ENGINEERING MONOGRAPHS

United States Department of the Interior BUREAU OF RECLAMATION

Reinforced Concrete

DESIGN DATA

By L. W. Crandall, George Evans,

Erik Sollid, H. L. Neve, and Samuel Judd

Denver, Colorado

No. 10

October 1952 (Revised and reprinted November 1965) 81.35

United States Department of the Interior

OSCAR L. CHAPMAN, Secretary

Bureau of Reclamation

MICHAEL W. STRAUS, Commissioner L. N. McCLELLAN, Chief Engineer

Engineering Monograph

No. 10

Reinforced Concrete DESIGN DATA

by

L. W. Crandall, George Evans, Erik Sollid H. L. Neve, and Samuel Judd

Engineers, Structural and Architectural Division Branch of Design and Construction

Technical Information Office Denver Federal Center

Denver, Colorado

Copyright, 1949, by Samuel Judd All rights assigned to the United States

Department of the Interior, Bureau of Reclamation

ENGINEERING MONOGRAPHS are published in limited editions for the technical staff of the Bureau of Reclamation and interested technical circles in government and private agencies. Their purpose is to record developmentS, inno­vations, and progress in the engineering and scientific techniques and practices that are em­ployed in the planning, design, construction, and operation of Reclamation structures and equip­ment. Copies may be obtained from the Bureau of Reclamation, Denver Federal Center, Denve!", Colorado, and Washington, D. C.

First Printing, October, 1952 (Revised and Reprinted, September, 1959) (Revised and Reprinted, November, 1965)

ACKNOWLEDGMENTS

"The Modified Slope-Deflection Equations," by L. T. Evans, a paper which appeared in the October 1931 Journal of the American Con­crete Institute, is the source of the influence lines for beams with sym­metrical straight haunches, pages 38 and 39. Other diagrams taken from this paper are those giving slope-deflection coefficients for beams with a full straight haunch, page 50, and those giving slope-deflection coeffi­cients for beams having an infinitely deep section at one end, page 52. The diagrams giving slope-deflection coefficients and end-moments for uniformly loaded beams having symmetrical straight haunches, pages 36 and 37, and slope-deflection coefficients for beams haunched at one end, pages 40, 41, and 42, and the influence lines for end moments on a beam having a full straight haunch, page 49, have been recalculated, rearranged, and extended beyond those published by Evans in the ACI Journal.

The influence lines for end moments on beams with a straight haunch at one end, pages 45, 46, and 48, appeared in "Continuous Frames of Reinforced Concrete," by Cross and Morgan, published by John Wiley and Sons, Inc. , and are reproduced by permission. Diagrams for uni­formly loaded beams with a straight haunch, and diagrams for carry­over and stiffness factors for beams with infinitely deep sections at both ends are also included in the book by Cross and Morgan, but the similar diagrams appearing in this monograph have been recalculated, rearranged, and extended.

Figures 93(a), page 105, to 93(i), page 109, inclusive, showing values of k and j for rectangular beams reinforced for compression, are taken from curves prepared by the Public Roads Administration, Fed­eral Works Agency. They are used by permission.

i

CONTENTS

~ IN'TRODUCTION . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . xiii

NOMENCLATURE . . . . . . . . . • . . . . . . . • . . . . • . • . . . . . . . . . . . • . . . . . . -m

CHAPTER I

DETERMINATION OF BEAM CONSTANTS . . • . . . . . . . . . . . . . . . . . . . . . . . 1

1. General Definitions • . . . . • . . . . . . . . . . . . . • . . . . . .. . . . . . . . . 1

2.

3.

~al Initial End Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 b Carry-over Factor .. ~ . ~ . . . . . . . . . . . . . . . . . . . . . . . . . 2 c Stiffness Factor or Modulus of Stiffness . . . . . . . • . . . . . . . . · 2

Beam Constants for Members of Constant Moment of Inertia

Example

Example

Example Example

Example

1 - Beam Constants fof Up.iform Section--Un,ifor~ Loading . ~ . . . . . . . . . . . . . . . . . . . . . . .

2 -Beam Constants for Uniform Section--Concentrated Loading· . . . . . . . . . . . . . . . . . . . .

3 - Use of Influence Lines for Initial End Moments ..... 4 ... Use of Influence Lines for Initial End Moments--

Moving Loads . . . . . • . . . . . . . . . . . . . • . . . . . . • 5 -Use of Influence Lines for Initial End Moments-­

,couple Acting at Any Point in a Member .......•

Beam Constants for Members of Variable Moment of Inertia

Example 6 - Beam Constants for a Member of Two

2

3

4 6

6

7

8

Prismatic Sections--Uniform Loading . • . . . . . . . 8 Example 7 - Initial End Moments for Member of Two

Prismatic Sections--Eccentric Loading. . . . . . . . . 9 Example 8- Beam Constants for Haunched Members • . . . . . . . . 10 Example 9 - Beam Constants for T -section . . . . . . . . . . . . . . . . 14 Example 10 - Modulus of Stiffness in Torsion--Member of Two

Rectangular Sections . . . . . . . . . . . . . . . . . . . . . 16

CHAPTER II

MOMENT DISTRIBUTION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1. Discussion . • . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 2. Sign Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3. Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 4. Principles of Moment Distribution . . • . . . . . . . . . . . . . . . . . . . . 57 5. Procedure of Moment Distribution . . . . • . . . . . . . . . . . . . . . . . . 57

Example 11 -Moment DiStribution--Continuous Beam . . . . . . . . . 58 Example 12 - Moment Distribution--Rigid· Frame Without

Sidesway . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Example 13 - Moment Distribution- -Symmetrical Frame

Symmetrically Loaded . . . . . . . • . . . . . . . . . . . . 63

6. Sidesway Correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Example 14- illustration of Simple Sidesway . • . . . . . . . . . . . . . 64 Example 15 -Temperature Effects on a Rigid Frame .. --. . . . . . . . . 67

iii

7. Moment Distribution for Frames Having Members of Variable Moment of Inertia. . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Example 16 -Horizontal Loading--Frame Having Variable Moment-of-Inertia Members . . . . . . . . . . . . . . . 81

Example 17 -Eccentric Loading--Frame Having Variable Moment-of-Inertia Members . . . . . . . . . . . . . . . 85

Comments on Example 17. . . . . . . . . . . . . . . . . . . . . . . . . . . 87

8. Moment Distribution with Torsion . . . . . . . . . . . . . . . . . . . . . . . 88

Example 18 - Moment Distribution with Torsion . . . . . . . . . . . . . 88

CHAPTER III

DESIGN DIAGRAMS 98

APPENDIX A

DERNATION OF EQUATIONS

General Equations for Initial End Moments . . . . . . . . . . . . . . . . . . . . 117 Specific Equations for Beam Constants for Given Methods

of End Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Equations for Plotting Curves of the Slope-deflection

Coefficients C 1' C2, and C3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Equations for Plotting Curves for Initial End Moments . . . . . . . . . . . . 122 Equations for Plotting Curves for Torsional Stress and Stiffness . . . . . . 123 Check on Moment Distribution ........................ ·. . . . . 124

APPENDIX B

DERIVATION OF EQUATIONS FOR PLOTTING, BENDING, AND DIRECT­STRESS DIAGRAMS FOR REINFORCED-CONCRETE MEMBERS

Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . • . . . . . . 126 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

Case I -Steel in One Face of Section--Direct Tension and Bending . . 126 Case II - Symmetrical Steel in Two Opposite Faces of Section--

Direct Tension and Bending. . . . . . . . . . . . . . . . . . . . . . 127 Case III - General Case ......................... ·. . . . . . . 127

Effect of Location of Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

Example 1 - Symmetrical Beam--Bending and Direct Tension . . . . . . 132 Example 2 - Symmetrical Beam--Bending Only . . . . . . . . . . . . . . . . 133 Example 3 - T-Beam--Bending and Direct Tension . . . . . . . . . . . . . 134 Example 4- Unsymmetrical Beam with Several Rows of Steel--

Bending and Direct Compression . . . . • . . . . . • . . . . . 136

iv

Number

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. .

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

LIST OF FIGURES

Loading, deflection, and moment area diagrams .............. .

Illustration of carry-over factor ..•......................

Illustration of stiffness factors ......................... .

Beam and loading diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Simple beam moment diagram

~ . -I- diagram . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Beam and loading diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Simple beam moment diagram

; diagram ..................................... .

Beam and loading diagram

Beam and loading diagram

Eccentrically loaded column ........................... .

~eaction diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._ .

Moment diagram . . . . . . . . . . . . . . . . . . ................ .

Beam and loading diagram ............................ .

Eccentrically loaded member . . . . . . . . . . . . . . . . . . . ....... .

Haunched frame and loading

Beam of frame in Figure 17

Hailllched column with triangular loading . . . . . . . . . . . . . . . . . . . .

~ diagram for external loading of column of Figure 19

~ diagram for mAB = 1 ..... · · · · · · · ; · · · · · · · · · · · · · · · ·

~ diagram for ~A = 1

Beam with two T-sections

2

2

3

3

3

4

4

5

6

6

7

8

8

9

10

10

11

11

12

13

13

15

24. Beam with two rectangular sections . . . . . . . . . . . . . . . . . . . . . . . 16

25. General equations for initial end moments, carry-over factors, and stiffness factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17-18

v

Number Page

26. Initial end moments for beams of constant moment of inertia . . . . . . 19-23

27. Influence lines for initial end moments--Beam with constant moment of inertia . . . . . . . . . . . . . . . . . o • • • o • • • • • o o • o • • • • • • • o • 24

28. Initial end moments for an applied couple--Beam with constant moment of itlertia . o o ••••• o ••• o o •••• o •••••••••• o • • • • 25

29. Moment and deflection diagrams for an applied couple--Beam with constant moment of inertia . . . . . . o • • • • • • • • • • • • • • • • • • o • •• 26

30. Slope-deflection coefficient c1 --Beam with two constant sections 0 • • 27

31. Slope -deflection coefficient C 2--Beam with two constant sections . o • 28

32. Slope-deflection coefficient c3--Beam with two constant sections . . . 29

33. Initial end moments uniformly distributed load--Beam with two constant sections . o • • • • o • • • • • o • • o • • o • • o • • • • o • • • • • • • 30

34. Influence lines for initial end moments--Beam with two constant sections o ••••••••••••••••••••• o •••••• o • • • • 31-32

35. Influence lines for initial end moments DBA --Beam with two constant sections ..................... o o 0 • • • • • • • • • • 33-34

36. Initial end moment, CAB' for an applied couple at change in section--Beam with two constant sections ... o • • • • • • • • • • • • • • • • • • • • 35

37o Initial end moment, CBA' for an applied couple at change in section--Beam with two constant sections . . • . . . . . . . . . . . . . . . . . . . . 0 35

38. Slope-deflection coefficients--Beam with symmetrical straight haUilches . • o • • • o • • • • • • • • • • • • • • • o • • • • • • • • • • • • • • • • • 36

39. Initial end moments, Uiliformly distributed load--Beam with symmetrical straight haunches . . . . . • . . . . . . . . . . . . . . . . . . . 37

40. Influence litles for initial end moments--Beam with symmetrical straight haUilches ............. o ••• o •••• o • • • • • • • • • • • 38-39

41. Slope-deflection coefficient Cl' c2, c3--Beam with straight haunch at one end . . o • • • • • • • • • • • • • • • • • • • • • • • • • • • • • o • • • • • • 40-42

42. Initial end moments, uniformly distributed load--Beam with straight haUilch at one end . . . . . . . . . . . . . . . . . . o • • • • • • • • • • • • • • • 43

43. Influence lines for initial end moments--Beam with straight haunch at one end .......... 0 •• o o • • • • • • • • • • • • • • • • • • • • • • • • 44-48

44. Influence lines for initial end moments--Beam. with full straight haunch . . • . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

45. Slope-deflection coefficients--Beam with full straight haUilch . . . . . . 50

46. Slope-deflection coefficients--Beam with itlfinitely deep section at both ends .......... 0 o o • • • • • • • • • • • • • • • • • • • • • • • • • 51

Vi

Number

47. Slope-deflection coefficients--Beam with infinitely deep section at one end . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . 52

48. Slope-deflection coefficients--Beam with symmetrical parabolic haunches . • . • . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . 53

49. Torsional stiffness of rectangular beams . . . . . . . . . . . • . . . . . . . . . 54

50. Moment of inertia ofT-sections about the centroidal axis . . • . • . . . 54

51. Areas and centroids of moment diagrams . . . . . . . . . . . . . • . . • . . 55

52. Slope-deflection coefficients--Beam with variable width, depth constant . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . 56

53. Beam and loading diagram. . . . . . . . . . . . . . . . . . . . . . . . . • . . • 58

54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

69.

70.

71.

72.

73.

74.

75.

Free-body diagram of beam AB in Figure 53

Free-body diagram of beam BC in Figure 53

Frame and loading diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Free-body diagram of beam AB in Figure 56

Free-body diagram of beam BC in Figure 56

. .............. .

Free-body diagram of column BD in Figure 56 ......•..••....

Frame and loading diagram .................•..........

Free-body diagram, columns--simple sidesway .............. .

Sidesway deflection of frame . . . . . . . . . . . . . . . . . . . . . . . . . . .

Free-body diagrams of columns showing unbalanced shear ...... .

Free-body diagrams of columns showing effect of side sway on moments .................................... .

Frame with 3 points of sidesway • . . . . . . . . . . . . . . . . . . . . . . . .

Frame showing values of K • • . • • • • . . . . • . . • . . . . . . • • . . • • •

Frame showing distribution factors S ........ ~ ...•........

Deflection of frame due to temperature rise ................ .

Unbalanced shears due to temperature loading . . . . . . . . . . • . . . .

Totals of unbalanced shears from Figure 69 . . . . . • . . . . . . . . . . .

Moments due to sidesway of top story . . . . . . . . . . . . . . . . . . . • .

Shears due to sidesway of top story ...................... .

Moments due to sidesway of beam EF ..................... .

Shears due to sidesway of beam EF ..................•....

Moments due to sidesway of beam GH . . . . . . • • . • . • . . • . • • • . .

vii

60

60

61

62

62

63

63

65

65

66

67

67

68

68

69

'7o

70

71

72

76"

76

77

Number Page

76. Shears due to sidesway of beam GH . . . . . . . . . . . . . . . . . . • . . . 77

77. .Frame with variable T-sections • . • . . . . . . . . . • . . • • . . . . . . . 81

78. Free-body diagrams of columns of frame in Figure 77 . . . . . . . • • 83

79. Sidesway deflection of frame in Figure 77. . . . . . . • . . . . . . . . . . 83

80. Free-body diagrams of columns of frame in Figure 77 •. ~ . . . . . . 83

81. Stepped columns eccentrically loaded . . . . . . ~ . . . . . . . . • . . . . 85

82. Free-body diagram, left column of Figure 81 . . . . • . . . . . . . . . . 86

83. Sidesway deflection of frame in Figure 81 . . . . . . . . . . . . . . . • . 86

84.

85.

86.

87.

88.

89.

90.

Free-body diagrams of both columns of Figure 81

Free-body diagrams of both columns of Figure 81

Frame in torsion ....•.....•................... · •...

Shears due to a load of 500 lb. per lin. ft. on beams FB and CE . . .

Shears for joint B deflected

Shears for joint C deflected

Bending and direct tension--Reinforcement on one face of section .•

86

87

89

91

93

93

99

91. Bending and direct tension--Symmetrically reinforced section . . . . 100-103

92. Bending and direct stress in reinforced concrete members . . • . . . 104

93. Rectangular beams reinforced for compression- -values of k and j for given values of pn and p'h. Ratios of d' /d from 0.5 to 0.30 . . . 105-109

94. Torsional shear in rectangular beams . . . . . . . . . . . . • . . . . . . • 109

95. Apyroximate values for allowable combined axial and bending stress--tied columns . . . . • . . . . . . . . . . • . . . . . . . . . . . . • . 110-113

96.

97.

98.

99.

100.

Sectional areas and perimeters of American welded wire fabric . . .

Sectional areas and perimeters of reinforcement bars

APPENDIX A

Beam with two constant sections . . . . . . . . . . . . . . . . . . . . . . • .

APPENDIX B

Effect of location of steel in beams . . . . . . . . . . . . . . . . . . . . • .

Unsymmetrical beam with several rows of steel--Bending and direct compression ......•..............•........

viii

114

115

122

131

136

TABLES

Number

I. Calculation of areas and centers of gravity for M

X d' . 5 -I- 1agram .................................... .

M II. Calculation of I x diagram for external loading . . . . . . . . . . . . . . 12

X

III. M

Calculation of Ix diagram for mAB = 1 and mBA= 1 X

IV. Moment distribution of beam in Figure 53, initially assuming

13

end A as fixed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

V. Moment distribution of beam in Figure 53, treating end A as hinged . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

VI. Moment distribution in frame of Figure 56 . . . . . . . . . . . . . . . . . . 62

VII. Moment distribution in frame of Figure 60 . . . . . . . . . . : . . . . . . . . 64

VIII. Moment distribution in frame of Figure 60 with only left half loaded . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

IX. Moment distribution for sidesway . . . . . . . . . . . . . . . . . . . . . . . . 66

X. Moment distribution for temperature rise . . . . . . . . . . . ; • . . · . . . 71

XI.

XII.

XIII.

Moment distribution for sidesway of top story

Moment distribution for sidesway of beam EF

Moment distribution for sidesway of beam GH

73

74

75

XIV. Solution of simultaneous equations . . . . . . . . . . . . . . . . . . . . . . . 78

XV. Side sway correction of moments in foot kips due to P 1 = -1,000 kips. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

XVI. Sidesway correction of moments in foot kips due to P 2 = -1,000 kips. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

XVII. Sidesway correction of moments in foot kips due to P 3 = -1,000 kips. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

XVIII. Sidesway correction of moments in foot kips due to P1 = + 124.66 kips, P2 = + 12.09 kips, and P 3 = - 113.40 kips. . . . . 80

XIX. Final moment~ in foot ki:RS due to 400 F temperature rise on beams A to D and 20° F temperature rise on beams EF and GH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

XX. Distribution of moments in frame of variable T-sections loaded as shown in Figure 77 . . . . . . . . . . . . . . . . . . .. . . . . . . . . • . . 82

ix

Number Page

XXI. Distribution of sidesway moments of frarne in Figure 77 . . . . . . . . . 84

XXII. Distribution of moments in column in Figure 81 assuming column fixed at both ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

XX!ll. Distribution of moments in columns of Figure 81 due to sidesway . . . 86

XXN. Moment distribution due to loading shown in Figure 86. No deflection at joints B and C . . . . . . . . . . . . . . . . . . . . . . . . . . 91

XXV. Distribution of moments in direction AD. Joint B deflected . . . . . . 92

XXVI. Distribution of moments in direction FBo Joint B deflected .. 0 • • • 93

XXVII. Distribution of moments in direction AD. Joint C deflected .... • . 94

XXVIII. Distribution of moments in direction FB. Joint C deflected. . . . . . 94

XXIX. Solution for correction factors . . . . . . . . . . . . . . . . . . . . . . . . . . 95

XXX. Correction in bending moments due to + 1,000 lb. at joint B and no load at joint C ........................... ~ . . . . . . 95

XXXI. Correction in torsional moments due to + 1,000 lb. at joint B and no load at joint C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

XXXll. Correction in bending moments due to + 1,000 lb. at joint c and no load at joint B . . . . . . . . . . . . . . . . . . . . . . . . 0 • • • • • • • • • 96

XXX!ll. Correction in torsional moments due to + 1,000 lb. at joint C and no load at joint B ..................... 0 • • • • • • • • • 96

XXXN. Final corrected flexural moments due to loading shown in Figure 86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

XXXV. Final corrected torsional moments due to loading shown in Figure 86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

APPENDIX B

XXXVI. Tabulation of computations for constants, symmetrical beam. . . . . . 135

XXXVII. Tabulation of computations for constants, unsymmetrical beam . . . . 136

X

Reinforced Concrete DESIGN DATA

INTRODUCTION

The Bureau of Reclamation,' in common with other engineering organizations, is frequently required to design concrete frame structures, such as powerhouse superstruc­tures and valve houses, in which heavy­duty overhead cranes pose difficult design problems. Practically all such structures are designed as rigid frames, and are stat­ically indeterminate.

This monograph is confined to deriving data necessary for designing reinforced­concrete structures of this type, but the method can be extended to apply to any statically indeterminate structure. Data contained herein were first published by

xiii

the Bureau as Technical Memorandum No. 629, and were later revised and issued as Technical Mernvrandum No. 637. Issuance of this material in the Engineering Mono­graph series is prompted by the interest shown in the preceding editions. No sub­stantial changes have been made in the text for the present edition.

Chapter I discusses the determination of beam constants for frequently occurring variable sections, Chapter II discusses mo­ment distribution, including side sway, and Chapter III contains a collection of design diagrams for reinforced-concrete members. Derivation of equations is shown in Appen­dices A and B.

E

I

NOMENCLATURE

a beam or member, A to B.

area of Mx diagram, Mx Ix

being the simple beam mo­menf at any point x due to an applied end moment, mAB = 1, at end A.

area of Mx diagram, Mx Ix

being the simple beam mo­ment at any point x due to an applied end moment, mBA = 1, at end B. ·

area of Mx diagram, Mx Ix

being the simple beam mo­ment at any point x due to

··the applied loading.

fixed end moment at end A of member A-B with both ends fixed.

fixed end moment at end B of beam A-B with both ends fixed.

slope-deflection coefficients.

moment at end A of beam A-B with end A fixed and end B hinged.

moment at end B of beam A-B with end B fixed and end A hinged.

Young's modulus of elas­ticity.

moment of inertia about the center of gravity of bean1 of constant section.

moment of inertia about center of gravity of shallow­est section in a beam of variable section.

moment of inertia about center of gravity at any point x in a beam.

stiffness factor at end A of beam A-B.

XV

p

R

stiffness factor at end B of beam A-B.

length of beam.

moment at end A of beam A-B.

moment at end B of beam A-B.

the moment in a simple beam at any point x due to the applied loading.

unit moment applied at end A of beam A-B.

unit moment applied at end B of beam A-B.

concentrated load. ~ r:-·

carry-over factor at end A of beam A-B.

carry-over factor at end B of beam A-B.

distance from end A to centroid of AA.

distance from end A to centroid of AB.

distance from end A to centroid of A

0•

relative linear deflection between ends A and B in a beam A-B, measured normal to the initial posi­tion of the beam.

change in slope of the elastic curve, at end A, from its initial position.

change in slope of the elastic curve, at end B, from its initial position.

Other notations are defined where first used. Some of the nomenclature given above is illustrated in Figure 1(a) to 1(h).

(a)

(b)

(c)

(d)

(f)

(g)

(h)

Any loading---->-01TIIIIID1 ·

I-*- ______ X _____ >1 ._ L-X · Lix

~- -----------Xs ------ ---->! 1 x ->:tc:dx 1 I -· 11 I : ~--~j-------1l Ae, oc.g~

~--------x~--->1·

Deflection· diagram:

Moment diaqramfor · applied load consid­erinq A and B free.

Moment diagram for mAe= I consideri nq A and B free.

FIGURE l - Loading, Deflection, and Moment .Area Diagrams.

xvi

CHAPTER I

DETERWNATION OF BEAM CONSTANTS

Usually the quickest way of solving the problems involved in statically indetermi­nate structures is by the method of moment distribution. However, before the method can be applied, constants must be deter­mined. These constants are. influenced by the loading and the properties of the struc­ture. The methc d of moment area has been found satisfactory by Bureau designers for determining these constants and is used in this discussion.

This monograph deals only with struc­tures of constant modulus of elasticity E throughout each member. Since E usually

divides out, the Mx diagram rather than M IX .

the __.!. diagram may be used. Whenever Eix

E has any effect on the final results, as in the case for stiffness factors and moments caused by linear movements of joints, it has been included in the final equations.

In order to reduce the work of computing beam constants, simplified formulas for members of uniform section and general formulas for members of any shape are included. Diagrams of beam constants for frequently occurring variable sections are presented. No attempt is made to desig­nate a sign system for initial end moments and carry-over factors. All sign systems are arbitrary, and the signs must be deter­mined from the conditions of the problem, based on the sign convention that the de­signer has assumed.

1. General definitions

The following beam constants must be determined before analyzing a frame by the method of moment distribution:

(a) Initial end moment

(b) Carry-over factor

(c) Stiffness factor or modulus of stiff­ness

1

(a) Initial end moment. The moments. at the endS of a member due to external load­ing, rotation of the ends, and relative deflec­tion of the ends are eXpressed by the follow­ing two equations:

MAB ~c [ Cl eA + C29g

- (Cl + C2) d +CAB

MBA E~c [ C29A + C39g

- (C2 + C3) t] + CBA

Here CAB and CBA are called fixed end moments and are defined as the moments at the ends of a member caused by the applied loads when both ends are held fixed in their original positions. However, in the method of moment distribution, any combination of the terms on the right side of the above equations thatis convenient or necessary for starting the distribution of moments in a given problem is called a fixed end mo­ment. In this monograph the term ''fixed end moments'' will be restricted to mean the quantities CAB and CBA only.

The moments used in starting moment distribution will be called ''initial end mo­ments" or "initial moments" and are simply the moments at the ends of a member as defined by the two equations for ~B and MBA· Thus, the distinction between initial end moments and fixed end moments is that any term or terms of the equations for MAB and MBA. may be called initial end moments, but CAB or CBA will designate fixed end moments only.

· (b) Carry-over factor. The carry-over factor in bending is ilhlStrated in Figure 2. When er1d A of beam AB is subjected to a moment ~B' a moment MBA is induced ~t end B. The carry-over factor is defined

. M as the ratio BA • MBA is caused by MAB

MAB only. This ratio, for any given beam, varies with the degree of fixation of end B. For the present purpose the carry-over factor

M will be limited to mean the ratio BA when

MAB end B is fixed. The case when end B is hinged will be used, but for this condition the carry-over factor is zero, since obviously MBA = 0. The carry-over factor in torsion .is defined in example 18 of Chapter II.

r. -MI.!. ••· Mu

A~ ---- MBA-~~ ~----- Fixed_.. ~ B

FIGURE 2 - Illustration of Carry-over Factor. (For actual values see Figures 25(a) and 25(b).)

(c) Stiffness factor or modulus of stiff­ness. In bending, the modulus of stiffness of a member is the bending moment required to turn the hinged end of a member through a rotation of one radian. The other end may or. may not be completely fixed. For the present purpose, the most useful cases are illustrated in Figures 3(a) and (b). This definition of stiffness is used without modi­fication throughout this monograph. It re-sults in a value of 4EI for the stiffness fac-

L tor of a fixed beam of constant section in-stead of the value of EI or I as usually · L L----given. The reason for using actual values

"Qi'""'S'ffffness, as defined above, is to elim­inate a source of error in computing moment distribution factors when members of both constant and variable moments of inertia are present in the same fram·e and also when members in torsion are present All dia­grams in this monograph for determining stiffness factors give actual stiffnesses KAB and KBA' and the equations for KAB and KBA on each diagram refer to a fixed beam. In torsion, the modulus of stiffness of a member will be defined as the torsional moment required to twist the free end of a member through a rotation of one radian

2

with the other end fixed. Whenever the un­balanced moment at a joint is distributed in bending only to other members of the same material, the modulus of elasticity may be omitted when computing the stiffness fac­tors. However, when members in torsion also frame into the joint, the modulus of elasticity must be included, since it is dif­ferent for bending and for torsion. This, of course, is also true when members of dif­ferent materials frame into the same joint.

Mu = KAe :'Fixed

A~~-:~----=:;;.:------ .,li la) .

~~~-:-~~~--,.---------------------------·H•noed A -- ••• _ '. e

··e= 1 (b)

FIGURE 3 - Illustration of Stiffness Factors. (For actual values see Figures 25(a) and 25(b).)

The values of case I in Figure 25(a) can usually be used in the solution of rigid frames. Where applicable, cases II and TII in Figure 25(a) lead to short cuts.

2. Beam constants for member§ of constant moment of inertia

For beams o! constant moment of inertia, the work of computing initial end moments is greatly simplified by the use of special equations which have been derived for vir­tually all types of loading and end supports. However, loadings are occasionally encmm­tered for which special equations are not obtainable. The general equations given in Figures 25(a) and (b) will greatly facilitate the calculation of beam constants when special equations are not available. Ex­amples 1 and 2 illustrate the use of these equations. Special equations for initial end moments for various types of loadings and end supports are given in Figures 26(a) through (e).

Influence lines are usually very useful for obtaining initial end moments, and illus­trations of their use are given in examples 3, 4, and 5.

~~~~~l=~~~~~~~~~~~~~~~ ~e~Uo.D.= JJ.niio.r!lllo,adj~.

Given:

A beam of constant section 1. 0 foot wide, 4. 0 feet deep, 20. 0-foot span, carrying a load of 100 pounds per linear foot, as shown in Figure 4.

.f'.IGURE 4 - Beam and Loading Diagram.

Required:

Stiffness factors K, carry-over factors r, and initial end moments which, in this case, are the same as fixed end moments CAB and CBA·

Solution:

1 ~243 = 5.33 ft. 4

From case I in Figure 25(a) the stiffness factors are

4EI ~A= L

L066 E

4 X 5.33 E 20

Where all members of the structure are of the same material and subjected to bending only, E can be cancelled, leaving

Further, from case I,

1 rAB = rBA = 2

and

(r is always equal to ~ for 2

beams of constant section)

Xo I A (4- 6-)-o L L

Xo I A(- 2+ 6-)-

o L L

3

As shown in the nomenclature, A0 is the area of the Mx diagram for the loading

I . under consideration, and X 0 is the distance from end A to the centroid. of A0 • The mo­ment diagram for a simple beam having a span of 20 feet, loaded with 100 pounds per linear :foot, is a parabola having a m::txi.mum ordinate at the center equal to .l wL2,

. 8 .

1 2 1 M = -wL' = -x100x202 8 8

5,000 ft. -lb .

FIGURE 5 - Simple Beam Moment Diagram.

Since the moment of inertia, I, is con-4 M stant and equal to 5. 33 ft. , the 2 diagram

I is also a parabola having a maximum ordi .. nate at the center equal to

Mx = 5,000 = 937 lb./ft. 3 I 5.33

M FIGURE 6 - ~ Diagram.

I

From Figure 51, the area of a parabola equals

A = .1_ ML 3

where M is .the value of the middle ordinate and L is the length. Therefore,

A0

= ~ X 937 X 20 = 12,500 lb./ft..2

The center of gravity is at the center of the span and is therefore 10 feet from end A.

L x0 = 2 = 10.0 ft.

Substituting these values gives

12 500(4- 6 1b) 5·33 ' 20 20

3,330 ft. -lb.

CBA 12 500(- 2 + 6 10 ) 5·33 ' 20 20

3,330 ft.-lb.

These values could, of course, have been found simply by applying the formula for initial end moments for a uniformly loaded beam, as given in Figure 26(b),

CAB = CBA = 1~ w L2 = 1~ x 100

X 202 = 3,330 ft.-lb.

Given:

A beam of constant section 1. 7 5 feet wide, 4. 5 feet deep, 30.0-foot span, and loaded with two· concentrated loads, as shown in Figure 7.

apoo.., to,ooo• r·o· -t u 150' ----t-•o'l A· Y X 8

' ----------- L=30.0'-----------

FIGURE 7 - Beam and Loading Diagram.

Required:

Stiffness factors, K, carry-over factors r, and initial end moments.

Solution:

1 3 I= 12

x1.75x4.5 13.3 ft.4

4

Referring to case I in Figure 25(a),

4EI ~A= L

1.77 E

1 2

4 X 13.3 E 30

ln order to evaluate the end moments, first calculate the simple beam moments and then find the area and the center of gravity of the corresponding Mx diagram as follows·:

I -

The reactions 'for a simple beam are

6 x 1o,ooo + 21 x 8·,ooo 30

7,600 lb.

9 X 8,000 + 24 X 10,000 30

10,400·lb.

The moment for a simple beam at the 8,000-pound load is

M8,000 = 7,6oo x 9 = 68,400 ft.-lb.

and at the 10,000-pound load

M 10,000 = 10,4oo x 6 = 62,400 ft.-lb.

# 0 0 <D f'-!"

FIGURE 8 - Simple Beam Moment Diagram.

II ID

a:::

'Then since I = 13.3 ft. 4, the Mx value I

9 feet from A is

Ma,ooo = · 68,400 ,= s I lS. 3 5,140 lb./ft.

and 6 feet from B

M1o,ooo 62,4oo 4 69 1 3 I 13.3 ' 0 lb. ft.

Mx FIGURE 9 - -I- Diagram.

Divide the Mx diagram into convenient I

sections - ,1, 2, and 3. Table I shows the calculations for areas and center of gravity of these sections.

Then,

A0 ~A = 110,900 lb./ft. 2

and

1,714,600 110,900

15.46 ft.

Substituting these values in equations of case I in Figure 25(a) gives

110 900(4 - 6 15·46 ) J-3·3 ' 30 30

44,900 ft.-lb.

110 900(- 2 + 6 15.46 ) 13·3 ' . 30 30

53,600 ft.-lb.

These values could have been found by the formulas given in Figure 26(a) for similar loading.

1 2 1 CAB'=.- ~Pab = - (8,000 L 2 302

X 9 X 212 + 10,000 X 24 X 62)

44,900 ft.-lb.

12 ~Pa ~ = _1_

2 (8,000

L 30

X 9~ X 21 + 10,000 X 24 2 X 6)

53, 500 ft. -lb.

If the above beam were hinged at end B, the value of DAB from case II in Figure 25(a) would be

*DAB = CAB - ~ CBA = 44,900 + -t X 53,500 = 71,600 ft.-lb.

*In the equation for ·DAB or ~A the sign of CAB and CBA must be taken into acco~t.

TABLE I

SEC-TION

I

2

3

Total

M Cal.culation for .Areas and Centers of Gt-avi ty ,+ Diagt"am

·D I.STANCE DISTANCE FROM LEFT FROM

-

AREA OF SECT I ON EDGE OF SECTION TO C.G A TO C.G.

OF SECTION

; X 9 6.00 6.00 ~ x9x5140= 23,100 !5140t2X4690 x~"5 =7 39 5140 + 4690 3 . 9+ 1.3 9 = 16.39 r\?14o ~ 4690)15=73,700

I 3 X 6 2.00 24+2:26.0

I . 2X6.0x4690= 14,100

l A= 110,900

5

MOMENT ABOUT END "A"

138,600

1,210,000

366,000

1,714,600

or, by formula in Figure 26(a),

DAB. = 2L12 ~Pab(b + L) = 1 2 X 302

Given:

[8,ooo x 9 x 21(21 + 30)

+ 10,000 X 24 X 6(6 + 30~

71,600 ft.-lb.

The same beam and loading as in ex­ample 2. . ~

0 0 0 o .. ~ Q

------~-~-~----~---- --oL-- --------

A ~------~.~----------~~----~-8 I 1 I I I ---9.0 --*·-----1s.o ------+-s.o-

- -------------L =30.0~-------------

FIGURE 10 - Beam and Loading Diagram.

Required:

The initial end moments CAB and CBA.

Solution:

Influence lines are plotted on Figure 27 for a beam of unit length and constant mo­ment of inertia carrying loads perpendicular to the axis of the beam. In order to facili­tate the use of these curves, the values of both CAB and CBA have been plotted For a beam of constant section they are, of course, similar curves but opposite hand. On the same page is also a curve for DAB which is the moment at A with the end A fixed and the end B hinged.

For the 8,000-pound load,

a = ~ = 0.3 30

and for the 10,000-pound load,

24 a = 30

= 0.8.

6

Then from the curve for CAB in Figure 27 the coefficient ~ for the 8,000-pound load, (a = 0.3), is 0.146 and for the 10,000-pound load, (a = 0. 80), 0. 032• Therefore,

CAB = (0.146 x 8,000 + 0.032

X 10,000) 30 = 44,640 ft.-lb.

Similarly, ~ for CBA for the 8,000-pound load is 0. 062 and for the 10,000-pound.load 0.128. Therefore, ·.. . .

CBA · = (0.062 x 8,000 + 0.128

X 10,000) 30 = 53,300 ft.-lb.

These values check the previously calculated values reasonably well. ·_ . . ·

Given:

A beam of constant section on a 30-foot span, loaded with two moving loads 3 feet apart. The larger load is 10,000 pounds and the smaller load is 8,000 pounds.

# # 0 0 0 0

~I ~I A~------+~-+-~ -----~ B

I I

-----X=oL----~3.01 !<--- (L-X-3)---" --------------L =30.0'-------------

FIGURE 11 - Beam and Loading Diagram.

Required:

The maximum fixed end moment CAB at A and the corresponding fixed end mo­ment CBA at B.

Solution:

The distance between the loads is ..! 30

= 0.1 of the span length. Several trials indicate that the largest value of the fixed end moment at A (CAB) will occur when

the 10,000-pound load is at a= 0.30 and the 8,000-potmd load at a = 0.40. With the loads in this posit\on the coefficient ~ at the point a = 0. 30 is 0. 146 and at the point a= 0.40, 0.144. Therefore,

CAB = (0.146 X 10,000 + 0.144

X 8,000) 30 = 78,400 ft. -lb.

Similarly, the corresponding fixed end moment at B is

CBA - (0.061 x 10,000 + 0.095 .

X 8,000) 30 = 41,1.00 ft.-lb.

Given:

A colunm 27 feet long loaded with a ver­ticalload of 30 kips 1. 25 feet from the center line of the column and applied 20 feet from the base, as shown in Figure 12(a).

A

B

(a)-Actual ~oading

).""T" , ·o

~~ ' 0 " : l---·

-~ ! ~ q. ... 0 _JN

' " :':J ' 0

~~ o I

' ' : !

~--±-

A

B

Couple =30x 1.25 =37.5' K

(b} -Equivalent ~oading

FIGURE l2 - Eccentrical.ly ~oaded co~umn.·

Required:

Initial end moments at A and B.

Solution:

This problem can be solved by the use of the special equations in Figure 26(e) or the general equations in Figure 25(a). How­ever, the influence lines in Figure 28 give a much quicker solution of this problem. The diagrams can be read accurately enough for

7

most designs. Determination of the signs of the end moments and the- shape of the moment diagram is difficult; since it varies. with the location of the point of application of the couple. Figure 29 has been prepared to facilitate the determination of the signs of these moments.

The column as loaded in Figure 12(a) may be replaced by a column with a centri~ load of 30 kips and a couple of 37. 5-foot kips, as shown in Figure 12(b), The con­centrated load has no effect on the moments; so the problem reduces to one of finding the moments for .a couple of 37. 5-foot kips.

First find

7 a = 27 = 0.259.

Then from the influence line, Figure 28, the coefficients ~ for CAB' CBA' and RA (the reaction at A) are 0.17, 0.31, and 1.16, respectively.

Therefore,

CAB 0.17 X 37.5 6.4 ft. -kips

CBA 0.31 X :17.5 11.6 ft.-kips

RA 3~75 X 1.16 = 1.61 kips

Since no transverse loads are applied, the reaction at B is also 1. 61 kips. The re­actions are always in such a direction as to oppose the applied couple.

The top deflection diagram in Figure 29 shows how the beam is stressed when the couple is applied between a = 0 and a = 0. 33. Similarly, the second and third diagrams show how the beam is stressed for the couple applie<d between a = 0. 33 and a=0.67;and a=0.67 and a=l.O, respectively. Naturally, if the couple,,ts reversed, all stresses reverse.

In the present example a= 0.259<0.33. Therefore, the shape of the deflection and

. the moment diagram will be as shown by the top diagram in Figure 29. This enables the reaction diagram, Figure 13, and the moment diagram, Figure 14, to be drawn. Since ten­sion exists on the right side of the column at the top and the bottom, the direction of end moments will obviously be as shown in Fig­ure 13. The reactions are always equal and opposite and in such a direction as to oppose the applied couple.

The moment on theright side at the application of the couple will be 6.4 + 1.61

0 r---.: N

I I

0 0 N

I v 1 J \ _v ____ ~ R8 =1.61K CaA= 11.6ft.K

8

FIGURE 13 - Reaction Diagram.

x 7 = 17.7 ft -kips, and on the left side 11.6 - 1. 61 x 20 = - 20.6 ft. -kips.

The sum of these two moments should equal the applied couple of 37.5 ft. -kips, 20.6 + 17.7 = 38.3 ft-kips. The discrepancy is small and due to inaccuracy in reading the curves.

3. Beam constants for members of variable moment of inertia

General equations solving carry-over factors, stiffness factors, and initial end moments for members of any shape are given in Figures 25(a) and (b). These equa­tions express the above constants in terms

of properties of the Mx diagrams. To facil-Ix

ita.te the work of computing these beam con­stants, diagrams are included in this chapter

8

CAB= 6.4 ft. K--- .. .. ,, . tc-.-'~->

A: , - 'A- ----- --------- -.--· -,.

I

-0

r:--: f<- 20. 6 ft. K -~---·~r-------+----~

I I

'-.---~-.,-.-177ft. K ~

8 !-, --~ I I I I

~<---X- ->4

CaA ~ 11.6 ft. K-'

FIGURE 14 - Moment Diagram.

covering the most frequently occurring shapes of members. Special attention has been given to members composed of two prismatic sections because this type occurs as a crane column in practically all power­house buildings. Although these curves and equations are self-explanatory, their use is illustrated in examples 6, 7, 8, and 9.

Example 10 illustrates a method of de­termining the torsional stiffness of a mem­ber made up of two rectangular sections.

~~:mmEL EL -_Jieg_rn Q..o.nsJa.llt§ ..fo.r ..e: w,grnb.§r Qt~o_p.riJ;!ID~i..Q ~~ctj.Q...n~ -U:.@Q.r..rn_lQ.a..Q]J)g.

Given:

A member of two prismatic sections, as shown in Figure 15, loaded with a uni­form load of 100 pounds per linear foot Both sections of the member are 1. 25 feet wide.

Required:

Initial end moments, CAB and CBA Stiffness factors, KAB and KBA Carry-over factors, r AB and rBA

Solution:

10 a = 30

1 3

m _bQ_ = 2 1.25

1.25 X 1.253 12 3 30 = 0.00678 ft.

The moment of inertia Ic is based on the smaller section. For the meaning of a and m, see Figure 15. By the diagram in Figure 33, for a =.! and m = 2, the

3 value of ~ for _CAB is 0.0525 and for CBA 0.115. Therefore,

cAB o.o525 x 100 x 3o2

4,730 ft.-lb.

cBA o.115 x 100 x 3o2

10,350 ft.-lb.

From the diagrams in Figures 30 to 33 are obtained values of the three coefficients C1, C2, and C3; C1 = 5. 5; C2 = 6.15; and c 3 = 26.o.

Then, by the equations shown on the same diagram,

K AB

I C ...£E

1 L

0.0373 E

Ic c3LE

0.176 E

c2 6.15 c1 = 5.5

c2 6.15 c3 26

5.5 X 0.00678 E

26 X 0.00678 E

1.118

0.237

9

,··100./ft. y

A B

·oL=IO.O'-.ic ...... (L-oLl=20.0' .... . ~ ............... L =30.0' ............ .

FIGURE 15 - Beam and Loading Diagram.

If, in this example, end A were hinged, the coefficients would be as follows:

By the equations given in case III in Figure 25(a), · ·

c2 *DBA = CBA - C1 CAB = 10,350

+ ~.155 4,730 = 15,650 ft.-lb.

r BA 0; r AB and K AB remain

the same.

2 = (c3 - c2 ) Eic

C = (26.0

1 L

-65~~2 ) 0.00678 E = 0.129 E

E~w~1~wu~L~~~~~m~w~m~~~ .Q.f.J~o_p_ri§mg_tli! ..§~CYQ11.2 -.E ~c~ll!;rjc_l.Qa..9wg:.

Given:

A member of the same dimensions as the one in example 6, loaded with an eccen­tric force of 1~000 pounds applied as shown in Figure 16(aJ. End A is not fixed against vertical movement.

Required:

The initial end moments due to this loading.

*In the equation for DBA or DAB the signs of C BA and CAB must be taken into account.

·o 2 ~ 0 I ~ ....,..

A

i~~ISII.: B

(a)-Actual ~oading

B

(b )-Equival.ent ~oading

FIGURE ~6 - Eccentrically ~oaded member.

Solution:

The loading of Figure 16(a) may be represented by a centric load of 1,000 polUlds and a moment of 630 ft. -lb. as shown in Figure 16(b). The centric load will have -no effect 011. the end moment. Then by the diagrams in Figures 36 and 37, the values of CAB and CBA are obtained for an applied moment of unity. Multiplying these values by the applied moment of 630 ft. -lb. gives the moments sought. Thus, for a= .l and m = 2, 3

CAB = + 0.19 x 630 = 119.7 ft.-lb.

-,.---------I I I I I

0 <X) II

.c. I I I I I I

j_ #-------:-.~ ~~~~

I ->'

I

and

CBA = + 0.275 x 630 = 173.3 ft.-lb.

Stiffness and carry-over factors remain ~e same as in example 6.

Given:

A frame having a uniform width of 1. 25 feet, with dimensions and loading as shown in Figure 17. .

Required:

(a) Initial end moments, carry-over factors, and stiffness factors for the beam.

(b) Same for the columns.

Solution (a) - Beam:

The length of beam iS considered to be the distance from center line to center line of colunms as shown in Figure 18. Note that the haunches are also extended to center lines of columns.

p = 1000#

~kL=5.0' -~ I

0.67'-

'

· FIGURE 17 - Haunched Frame and Loading.

10

A

' I md =,4.0'

I I

, -----r-----------------~ .I I

jo<- ).-·->t . I

P=IOOO# rkl = s.o·-

. I I I I : '---·a L-= 2.33

:<- · ---- -- ------ ------ L = 2 o. o' • I

-- ~ -- - ----------- -> I

FIGURE J.8 - Beam of Frame in Figure ~ 7.

The ratios, a, k, and m are

a = 2·33 = 0.117 20

k 5 20 = 0.25

m 1~6~ = 2.40

1.25 X 1.673 0 485 ft 4 12 = . .

From diagram in Figure 3 8 for a = 0.117 and m = 2.4,

3.9

K. C IcE 6 5 0.485 E AB 1 L = . 20

0.158 E

.... ue to symmetry, KBA = 0.158E, and for the same reason,

Values for finding initial end moments are obtained from curves in Figure 40(a). Thus, for k=0.25,m=2.4,and a=0.1,·~is· found to be 0. 036, and for the curve a = 0. 2,

11

~ = 0.034. Interpolating to the value a= 0.11'i of this problem, ·

and

~ = 0.036 - 0.002 X OQ~i 7 = 0.0357

0.0357 X 1,000 X 20

714 ft.-lb. t

Similarly, by taking k from .the other end of the beam, it is found that for this new k = 0.75, ~ = 0.171, and

CBA = 0.171x1,000x20

3,400 ft. -lb.

Solution (b) - Columns:

The height of column is taken to center line of beam and the haunch is extended to this line also, as shown in Figure 19.

l<md=3.~~ :A : l ·~ -r- Point A _ x

Lt---,--' ---r- II - I --::=t---A.-------

,.,__~--=~==-=--=--- II -2 --==t-0 w-..f=--===-=-==-- 11-3 --l aS r---+---'----- .. -4 _-x II • s:. L.-. ...t--+--=--::::-=:----11 -5 _-. : r---'1--+-~~-:=-=:;--- 11-6 --* : J.---,..r-t------ H -7 . _-t

y__ ,,_ 8 _---=.L I I B

· >1 :<·aoo#/lineor ft.

FIGURE ~9 - Haunched Co~umn with Tri­angular Loading.

The moment of inertia Ic is always based on the shallowest section. In this case, then,

m

and

a

1.25 X 0.673 12

3.5 0.67

2.83 -8-

5.22

0.354

0.0313 ft.4

For these values of a and m the dia­grams in Figures 41(a), (b), and (c) give

cl 15.3

c2 6.2

and

c3 5.5

Then

KAB IcE

elL

0.0599 E

K_sA IcE

C3--y:-0.0313 E

5.5 8

0.0215 E

rAB c2 6.2 0.405 Cl= 15.3

rBA c2 6.2 1.127 = c3 = 5.5

Note that rBA is greater than one.

Both colunms of Figure 17 are identical and therefore the stiffness factors and the carry-over factors are identical The right column supports no direct load. It there­fore has no initial end moments. The initial end moments for the left column could be found by 'USing the influence lines in Figures 43(a), {b), (c), and (d), but, for the sake of

12

illustration, the general equations of case I in Figure 25(a) will be used.

For convenience, the column is divided into eight 1-foot long sections as shown in Figure 19. First, compute the Mx ordi-

Ix nates. This is done in Table II. The corres­ponding diagram and its area A0 = 445,630

lb. /ft. 3 and the distance to the centroid Xo = 4. 78 ft. are shown in Figure 20.

TABLE II M

· Calculation of ~ Diagr:-am

for Ex:ternal Loading

DISTANCE s:6~~ DEPTH OF Ix ~ POINT FROM SECTION FT.4 END"A" Mx FT.• D. FT. Ix

A 0 0 3.5 4.466 0

I 1.0 1050 2.5 1.628 645

2 2.0 2000 1.5 0.352 5,682

3 3.0 2750 0.67 00313 87,Bro

4 4.0 3200 0.67 0.0313 02,24C

5 5.0 3250 0.67 0.0313 103,830

6 6.0 2800 0.67 0.0313 89,46C

7 7.0 1750 0.67 0.0313 55,910

8 8.0 0 0.67 0.0313 0

II :89,460---- --------- --------------6

II : 55,910---~---- ----- ---------------7

II :0------------------- ---------------8 8

MX FIGURE 20 - T Diagram for External

Loading of Column of Figure 19.

Table III shows calculation of Mx dia­Ix

gram for unit moments applied at ends of column.

TABLE III

Calculation of i Di&gt"am for m.AB = l. and~= l. X

POINT Ix Mx FO~ Mx FOR MJr FOR ~FOR mAe= I maA= I

Ix Ix mAs-= I maA-= r

A 4.466 1.000 0.0 0.224 o.o I - I. 628 0.875 0.125 0.538 0.077

2 0.352 0.750 0.250 2.13 0.71

3 0.0313 0.625 0.375 20.00 12.00

4 0.0313 0.500 0.500 16.00 16.00

5 0.0313 0.375 0.625 12.00 20.00

6 0.0313 0.250 0.750 8.00 24.00

7 0.0313 0.125 0.875 4.00 28.00

8 0.0313 0.0 1.000 0.0 32.00

The Mx diagrams for mAB = 1 and for mBA= 1 are shown in Figures 21 and 22, lx respectively.

~ FIGURE 21 - I;" Diagram for m.AB= l..

13

M

II : 24.00

II : 28.00 II : 32.00

FIGURE 22 - ~ Diagram for mBA = l. •

Now, by equations of case I in Figure 25(a).

Ao{XB - Xo)

CAB = A A {XB - X A)

···K . AB

445,630(5.69 - 4. 78) 62.8{5.69 - 4.22)

4,390.ft.-lb.

Ao (Xo- XA)

AB(XB- XA)

= 445,630(4.78 - 4.22) 116.8{5.69 - 4.22)

1,450 ft.-lb.

A AX A 62.8 x 4.22 ABXB = 116.8 x 5.69 =

AB(L- XB)

AA{L- XA)

= 116.8,(8.00 - 52.629)) = 1.14 . 62.8 8.00 - 4.

XB E

0.40

AA(XB- XA)

5.69E = ~~~~~~ = 0.0616 E 62.8(5.69 - 4.22)

(L- XA) E

AB(XB- XA)

(8.00 - 4.22) E 116.8(5.69 - 4.22)

0.022 E

The values for carry-over and stiffness factors differ by less than 6 percent from those obtained previously by use of dia­grams. {See page 12 . )

The values of the coefficients C1, C2, and c3 may be obtained also by the equa­tions at too of Figure 25(a).

L XB . C 1 = Ic A A {XB - X A)

8.00 5.69 = 0.0313 62.8{5.69 - 4.22)

15.75

14

_L (~- XB)

Ic A~ {XB - XA)

8.00 (8.00 - 5.69) 0.0313 62.8(5.69 - 4.22)

6.40

L {L- XA)

Ic AB(XB .. XA)

8.00 (8.00 - 4.22) 0.0313 116.8(5.69 - 4.22)

5.63

Since accurate values for cl' c2, and c3 are now available, the initial end moments for any loading could be found by the second part of the equations of case I in Figure 25(a).

*C1, C2, and C3 as evaluated from curves . are generally too inaccurate to be used for finding initial end moments.

Ao [c1 -X~ (C1 + C2~ I~ = 445,630 [15.75- ::~ (15.75

J 0.0313 + 6.40)J 8.00 = 4,394ft.-lb.

CBA Ao [- C2 + ~ (C2 + C3~ I~

Given:

445,630 [- 6.40 + ::6~ (6.40

11 0.0313 + 5.63)J 8.00 = 1,378ft.-lb.

A member with two T-sections, as shown in Figure 23, loaded with a concentrated load of 1,000 pounds 6.0 feet from end B. The member is hinged at A and fixed at B.

SEC. A-A

FIGURE 23 - Beam with Two T-sections.

Required:

Initial end moment DBA

Stiffness factors K AB and KBA

Carry-over factor r AB' r BA = 0

Solution:

In order to find the ratio m necessary for the use of the design diagrams for mem­bers of variable sections, irregular sections must be changed into equivalent rectangular sections of the same moment of inertia. Further, for the curves to apply, the equiva­lent rectangular beam must be of the same shape as the beam shown on the diagram to be used. The width of the beam is chosen arbitrarily, but it must be constant for the same member. Therefore, the depths of the equivalent sections are proportional to the cube root of the inertias of the original sections, and

m =\3fTL VT; where IL is the largest value and Ic the smallest value of the moments of inertia of the beam.

By use of the diagram in Figure 50, the moment of inertia of the smaller section of the beam in this example is

b'h3 ~12

1.55 ft. 4

1.5 X 23 1.55 12

15

and of the larger section the inertia is

b'h3 ~ 12

1.5 X 33 1.51 12

5.10 ft. 4

Now by the diagrams in Figures 30, 31, and

32 for a=~= 0.4, and

m = ~ =1.49,

the slope deflection constants are

c1 4. 75

c2 3. 70

and

c3 = 11.8

Then, from case III in Figure 25(a),

c1 E~c = 4.75 1i~5 E

0.491 E

rAB c2 3.7o = o 78 C1 = 4.75 .

By the diagram in Figure 35(d), for

k = 6ig = 0.4, a = 0.4, and m = 1.49, the

moment at end B is

DBA = 'P PL = 0.224 X 1,000 X 15

= i 3,360 ft.-lb.

Given:

A member composed of two rectangular sections, as shown in Figure 24. Both sec­tions are 1. 25 feet wide.

[----- -Po•t I · -- -----.;.------ Po•t 2 - -------><

A -- ______ t, ______ l_ _______ -!a' _______ _j 8

FIGURE 24 - Beam w1 th Two Rectangular Sections.

Required:

Modulus of stiffness K in torsion.

Solution:

Referring to case VII in Figure 25(b), the stiffness of this beam in torsion is

1 KAB = ~A = _1_ + _1_

~1 Kt2

where Kt1 and Kt2 are t~e stiffnesses of~ parts 1 and 2, respectively. For part 1, h =· 2. 0 ft. and b = 1. 25 ft. Then,

h 2.0 1 6 b 1.25 = .

16

Now from the diagram in Figure 49, p = 0. 204. Take Poisson's ratio ~ for concrete to be 0. 25. Then,

~ b3hE L2(.1 + ~)

0 204 1.253 X 2.0 E • 10 X 2(1 + 0.25)

0.0319 E

For part 2, since h is always the larger side, h = 1. 25 ft. and b = 1. 0 ft. Then,

...!!. = 1. 25 - 1 25 b 1.0 - .

From the diagram in Figure 49, p = 0.175, and

13 x 1.25 E 0•175 10 X 2(1 + 0.25)

0.00875 E

The composite stiffness for the whole beam from A to B is then

1 1 1

0.0319 E + 0.00875 E

0.00687 E

Now if the modulus of elasticity E of concrete be taken as 432 x 106 pounds per square foot, and assuming that the beam w:ould not rupture, the actual moment re­quired to twist the end of this beam through an angle of one radian would be

KAB 0.00687 x 432 x 106

2,968,000 ft.-lb.

General slope- deflection equations r-

MAe= E(c [ c,eA + Cz9e - ( c, + Cz) t] + C~a

MsA = E[c [ Cz9A -t C38s- (Cz + C3) t] + CsA

C _ L x Xs C _ L x ( L- X.s) 1

- Ic AA ~Xs- XA), 2.- Ic AA (Xs- XA)'

METHOD OF END SUPPORT

VARIABLE MOME.NT OF INERTIA CONSTANT MOME.NT OF INERTIA

Any loadinq

A J:--ti--i B

I I

!<----- L ----~

CASE I

Any loadinq

A ~----Ll ~--~ B

CASE ll

Any loadinq

Ar U -1w ~- ----L -----~ "

CASE ill

D _ AoXo _ c· _ Cz C

SA- AaXs - aA ~ AB

r 0 r = AA XA = Cz BA = , I AB A B Xs C'

K _ Xs E _ C Eic. AB- AA (Xs- XA)- 1 L

K = J:l:_ = (C _ Ci) Elc BA As Xe 3 C, L

CAB= Ao(4--6 z) t CsA = Ao(-2+6 ~) ~

2 I rAB = 4 =y= reA

K _ 4-EI AB- L

K _ 4.EI BA- L.

DAB = Ao{3-3 2) t - c· 1 c· - A& -z: &A

r: -o· r - 1 AB - , BA- I

3EI KAe-= T

4E.I KeA=L

DsA = 3A. t· x t = Ce~tcAa

I reA : Q; rAB = z K

_ 4E.I AB- L

K _ 3EI

8A- L

~:~ C,, C2 and~ as evaluated from curves are qeneralfytooinaccurate to be used for computinq CAs and CeA

SHEET I OF 2 X- 0·1421

FIGURE 25(a) - General Equations for Initial End Moments, Carry-over Factors, and Stiffness Factors.

17

General slope- deflection equations :-

MA8 = ECc [c,eA + c1ea- (c, +Ctl t ]+ cAB

Mu = E.~c [cz eA + c3e8- (Ct+C3) t] + CsA

C L x X a . C LX L -Xe . ,= Ic · AA (X.-XA \, 2 = Ic AA ~-XA)'

METHOD OF E.ND SUPPORT

->f .. l\ :.C­. .iL.ll

B' .I ~

I I . 1 I I I

CASE N \At

VARIABLE MOMENT OF INERTIA CONSTANT MOMENT OF INERTIA

Stiffness and carry-over factors as for Case I :.Ct.>\. a·:· . _" ___ ., ~p

I I. I I I

I

...J I I I I

CASEV: . . _y_

I I I I I I I I

a~~(U' --f- ' : ~\ ..J : : t I t I I 1 . I

PX ·.: A CASE ID ~~>1

3EI ~ MAB = PL=- L2

PL3

~=--3EI

M _ PL- -3EI~ BA- - L 1

PL3

l:l. =- 3EI

Stiffness and carry-over factors in torsion. Rectangular sections only. Note: .. h" is the (arqer side.

CASE 1lii

If the beam has two or more rectan- b3 hE qu lor parts, the stiffness is" one.. KAa = KaA = /.3 L z(I+U) divided by the sum..of the reciprocal stiffness of each part. For values of /3 see

d iaqram on paqe 54. I

KAe = KsA = 1 1 - +- + ----Ktl Ku. rAB =reA= -1

fAB : reA = -1 SHEET ~ OF 2 X- D-1421

FIGURE 25 (b) - General Equations for Initial llhd. Moments, Carry-over Factors, and Stiffness Factors.

18

Beam fixed at both ends Left end fixed Left end hinqed f?iqht end hinqed f?iqht end fixed

p p

f-a -r----b----~ r-. a-r--.--b ----.. ~. A- y iBAi y ·.-8

-- ---- L - - - --->I F< ~--- -- L - - - ---

· Pab ( ) DAB= 2L2 b+L

/r< - - - - - a3- ---~ b3 1<1 - t< - - - - - a - - -·1 b ~-1 . ,3 3, /f<- - a -->t< - b - - -- ~ r< - - at ->t< ~ - bt - - > / -~ a, rr:R-~- -1 ~b,_z_- ~ ~ -:...; a,rrR- ---1 ~b.--- ;>~

A; v • v•2. yr3 1 B A 1 v 1 vr2 ·. vr!J ,B /f<- - - ---L ~ - - ---~ i<- -- ~ - -L -- - - -->

DA 8 = 2 ~2. [ Pab (b+L) D8A = 2~ I:Pob(a+L)

3 DAB = 16 p L

- I DAa-3PL

SHEET I OF 5

3Pa ( ) DBA= 2L a+c

X-D-1422

FIGURE 26(a) - Initial End MOments for Beams of Constant Moment of Inertia. ·

19

Left end fixed Beam fixed atbothends

Right end hinged p p p

, -'=-th_ t-'=-t1-4 4 . 4 4

(n-1) equal loads (n-l)equalloads

A +.· .....J.-.1~---L...L.....&........r/8 A 8

n equa I divisions W =Total load

CAs= CaA=(n~l) ,'2 WL

----- --L------- -->j n equal divisions W =Total load

DAa=(n~l) T WL

Left end hinged Right end fixed

{n-l)equalloads

n eq ua I divisions W =Toto I load D =(.!1±.!) .!. W L BA n 8

DaA-= k2J: yx (L-x)(2l-x)dx

3lL I DaA= 1.? M (L-x) dx .0

M1 =Simple beam moment M1=Simple beam moment

D _J.. W'2 AB- 8 '-

SHEET 20F5

D I 2 BA= 8 WL

FIGURE 26(b) - Initial l!hd Moments for Beams of Constant Moment of Inertia.

20

X-0-1422

Left end fixed Beam fixed at both ends

Right end hinged Left end hinged

Right end fixed

A ,~.,.."---------+' 8 A B A 8 - - - '- ""' - - L -- ..;. - - - >! - - - -·- - - L - - - - - --

A~~---~~B A 8 A 8 -------- L------~ k----..,.- -L-------

I

t<-a ->j I I

I a I

;- ->j w lb/ft ~~..............- I

I : ~~~~~

A-1--....-....~ ..... ~---r/8 A ·;--_..,.,.......,...._...__-¢1 8 A <>-1 ----!;li~..,...~---v 8 - - - - - - L - - - - - - ->1 f<------ L ------

CAs=CsA= I~L (~-6a2 Lt4cf) DAs=~L(L3-6a2 L+4a5 )

---- 0 I - - - ->J 'il2~ WI I b;tt

I I

~.......,~......,~...__-o;Q8 A ~~.....,.~~"----Y8 ------L ------->1

SHEET 3 OF 5

FIGURE 26(c) - Initial l!hd·Mom.ents for Beams of Constant Ma.ment of Inertia.

21

X-D-1422

Left end fixed Left end hinged Beam fixed at both ends Right end hinged Right end fixed

;t:.Y_ :da

1.-:~~~~::...~...--v B A ------L------

c =Wa2

(IOt.2-IOoL+3a2 ) AB 60L2

CeA= W a~ ( 5 L- 3 a ) 60~

------ L --~---Wa2 2 2

CA8 =3ol.? (10l.:-150L+6a )

C = w a3 ( 5 L- 4 a ) BA 20L2

-T ';3 _ _tN ~1-~~._~~~~-r

-- - - - --L- --- - -~ B

----a ----->l

wa2 2 2) DAe= 120~(40L.:-45aL+12a

SHEET 4 OF 5

8

D Wa2 2 2) BA = 30 L2 ( Sl..! - :3 °

D -~w'2 BA -64 ._

X-0-1422

FIGURE 26(d) - Initial End Moments for Beams of Constant Moment of Inertia.

22

Beam fixed at both ends

- a ->+<- - - - b - - -: W:: Total load

CAB= 3~t.?[2a2(a+4b)+3b2(4a +b)]

C8

A= 3~@[3a2(0+4b)t2b2(40+b)]

W=Totalload

CAs= CsA=l WL 48

>( 'i ~ ~

a A ~ ........... ........_......._.~~·-~8

----.-L----

Left end fixed Left end hinged Right end hinged Right end fixed

-a ->k- - - - b - - - ->1 k- a ->+<- - - - b - - - -~W=Totalload: l :w:Totalload

I I ~...._.~....,......_~~-oB A ~........_~......._. ........ ........,~-vB ------L-----~

A~B W =Toto I load

DsA = ~2 WL

IX

.Y~ 3

A ~f< ....... - .... _ ..... _ olio..-_ ... Lo....lo-....... _ .... _ .... _~_"¥--~ 8

2 2 2 W=Totalload = 3 Lw MAX. W=Totalload = 3Lw MAX. W=Total load= 3Lw MAX.

CAs= GsA= 16 WL DAB= io WL DBA= io WL

n equal divisions W =Total load

CAs= GsA= (I- .l2) j_ W L n 12

n equal divisions W =Total load

DAs={l-.l)lWL nz 8

n equal divisions W =Total load

I I D = ( I - n2 ) a W L

FOR TYPICAL CASES SEE SHEETS X-D-1424 AND 1425

SHEET 5 OF 5

FIGURE 26(e) - Initial End Mo.ments for Beams of Constant Mo.ment of Inertia.

23

X-D-1422

~ ..... 0 C/) L&J :J .J c{ ::;.

C!l.. u.. 0 (/)

w :::::> ...J ex: >

0.14

~ ~ rr v I ...........

r-..... / "' <-- -/- ~--'+ ' ~.333/ v ~- ~ --0.333-----0.333-- -·- --- -- 1--- ->

0.12 I )K \ 1/ 7 \ \

0.10

0.08

0.06

I I \ r\ I "7 ' \ I.

p ' I l <-aLl ~\ \ r1 I ~ ~-

-CAB

A y ~8 \

I 7 ~ \ \ l CeAp <-- -- L -- --~~ \ \ ~

0.04 I J \ \

I I CAB =/3 PL \ \ 0.02 I 7 CsA=f.3PL "' \

I / " \ 0

/....-': / '-.... •.•.. ...\ 0 o.r 0.2 0.3 0.4 0.15 0.6 0.7 0.8 0.9 1. 0

VALUES Of Q __.,- ......,....,..

0.18 / I ..... ~ I < -- --- - 0.4'2.3 -----; -- -- >\< --- ---

,_ -- - 0.517-- -- -- -- ->

0.16 v ' I ~

I \

0.14

7 i\ i

I ' x--DAe

0.12 v p \

I ~ f<-aL -~ \

0.10 I A

~ B \ I ~ f<-- --- L -- __ ,.. ' ~

0.08 \ I DAa =/3PL \

-· 1\

0.06 -- I \ - ~- \

0.04 I ~ I \

O.Ol I \

I \ I \ 0

0 0.1 0.2 0.3 0.4 0.5 0.6 OJ 0.8 0.9 1.0

VALUES OF Q X-0-1423

FIGURE 27 - Inf1uence Lines for Initial End Mo.ments-­Beam with Constant Moment of Inertia.

24

0.4

0.2

0.0 y \

0.2 \ \

~ 0.4 0 z <t 0 6 ~· I.L 0

0.8 (/)

lLI I :::> ..J <t 1.0 I >

1.2

1.4

0.2

~-0.0

1\ '\

0.2

0.4

(Q 0 z 0.6 <t

CQ I.L 0.8 0

I (/)

LIJ 1.0 :::> I

..J <t >

1.2

1.4

1.6

v ..... I> < ......... ~~ /

/~ 7 "' .......

' --- -- a3r3 --- ->rc /-- --- 0.~33 ----~ H<- --- - -q.333 -- ---I

.I . 2 .3/ .4 .5 .6 "".7 .a .9

/ VALUES OF a '\ / / I _J " o<.-OL-

\ J ~M Ra \ I \ ,{_cAB

A L>-

B ,'~ j ~ RA GsA-

\/ ~ ~- --- -L --- ---- -> 'v ;'\ ~

/ j', I \ CAs= /3M I \

\ R =/31 ~ I

\ CaA=/3M I \ I

\ I' '\ ~- RA =- Re /

/

"-.... ~/ ~ --............ ........

i'--... / 7 v ~ ~-DAB

/ v

v --- ---- 0.4,23 --- -- ---~

,/-_ --- -- --- 0.577 -- --- -- ---

.I .2 .3 7. .5 ·? .7 .8 .9

\ /v VALUES OF a

\ v /

/ <.-- aL ~

!'\ /v A •fi¥

\. / RA L:!-/' '

<-- --- L ---

1/ "~ I "~ DAe=/3M

/ "~ 1M

RA=/3 T

"'" "' v RA= -Re .... ,

"" ......... ......... !'....

............ ....._ ~

FIGURE 28 - Initial End Moments for an Applied Couple-­Beam with Constant Moment of Inertia.

25

~ j I • 0

I v

\ \

->-

L. 0

! T I I

--- ~

X-D-1424

0

I

: I

DEFLECTION DIAGRAM

------- Q = 0.33 to 0. o 7 · -----~

MOMENT

I I

I I

! I I I I I

MOMENT DIAGRAM

DIAGRAM I I I I I I I

! I I I I I I

---------------------q: 0.67to 1--------------------.:::>-j

DEFLECTION DIAGRAM

B

I I

FIGURE 29 - Mo.ment and Deflection Diagrams for an Applied Couple--Beam with Constant Mo.ment of Inertia.

26

B

X-D-1425

\3rrrr~TTTTTITIII~II,~~~~nr----------------------------------

' _y___A B

u I Ic 1------~~.----_=ofr -f- -: I . I • 1 :<--aL-->: : f<----------L --------->l ' 12 \ \

--' MAs= E Ec[ceA+C28a-(C,+Cz.) ~]+c"s

' ·; M6.: E [c [c2e.•c.e.-(C,• C3l t]+c •• ~

1\

\ l""o ' \. { rAB::: ~ ~1\+++4\~~· ~++~~~++~~·~~++~,rl-~++~~~~carry-over c·

\1\ r - ~ \ \ 1\ \ BA - C3

u \ \ :? t-+-+-1--J.-+---l--1

~ 9H\~~44~~~+4~~~+44-~~++~~~.~~~ I- \ \ i\ : (..-?,, ~. t-++-++4--+-iH---H-+4--+-iH-~++~~t-+++~~ z ~~~~~~~~+4--+-iH---H-+44-H-~++~~'~l~~++~~t-+++44~1~+-+444-~~++~~ w \ \\ ~~

u \ \ .... """'... +-t--t-f---1{~\fl\.ld---+~H-+-+-++~~+-+-++~~1-+-+++-+-H~ ~ l--+-l\'+-l1\H-'.I' ~~++44-+-t-+++44-+-+-+-+-""'~0 ~ A'-+=t~~s~t;t-t-t-t~-::t~t=tt-t-t~-::t~~tt-t-t-tj~~~ttt-tjj ~ 1--+~~~~~+-+-++44-+-1--+-+-+44-+-H-+-+++~-~~}+-+-++~~~+-+-++~~H-++++-+-H-+-+++~~~ 0 8~~~~~~,~~~++44~1--+-+-+44~H-+-++4~~~~+++4~~~+44-~~+4~~~++~4-~~ u ' 1\.

~ ~I\~++~IH-\~~~~~~~44~1--+-+++4-+-H-++++~~~++~-P'~~~~"~~~++~~~++~4-~~ :J \ 1\ ~ ~~ ..... ...J 1-t\: ,++-+-+---t-\:1-+--M-+-+-+--l~ .... +-+-++~-+ m = 3,-:. "

>~ 7 ~ 1\. v+44-~+-++-+-HH-~+41~~~M-I-+-~+-+-HH-~++44~

~~- ~~~

1\

1\ ~ .... .... ..... ~

~++~~,~++~~~~l""o~~ ... ~~~-m=ZS'~ +-+-+-+-+-t-++++-+-+-i-+--+-++'~ .... -+-f-P~~~~~~~~~~+-++-+~ \. ~

6~+++4-+-~++++~~~++~~H-+++44-~~+44-~+++ ... ~ ... ~~ ... H-++~~~~ .... bP~~~~~~~~ ' ' ~ ~ ~~ ' ' l""o.._ .... ~~

H-+++44-H--~'~,-+-+4-H--+-++'~~2-~-+--~--+-+-r·H-+++4+~-+--~~++~~t-+++'~ .... ~~,~:~~~~~~ ~ ~~

I I I I I - .....,, '

4-1-+-H-t-+-+-+--+-++++-++++-H-H-+ m = ,. )r 0.1 0.'2. 0.3 0.4 0.5

VALUES OF a

FIGURE 30 - Slope-deflection Coefficient c1-­Beam with Two Constant Sections.

27

0.6 0.1

X-D-1426

N u V)

1-z u.J <..)

LL LJ.. LaJ 0 u u.. 0

CJ)

uJ ';:) ...J ct >

00 ~

9

f{

4-l

I"' 3..,

v ~

--l..l

':Ill 8

7

6

5

"

~

? ~.I

j,.o ....

l..oo v

"""

0.'2.

A B

1:1 Ic [j I I I 1 I

I ~--a L- >l I I

!<----------- L - - - ------~

MAs- E Ec [c.eA + c2.ea-(C,+Cz) t~+CAB

MsA = E [c [cz8A+C38s-(C2+C3)~+CsA

""' 0 Sfff t•s = C, Ec E

1 ness I KaA=- C3LE

·~~ <' rs= ~· N ~ " ~ I' Carry-over c' I"~

r-.0''.:;: I' r BA = c~ .... ~ .. ' I"~ "'

""' ~

~ "

f!. 1 """ r""'loool

I' r-. ~."" r"'""

;... l"ooo

~'""" """ ;..... " rn_ •

rn::: l a

rr ,t.S

.fTl .l:v. 0.3 0.4- 0.5 0.6 0.1

VALUES OF a

X-D-1427

FIGURE 31 - Slope-deflection Coefficient c2-­Beam with Two Constant Sections.

28

10 9

00 00 00 8

1 00

6 ()()

5

(f)

u

4

3

00

00

00

00

cnl 1-z u.J

00

u LL. LL. LLl 0 u u. 0 \/) uJ ::J .J <(

>

90 80 10

60

50

,.,

30

zo

10 9 8

1

6

5

4! O.t 0.2

¥frc B

[} I I

~- -al -->i I I I

:.<------ ----- L ----- ----~

"0 MAs= E [c. [cleA+Cz9a-(C+CJt]+cAa ~ . 6'.

0~ s. L.-

MeA= E Ec:[cz.9A+C39a-{C2+C~~]+CaA .... " {K oC{<E '' Stiffness As ' 1 I

07::::4. ' KsA:: c3Lc E ..... -. " r -~2 .... ' I" ~

i"'- ~['. Carry-over AB- c, "''

t\_l'lo r - c2. ..... I' BA- (3

rn:::3-· - v --

m =zs-· l"ooo ..... ~ ..... '

i""' roo. ~-...

m-2·v

.... .... ..... .... 1-o. ......

"""""" .... l"""llllo. r""- ..... ~ ~ ..... ~ ~

m=t.s·~ ..... , .... ~~ ~ ..... """' ......... I"

r--~-~- """~

-

rn~= t-v. 0.3 0.4 0.5 0.6 0.1

VALUES OF 0

X· 0-1428

FIGURE 32 - Slope-deflection Coefficient C --Beam with Two Constant Sections. 3

29

.2.5

CQ LL

.2.0 a Ill' Ul ::I ..J <t >

.15

.10

co_ LL

.OS 0 Cf)

LLI ~ ..J ~

00 0.1 0.'2. 0.3

m= 4 ..

T3 .. (\\"'

m=rv

m = 2-~­m=-3·:, m- 4-"

0.4 0.5 0.6 VALUE.S OF' a

0.1 o.s

FIGURE 33 - Initial End Mo.ments, Unifor.mly Distributed Load-­Beam with Two Constant Sections.

30

0.9 1.0

X-D-1429

0.2

0.1

0 1.0

0.1

u.. 0 1/) uJ :::> .J

~

IJ... 0

Ia :::> ...J

o:a ~

0.

QS

0.2

tQ. u. 0

o.r 1./) u.J :::> ..J ct >

0 1.0

0.2 <o. LL 0 1./) LLI :J .J

0.3 ct >

0.4

0.5

0.9•

0.9

m=l''

0.8 o.1 H+t o.s 0.5 0.4 0.3 0.'2. 0.1

VALUES OFk

p

I r--kL --, ·f$=1 i I·

~<-aL --~ 1

~------~----L _________ J CAs= ~PL C8A=)3'PL

SHE.E.T I OF 4 x-o- t430

FIGURE 34la) - Inf~uence Lines for Initial End Moments-­Beam with Two Constant Seetions. a= 0.1.

m= , ...

0.8 0.1 O.G 0.5 0.4 0.3 0.2 0.1

VALUES OF K

"'!··.

m,;J! 07:::3··

p

I t-kL --, ·~ 1 IB

~-aL-->: I 1<---- .. ----- L ________ .....,

SHE.E.T 2. OF 4- X- 0-1430

FIGURE 34(b) - Influence Lines for Initial End Moments-­Beam with Two Constant Sections. a = 0.2.

31

0

0

J.l

0.1

0.

0.2

tQ u. 0 en u.l :::> .J ct >

u.. 0

t3 :::> .J

0.3 ~

0.5

0.1

u. 0 VI

0.1 uJ :::> -' c:x: >

0 1.0

0.1

0.2. CQ. u. 0

Vl uJ :::> ....1

1).3 ~

0.5

0.9

0.9

0.8 0.1 0.6 0.5 0.4 0.3 0.1 0.1

VALUES OF 1<.

rn,;-.

rn, 2·.

p

I r--kL -->:

Af$=1 1 le ~-aL-~ : ~ ----------L ----------!>"

CAs= ;3PL CaA= fj'PL

SI-IEE.'T 3 OF 4- X-D-14?10

FIGURE 34{c) - Influence Lines for Initial End Mo.men Beam with Two Constant Sections. a= Oo3.

m= /··

0.8 0.1

07"'.3··

0.6 0.5

VALUES OF k

0.4 0.3 0.'2. 0.1

A~ j Ia ~-al-~ : ~--------- L --------..l

CAa = ,BPL CaA = j3'PL

SHEET 4 OF 4 x-D-1430

FIGURE 34(d) - Influence Lines for Initial End Moments-­Beam with Two Constant Sections. a = 0.4o

32

0.6

0.5

0.4

co. u. 0 VI w 0.3 ::> _J

~ >

0.2

0.1

9.o 0.9

p

r--k L-~ i'

A ~--Hinqed ~ ···"> t f. 1

.~ Fixed-

jot---aL --->l ~----------- L -----:-

DsA==;SPL

~ .. ~· ;..?{>-~·

~tt\,. ·'

0.8 0.7 0.6 0.5 0.4 0.3 0.2. VALUE.S OF k

FIGURE 35(a) - Influence Lines for Initial End MOments DBA-­Beam with Two Constant Sections. a = O.l.

DAB

0.1

-:-..;>:.

=0

6

0

r--kl- ... 0.6---~Blp r i

0.4

0.3

0.2

co. u. 0

Ill Ul ::J _j

~ >

I ')··

rn"'"' H-+++-+-.IH-hof'+++-+++-H-t.lo"F--+ ~"' ,...

0.1 ___ _

q,o 0.9 0.8 0.1 Q6 ~5 Q4 0.3 0.2 VALUE.S OF k

FIGURE 35(b) - Influence Lines for Initial End Mo.ments DBA-­Beam with Two Constant Sections. a = 0.2.

33

0.1 0

0.4

CQ. u.. 0

0.9 VI uJ ::J .J c:(

>

o:z

0.9

A ~-·Hinqed ~ 1 ~ ! ~ Fixed----:.. 8

><---aL --->1 ~------------ L -- ------>

DsA = ~ PL DAB= 0

0.8 0.? 0.6 0.5 0.4 0.3 0.2 0.1 VALUES OF k

FIGURE 35(c) - Influence Lines for Initial End Moments DBA-­Beam with Two Constant Sections. a = 0.3.

0

o.s'BIE!EfEIHEffEIEIEIHIHEffEIEIBfHEffEIE±EIHEJEffEIHf ____________ -pp----~ r--kL-->

0.4

c:Q. u.. 0

0.3 ~

0.2

0.1

.0

::J ..J c:(

>

0.9

m=4··

m=3··

0.8 0.1 0.6 0.5 VALUES OF k

0.4. 0.3

A c}<--Hinoed ~ , B 1 ~ ! ~ Fixed----:..

~< --aL- ~ :.. -------- -- L ________ ...,

DaA = f3Pl DAB = 0

0.2 0.1

~IGURE 35(d) - Influence Lines for Initial End Moments DBA-­Beam with Two Constant Sections. a = 0.4.

34

0

+0.4

i-0.3

+0.2

+0.1

0

C!:l. -0.\

LL. 0 -o.z r.n u.J

3-0.3 <{

> -0.4-

-0.5

0.'2

<.:l ,,·

-0.3 .4 0.5 0.6 VA LUE.S OF a

NOTE

0.1 o.s

A~M r-r--1>-~ al ->i !<---------- L --

CAr.= /.3M

<>;

0.9

1 ,. ------ >{

-0.6

For an applied moment "M" as shown (counterclockwise) the + sign of the coefficient for CAs means that the mo­ment CAs tends to turn joint A, in a clockwise direction. Rever sa I of "M" reverses CAs

MAs = E. tc. [ c,eA + Cze!l- (C+ Cz) t1 + CAe

MsA = E. tc [ c2.eA + C.s8s-(Cz+(~ ~ + CsA -0.1

-o.s

-0.9

-().4

-0.3

-0.2

-\.0

FIGURE 36 - Initial End Mo.ment, CAB, for an Applied Couple at Change in Section--Beam with Two Constant Sections.

MAs= E. t [c,eA~ c2.es-(C,+Cz)t]+CAs

MsA= E. ~c[49A+C39a-(Cl+CJ~+C6A -0.'3

NOTE ~-o.s For an applied moment "M"as

shown (counterclockwise) the + siqn of the coefficient for C!IA means that the mo­mentCsA tends to turn joint B, in a clockwise direction. Reversal of "M" reverses ( BA·

0

:3-0.1 ::J -l :; -0.6

-0.5

;j S:

;5 ~

~- ;;,· ~ ~

9 ~

t.'-~

X-D-1432

1.0

'ill. -0.1 VAL a LL 0

~ 0

3 +0.1 ~ >

+0.2

+0.3

+0-4

+0.5

0.1 0.'2. .3 0.4 ~ 0.5 0.6 07++- 0.8 0.9 l.p

A,Ic f<t~ j ~e ~ al -~ , ~ - -------- . L-- - - --- --->l

CaA ::. /d'M

x-D-1433

FIGURE 37 - Initial End Mo.ment, CBA' for an Applied Couple at Change in Section--Beam with Two Constant Sections.

35

3'2

30

28

J''Z.6

ttl 112.4 <.)t.J

o E2'2

~ g2o .0

u_ts LL.. s 16 0 'i:

-+-C/) Ql 14 LLI E ::::> E 12. _) ~ <t: 0 10 > I-

& 8

6

4

0.0

I 30

28 j_ I

L I

?/;,' -<-

~~-= ~ ~-II '' J Ej r:-t l1 'o ~

1 ('v .f-

1/ I"~ j_ ~~ Ill/

~ IL ~'\· I / ~ /~ -'--'- .i....t. v 'l.;·

if L v~ lL L'f'. / // ~..~~ iL

..1 ~~ v ""I.--~ /~ ~ ~~

~~ ~ .... I--' - m~ 1---"~ ~~ - -~ m-=L'2__::: ~ ~- ...... .....

m = 1.0-

2.6

24

22

'2.0 N

c.) 18

u.. 0 16 IJ) 14 uJ :J 12 _J c:!

10 > 8

6

4

2

IL

J

~~~ r-

€~r t-

"~ t-Ill ~

/I I y <"\,~ t-

1/,;~ I j <S: ~'\..·

v~~ 1/ I ) vv~ o-

I' v~ vv v:,.r v v:~ " ~{1\~ ~~ ~~ ~ ,~

~

~~ 1--'~ ..... m-=1.4...,

m=l.2-~ ~ ..... - m= 1.0-

0.1 0.1 0.3 0.4 0.0 0.1 0.'2. 0.3 0.4 VALUES OF a VALUES OF a

A c=;:--l,..._, -===~Ic_t __ -_-_-__,-N~~] I I I 1

~--a L -->l ~ --aL --~ I I

~------ - ------- - L --- ---- ----- -- ->l

FIGURE 38 - Slope-deflection Coefficients--Beam with Symmetrical Straight Haunches.

36

x- o-1434

0.116

0.114

0.112

'0.110

0.108

0.106 /0

/ ~ .r ~ / '/

/ ./ v ~ / / /

t/ ~ / ./

v / ,.-

./ CQ. 0.104

LL 0.102 0

0.100

/.~ ~ ~ / ,.,. ...-/

Ul w :J 0.098 ...J ct 0.096

> 0.094

~

A~

~ / v::v ~~ ~ ~ v

/~ ~ v v v

~ ~ v v v v

? / / / ,_.,-

v v , v ~

, / ~ _....

/ ~ -.,... '/

/I"""

~ ---~ ~ 0.092

0.090

0.088

0.086

0.084

~ ~ ~ V' ---~~-lL: ~ v ~ ~ ~ v ._

o.oazo.o

" ....... ~

0.1

I I I I I

~ ~

I<-a L ->l ~-a L ->1 k---------- L ------------~

0.2 0.3

VALUES OF a

- m ::3 s--.:.. ·.,,

v / IT] ~3. b- .........

v ~ m~ c .,..,...-- m~~ ~ .,..,...- ~ ~ m.::: 2.4-, ~ ...- m==2.?:t b

~£~ r--.. -~ I--

I~ IT7::::18 ~ ~ ~

m::::,61 ~ ~--"' rot I--

r--.. m-Il,. - . --~ -I I .I m = 1.2-~

m = 1.0 -, 'i'

0.4 0.5

x~ o~t435

FIGURE 39 - Initial End Moments, Uniformly Distributed Load-­Beam with S;ymmetrical Straight Haunches.

37

C!l c.-

0.16

0.11

0 0.08 VI Q)

::J

;g

h ~~

jj '/ /II

~r; 0.04

f 0.0 ' 0.01

1.0

0.'2.4

O.'ZO

0.16

CQ c.-0

Ul0.12 Q) ::J

~ 0.08

0.9

II v~

!J. v II

A' 0.04

I v 0.0 1.0 0.9

- ~ m= ~.6

1/ v - ....... ~ v -....... m= 1.8

~ ~ ~ R ~ R': 1.4 -~ -r--... r: / I' ~ ~m= 1.0

/ "~~ ~

' ~ 'g m=

m= ~.6· ~ ~ m= /.8 ~

0.8 0.1 0.6 0.5 0.4 0.3 0.'2. Values of k

VALUES OF f3 WHEN a=O.IO

'/ - ~ m= 2.6 , ..

/ / --~ " .. -rn= /.8

I '/ - ........ ~"\ m= /.4 v --.. ~

v~ '/ - ""' ~ ~ l/ r-- r--..... m::. 1.0

~ / !'-. ~ ~ v ""' ~ ~

~

'~ ~ ~ m= m= ~6·~ ~ ~

m:. l8 ~

0.8 0.1 0.6 0.5 0.4 0.3 0.'2. Values of k

YALUE.S OF {3 WHEN a: 0. '2.0

p r- kl ->! j---p- 'j'Ft Initial E.nd Moment =-CA.s=/3 PL

-u A __:--o~ -_-· 6 E ' • x-- : ! ' !<-aL .,..: ~-cl . .,..,

I t

k------- L -------~

. SHEET I OF 2

/.0

m= 1.4

~ t:::---

0.1

/.o

m= 1-4

~ 0.1

FIGURE 4o(a) - Influence Lines for Initial :End Moments-­Beam with S~trical Straight Haunches.

38

0

0

X-D-1436

0.1.4

0.'2.0

0.16

C!l.. ~ lllO.I'Z. G) :J ;g

0.08

0.04

A &

!J. 'f/ /It v

~ r u j,

I

k-': ::......... I!. m::

/; ~ ............ ~ w~ ;; / ~ ......:: ~ ~ /

~ --.... ......... ~

'/ - '<: ,.,.,- -..... v .... ,

~.6

m• ~2

-m -1.8

~ m: 1.4

~~ ~~ m= /.0

.t'>\ ~ " ~ ~ ~

~ ~m:; 1.0

~ .. -m= 1.4

m: ~ ~.2.: ~ ~ m= 1.8

v o.o 1.0

m•'l.6-~~~ 0.9 o.e 0.1 0.6 o.s 0.4 0.3 O.'Z. 0.1

Values of k VALUES OF~ WHEN a:: 0.30

0.1 0.6 o.s 0.4 0.3 O.'Z. o.l Values of k

VALUE.S OF f3 WHE.H a~0.40

1 P~-~kL ~ Initial

1---A t?u~~E-;:~ B _A____ : : · :<QL ~ ~ol >l

I I :<------ L ------>,

SHEET 2 OF 2

FIGURE 40(b) - Influence Lines for Initial End MOments-­Beam w1 th Symmetrical Straight Haunches.

39

0

0

X-0-1436

-70

60 .....

so

40

~~- ~.- m~~, ... . m=- .4

8

7

5

4 0 0.1 O.'Z. 0.3

~<-md -~ : -~~:~~:i-~<t_ of connecting A 1

: member .C. I o:

_____ }'_ I

..c

0.4 0.5 0.6 VALUES OF a

SHEET I OF !

0.1

m .3_

0.8

FIGURE 4l(a) - Slope-deflection Coefficient c1-­Beam with Straight Haunch at One End.

40

0.9 1.0

N u u. 0

(.{) uJ ::J _j

~

40 __ _

30 ,I

10

9

8

1

6

5

4

0.'2.

I

0. 3 0.4. o.s 0.6 VALUE.S OF a

SHEET 2 OF 3

fT) :LE

.m= .4 v

I I I

0.1 0.8 0.9

FIGURE 4l(b) - Slope-deflection Coefficient c2-­Beam with Straight Haunch at One End.

41

1.0

x- o-1437

\8

11

16

15

14

\3

tJ u. 12 0 (/)

ul \\ :::> .J <{

10 >

9

8

7

6

, L..<

"""' 5 ~~

""' io"': [; .....

~ I""

I' .....

4 0 0.\ o:z. 0.3

r<-m d -::; __ ,. . I

1--K :i· ~ of connect1 nq

I

I ->t

I A ' 1 member .r::,l a: _____ -{ I

..c.

f'

..; 1.1

~

..; ..... ..; ~

""' """ , ""' --'""~""'

""" -~""' I;' ,.-r"' -~""' ..... -~ .... ..-

-~---

0.4 0.5 0.6 VALUES OF a

SHEET 3 OF 3

0-(J;)-i

El I

L\ I

1/

1

!/ I

I I

1/ f' I

I l

~ rlc,-1-1/ ~-il :~

(!' ~ r'? t ~ .- r- II ..; , ..; f~'-t

)I' I -

T ~

""" v

""' / II'

..-~'\. ~ .. /~f--7

,~~ £ , .... -~~; -~ """ ~'1., \· ..- -~' ~ ..- , •• ,L.

""~,~ ----- ~--~--- -~""' ·~~ lv I I I I I 1 t 1 l I

0.1 0.8 0.9 1.0

X- D-1437

FIGURE 4l(c) - Slope-deflection Coefficient c3-­Beam with Straight Haunch at One End.

42

I

0.2. I

0.20

O.f9

0.18

0.17

0.16

0.15

~ 0.14

u. 0 0.13 (/) LL.l ::J 0.!2 _J

<! ::;- 0.1 I

I I I

0.10

0.09

{------: 0.08 I

~0.01 11. 0

(/) uJ

0.06

:J 0.05 .J c::r > 0.04

/) !IJ

/'/~ /i v

IF ~ ~ ~

'

J

I II il

II If

// / II / 'I /

/ / ~/

~

r-..... ~ .......... r--~ ......... r-...... ~ ~

0.1 0.'2.

;..:..md-~ I I

I I 1

L-r--~--

...c I

C5 : _______ '{_ :

---.... / ...........

v ' / '\ ,-.. ~.,/ ' ~'Y " i)C- ~~Y>

, ......... ,..........

-'/y ' "" v ~ "' "\ I 7

' /

I / ~ ....... J ............

/ ' o' ~~,

~-:-~· -..... ~-- ......... / ............ .........

/ ............. -- ...........

ro -;.I. s ··v -~- --.. -- -- --

m =- 1.0·~·, I I

1--. m =- 1.5 ··v I - -lo...

.......... r-- ~__:2.o·-v r-... r---" ............ !!1~3a-- -.......

........... '~ 1--..

'~o- -,.., r-- r--... I ~ ~ ....... -~

I .... ......_ 0.3 0.4 0.5 0.6

VALUES OF a 0.1 0.8 0.9 1.0

lnitiat end moment CAe = AWh2

Initio I end moment CaA = j3'Wh2

~ ..c

'~ I

-m,?7;;tn~, . - - - - - - - _y-

FIGURE 42 - Initial End Moments, Uniformly Distributed Load-­Beam with Straight Haunch at One End.

43

X- D-1438

\

I

I I I

co.. u. 0

ell u.J ::> ...J <( ::>

I I I

--t--I

C!?.. I.L. 0

ell uJ :::> ...J <( >

10 09 08 01 06 05 04 0~ 01. 01 00 0;3'2

0.:30

0.'2.8

O.'l6 [J...: -,

~ t:::=-r-.. ~0~ ~"~o A ~ ~ "'"- ~~/ --~

~ ~..._;.; ~rr;:t~ I~ - ~-~~~ ./ -

~ v/ lo"' N'' ~ ''<: 1(/j - J ~0~/ J"-~'~ 0 __, --

~ v v ~ "~\~ ~

"""' ~ v ' ~ " ~~ !J v "' "' ~ 1/ " ."\ ~ ~ " ~ ~ j

I ........

~ v ~ b...

0.'2.4-

0.2.2

0.'2.0

0.!8

0.16

0.\4

0.1'2.

O.lO

0.08

0.06

0.04

0.02

0.00

!": ~ ~ ~ I """ ~ ~ ~ ~' IJ

' ' R ~a ,-·1/J ... j ~ .~ . ..... < If -'0~ ~·S

"'< ~~~ ~"0~;0 ~

·~~ ~~~ ~¥r ~, ~'-..... r-.... ~

...... ~ ~

0.02.

0.04

0.06

0.08

0. tO

o.tz 0.14

0·16to 0.9 0.8 0.1 0.6 0.5 0.4 0.3 0.'2. 0.1 0.0

VALUES OF k

Initial end moment CAB = ,BPL lnitia I end moment CsA ::. ,B'PL

)'_ ______ ·~~~

B SHEET I OF 5

FIGURE 43(a) - Influence Lines for Initial End Moments-­Beam with Straight Haunch at One End. a= 0.2.

44

I

I I I I I I I I I

t!l_ LJ... 0 lf)

LJ.J :J _J

<( ~

I

I I

-*--

~ u.. 0

(/) uJ :J ..J

§

I 0 0 9 0 8 01 0.6 0.5 0.4. 0.3 0.'2.. 0.1 0.0 0. 3 2 ~~...-..,.........::.r--r---r--r---r----r--~-r----r--~~---r-~---r----r----r----,

0. 3 0 1---~+--+--+---+---1--+-+-+-+--+---+--+--+--+--+--+---+--+--i

0. '2. 8 1---~+--+--+---+---l--+-+-+-+--+---+--+--+--+--+--+--+--+----1

0.2.6

0.'2.4

0.22

0.2.0

o. 18

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00

0.02

0.04

0.06

0.08

0.10

0.12

'~~~~~ /~ 1---t---+---t---+--t---+---+--~~ ~ ~s I '~... .../th'

~,r---..~ --~7 0. 14 l---+--+--+--+--+---l--+-+--+--+-~ ......... --+--+---ll...,...oo"'i----::::3~1--+--~+--l

0. I 6 L__L.._.L__.L__...l.....-..L.-..L.-....L..-...J...._-'---'--.....I...--1....-......L-......L----L----L----L....--L...--'----J

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 VALU E.S OF k

Initial end moment CAs= j3PL tn i tia I end moment CaA = f3 PL

y_- -/_:/,%-~~~

B SHEET 20F 5

FIGURE 43(b) - Influence Lines for Initial End Moments-­Beam with Straight Haunch a.t One End. a= 0.25.

45

X·D-1439

0.321.0 0.9 0.8 0.1 0.6 o.s 0 4 0 3 0 2 0. t 0

0.30

0.2.8

0.'2.6

0.2.4

0.1'2.

0.2.0

0.18

0.16

CXl. 0.14

LL. 0.12. 0 lf) 0.10 uJ ~ 0.08

§ 0.06

: 0.04 I I : o.oz

-l-- o.o--1

: 0.02. I I

- 1 0.04 C!l. ..,_ 0.06 0 (/) 0.08 UJ ::J 0.10 .;..1

~ 0.12

0.14

~ lj,

II '/. II! /

;/1/ !IJ Iff

!J {

~ ::::::::

"

,~V ..._~

I r\-?t; / [....--r--....... ~~

v / "' ~T \ ~

I v ~ \ o.

'I / ~ ~0" N \

/ 1/V Ks-.. " 7 - ~........_...._ ~ _.--

/ ..... ~01 ~ ~/

~-

' ::::::-- ..._..._

t' ~ ......... ~~ "'-

"~ ~ ~ " ~ "'-'~ " ~ r--..~

"' ~~o .. (?) .. ·O.N I': '/ ~ ~~ ~

')(g .~r---.... ""'-

\

\ \ \ \ \

" \ l

" ~ " ',\ \ " ......._\ ~\ '~ ~ ~ ~ :::.....

I )f

IJ A'

~ w ~ 1-- ..---; ~ r--.._ 'i ~ r---- -~ ~

o.rs~,0 0.9 0.8 0.1 0.6 0.5 0.4 0.3 0.'2. 0.1 0

}-

' I I I I I I I I I I

..J

I I I I

A a= 0.35

0

VALUES OF k

Initial end moment CAs = ~PL Initial end moment CaA = /3'PL

t----~~~ B

SHEET 3 OF 5

FIGURE 4J(c) - Influence Lines for Initial End Moments-­Beam with Straight Haunch at One End. a= 0.35.

46

X-D-1439

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.'2. 0.1 0.0 0.3'2.

0.30

0.'2.8

0.1.6

0.'2.4

0.'2.2

0.'2.0

0.18

0.!6

0.14

CQ_ 0.1 'l lJ..

0.10 0

V) 0.08 u..J ::::> 0.06 ...J <( 0.04 >

I 0.0'2 I

I

-i--- 0.00 I

(Q u. 0 (/)

IJ..I ::J ...J <( :>

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

~

~ h v

)(/, v 1/ v

~'I v ~ !J '/

(// 1-i.o""""

~ r/ /

~ / A v

If J'

L I

"'-' l"':: ~ ~ """"' "' ' ~ ~ '-

""' ~ i'-

"'

..... ~ ' ~

~ -,. ~ ~ J ~ ~ ,--..... ~~, ·.$'

'~~~\ \ '\rO \'~ - K'

--~ 21~ \ \ ,\ "/

" \\ ~.s -.... ~ " '" ~' ~\ ~~/)')~/ '\ \ ~ ,p 1'\ :" ~ ~

'" 1\. '~~ ~ ",\ ~

' ' ~ ~ ' ' ~ ~~ ~

j ::::::~,A0~2 I '="~7 ., ) A '" '- ~ 0 I

''~, ,

jf I

~~ ~ ~V/

' ~ lo... ~~ ~~. ·S_ ' ~

·;~I ......._ .., . ~ 'i'-.. ~ r-

~ ,

........... -r--

1.0 0.9 0.8 0.1 0.6 0.5 0.4 0.3 0.1 0.1 0.0.

I I I

A a == 0.40

: -,-P I ~ I .:::£. I I

y_- -~~~--Y. B

VALUES OF k.

Initial end moment CAs =;BPL Initial end moment CsA =)3'PL

SHEET 4 OF 5

FIGURE 43(d) - Influence Lines for Initial End Moments-­Beam with Straight Haunch at One End. a = 0.40.

47

X- 0·1439

I I

_.J

I I I I I I I

-CQ u.. 0

V'J w :::> _j c:{

>

A a= 0.5

·Initial end moment CAs = ~ PL Initial end moment CaA = f3'PL

v..-----~~w B SHEET 5--0-f:-5

FIGURE 43(e) - Influence Lines·for Initial End Moments-­Beam with Straight Haunch at One End. a = 0.5.

48

X- 0-1439

1.1. 0

1.1) w ::) ..J <(

> I I I I I I

1.0 0.9 0.32

0.8 0.1 0.6 0.5 0.4 0.3 0.2 0.1

0.30

0.28

0.26

0.24

0,2.2

0.20

0.18

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

~ A~

& I'

J'r II

l' I

b h '// f/ / 7 !...."'

/

-/"

/ /"" r--...... 7 --/ ......

"' ,..- -"' ...... ' - ._

...... !'... ......

'

'< ~\~ ~ r-.'R.~

....... ' '5(

K.~ ·o

~ ~ ~',,T ~ ' \ ~· ,~~1\ \

.......

~~ 1\ \ ~ '<:~ ' .s \ \ ~~~ " \ \

<o " '\ ' \ \\ ....

"' \

'" .\'\ ........

~~ ~ ~ -.......::

0

I ~ ~ -1

-l- --0.0-

CQ.

LL 0

V)

IJ.I ::) ....J <(

>

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

"'r-:::::: "

~- .........;;;:

~ ~

'" ~ -r-...... ~

~

~"..::: ~ r--.. r-........ ...........

!'... ~ ~ r-..... ....

!'-.. ""' ~ " ..........

~

' ~ "'!'-..

,~ ... If ~ ~0 ~ ..,. ~ ~ w -' ~z.o v ~

K r:.:.m. = I. 5 """""' ~~

~-I I ./~

~m =I .g_. I.T

1.0 0.9 0.8 0.1 0.6 0.5 0.4 0.3 0.2. 0.1 0

A

->~ d 1<-

VALUES OF k

Initial end moment CA 8 =,8PL Initial end moment CaA=f3'PL

FIGURE 44 - Influence Lines for Initial End Moments-­Beam with Full Straight Haunch.

49

x- o-.1440

45

40

35

30

'2.5

2.0

(f)

u 15 a z <t 10

N u

u Ll.. 0 (/) w :J _l <l: >

9

8

1

6

5

4. ~ 3

2 __....,-

0 1.0

/ / ---~ ~ ~

1.2.

{

r _ ~2 lAB -

Carry-over ~· reA=~

I I

/ I

_,_..,.... -~ ~

1.4

/

./ /

./ v C( ~ ~

~ .,..,...

~

~

,_..,.... ::::::1 / ,_..,....

~ >---____...

/" /

/ ____... .....-- /"" .Ct' ·v

~ ~

~

1.6 1.8 z.o '2.2 2.4 VALUES OF m

FIGURE 45 - Slope-deflection Coefficients--Beam with Full Straight Haunch.

50

v

~

2.6

X- D-1441

AS

46 44

42 40

38 36

34 -32 u

30 u. 0 zs (/) ~6

~ ?.4 .J 22 <(

> ?.0 18

16 14

12

10

8

6

L..o

"""

"""

.... a.-1-"" ....

~

~ v ~

1.-" ~

'"" i,.ool""'

~

""" l...ol-

""""" """

l.olo- ....

I

I

I

II

II

IJ'

~C) 0)

II

o-, 1/

'\,<, ll' ~0;.

0./

1...1'

..,Q~ ""' o:...r·1 j,;f I I I

I l I I I I \S

""" a-:::.0~ ~II

n~ojg .... ~~o~os .... o=O.O

I I

tO

50 48

46 44 4'2.

40

38

36 34

u 32

1.1.. 30 b 2.8

(/) 2.6

~ Z4 _J 22 cd: > 2.0

18

16 14

12 10

L..o

l,.oo """ L.oofoo"'""'

""'~ .:::;;;; ...

v -;

L.; l.i l.i

, ....

" r.,.. l..,, l..,,

""' ~ ""' ""' V'l~ !;;;; [... J;;..o

v [., ~ ""' I"""L.. 1.-" 1.-" .... c;o. l,.oo ~ ~r;.

I

I II

11 II

~~

1/

I I !/

J 1/ rr

IJ IJ' II

~ j ,., (;)?/ rr ,, ~

~~ ,r~ i) (;::)'> 1-~ 1-

7p:l) (/ ~11 o)~ \)~ J-

1/ Ll q ~ l.i 0:~

~ ~· l/~ ~ L;l 0 ~ v .-.,

v [...) 1-1-

I..., [.,

""'I....

4 0.05 0.10 0.15 0.'2.0 0.'2.5 0.'30

8 6 4

44

42

40

38

36 34

32 30

t,)'2.8 2G u.

0 24

(/) 22 uJ :J 20 -j_ I 8 > 16

14

1?. 10

8 6

4

~ j.,oo

t.... ~

"""' """ """ 1-'

'"" '-"""

VALUES OF b

1/

I

1/

r$?1 o· 117

'(I

~ J

,., 1/

~.I' II' Cv1

1.1 ""ill I ~ o.i'l t J

I~ l.il '\..() II" II' ~<¥1' I~ ~

""" ~II

v I""' V'1 ... 'c~ ~ ~

"' """' ()y

......... i.o" ~~o.\C ~ 1.-" _,I"" t....l""

"""' ~)~~ ~ ... ~ '-"""" l,.ooi""" l,.oo .... ~:::.o.o l...oi-'1"""

~ ... io-1-' a.-~-,

~ '-""' I-I"" I I I

I

:I

IL

IJ'

r.,..

o.os 0.10 0.15 0.20 0.2.5 0.30 VALUE.S OF b

B --.!.---! -----

\ c ...&....4-1---'-. • -I~ I n f i n i ty --- ·..L.-~'--'t.-......_

MAP,= ECc [ c,eA + c~es -(C,+Cz) ~]+cAs

MsA = E.Cc [ Cz.9A +C39s -(Cz +C3) t ]+CsA

St 'ff { KAa = C, -fc E. t ness I

KsA = C3T E.

{

fA& =.9. Carry-over ~

rsA : C3

I I ~.OS 0.10 0.15 0.'20 O.'Z.S 0.30

VALUES OF b

FIGURE 46 - Slope-deflection Coefficients--Beam.with Infinitely Deep Section at Both Ends.

51

X- 0·1442

100----~~~--~--~~--~~--~--~~-~,--~~

90~~--~~--~--~-+--+--4--~--~-++1-+--4-~ I

80--

10 r----

60--

sor---

40....__ -

30~

-28-

-

A[fQ) ·T1=t; ~

I I I I :

I ' i<-- d> I ~

.c:-t: I • S: I (+...

11.! : c::

: t ____ y_ ~

Bl

c=fi

II I

I 1

I v

I I

/

/ I I

I rS ?.6- I I .. 0 z~r--r--+-~--,_--~-+--~~'---r--~-+-+1+--;--~

I I u I I

22~-+--+--4---r--r--+--~-4---r--r--++-4-~--~ ~ I I 0 J I ~ 20r--r--+-~--,_--~-+~,~-;---r--~-Jr--+--,_~

~ I I ~ 18r-~--~-+--+-~--,_,~~~--~-r~,~--+--+--~

~ ,~1 I 16~~--+--4--~--~:~j+--+--4---~-j~-+--+--4--~

<..Jj I 14r--r--+--+--+-~-~~r--r--r--+/~+--+--4--;--~

t2~-+--+--4--~~~v--+--4--~1~~d~~--+--+--4-~ / ,~ 'l v ~v 10~-+--+--;--~~~-r--+-~--/~--~-+--+~--~~--~

ar-~--r-~~~-r--+--+--~~~+ +-=~~~+-~--+-~ /~ / .. c~_v-

·--- --- +----l---------+-·-+--4-----1~~--+---+--4-----1~~--+----f

0.6

FIGURE 47 - Slope-deflection Coefficients--Beam with Infinitely Deep Section at One End.

52

0.7

X- D-1443

I I J I I I I VJ/.

:::~ It);)' It/ ,

S",+ /0 f-I J I '?' 1-'\ ~~ I It"~ f-

i/ ""' ... , "\.,~~ , ~!~--~-v

/""!: 'I I I "' .,. ~ ~ /

""" ~~-r~ -f-

""' ""ij

-f-

~ .... - I I -~-;:;a:~ .... ,....-

,;..o]!:=;; ....... - {11:::\.S -r-

- m= 1.0 --f-I I I I I I I

0.1 0.2 0.3 0.4 0.5 VALUES OF Q

• ~------1 L/-i u~ j Ic ~ :::;j __ ~_J ~--al-~ ~--al -->l I I ~-<------------- L -------------.>~

A

MAs= Ecc [cleA+C2 9s-(CI+Cz)~]+CAs MsA = E~c [Cz9A+C39s-(Cz+C3) t]+CsA Ic =Moment of inertia at <E

St:ff {KAs:: cl Ec E 1 ness K _ C .k E

BA- 3 L

Carry-over{rAs: ~ rBA - Cs

FIGURE 48 - Slope-deflection Coefficients-~Beam with Symmetrical Parabolic Haunches.

53

/

1/

X·D-1444

3. 0

2.. 5

0

I. 5

Fixed end.

~ ~.,

~/

.11J..1----------~----1i

I I I

' I I

h

Mody(us of Stiffness "Kt" in torsion eq~.~als the moment required tc twist the section where the r."~ment is ap­plied thro1.1qh Qn angle of one rodian.

K .t.:~~G t"'~ "'Anqle of twist per t ~-' L b h G/3 unit lenqth

I

/ /

}--------------I

,~--~ __ )-

. . . E G=Modui!As of Elast1c1ty 1n sheQr" '2.(ITJ..I.)

1 1'-.. :Q ~---~---------L-----------~ ~-

E•Youn9's Modulus of Elasticity ~" Poissons ratio. h ·The larger side Mt • Torsional moment

4.1) 5.0 &.0 Valuer. of~

1.0 8.0

TORSIONAL STiffNESS OF RECTANGULAR BEAMS

9.0 10.0

X-D-144-5

FIGURE 49 - Torsional Stiffness of Rectangular Beams.

,. ....... I ,.~"'

S'[W"" ~~B~O- , ~

~~~~ 1--,...... o.A: 1---r-~I--""'

_ ....... r-"" 0.4-'jC 1-- ....

....... ~--""' ~~ ----f-"" 0.~·-.'i.

............ ,......1- -1-- -~ ..... I"""' 0 2-· --1---

~"""" 1- ....... ,. -...... ====~=""' - d::"tT

....... ~--"" -~ -""'"Ls-=O.I·· ~~ ~:;:;;--- -~- h

k;::; ~~F"" -~----~~~ ,.~ f-""

~~ _,.f.-~~ ,......r-""

~'r' ....... ~-""" v v v

~ 1.1.0 6.0 7.0 10.0 11.0 z.o 8.0 9.0 3.0 4.0 5.0 12.0

I*---- b ---- ->i

~-1:;-~

-;_~~---~Centroidal axis

b'h~ I = ..8 IZ:""

Where /3= I+(A-1)83 + 3(\-B)z. B(A-1) t·~B(A-1)

VALUE.S OF ~=A

FIGURE 50 - Moment of Inertia of T-seotions about the Centroidal Axis.

54

X-D-1446

I I I I I I ~-<-! L->~ I 3 I I< ------L --- -- ->-~

A=! ML

I I 1 I rc=- K L ->+<1-(1 ~K) l:-~ ri(I+K)L>+<! (2-K)L~ t<------L ------>-~

A=! ML

M .. 'M .. '

I 1 I I 1 1 I I ' I I I ' I ·z I I I J(2M + I l(2M+M~' 1<-- L ->t<--.; L ----->I ~ 1 3 ~ I· I I

I 3(M+M 3(M+M' I t<---~ L --->+<-- L~ I I I ~ 3 I 1<------L ------>~ 1<------L ------->-~ A =l (M+M') L A= 1 ML+l M'L

I' .......

y-4 '~,, ~~----~--'':::::.. .....

I 1 I I I I 1 I I I I I I<- - b - ->I I I I I : r<-~---- a----->l I '<J_ >1 I

! :,_ (a-b)(o + 2 b) ! : 3(a+b) : I I 1<------ L ------~

y(a2-b2) A= 2L

*ur =Load per unit lenqth

I I

I I I I

--L--~-'-1 I I I : I I I I I

1< - - --: - a -- >+< - b ->~ I I I 1 I I I I I ,, 'I I 1<-- a ->+<- - a ->+c; b*J b>i I 2 . 1 ~ 2 1 f I I I t< - - - - - - L - - - -- ->-~

A = ura a 'rr A = urb 3

1 12

I I I I I ~<KL -,->+<--~-(1-K)L->1 I_, I I 2 I ~KL>~<-lL--~(1-K) L>~ 1 3 3 3 1

t< --- --- L ----- ->t

A=~ M K L + ~ M ( l-K) L

MBM' I ljll i ~-<- a ->K - b .- >+<- c ->-~ I I I 1 ~<-a +L b>~<--c + 2 b ->-~ 1 .;, I '! I

~-a+ J b -->4<-c +i b->: ~-<--- --- L---- ---->-~

A=! Mb + 1 M'b

13 I I ->4-x~<- 1

110 ·I t<3 X>-~ I 5 I I<-- X- ->-l

X· D-1447

FIGURE 51 - Areas and Centroids of Moment Diagrsms.

55

"'"'--I

s 0 :z <!

('I

u

u u. 0

(/') LLl :J _l

<! >

2 0

~-8

1

6

5

,,

3

1..;

1..;1

2/

I

~nb ~ I

... .. ....

"""' (?i ~ io"" ~

-~ -\., I' f../

lr'l 1 ~

l.oo' ,

'2. 3 4 5 VA LUES OF n

MAe= ECc [C,8~+C29a-(Ct+Cz) t]+ CAa

. MeA= E.Cc [c2eA+ c3ea- (Cz+ C3) t] + cBA

. { K"s = C, tc. E { St1ffness 1

Carry-over KsA = C3 Lc E

Iii"

A ,.--Ic y ____ I • ../ d=constant

-~b ~-

FIGURE 52 - Slope-deflection Coefficients--Beam with Variable Width, Depth Constant.

56

X-0·1456

CHAPTER II

MOMENT DISTRIBUTION

1. Discussion

Moment distribution, as developed by Professor Hardy Cross, is essentially a ~ethod of solving slope-deflection equa­tions by successive approximations. These approximations are accomplished by first computing initial end moments, due to the applied loading, considering all joints as fixed in a given position. Then by releasing each joint inturn, the unbalanced moments are distributed throughout the frame until all joints are in equilibrium. During this proc­ess all joints are considered held against lateral movement. In frames where lateral movement (deflections) will occur, sidesway corrections must be made in order to bal­ance shears as well as moments. No dis­tribution of moments takes place at joints permanently fixed; hence they are notre­leased. Short cuts are possible by treating hinged joints as such to begin with. These joints, after the beam constants have been computed, may simply be disregarded. This will be illustrated in the examples.

2. Sign convention

Up to this point it has not been necessary to designate a sign system. Hereafter, for the purpose of illustrating moment distribu­tion, the following sign system will be used. If the end moment in a member tends to cause clockwise rotation of the adjacent joint, that end moment will be called posi­tive (+). When the tangent to the elastic curve of a member has rotated in a counter­clockwise direction from its initial position, the change in slope will be called positive. The deflection ~ with respect to a member will be called positive if it rotates a line joining the ends of the members in a coun­terclockwise direction from its initial posi­tion. Reversal of rotation in each definition above causes reversal of signs. Positive thrust causes compression in a member and negative thrust causes tension in the member. The directions of shears are best visualized from free-body tiiagrams. For the· purpose of correcting for sidesway in moment distribution, a shear, at the end of a column, will be called positive (+) if it tends to move the beam into which it frames toward the left and negative (-) if it tends to move the beam toward the right. For cases where this definition is not adequate,

57

'the designer must define a system of signs that is consistent throughout the problem.

3. As.sumptions

It is assumed that:

(a) The ends of all members meeting at a joint rotate through the same angle when the joint turns.

(b) Axial loads or stresses will not change the length of members.

(c) Deformation due to shear will.be disregarded.

4. Principles of moment distribution

(a) An external moment applied to a joint in a structure is resisted by all members meeting at that joint in proportion to the stiffness factor K of each member.

. (b) When a joint is rotated, the moment mduc:ed at the far end of a connecting mem­ber 1s equal to the moment distributed to the member at the joint times its carry­over factor. This induced moment is called carry-over moment.

5. Procedure of moment distribution

(a) Compute and record initial end mo­ments, carry-over factors and stiffness factors of each member in the structure.

(b) For convenience, compute distribution factors S for each member at a joint. There cannot be any distribution of moments from one member to another throug_h a hinged joint. Therefore, any initial moment at the hinged end of a member, resulting from asswning the joint fixed initially, distribUtes to that member only and its distribution factor S = 1. At permanently fixed ends no moment distribution takes place; hence no distribution factor is required, or, S = 0. For other joints, the distribution factor S for each member is proportional to the stiffness factor K of that mernber and is computed as follows:

SAB = 2: K of all members meeting

at the joint

(c) Sum up algebraically all moments at each joint. If this sum is not equal tu zero, the joint is said to be unbalanced by the amount this sum differs from zero. This difference is called the unbalanced moment.

(d) Release each joint in turn and dis­tribute the unbalanced moment to the mem­bers by multiplying the unbalanced moment by the distribution factor S of each mem­ber. The sign of the distributed moment is always opposite to that of the unbalanced moment. After each distribution the joint is again held fixed.

(e) Carry the distributed moments to the other end of each member by multiply­ing the distributed momerit by the carry-over factor. The sign of the carry-over moment is the same as that of the distributed mo­ment.

(f) Repeat steps (d) and (e) until the carry-over moments are small enough to be disregarded.

(g) Sum up algebraically, at the ends of each member, the initial end moments, dis­tributed, and carry-over moments. The results are the moments existing in the structure due to the applied loads or de­fle~tions. At a permanently fixed end, no distribution takes place; hence the summa­tion of moments includes only initial end moments and carry-over moments. Free ends, when treated as such initially, are omitted from the moment distribution.

(h) Shears and thrusts may now be found by statics.

This procedure is illustrated by the following examples.

-0 ~

E)@IDIDEi...lJ_-_M.o.w~n_t gi~t.I:i.l21!1;i.Qn_ -_Q._o.n­tillllQ..~ .Qe..am.

Given:

A continuous uniform beam, hinged at A, simply supported at B, and fixed at C, as shown in Figure 53. The beam is 1. 0 foot wide, 4.0 feet deep, and supports a load of 1,000 pounds per linear foot.

Required.:

Moments, shears, and reactions due to a load of 1,000 lb. per linear ft.

Solution:

In order to show the saving in time made possible by treating free ends as such, this problem will be solved tw·ice; first, treating end A as fixed initially, and second, treating end A as free initially.

Solution treating end A ~ fixed

Referring to case I, Figure 25(a),

KAB K _ 4EI BA- L

4 X 5.33 E 20

1.066 E

KBC 4EI 4 X 5.33E

KcB = T 30

0. 711 E.

rAB rBA rBC rCB 1

2

--~--~A +8 · I I ----------20.0 ----- --------------30.0

FIGURE 53 - Beam and Loading Diagram.

58

By equations in Figure 26(b), the initial end moments due to a uniform load of 1 000 lb. per linear ft. are: '

and

wL2 -c *--­BA - 12

1 ,oo~~ 202 = 33,300 ft.-lb.

wL2 -c *=--CB 12

1 z 00~~ 302 = 75,000 ft.-lb.

*The moments at the ends are equal but of opposite sign.

Referring to step (b) on page 57, the distribution factors are:

and

~A+ KBc 1.066 E

1.066 E + 0.711 E 0.60

KBA + KBc 0.711 E

1.066 E + 0.711 E = OAO

The moment distribution is shown in Thble IV. The line of initial end moments is marked I.E. M Lines of distributed mo­ments, carry-over moments, and final mo­ments are marked D. M., C. M., and F. M., respectively. HeaVY lines are drawn under the distributed moment.s to signify that the joints are in balance at that point. Distri­bution factors are shown in boxes at the end of each member. with the carry-over factor at the end of the small arrow.

Solution treating end A as ~ free end initially.

Referring to case III, Figure 25(a).

KBA 3~I 3 x2~33 E = 0.8 E

r·BA = 0

and from Figure 26(b),

wL2 = 1,000 X 202 DBA 8 8

50,000 ft.-lb. TABLE IV

Moment Distribution of Beam in Figure 5:3, Initially A.ssuming End A as Fixed C A ~=r-<-i 0.6 I B I o.4 r>r=~ rol

I.E.M - 33,300 + 75,000 - 75,000 D.M. - 25 000 - 16,700 C.M. 0 ~- 8,350 D.M. + + C.M. )...+ 3,325 D.M. C.M. ~- 1,250 D.M. + + 1,500 + C.M. t 938 500 D.M. 563 • 375 C.M. 28 2 .-£ 375 0 ~- 188 D.M. + 282 + 225 + 150 F.M. 0 - 63,225 + 63,225 -80,960

Note: No distribution made at joint "C" (a permanently fixed joint).

59

,I

TABLE V

Moment Distribution of Beam in Figure 53, Treating End A as Hinged c

A 0.= r~ 0.53 8 0.47 ~r ="2' ~ I.E.M

D.M C.M. F.M

~o

0.-t!..--

0

-50,000'

----- 13 250

- 63,250

\ + 75,000 - 75,000 ~ -II 750-......._ ~ %

-~- 5,8 7 5 ~ . + 63,250 -80,875 ~

~

Initial end moments, stiffness and carry­over factors for span B-C remain the same as before.

The moment distribution is shown in Table V. Once the moments have been found,

shears and reactions are obtained by statics as follows: Draw free-body diagrams of each span, as shown in Figures 54 and 55. Positive moments, as they have been defined, tend to rotate the beam in a counterclockwise direction.

In this case distribution is required around joint B only. Distribution factors are:

The shears at each end may now be found simply by taking moments about the opposite end. Thus,

0.8 E + 0.711 E

KBC

0.53 20,000 X 10 - 63,250 20

6,840 lb.

KBA -r KBC 20,000 X 10 + 63,250 20

. 0.711 E = 0.47 0.8 E + 0.711 E 13,160 lb.

1000#/ft.-= 20,ooo#·total

I I * ~ ~ ~ ~ ,1 ~ ~ ~ ~ * * i \ _ ·~ L ~B.JM8A-63,250 )I I y I :

I I I

<---------------20.0 ------------~ # VAe = 6,840* V8A= 13,160

FIGURE 54 - Free-body Diagt'am of Beam AB in Figure 53·

•# M8 c;= 63,250

1000*/ft. = 30,000#totol

t ! ! ! ! ! ! ! ! ! ! 1 l * ! l ! l l t \ ' (B C 1 Mea= 80,875 # ~}.., : ~, }'

I I t I I I I<-- - - --- - - - - -- --- --- -3 o. o ------------- -- -~

V, = 14 41·o• V.c8 = 15,590# BC ,

FIGURE 55 - Free-body Diagram of Beam BC in Figure 53·

60

30,000 X 15 - 80,875 + 63,250 30

14,410 lb.

30,000 X 15 + 80,875 - 63,250 30

15,590 lb.

RA VAB

RC VCB

RB = VBA + VBC = 13,160

+ 14,410 = 27,570 lb.

The directions of the shears are best visualized from the free-body diagrams.

EX..2-IDQ..l~ l.2_-_1\iOJP.~Ilt .Q,i.Qt.r.i:Qt!ti.QIL ·:J3J~q i.ra.:rne_~tlJ.Q.Ui sj.Q.e~'@.Y..

Given:

A frame 1. 0 foot wide with supports, loads, and depths of members as shown in Figure 56.

FIGURE 56 - Fr&llle and Loading Diagram.

Required:

Moments, shears, and thrusts due to the loading shown in Figure 56.

Solution:

Note that the center lines of beams AB and BC do not coincide. In this case the working line for the frame was arbitrarily chosen midway between them. It is neces­sary to approximate the location of the work­ing line in cases of this kind. However, minor inaccuracies in the location of this line will not appreciably affect the results.

61

The moment of inertia of each member is

1AB

1 X 33 2.25 ft.4 12

1BC

1 X 23 0.667 ft.4 12

1BD = 1 X 3.53 3.575 ft.4

12

Referring to cases I and II in Figure 25(a),

KBA = 4~!

0.45 E

4 X 2.25 E 20

3EI L

3 x ~5667 E = 0.133 E

KDB

0.953 E

4EI T

4 X 3.575 E 15

Since joints A and B are permanently fixed and C hinged, moment distribution is required around joint B only. Distribu­tion factors around joint B are:

SBA = KBA + KBC + KBD

0.45 E 0.45E + 0.133E + 0.953E

0.293

KBA + KBC + KBD

0.133E 0.45E + 0.133E + 0.953E

0.087

KBA + KBC + ~D

0.953E 0.45E + 0.133E + 0.953E

0.620

A c v ~ v

I.E.M.+3333 -3333 0

D.M. + 34

C.M. + 17

F.M. +3350'# -3299'# + 2232'# 0

By the equations in Figures 26(a) and (b),

wL2 100 X 202 CAB = - CBA = 12 = 12

3,333 ft.-lb.

Pab (b + L) 2L2

1, 000 X 1 0 X 5 ( 5 + 15) 2 X 152

2,222 ft.-lb.

871 ft.-lb.

D 'Y

-871 +35

I.E.M. C.M. F.M.

500 X 82 X 7

152

996 ft.-lb.

The distribution of moments is shown in Table 'VI. For clarity this table is shown on a diagram of the frame. Arrows indicate the members to which the moments apply. This system will be used in all of the fol­lowing examples.

In order to find the shears, draw free-body diagrams as shown in Figures 57, 58, and 59.

1004/tt. = 2000# total

AJ!J!!ll!ll!lllii!!I~L!lllB) MAs= 3350'# ( ~I j, Mu= 3299'#

I k-- ----------------20. o· -----------------~I VAe =1000# V8A=IOOO#

FIGURE 57 - Free-body Di&gr:'am of Beam AB in Figure 56.

1000#

B ~-------------10.0' --------t----5.0' -->i C

Mac= 2 2 32'# ( J Y ll. ) Me's = 0

~ I I I I I

~-<.----------------15.0- ---------------~

V8 c = 480* Vee= 520** FIGURE 58 - Free-body Di&gr:'am of Beam BC in Figure 56.

62

# M80 = 1067'i V80-=280~-x-

B , 0 !'=

;-sao# '

0 a)

D I - ~_.)' __

Voa-220 Mos-=836'#

FIGURE 59 - F.ree-body Diagram of Column BD in Figure 56.

As explained in example 11, the shear at one end of a member is found by taking moments about the other end. Thus,

2,000 X 10 + 3,350 - 3,299 VAB 20

1,000 lb.

2,000 X 10 + 3,299 - 3,350 20

1,000 lb.

1,000 X 5 + 2,232 15

480 lb.

1,000 X 10 - 2,232 _ 520 lb. 15 -

500 X 8 + 1,067 - 836 15

280 lb.

500 X 7 - 1,067 + 836 15·

220 lb.

I'he directions of the shears, hence the direct stresses and reactions, are easily visualized from the free-body diagrams. The thrust in member AB, assuming the hinge at C cannot support horizontal loads, is equal to the shear VBD = 280 lb. The vertical thrust in column BD is the sum of the shears VBA and VBc = 1,000 + 480 = 1,480 lb.

~mill§. l.3_--~o.rn.§m .Qilitrtingwu. =..Sy.IP­metti~l :f.ra.ru.eJ>.i::Wlll~­<l.alllL J.Qq_dgd.

Given:

A frame 1. 0 foot wide with load and depth of members as shown in Figure 60.

63

A

---. ------15.0' ------

' I

·2.0'

~~~'7/'/,.- y_

C 0 FIGURE 60 - Frame and Loading Diagram.

Required:

Moments at all joints A, B, C, and D.

Solution:

References to equations for beam con­stants are the same as for previous prob­le~. Refe~ences, therefore, will no longer be g1ven. The equations only will be shown and computations performed.

wL2 - CBA = 12

10,000 X 152 12

187,500 ft. -lb.

bd3 12

1 X 2.53 = 1.3 ft.4 12

KBA =

0.347E

KBD =

0.133E

1 X 23 = 0.667 ft.4 ---rr-4EI L

4EI L

4 X 1.3 E 15

4 X 0.667 E 20

KAB SBA = KAB + KAC

0.347E 0.347E + 0.133E = O. 723

S = KAC BD KAC + KAB

0.133E 0.347E + 0.133E = 0~ 277

r =- ~ for all members.

The distribution of moments is shown in Table VII.

TABLE VII Moment Distribution in Frame of Figure 60.

I.E.M. C.M C.M. C.M. C.M. C.M. C.M: ,.M . . M.

- 24,500 ... 17 700 - 8,850 ... 6 400 ... 2 450

... 3 400 + I 225 • 450 ... 160

Sin.ce the final moments are the same but of opposite signs, the shears at the ends of the columns are equal and opposite, and therefore no sidesway takes place.

6. . Sidesway correction

In the discussion on page 57, it was stated that moment distribution is, essen­tially, a method of solving slope-deflection equations by successive approximations. These approximations commonly do not take into account the effect of lateral movements of the joints that usually occur due to the applied loading. Since the effect of lateral movement is not included, it implies that an imaginary force is employed to hold the frame in place, or, in other words, it is simply assumed that the joints of the frame do not move sideways while the moments, due to loading, are being distributed. This gives rise to unbalanced shears that must be talanced by changing the moments. These changes in the moments are commonly called sidesway corrections. The amounts of these corrections are dependent on the magnitude of the deflections that the structure would undergo due to the loading if it were allowed to deflect freely. Since these deflections are not known, a direct evaluation of the sidesway corrections cannot be made. A

64

convenient procedure for finding these cor­rections is to subject the structw-e to known arbitrary deflections and obtain the resulting shear forces in terms of unknown ratios be­tween the unknown deflections, due to load­ing, and the arbitrarily assumed deflections. These ratios are then found from the condi­tion that shears due to the arbitrary deflec­tions must be equal and opposite to the un­balanced shears due to loading. Now, as shears and moments are proportional to deflections, the values of the sidesway cor· rections can be obtained by multiplying the moments obtained from the known deflec­tions by these ratios. The following ex­amples illustrate and explain this procedure in detail. ·

Given:

The same frame as for example 13, Figure 60, but with only the left half of span AB loaded with 10,000 lb. per ft.

Required:

. Moments at all joints A, B, C, and D, including the effect of sidesway.

Solution:

129,000 ft. -lb.

1 ~ 2 wL2 = - 1 ~2 X 10,000

X 152 = - 58,600 ft.-lb. The other coefficients remain the same as in example 13. Moment distribution is shown in Table VIII. TABLE VIII

Mo.ment Distribution in Frame of Figure 60 with only Left HaJ.f Loaded

I.E.M. 0 D.M.-35 800 C M. D.M.- 5 900 CM. D.M.- 4 650 C.M. O.M - 765 C.M. O.M.- 610. C.M. 0 M.- I 02

. M. -47,830

c

- 2 950 2 325 + IPSO

- 382 + 850 - 305 + 139

~ 110 - 5 I +16,710 -23,915

The free-body diagrams of the columns are shown in Figures 61(a) and 61(b).

MAC= 4 7,830'1 M80= 33,420'1

VAc=-35901 V10=•2~ T~ ,;-··p-·r.'

i A -6 8

Unbalanced shear 2 ·a equals ! 0 -3590+-2500=-1090 .. •·

~.· . la) : (b) . : C : D

.'l~--U ~ Y ... ~ Vc4=+-359~ V08=-2500

Mc4 =23,915'# M08 = 16,710'#

FIGURE 61 - Free-body Diagram, co1umns--Simp1e Sidesway.

Taking moments about joint C, Figure 81(a) gives:

47,830 + 23,915 VAC = 20 3,590 lb.*

Taking moments about joint D, Figure 61(b) gives:

VBD = 33,420 2~ 16,710 2,500 lb.*

Since no external lateral force is acting on the frame, equilibriam requires that V AC and VBD be equal in magnitude and opposite in direction; that is, the sum of the horizon­tal forces acting on member AB must equal zero. This is not the case, and the frame is out of balance in shear by -1,090 po1.mds. Therefore, a reaction of +1,090 pounds would be required to bring the frame into equilibrium. This reaction, which is equal and opposite to the unbalanced shear, will be called balancing force.

Since this force actually is not present, the frame will deflect horizontally, or side­sway, until the shear caused by this deflec­tion is equal to the balancing force. In other

*Note that the sign of the moments does not determine the signs of the shears. The direction of the shears is visualized from the free-body diag!'ams and then signs are given in accordance with the definition given previously.

65

Nords, the frame will deflect exactly as though it were acted upon by a force equal in magnitude and direction to the unbalanced shear. The change in each moment caused by this load is commonly called sidesway correction. To find this correction, first give the frame a deflection A, holding all joints rigid, as indicated in Figure 62.

->. ~t<-1 I

FIGURE 62 - Sidesway Deflection of Frame.

Then from case IV in Figure 25(b), the end moments of column AC due to A are

Due to symmetry, the end moments of col­umn BD are equal to MAC' If it were known how much of a deflection A to give the frame in order to balance the shear of -1,090 pounds when joints A and B are released, the problem of sidesway could be solved simply by distributing the initial end moments caused by A and adding the final moments to the moments of Table VIII. Since the correct value of A is not known, the frame is given any arbitrary deflection A' corrected by a factor X. That is A = 11 1X. How to evaluate X, called the correction factor, will be shown later.

Since ll' may be given any arbitrary value, it is obviously convenient to give it such a value that it divides out all common factors in the equation for end moments. In this case let

L2 A = A I X = -1,000 6EI X

*The sign of the deflection should not be introduced before it is used in obtaining the moments for a given member, since in multistory structures, the same deflection may be negative for one member and posi­tive for another member.

Using this value in the equation for moments gives

MAC = MCA = MBD = ~B

_6EI(-1000 L2)X L2 ' 6EI

1,000 X

No unit need be specified for this moment since it would merely change the value of X. The moment distribution is now performed in the usual manner, as shown in Table IX. By inspection of this table, it is obvious that the correction factor would not have to be shown except in the final moments. This will be done in subsequent examples.

TABLE IX

Moment DistributioL for Sideswa.v

I.E.M.1- 1,000 X ::;; O.M.- 277X d C.M. O.M.f. IOOX C.M .

. M.- 3.6X . C'.-M.

O.M.+ 13X C.M. O.M.- 5X F.M. +795X

C.M. - 18X C.M. + 6X C.M. 3X F.M. + 896X

+ 50X - 18X + 6X

3X + 896X

The free-body diagrams of the colunms are shown in Figures 63(a) and 63(b).

MAc= 795X

AK?.~l-+84.5X

: I I

M80=795X BJ.-.... --:::....,,Vao~ + 84.5 X

<

0 (a) 0 (b)

(\J

I I I

C : D > -.r-Y--.: ..........:::;:7 .... ~-f

VcA ;-84.5X V08 =-84.5X McA=896 X M08 -=896X

FIGURE 63 - E)-ee-body Diagram of Columns Showing Unbalanced Shear.

The shear at top of column AC is

795X + 896X VA C = 20 = + 84. 5X

66

Due to symmetry V AC = V BD = + 84. 5X

Now, for equilibrium these shears added to the unbalanced shear of -1,090 pounds, due to external loading, must equal zero. Therefore, 84.5X + 84.5X - 1,090 = 0, and X = 1,090 = 6 45

169 . .

If the frame had been given an arbitrary deflection A' in the opposite direction, the moments of Table IX would have had opposite signs, but X would then have been negative. tJ' may therefore be given any positive or negative value. The corrections in moments due to. sidesway are now found by carrying out the multiplications with X in Table IX. Thus, · ·

MAC MBD = + 795 ·x 6.45

+ 5,130 ft.-lb.

~B = MBA = - 795 X 6. 45

- 5,130 ft.-1b .

MCA MDB = + 896 x 6.45

+ 5, 780 ft. _;lb.

Adding these moments to those of Table VIII gi~es the correct moments, including side­sway. Thus,

MAC - 47,830 + 5,130

- 42,700 ft. -lb.

MAE + 47,830 - 5,130

+ 42,700 ft. -lb.

MeA - 23,915 + 5,780

- 18,135 ft. -lb.

MBD + 33,420 + 5,130

+ 38,550 ft. -lb.

MBA - 33,420 - 5,130

- 38,550 ft. -lb.

MDB + 16 '71 0 + 5 '7 80

+ 22,490 ft. -lb.

The free-body diagrams of the columns are shown in Figures 64(a) and (b).

MAc= 42, 700#

....-;: _.~c=-3042#

A

lo)

c· VcA=+3042~

MeA= 18,135*

Meo=38,550# V80=+3050# ~

8

(b)

D ~V08=-3050#

M08 = 22,49~ # FIGURE 64 -·Free-bOdy .D1a8rams ot Co~umns Showing Effect of Sidesway on Momenta.

Shears as well as moments are now in eqUi­librium. For a complete check on the re­sults, see the discussion "Check on moment distribution" in appendix A. ·

~W.~ lJ)_-~m.Ii.el:a.llU'~ ~~_rigid :fra.m_e.

Given:

A reinforced-concrete frame of the dimensions shown in Figure 65. All mem­bers in this frame are 2. 0 feet wide.

A to D and a temperature rise of 200 F on beams EF and GH.

Solution: In this case, since the center lines of

members do not coincide, an assumption must be made as to the location of working lines of the frame. The lengths of members will be assumed equal to distances between working lines as located in Figure 65. Mo­ments of inertia of members will be taken about the centroidal axis of the gross sec­tion, ignoring the effect of the reinforcement.

Required: The stiffness factors of all of these Initial end moments throughout the frame beams or columns are found by case I in

.due to a temperature rise of 400 F op. beams Figure 25(a).

B ·o .• C ·a ~- q- ----40.0' - --------~-- ~ --30.0 -- ---=-,<- ~ --20.0 -->t I \t) 1 I I

A I : I : D

r if=;= ==;t· ·~---~

~ -1f~." - -~! 1-3.o' ir2.5 __ ·-.... r;;2_._5· ...... · -~

ON:,.-- -t= -s -·~ G ~ ,···--tH II. _ ·4.5' 2.75'· 2

I t I --. t_ b= 2.0

1 K L

FIGURE 65 - Frame with 3 Points of Side sway.

67

A ·e G D lLJ (!) N.

2.08 E 1.42 E UJ 0.90E UJ w en en 0

<D '<D N 0 0

E F G H 5.72 E 0.90E

w v 0 rr>

w UJ UJ v en en 0 <D rti <D

0 0

J K L M

FIGURE 66 ... Frame Showing Values of K.

K = 4EI L

Since all members have rectangular sec­tions and the moment of inertia is taken about the centroidal axis,

I = bd3 12

and

K 4Ebd3 12L

Values of K for all members are shown in Figure 66.

A 8

J K /

The boxed-in numbers in Figure 67 are distribution factors. For example, S for member AB is

2. 08E + 1. 20E 0.63

In solving this problem it is convenient first to consider members A to J and G to L held in their original positions, let­ting the expansion due to temperature rise take place in one direction with all joints held rigid, as shown in Figure 68.

G D

L M

FIGURE 67 - Frame Showing Values of Distribution Factors S.

68

0 0 N --------40.0 1

-------1 I I 'j __

FIGURE 68 - Def~ection of Frame Due to Temperature Rise.

Taking the coefficient of expansion of concrete to be 0.000,006,5 and referring to Figure 68,

!1 0.000,006,5 X 400 X 401

0.0104 ft.

A2 0.000,006,5 X 400 X 70 1

0. 0182 ft.

A3 0. 000,006,5 X 40° X 90J

0.0234 ft.

A4 0.000,006,5 X 20° X 40 1

0.0052 ft.

A5 0. 000,006,5 X 20° X 20 1

0.0026 ft. The initial end moments in the columns for these deflections are given by case IV in Figure 25(b) to be

_ 1.5A K L

Values of K are obtained from Figure 66. Assuming E for concrete to be 432 x 106 pounds per square foot and giving proper signs to the deflections, the moments are

1."5 MBF* = MFB = 11) (0. 0104

- 0.0052) x 432·x 106

x 1. 20 = 270 ft.- kips

69

MGC = MCG = \~ x 0.0182 x 432

X 106 X 0. 69 = 543ft. -kips

MHD = ~·; (0.0234

- 0. 0026) X 432 X 106

x 0. 69 = 620 ft. -kips

~K ~F = ~ 0.0052 x 432

x106x3.04 = 511ft.-kips

MHM MMH = ~g x 0.0026 x 432

X 106 X 0.69 = 58 ft.-kips

The moment distribution for these initial moments is shown in Table X. The shears in the columns are now found as shown previously. For example, the shear at the top of column AE is

V ._ v - 21 + 32 AE. - EA - 15 3: 53_kips

The directions of the shears are best visual­ized by drawing free-body diagrams of each column. The shears at the ends of the col­umns and their direction of action on the corresponding beams are shown in Figure 69.

*Note that ll to be used for MBF and MD H

is the difference between the top and the bottom deflections.

A 8 c 0 +3.53~ +22.13-< +52.8()< +46.20<

~-3.53 ~-22.13 ;,..-52.80

+5.35-< :>-9.25

J K L M

FIGURE 69 - Unbalanced Shears Du.e to Temperature Loading. (All shears are in kips. ) .

By inspection of Figure 69 it is seen that the forces on the beams, resulting from

. the shears in the columns, do not sum to zero. The totals of the unbalanced shears .are shown in Figure 70. It is seen that the beams AD, EF, and G H may move inde­pendently of each other in a horizontal di­rection. It may be said, therefore, that this frame has three points of freedom. Since no balancing forces are present, the frame will deflect at each point until the shears caused by these deflections are equal and opposite to the unbalanced shears Pv P2, and P3; that is, the frame will deflect as if acted upon by forces equal in magnitude and direction to Pl' P 2, and P 3 of Fig-ure 70. The correction due to sidesway in this case is found in a manner similar to that employed in example 14. The exact deflections are not known. Therefore, as before, each point of freedom is given any convenient arbitrary deflection multiplied by an unknown correction factor. Barring

coincidences, this factor will be different for each point. Since an unbalanced shear at any point of freedom affects the unbalanced shear at·all other points of freedom it is necessary to treat each point separ~tely obtaining an equation for each point. Th~ an equation is obtained for each unknown correction factor, making it possible to solve for these factors. This will be made clear by the following procedure of the present example.

Give the top story a deflection A holding. all joints rigid, as shown in Figure 71. From case IV in Figure 25(b), the equation for initial end moments is

or, since K = 4EI , L

A B C D ...---------...-----------.,...._----.<·-~: t 124.66

..,.E ________ F-k-_P2= +12.09 -->t_G ___ H-t

J

P3=-113.40

K L

FIGURE 70 - Totals of Unbalanced Shears from Figure 69. (All shears are in kips. )

70

M

A ro:63l

r@ 0 ~ 0 0 0 0

0 -59.4 +22.0 + 37.4

+13.6 +38.4 -19.2 -32.8

- 1.3 -19.9 + 7.8 + 13.4 + 2.1 + 10.3 - 4.6 - 7.8 - 0.5 - 4.3 + 1.8 + 3.0

+ 0.4 + 2.0 - 0.9 - 1.5

+21 -21

E \oJ2 rom t W1 ~

0, 0 0 0 0 0 0 0 -226.5

+27.2 +68.0 + 131.3 +11.0 0 + 10.2 - 2.5 - 6.4 - 12.3

- 9.6 0 - 25.7 + 4.2 + 10.6 + 20.5 + 3.9 0 + 5.2 - 1.1 - 2.7 - 5.3

- 2.3 0 - 4.7 + 0.8 + 2.1 + 4.1 + 0.9 0 + 1.5 - 0.3 - 0.7 - 1.4

+32 + 71 -103

;~ 0

+34.0

- 3.2 + 5.3 - 1.3

+ 1.0

- 0.3

+36

re-t:,. I I

TABLE X Moment Distribution for Temperature Rise

(All moments are in foot kips)

B ro.44 jQ.30j

f S1 {

0 + 270.0 0 -118.8 - 70.2 - 81.0

0 - 46.9 - 127.6 + 76.8 + 45.4 + 52.4

+ 18.7 + 2.1 + 69.5 - 39.7 - 23.5 - 27.1 - 16.4 - 5.3 - 25.2 + 20.6 + 12.2 + 14.1

+ 6.7 + 1.1 + 11.5 - 8.5 - 5.0 - 5.8

- 3.9 - 0.9 - 4.4 + 4.0 + 2.4 + 2.8 + 1.5 + 0.3 + 1.9 - 1.6 - 1.0 - 1.2

- 61 + 181 -120

F IQ.58l lo.i2l

¥ ~ ~ 0 +511.0 +270.0

-453.0 -234.3 - 93.7

0 0 - 35.1 + 20.4 + 10.5 + 4.2 + 65.7 0 + 22.7 - 51.3 - 26.5 - 10.6

- 6.2 0 - 11.8 + 10.4 + 5.4 + 2.2 + 10.2 0 + 6.1 - 9.5 - 4.9 - 1.9

- 2.6 0 - 2.5 + 3.0 + 1.5 + 0.6 + 2.0 0 + 1.2 - 1.9 - 1.0 - 0.3

-413 +262 + 151

s-;; K ~ y

+511.0 -117.2 + 5.3

- 13.3

- 2.4 + 0.7

- 0.5

+386

->l I I

[Q.471

¥ 0

-255.2

- 40.5 + 138.9

+ 26.2 - 50.4

- 13.6 + 23.0

+ 7.0 - 8.9

- 2.9 + 3.9 ...j. 1.4 - 1.6

-173

c

~ +543.0 -124.9

- B 1.5 + 68.0

+ 29.7 - 24.7

- 12.5 + 11.2 + 4.2 - 4.4

- 1.8 + 1:9

+ 0.7 - 0.7

+408

ro:w r +543.0 -162.9

- 62.5 + 59.4

+ 34.0 - 24.9

- 12.4 + 8.5 + 5.6 - 3.7

- 2.2 + 1.4 + 0.9 - 0.6

+384

L ~

loTol t

0 - 162.9 - 173.6 + 88.7

+ 51.3 - 32. I - 22.8 + 14.7 + 7.8 - 5.7

- 3.5 + 2.4 + 1.2 - 1.0

-236

G

~ 0

-162.9 0

+ 59.4 0

- 24.9 0

+ 8.5 0

- 3.7 0

+ 1.4 0

- 0.6

-123

~ y 0

-81.5 +29.7 -12.5 - 1.8 + 0.7 - 0.3

-62

[D.4ol

~ 0

-217.2

-135.6 + 79.3 + 49.0 - 33.2

- 16.0 + 11.4

+ 6.8 - 5.0

- 2.4 + 1.8 + 1.0 - 0.7

-261

+4~3X

loAo1 {

0 -271.2 -108.6 + 98.0 + 39.7 - 32.0

- 16.6 + 13.7 + 5.7 - 4.8

- 2.5 + 2.1

+ 0.9 - 0.7

-276

D ro.56' ( ~

0 +620.0 -347.2 -272.6

- 81.5 -101.7 + 102.6 + 80.6

+ 44.4 + 36.8 - 45.5 - 35.7 - 16.1 - 12.0 + 15.7 + 12.4 ...j. 7.3 + 5.2 - 7.0 - 5.5

- 2.8 - 1.7 + 2.5 + 2.0 + 1.2 + 0.8 - 1.1 - 0.9

-328 + 328

H

to.30l ~ ~

+ 58.0 + 620.0 -203.4 -203.4

0 - 136.4 + 73.5 + 73.5

0 + 40.3 - 24.0 - 24.0

0 - 17.9 + 10.4 + I 0.4

0 + 6.2 - 3.6 - 3.5

0 - 2.7 + 1.5 + 1.6

0 + 1.0 - 0.6 - 0.6

- 88 + 365

r;;~ + 58.0 -101.7 + 36.8 - 12.0 - 1.8 + 0.7

~ - 15

"{ f t925X -228X -697X

v +998X

r----1 'V

-168X

v -315X

K L

-7

FIGURE 71 - Moments Due to Sidesway of Top Story.

71

A

;:.--II4.27X

- 16.30X

• +67.4QX-<

=>-129.80X - 19.15X

-53.20X -9.90

FIGURE 72 - Shears D\1e to Sides"W&.Y of Top Story.

In this case let

ll = 6 • X = _ 15,000 X - 1.5E

Using this value of 6 , and K from Figure 66, the moments are

MAE = MEA = - l·; 1. 20E

~H

L ·15,000 xl = 1,2oox [ 1. 5E j

1.5 . ~B = - 15 1.20E

L 15,ooo xl [ 1.5E J 1,200X

MGC = - i·; 0.69 E

[ 15,ooo xl 690x L 1. 5E J

M = -~069E HD 15 .

L 15,ooo xl 690x L 1.5E J

The moment distribution for this case is. shown in Table XI. The final moments, copied from Table XI, are shown in Figure 71. The corresponding shears at the ends of. the columns and their direction of action on the beams are shown in Figure 72.

72

Next, beam EF is given a deflection A hold­ing all joints rigid, as shown in Figure 73. It is convenient but not necessary to use the same numerical value of A' as was used for b·eam AD. The end moments in columns AE and BF will then be the same as before except for sig!'l.s and the correc­tion factor. Thus, using Y as correction factor and in accordancG with Figure 73, the moments in columns AE, BF, EJ, and FK are

- 1,200Y

M_ = M_ = -.12 1.20E --7B -~F 15

~K

[15,000 yl L 1.5E J - 1,200Y

MJE = - ~g 3.04E

[ 15,000 YJ = 2 280Y L 1. 5E '

~ = - ~g 3.04E

[- 15,000 y] = 2 280Y

1.5E '

Moment distribution is shown in Table XII. The corresponding final moments and shears are shown in Figures 73 and 74, respectively.

TABLE XI

MOment Distribution tor Sides~ ot Top Story

A B c [Q:§3] [Q.44 ro:3o1 .IM7l -o.30I

r@ t r ~ ~ ~ r@ t + 1200 0 0 + 1200 0 0 .f.690 0 - 444 -756 -528 - 312 -360 -324 -159 -207

- 72 -264 -378 - 72 -162 -180 -103 -193 + 124 +212 +269 ..... 159 + 164 -+ 224 + 109 +143 + 34 + 134 -+106 + 30 + 112 -+ 92 + 32 +- 58 - 62 -106 -109 - 65 - 74 - 85 - 42 - 55

- 12 - 55 - 53 - 15 - 42 - 37 - 17 - 33 ... 25 + 42 +- 48 ... 29 +- 33 + 41 +- 20 -+ 26

+ 6 ..... 24 + 21 +- 5 -+ 20 +- 16 + 5 ..... II - II - 19 - 20 - 12 - 14 - 15 - 7 - 10 ..,. 3 - 10 - 9 - 2 - 7 - 7 - 3 - 5 ... 5 ..... 8 + 8 + 5 +- 5 + 7 ..... 3 -+ 5 .j.. I -+ 4 + : ..... i -+ ~. + ~ +- l +- ~ - 2 3

+ 789 -789 -645 ..... 949 -304 -268 .f.528 -260

E F G fQ.i2 [Q.581 [Q.58] Q.i2l [QjO ro.4o1 ~ ~ { t ~ i ~ ~ ~

... 1200 0 0 0 0 ... 1200 -+690 0 0 - 144 -360 -696 -696 -360 - 144 -207 -207 -276

- 222 0 -348 -348 0 - 156 - eo 0 -138 + 68 ... 171 +331 +292 +- 151 +- 61 ... 65 ..... 66 + 87 + 62 0 + 146 + 165 0 ... 80 + 55 0 + 58 - 25 - 62 -121 - 142 - 74 - 29 - '34 - 34 - 45

- 31 0 - 71 - 60 0 - 32 - 21 0 - 18 + 12 +- 31 + 59 + 53 + 26 + II + II ..... 12 + 16

+ 12 0 + 26 -+ 29 0 + 14 + 10 0 + 9 - 6 - II - 21 - 25 - 13 - 5 - 6 - 6 - 7

- 5 0 - 12 - 10 0 - 6 - 3 0 - 3 + 2 + 5 + 10 + 9 + 5 ..... 2 + 2 + 2 +- 2 + 3 0 + 5 + 5 0 +- 3 + 2. 0 + 2 - I 2 - 5 - 5 2 - I - I - I - 2

+ 925 -228 -697 -733 -265 + 998 +483 - 168 -315

~~ r;~ ~~ 0 0 0

-180 -180 -104 + 86 + 76 + 33 - 3t - 37 - 17 +- 16 + 14 + 6

5 - 6 3 +- 2 + 2 + I

0

-113 -112 84

73

loAOl ~

0 -276 - 138 +- II. 6 + 44 - 36

- 22 + 19 + 8 - 7

- 3 + 3 + I - I

-292

0 !Cia (

0 -386

- 103 + 115 +- 72 - 65 - 27 ..... 23 + 13 - II

- 5 + 4 + 1 --369

~ 0

-207 0

+ 87 0

- 27 0

+ 14 0

- 5 0

+ 2 0

- I

-137

.r---; v 0

- 104 + 44 - 14 + 7

2 + I

0

- 68

~ +690 -304 -103 + 91 + 44 - 51

- 14 + 18 + 7 - 9

- 2 ..... 3_ + l -+~69

H roi01 t

+690 -207 -152 ..... 87 + 46 - 27 - 25 -+ 14 + 9 - 5

- 4 -+ 2 + 2 - I

+429

M ~

A Q.63l

~ ~ - 1200 0 + 444 -4-756

- 65 +264 - 74 -125 + 6 - 69 + 23 ... 40 + 5 + 20 - 9 - 16

0 - 6 + 2 + 4

0 ... 3 - 1 - 2

- 869 +869

E [Q.i2 fQ58l

r 81 ~

- 1200 +2280 0 - 130 - 324 -626 + 222 0 -313 + II + 27 .... 53

- 37 0 + 45 - I - 2 - 5 + II 0 + I - I - 4 - 7

- 4 0 - 3 + I + 2 + 4 + 1 0 + 2

0 - I - 2

- 1127 + 1978 -851

~~ + 2280

- 162 + 13

I 2

+ I 0

+2129

TABLE XII

Moment Distribution for Sidesway of Beam EF

B fQ.44 ro:30l r ~ r

0 -1200 0 + 528 ... 31.2 +360 +378 - 65 0 - 138 - 81 - 94 - 62 ... 10 - 42 + 41 + 25 + 28 + 20 ... 4 + II - 13 - 11 - II

- 8 0 - 5 + 6 + 3 ... 4 + 2 0 + I - I - I - I

... 753 -1004 +251

F .[0,5a] ~

~ ~ i 0 +2280 -1200

-626 - 324 - 130 -313 0 + 156 + 91 + 47 ... 19 + 26 0 - 40 + 2 + 4 .... 8

- 2 0 + 12 - 6 - 3 - I

- 3 0 - 5 + 5 + 2 + I

-+- 2 0 + I - 2 - I 0

-826 +2005 - 1179

~~ +2280

- 162 + 23 + 2

+ 1 0

•2143

ro.47l ~

0 0

... 180 - 85 - 47 .... 22 -+- 14 - II

- 5 ... 3 ... 2 - 2

.... 71

74

c

m 0 0 0

-41 0

+ II + 3 - 6

0 + 2 + I• - I

-31

fQ.30

t 0 0 0 0

-22 + 1 + 5 - I - 3 + 2 .... I

0

-II

L w;,

o:3ol t

0 0 0

-54 0

+14 + 1 - 7 - 2 + 2 ... I - I

-40

G

l§h 0 0 0 0 0

+7 0

-2 0

+I 0 0

+6

~' y 0 0 0

+4 -I

0 0

+3

IQ.40l

f 0 0 0 0 0

+8 6

-2 -2 +2

0 -I

+5

loAol v 0 0 0 0 0 0

+4 -4 -I +I +I -I

0

0 ro.ss t 0 0 0 0

-27 + 15 -+- 7 - 4 - 3 + 2 + I - I

-10

~ 0 0 0 0 0 0 0

-3 0

+I 0 0

-2

·~ y 0 0 0 0

-2 0 0

-2

@h 0 0 0 0 0

+ 12 0

....;. 3

' - I + 2

0 0

+10

H I0:3o1 ~

0 0 0 0 0 0

+6 -3 -I

0 +I -I

+2

M ~

TABLE XIn

Mo.ment Distribution for Sidesway of Beam GH

A B c D ro:63l fM4 IQ.30l IM7l loTol fQ56

r.@ ~ t ~ ~ ~ ~ l r @h 0 0 0 0 0 0 -690 0 0 -690 0 0 0 0 0 -1-324 ... 159 -1-207 +386 +304

0 0 0 0 + 162 0 + 26 + 193 +104 + 26 0 0 -71 -42 - 49 -103 - 50 - 66 - 73 - 57 0 -35 0 0 - 51 - 25 - 17 - 36 - 33 - 28

+13 -1-22 +23 + .13 + 15 + 37 + 18 + 23 + 34 + 27

0 +12 +II + 2 + 18 + 8 + 10 + 17 + II + 7 - 4 - 8 -14 - 8 - 9 - 16 - 8 - II - 10 - 8

0 - 7 - 4 0 - 8 - 4 - 2 - 5 - 5 - 4 + 3 + 4 + 5 + 3 + 4 + 5 + 3 + 3 + 5 + 4

0 ... 2 + 2 0 +- 2 + 2 + I ... 2 + I + I - I - I - 2 - I - I - 2 - I - 2 - I - I

+II -II -50 -33 + 83 + 226 -551 +325 +419 -419

E F . G H [Q.i2" IMsl ro.58l tQl2l [0:30 loAOl loAol 0301 r ~ ~ ~ fi! ~ r'

'( ~ ~ ¥ fi! ~ 0 0 0 0 0 0 -690 +517 0 0 +517 -690 0 0 0 0 0 0 + 52 -1- 52 +69 +69 + 52 + 52 0 0 0 0 0 0 + 80 0 +35 +35 0 +152 0 0 0 0 0 0 - 35 - 34 -46 -75 - 56 -·56 0 0 0 0 0 -21 - 25 0 -37 -23 0 - 28 0 0 0 +12 +6 + 3 ... 19 + 18 +25 +21 + 15 + 15

+7 0 +6 0 0 :+ 7 + 9 0 +10 + 12 0 + 14 -I -4 -8 - 4 -2 - I - 5 - 6 - 8 -10 - 8 - 8 -2 0 -2 -4 0 - 4 - 4 0 - 5 - 4 0 - 4

0 +I +3 +5 +2 + I + 3 + 2 + 4 ... 3 -1- 2 + 3 +I 0 +2 +I 0 .... I ... I 0 ... I + 2 0 + 2

0 -I -2 -I -I 0 0 - I - I - 2 - I - I

+5 -4 -I +9 +5 -14 -595 +548 +47 +28 +521 -549

~~ ~~ ~~ 1.-M '( ~~

0 0 +517 +517 0 0 + 26 ... 26 0 0 - 17 - 28 0 +3 + 9 + 7

-2 -I 3 4 0 +I + I + I 0 0 0 0

-2 +3 +-533 +519

75

'f -1127Y

,,__--,r--~

ti978Y

+2142Y

FIGURE 73 - Moments Due to Sidesway of Beam EF.

The sidesway procedure for beam GH is the same as for beam EF, this time using Z as correction factor. Deflecting beam GH in the direction shown in Figure 75 and us.ing the same numerical value for 6' as before gives

MGC = MeG = -1i; x 0.69E

[15,000 zJ = -690Z [ 1.5E

~ ~H=-i;x0.69E

[15,ooo z]

1.5E - 690Z

MGL MLG = - ~05 x 0.69E

[ 15,ooo zl L 1. 5E J

A

5172

~-133.07Y

MHM = MMH = - ~05 x 0.69E

[- 15,ooo zJ = 517Z

1.5E

Moment distribution is shown in Table XIII, and the resulting moments and shears are shown in Figures 75 and 76, respectively.

As stated previously, for the frame to be in equilibrium the shears due to side­sway must be equal and opposite to the un­balanced shears caused by external loading. The total unbak .. iced shear for each point of freedom is shown in Figure 70.

. Now, the summation of shears around any row of continuous beams in Figures 72, 74, and 76, plus the corresponding unbal­anced shear, must equal zero. Therefore, for the beams A to D,

364. 67X - 280. 60Y - 143. 06Z

- 124.66

-145.53Y

+2.80Y +0.4Y-<

FIGURE 74 - Shears Due to Sidesway of Beam EF.

76

-2Z +2l 'f

+548l

0

+517Z

FIGURE 75 - Moments Due to Sidesway of Beam GH.

and for beam EF,

- 279. 52X + 691. 20Y - 2.18Z

= - 12.09

and for beam GH,

- 142.85X + 2.15Y + 247.63Z

= + 113.40

These three equations furnish the neces­sary conditions for evaluating X, Y, and Z. However, solving the equations as written above would give a set of values of X, Y, and Z valid only for the unbalanced shears used in solving the equations. Since most structures are subject to various kinds of loading, it would be convenient to solve the above equations in such a way that the ef­fect of each unbalanced shear would be o b­tained separately. This may be done with­out much additional work, as shown m .Table XIV. In column 8 the equations are set equal

A 8 +1.0 i! <:

-0.3Z +0.35

to the condition of an unbalanced shear of -1,000 kips on beams A to D and zero un­balanced shear elsewhere; that is, P1 = -1,000 kips and P2 = P3 = 0. (P1 is shown with a plus sign in the table, since there it is to the right of the equal sign.) Hence, the values of X= +5.894, Y = +2.373, and .z = +3.379 in column 8, lines 16, 15, and 12 of Table XIV, give the effect of an unbalanced shear of -1,000 kips on the beams A to D. Columns 9 and 10, lines 16, 14, and 12, give the effect of an unbalanced shear of -1,000 kips on beams EF and GH, respectively. The corrections in moments due to P1 = -1,000 kips is now found from Figures 71, 73, and 75 by carrying out the multiplication, using the values of X, Y, and Z of column 8 of Table XIV. Similarly, the. correction in moments due to P2 and P3 being succes-sively equal to -1,000 kips is obtained by multiplying by the values of X, Y, and Z of columns 9 and 10, respectively. These computations are tabulated in Tables XV,

-3.132 -64.53i!

FIGURE 76 - Shears Due to Sidesway of Beam GH.

77

TABLE XIV

Solution of Simultaneous Equations

RECIPRO- X y l CALS I 2 3 4 5

I 0.002742 + 364.67 -280.67 -143.06

2 0.003578 -279.52 +691.20 +2.18

3 0.007000 - 142.85 + 2.15 +247.63

4 +I -0.7695 -o. 3923

5 -I +2. 4728 +0.0078 6 -I +o.o 151 + 1.7335

7 0.587036 0 +I. 7033 -0.3845

8 1.325557 0 -o. 7544 + 1.3412

9 +I -0.2257 10 -I + I. 7778

II 0 +1.5521 12 l

13 +0.2257Z 14 y

15 0.7695¥ 0. 3923Z

16 X

XVI, and XVII. By means of these tables the correction in moments due to any un­balanced shear may now be obtained. As shown in Figure 70 of this example,

+ 124.66 kips

+ 12.09 kips

and

P 3 = - 113.40 kips

These values are then used in Tables XV, XVI, and XVII, respectively, to obtain the values of Table XVIII. The final moments in the frame of Figure 65, due to a :empera­ture rise of 400 F. on the top beams and 20° F on the lower beams, are now the alge­braic summation of the final moments of Tables XVIII and X. These moments are shown in Table XIX. It is seen that the final moments sum to zero around each joint. Verifying that the shears sum to zero on any row of continuous beams merely checks the numerical work of correcting for side­sway, since this cond:ltion was used as the basis for finding the correction factors. Continuity at joint B between members BA and BC and at joint C between members CV and CD can be verified by the criterion derived on page125 in appendix A. Due to the limitation of the criterion, it cannot be used at any other joint in this case.

The beams at joints A, B, C, and D had no initial end moments, and, therefore, the

CHECK

6 7 8 9 10

-59.06 P,=+IOOO P.=O P,= 0 +413.86 P1 =0 P.=+IOOO P1 = 0 + 106.93 P.=O P.=O P..= +1000 -0.1618 -0.1618 +2.7422 0 0 + 1.4806 +I. 4806 0 +3.5776 0 +0. 7485 +0. 7485 0 0 -+7. 0004

+ 1.3188 + 1.3 188 +2.7422 +3.5776 0 +0.5868 +0.5868 +2.7422 0 +7.0004

+0.7743 +0.7743 + 1.6099 +2.1004 0 . +0. 7778 +0.7778 +3.6349 0 +9.2794

+1.5521 +5.2448 +2.1004 +9.2794

78

+3.379 +I. 353 +5.979

+0.7626 +0.3054 +1. 3495 +2.373 +2.406 +I. 350

+3.1516 +2.3822 +3. 3844 +5.894 +2.382 +3.384

changes in moments are equal to the final moments. Using the final moments of Table XIX in the criterion on page125 of appendix A gives for members BA and BC of joint B,

1 42-- 127 2

- 28 -l (- 29) 2

= = ~~: ~ = 1. 59

The ratio of the corresponding stiffnesses is

KBA 8

BA = 0.44 = 1.47 KBC = SBC 0. 30

The error is only 8 percent. This is con­sidered to be very good since the criterion is extremely sensitive to changes in the moments. Similarly, for joint C the error is 9 percent.

7. Moment distribution for frames hayin~ members of variable moment of inertia

The method of moment distribution applied to frames having members of vari­able moment of inertia is exactly the sai'I'e as for those having members of constant section. The carry-over factors, the stiff­ness factors, and initial end moments are, of course, different and must be determined in the manner shown in Chapter I.

TABLE XV

S1desway Correction of Moments in Foot IC1ps Due to P1 = -1,000 IC1pa

A 8

+4650 -4650 -3802 +5593 -2062 +2062 4- 1787 -238 ... 37 - 37 - 169 - 112

+2625 -262S -2184 +3099

E F

+54S2 -1344 -4108 -4320 -IS62 +S882 -2674 +4693 -2019 -1960 4-4757 -2797 + I 7 - 14 - 3 + 30 4- 17 - 47 +2795 +3335 -6130 -6250 +3212 +3038

+5047 1

+4462

- 695 +5082 + 7

+4394

-1791 --1580 + 59S ... 168 + 281 + 764 - 91S - 648

TABLE XVI

c 0

+3112 -.15 2

- 73 .;. 95 -1862 + 1098 +1177 - S29

..o201 I + 1852 + 159 + 95 + 1760. ,... 1855 + !JO + 876 -1686 .-1626 + 948 + 678

-·460 + s +1852

+1397

+.1747

+1380

Sides'WIQ' Correction of Moments in Foot IC1pi Due to P2 -1,000 ICips

A B c 0

+1879 -1879 -1536 +2260 - 724 - 639 + 12S8 - 619 - 879 + -2091 +2091 +I 811 -2415 + 604 ... 171 - 7S - 96 - 24 + 24 + IS - IS - 68 - 45 ... 113 + 306 - 746 + 440 + S67 - S67

- 197 + 197 "'" 207 - 200 - 7 - 162 + 437 - 275 - 336 + 336

E F G H

-27'11 +4758 -2047 - 1987 +4824 -2837 + 7 - 5 - 2 + 12 + 7 - I 9 - 805 + 742 + 63 + 38 + 705 - 743 - 501 ... 4210 -3709 -3721 +4200 - 479 + 320 + 356 - 676 - 658 + 374 + 284

y - 186

+5117 +5153 + 5 7 3 + 3 + 742 ... 670

+4881 4-4875 + 561 + 518

79

+2670 -1173 + 66 +1563

+ 30 +16-39

A

TABLE XVII

Sidesway Correction of'.M-nts in Foot Kips Due to P3 -1;ooo Kips

A 8 C

-2670 +1173 - 66 -1563

E

+2669 - 24 +1873

+-2871 12

+2527

-2358 -2480 -1148 -1115

6 + 54 -3512 -3541

-2183 + 1016 - 299 -1466

+2706 + 30 + 1839

+2891 + 12

+2504

+3211 -1-355 - 197 + 1659

F

+3377 - '1591 - 84 + 1702

-1028 - 907 + 1787 4- 339 + 96 - 42 + 496 + 1351 -3294

- 193 ~ 540 -1549

G P. = -1000 3 -

Kips

+1634 15

-3557 -19'38

TAB~xVIII

- 880 - 54 +1943

+1009

- 568 + 8 +3276 +2716

0 y - 264 + 3 +3276

4-3015

-1066 - 988 + 7 0 + 281 + 167 - 778 - 821

Sidesway Correction of' Moments in Foot Kips Due to P1

= +124.66 Kips, P2

= +12.09 Kips, and P3

= -11].40 Kips

B c

-1249 13

+2505 + 1243

- 464 3

-+-3115 +2648

'( - 206

4 +3091 +2881

D

D

+1249 + 13 -2505 ·-1243

H

+1452 + 3 -3282 -1827

...---l~:::..I.--------.J..::::.~h----....J...:*:.J...,....L..::::_;!.!...L...,----......j...::.:.;::~----------J~~-r'~.>P.'.'Ii+l~4.66 Kips

- 327 +- 2 + 177 - 148

- 348 + 6 + 186 - 156

+ '327

- 2 - 177 + 148

E

- 51 + 212 - 255

- 59 + 287 - 328

+ 764 + 779 + 45 + 45 - 398 - 402 + 411 + 422

+ 272 3

- 166 + 103

- 400

- 51 + 209 - 242

- 548 - 59 ± 284 - 323

- 386 + 114 + 3 0 + 188 - 22

- 195 + 92

F

- 379 + 6 + 193

- 180

+ 81 - 147 + 2 - 5 +- 61 - 175

+ 144 - 327

G P3= -113.40

Kips

- 101 4

- 220 - '325

80

+ 66 + 3 + 114

+ 183

-+ '308 + 195

+ 342 + 161

+ 210 + 203 8 + 8

- 88 - 9'3 +- 1'30 + 118

y + 98 - 98 + 4 - 4 + 141 - 141

+ 243 - 243

- 118 5

+ '300 + 177

172 6

+ 327 + 149

H

-- 295

TABLE XlX

0 Final Moments in Fgot Kips Due to 40 F Temperature Rise on Beams A to D

and 20 F Temperature Rise on Beams EF and GH

A

+ 21 - 21 -148 +148 -127 +127

-156 -124

E

-255 -184

+ 36 ....,328 -292

+411 +422 +308 ... 9

B

- 61 + 181 -120 +103 -195 + 92 ... 42 - 14 - 28

F

-242 -180 + 20 - 29

+386 -323 + 63

The following examples illustrate both moment distribution and the method of mak­ing the sidesway correction for frames · having members of variable section.

E~IDtU~ L6_-Jk>ti~oniaU®.,ging_ --Er~rne b..a.Y:i.ng_ v_e.tia.ble_:r.g_o lll.§.!lt­Q1 ::..in.e.I:tia_:r.g_e lllQe..r a.

Given:

A frame with variable T-sections and loading as shown in Figure 77. The total flange width of column AC and beam A~_ is 3. 75 feet and of column BD, 5. 50 feet. The width of the stem is 1. 0 foot for all mem­bers.

0

FIGURE 77 - Frame with Variable T-sections.

81

c

-173 +408 - 236 +144 -327 ... 183 - 29 + 81 - 53

G

+3.84 -261 -325 +195 +130 ~ 59 ... 72 ·-131

... 99

Required:

-276 +118 -15.8

-328 +328 ... 243 -243 - 85 + 85

+177 ... 89

+149 -+ 134

H

+365 -295 ... 70

Moments at all joints A, B, C, and D, including correction due to sidesway.

Solution: By use of the diagram on page 54 , the

moment of inertia of the s mailer section of column AC is

Ic 1. 75 1 X1 ~· 53 = 0.492 ft. 4

and of the larger section the inertia is

IL = 1. 72 1 ~233 = 3.87 ft. 4

Now in accordance with the explanations given in example 9 of Chapter I,

IL 3 3.87 w m = Ic = Va.492 = 1.98

Then by means of the diagrams in Figures 30, 31, and 32, for this value of m and a =

i'; = 0.30, the slope-deflection coefficients are

The stiffness and the carry-over. factors are

Elc elL 5.6 9· 49.~·· E

15

0.184E

c2 r AC = C1 ;:~ = 1.05

Similarly, for column BD

1. 98 l Xi~ 53 = 0. 557 ft. 4

1.96 1 ~233

2~.~ = 0.20

4.41 ft. 4

a

m

~D

"3/4.417 == Vo:557 2.0

6.49

5.3

27.0

6 49 0.557 E • 22.5

r 5.3 BD = 6.49 0.817

0.161E

For the load of 10,000 pounds applied to column BD 10 feet from end D the value

. 10 '· . of k 1s 22. 5 = 0.445 (see Figure 77). By

the diagram on page 31 for this value of K rand a and m equal to 0.2 and 2.0 respec~ tively, the initial end moments ~e ·

CBD = 0.044 X 10,000 X 22.5

9,900 ft. -lb.

CDB 0.17 X 10,000 X 22.5

38,250 ft. -lb.

For beam AB by the diagram in .Figti.:re 50 the moment of inertia is .

82

I = 1. 72 1 i223

= 1.146 ft 4

Then, by case I in Figure 25(a),

4E 1.146 2.5. 5 = 0.180E

The distribution factors are

S . = KAB AB KAB + KAC

0.180E 0.180E + 0.184E

KAB +KAC

0.184E 0.180E + 0.184E

KBA

KBA +KBD

0.180E 0.180E + 0.161E

KBD SBD = ...,K_B_A.;...+_K_B_D

0.161E 0.180E + 0.161E

TABLE XX

0.49

0.51

0.53

0.47

Distribution of Moments in Frame Having Variable T-section Loaded

as shown in Figure 17

The free-body diagrams of the columns are shown in Figures ·78(a) and {b).

MAc= 14301

• Mao= 4;9301

•

~---- e . A y---... V . .··x-- - AC

I

-~ ----- ---~ Va·o

I. I I I I _,

I I I

0 .o to)

I I I I (b)

~10,000• I I

_t __ ~--c ~-~ · VcA

~ N N

I j

·a ci "j'

MeA= 1soo·• I I 0 j .

· .-¥- •. ~~'-~Voi · M - 42 3101

• . ''DB ,

FIGURE 78 - Free-body Diagrams of Columns of Jrame 1J:l Figut'e ·71.

The shear at top of column AC is

and at top of column BD the shear is

. ·vBD = 4,930 + (10,000 X 10) - 421310 .

22.5

= ~ 2,783.lb.

The unbalanced shear for this case .is then

195 + 2, 783 = + 2,978 lb.

To correct for side sway, give the frame 9:. deflection as shown in Figure 79. · · .

->i~ ~ ~£\:c.

A ~: ~--~--------------~~~ • I

0 tri

~T.

.J I

"? N N ..

N ...J

I I I I

: D y ___ _

FIGURE 79 - Sides~ Deflection ot . Frame in F':l.gure 77.

The initial end moments for this frame are . obtained by equations of case IV inFigure . 25(b). By inspection of these equations it . is seen that the only common factor for both: ·

. co~~ is 'E. However, 6 ' may be chosen stt:ch·'that it divides out all the factors for at least one end moment of one of the col­umns. In this case, then, let

6 ·= .··A'''X = . 152 X 1,000 X . E 0.492(5.6 + 5.9) 2 <

. 15 .~- 1,000 X 11 . .5E 0.492

Using this value of 6 in the equations of <:~se IV.,in Figure 25(b) gives

.MAC. - ... E O.t92 (5.6 + 5.9) 15

2 . . X 15 X 1,000 X = - 1,000X

. 11. 5E 0.492

·Mc_A:· =.- E 9.~92(5~9+ 26.4) 15

X 152 X, 1,000 X - 2,810X . · · .. 11. 5E 0.492 -

·M,. = ... E 0.557 (6.49 + 5.3) BD 22.52

152 X 1,000 X = - 516X . X 11. 5E 0.492

~B. "' - E2~.~~7

(5.3 + 27.0)

152 X 1,000 X 1L 5E 0.492 X = - 1'413x

'!'he·lrioment distribution is shown in Table X:Xl, and the corresponding free-body: dia­

. gram is shown in Figures 80(a) and tb) •

MAc• 530X

. VAc•·I9.0XiJ. --·K ·. A ! . ·to) -~

c : ·· .... t

VcA=tt90X ~-=-McA•23t6X

M80 •379X

x--~~Veo=·75X ; 8 I

-.;, (b) ,_;

"' '

: D 1·--~ ""Voa•+75X

M08 •t301X

,ttWRE 8o - Free-body Diagrams Of ·· Columns ot F.rame · in Figure 77.

TABLE XXI Distribution of Sidesway Moments of Frame in Figure 77

0.49 r>r :.2 2=r-<1 0.53 1 l

{-I. E.M. -IOOOX D.M. C.M. D.M. C.M. D.M. C.M. D.M. C.M. D.M. F.M.

+ 510X

- 70X

+ 33X

- 5X

+ 2X -530X.

I.E.M. C.M. C.M. c.M. C.M.

·c.M. F.M.

A

-10

0

T L.. II

10 0 -

c

-t 0

t490X + 137X - 67X - 65X t 32X + 9X - 4X - 4X + 2X + 530X

., , ~ -2810 X + 536X

74 X + 35 X

5X + 2X -2316 X

The shear at top of column AC is

530 + 2,316 15

- 190X

and at top of column BD the shear is

V . _ 379 + 1,301 _ _ 75X BD - 22.5 -

The total shear due to A is

- 75X - 190X = - 265X

For equilibrium, this shear plus the un­balanced shear must sum to zero. There­fore,.

or

- 265X + 2,978 = 0

X 2,978 - 11.24 265 -

84

8 ~ f 0

,..... - 516X

+ 273X v + 243X

~ 245X 0

-130X q- - 115X - 34X L.;..

ISX II

+ 1'- + 16X t 16X CX)

- ax 0 - ax - 2X + IX + IX + 379X - 379X

~l, i I.E.M. C.M. C.M. C.M. C.M. C.M. F.M.

-1413X +199X - 94 X + 13 X

7X + IX -130 IX

Now, using this value of X in Table XXI and adding the resulting moments to the moments of Table XX gives the correct moments due to the loading shown in Figure 77, Thus,

MAC + 1,430 - (530 X 11. 24)

- 4,527 ft. -lb.

MeA + 1,500 - (2,316 X 11.24)

- 24,532 ft.-lb.

MAB - 1,430 + (530 X 11.24)

+ 4,527 ft.-lb.

MBA - 4,930 + (379 X 11. 24)

- 670 ft. -lb.

MBD + 4,930 - (379 X 11. 24)

+ 670 ft. -lb.

MDB - 42,310 - (1,301 X 11. 24)

- 5f3,933 ft. -lb.

Given:

A frame 2. 0 feet wide with load and depth of members a:s shown in Figure 81.

••· Pin connected o----.:::.~'--~~~~·· (both columns)

0 0 tf)

i 0 0

I 7 l : c y ____ v_m~•

D

FIGURE 81 - Stepped Columns ECcentrically Loaded.

Required:

Moments and shears in the columns.

Solution:

As explained previously, an eccentric load P may be represented by a centric load P and a moment Pe. In this case the moment is 100 .x 0. 5 = 50 ft. -Kips. The _proportionality constants for column AC are {see Figure 81):

a 10 30

1 3

m = 2.0 = 2 0 1.0 .

Now, from the diagrams in Figures 30, 31, and 32, the slope-deflection coefficients are

c 1 5.5

c 2 6.15

26.0

Assuming the column fixed at both ends, the stiffness factors are found from case I in Figure 26(a).

1 5· 5 6 x 30 E

The carry-over factor is

85

r . = c2 = 6.15 = 1.12 AC c1 5.5

The same constants for column BD a

m

10 25 = 0.4

2.0 = 2.0 1.._0

5.42 6.5 24.5

5.42 6 ;SO E

~ = 120 5.42 .

0.030E

are

From the curves in Figures 36 and 37, · c Ac* 0.19 ( - 5o,ooo)

- 9,500 ft. -lb.

CcA* 0.275 ( - 50.000)

- 13 '750 ft. -lb. T.ABLE :xxn

Distribution of Moments in Columns in Figure 77

.Assuming Col.umns Fixed at Both Ends

I.E.M. -13750 C.M. +10640 FM. - 31 10 ..

The moment distribution is shown in Table XXII. The distribution factor SAC= SBD = 1,

since joints A and B are hinged and no part of the unbalanced moment distributes to member AB. See discussion on page 57. The free- body diagram of column AC is shown in Figure 82. The shear at the top of column AC is

V - 50,000 + 3,110 1 770 lb AC - 30 = ' .

*In the curves in Figures 36 and 37, note· carefully that for an applied moment M (counterclockwise) at the change in sec­tion, the plus sign (+) of the coefficient for CAB or CBA means that the moment CAB or CBA tends to turn joint A or joint B, into which the beam frames,__in_a clock­wise direction. If the applied moment reverses, the end moments reverse, which is the case in the present example.

MAc= 0

f--- -;:-vAc=- 1110•

! M= so,ood• I

0 0 rt)

I I I I I

: I c :f ____ ..____,

""~ MeA-= 3,110 ft.lbs.

FIGURE 82 - Jree-body Diagram, Lett Column of Figure 81.

Since no external force is applied at the top, this is the unbalanced shear that must . be taken care of· by Side sway correction.

· This correction is found in the same way as it. was found in problem 14. Give the frame a deflection 6 as shown in Figure 83. In this case let

6 ::A' X

-~{l !<· I A I

_ -10,000 X Eic

I I

_I

0 0 ~ : : I I I

~~--Y..

! I I I

_1

0 It) N

I I I I I

"1ll ~-1 .• B I . . I I

.Y ... m7)')~

FIGURE 83 - Sides~ Deflection of Frame in Figure 81.

N. ow t by the equations of case IV in Figure 25(bJ, the initial end moments are

MAC = - Eic (5.5 + 6.15) ·10,000 X 302 Elc

= -+ 129X

- Elc (6.15 + 26.0) -10,000 X 302 Eic

= + 357X

86

MBD =- Eic2(5.42+6.5) -10,000 X 25 Eic

+ 190X

- Eic (6.5 + 24.5) -10,000 X 252 Elc

= + 496X

The moment distribution is shown in Table XXIII.

TABLE XXIII

Distribution of Moments in Columns of Ftgure 81 Due to Sidesway

A 8

0

II

~

c

I.E.M. +357 'C.M. -144 F.M. +213X

The free-body diagrams of the columns are shown in Figures 84(a) and (b).

MAc= 0

VAc ---·-r--

-0

0 ~ lO) I I I I I

t___ c .......___,

MeA= 213X

Vso

(b)

Mao=O

----,.-: I I I

0 It)

! I I

D : ~---~----V... ~

MBA: 267X

FIGURE 84 - Free-body Diagrams of Both Columns of F~e 81.

The shear at top of column AC is

213X 30

+ 7.1X

and at top of column BD the shear is

268X . VBD = ~ = + 10.7X

The total shear due to A is

+ 7.1X + 10.7X = + 17.8X

For equilibriurn, this plus the unbalanced sheq_r must sum to zero. Therefore,

or

17.8X- 1,770

X 1,770 17.8

0

99.4

Multiplying the final moments of Table XXIII by this value of X and adding them to the final moments of Table XXII gives the correct moments caused by the loading shown in Figure 81. Thus,

- 3,110 + 213 X 99~4

+ 18,060 ft~-lb.

MDB 0 + 268 x 99.4

+ 26,640 ft. -lb.

The free-body diagrams for these conditions are shown in Figures 85(a) and (b).

The shear at top of column AC, caused by the loading shown in Figure 81, is

V . - 50,000 - 18,060 1 065 lb AC - 30 = ' .

and at top of column BD the shear is

Comments .2!! example 17

Since the top of each column is hinged, it would not have been necessary to resort to moment distribution· to solve this problem.

87

·o 0 rt)

I I I I I I I

lo)

Mao=O

v. =+1065#~ BD ---1:..--

(b)

8 : I I I

I I

0 : ..__ _ ___,_ --_'{_ -! c _y_ ___ ..___ .....

~..(

~-~ M08 = 26,640

1#

MeA= 18,0601

#

FIGURE 85- Free-body D1agrams·of Both Co1umns of Figure 81.

With the top of the columns hinged and held against sidesway, the moment at the bottom of column AC is, by case III in Figure 25(a),

c2 DCA = CCA - c

1 CAC - 13,750

- 6· 15 (- 9 500) 5.5 '

= - 3,110 ft.-lb.

The unbalanced shear of - 1, 770 lb. could have been applied as a cantilever load at the top of the columns. The proportion of this load taken by each column may be found by case VI* in Figure 25(b). The deflection A is the same for both columns. Therefore, for ends C and D fixed,

Eic (c22 -C3\ C1 lj AC

PBD LBD3

EI (c22 - c~ c C1 ~ BD

*Note that cases V and VI are different for variable sections. To agree with the dia­grams from which c 1, c 2, and c 3 were obtained, case VI has to be used when end A is deflected and case V has to be used when B is deflected.

or

253 (W- 26.o)

303 (~:;~ - 24.5)

0.662.

The subscripts A. C and BD refer to the respeptive columns.

Now, from the two equations

PAC = 0.662 PBD

or

1. 510 p AC PBD

and

and

1,770 2.510

1,770

1,770 1.662

705 lb.

1,064 lb.

The cantilever moments at the base of each column are

MCA PAC LAC = 705 x 30

+ 21,150 ft. -lb.

and

= + 26,600 ft.-lb.

These are the moments due to sidesway, and added to the respective moments of - 3,110 ft. -lb. at bottom of column AC and 0 at

88

bottom of column BD, they give the final moments.

8. Moment distribution with torsion

The following discussion treats the dis­tribution of moments in frames having mem­bers in both bending and torsion Torsional stresses are developed in a member if it is subjected to a moment normal to its axis. In building frames, torsional stresses usually occur in marginal beams and in girders that support cantilevers. By inspection of Figure 86 it is seen that member AD will be in torsion and members BF and CE will be in flexure because of the applied loading. Slight torsional moments will occur in mem­bers BF and CE due to the curvature of members L\.D, but, as shown later, torsion in members BF and CE may be neglected. The moment distribution when torsion and bending are involved takes pl'3.ce in two di­rections perpendicular to one another. Mo­ments in other directions might be present, but that case will not be discussed here. The principle of distribution is the same as when bending only is considered. For the S'3.ke of clarity, the moments in each direc­tion will be distributed separ'3. tely and the results combined. Positive and negative moments are defined as before; that is, an end moment in a member, whether torsional or flexural, that tends to rotate the joint ciockwise is called positive, and it if tends to rotate the joint counterclockwise, it is called negative. A torsional moment induced at the end of a member is transferred di­rectly through the member without any re­duction in magnitude. Thus the moment at the other end is equal, but, since it is are­actionary moment, its direction is reversed. For this reason the carry-over factor is always equal to -1. From this it follows that all joints g,nd members have to be viewed in the same direction while distributing the moments. This will be made clearer by the following example.

g: ~awQl~ 18_ -_r:1 o_m~u_t _9i§trwu.ti9n. w:i.ih io.r.sjo,ll.

Given:

A frame, as shown in Figure 86, support­ing a uniform load of 500 pounds per linea:r foot on members BF and CE, and two con­centrated loads of 50,000 pounds at joints B and C.

Required:

Torsional and flexural moments at all joints in the frame.

---~---- 5.0'---------~ I I

~-?1

_.za.o·--/----- so,opot

FIGURE 86 - F.rame in Torsion.

Solution:

The two directions of bending in this case are parallel to the axis of beams BF and AD.

Case I - Bending in direction BF

For this case, BF and CE are in com­mon bending and AD in torsion. From case I in Figure 25{a)

4EI KFB =--y:-

0.1064E

4EI ~c = L

0.133E

4E X 1 X 23 25 X 12

4E X 23 20 X 12

The torsional stiffness of members AB, BC, and CD are found as follows: For all

• h 3 . members A. to D, b = 2 = 1. 5, and from the

89

diagram in Figure 49,' ~,;. 0.196. Then by case VII in Figure 25(b), assuming 1-L = 0.25,

b3 hE K AB = KBA = ~ L 2 ( 1 + ~)

23 X 3 = 0.196 10 X 2 (1 + 0. 25) = 0.188E

23 X 3 KBC = KCB = 0· 196 15 x 2 (1 + 0.25)

= 0.1256E

· 23 X 3 Koc = 0·196 5 x 2(1 + 0.25)

0.376E

The distribution factors S for case I, around joints B and C are now

0.1256E 8BC = (0.1256 + 0.1064 + 0.188)E

= 0.30

0.1064E 8BF = (0.1256 + 0.1064 + 0.188)E

0.25

0.188E (0.1256 + 0.1064 + 0.188)E

0.45

0.376E (0. 376 + 0.138 + 0.1256)E

0.59

0.133E (0. 376 + 0.133 + 0.1256)E

0.21

0.1256E (0. 376 + 0.133 + 0.1256)E

0.20

Case II - Bending in direction AD

For this case, members A to D are in common bending and Inembers BF and CE are in torsion From case I in Figure 25(a),

KAB

KBC

KCD

4EI = KBA =L

= 1.80E

= l<cB 4EI

=L

= 1.20E

= ~c 4EI

=L

= 3.60E

4E2 x 33 10 X 12

4E2 X 33 15 X 12

4E2 X 33 5 X 12

The torsional stiffness of members BF and CE are found as follows: For both mem-

bers, ~. = f = 2, and from the diagram in

Figure 49 ~ = 0. 228. Then by case VII in Figure 25(b), assuming ._.. = 0.25,

K - ~ b3 hE BF - L2(1 + ._..)

90

13 X 2E 0•228

25 X 2(1 + 0.25)

0.0073E

13 X 2E 0•228

20 X 2(1 + 0.25)

0.0091E

The distribution factors for c~se II around joints B and C are now

1.20E (1.20 + 0.0073 + 1.80)E =

0·40

8BF 0.0073E

(1. 20 + 0. 0073 + 1. 80)E 0

8BA 1.80E

(1. 20 + 0. 0073 + 1. 80)E 0.60

8cD

3.60E (3. 60 + 0. 0091 + 1. 20)E

0.75

8CE 0.0091E

(3.60 + 0.0091 + 1.20)E 0

1.20E 8CB = (3. 60 + 0. 0091 + 1. 20)E

0.25

Viewing the frame in the direction from D to A, the initial end moments in members BF and CE due to the load of 500 pounds per linear foot are

CFB _ C = 500 x 252 BF 12

26,000 ft. -lb.

· 500 X 202 - CCE = 12

16,670 ft. -lb.

Moment distribution is now done as before, with the exception that the carry-over fac­tor r for members in torsion is equal to -1. For this distribution, shown in Table XXIV, members AB, BC, and CD are in torsion. Members BF and CE are in bending.

The shears at the end of membets BF and CE, close to joints B and C, are as shown in Figure 87. The plus (+) sign sig-

TABLE XXIV

Moment Distribution Due to Loading Shown in Figure 86 No Def1ection at Joints B and C

B

A D

+9850 -11,700 -9,850

- 1,500 -4,600 + 560

- 700 - 590 + 90 + 50+ + 100 t 280

90 - 280 + 40 + 20 + 10 t 40

40 - 40 + 10 0 10 + 20

10 20

-14,040 +14,040 -18,200 +4,160 -4,160 -11,220 +15,380 -15,380

NOTE: Moments ore to the nearest 10 ft.-1 b.

+26,000 / + 3,2 50

F + 420 + 190 + 30 t 10 +29,900

nifies that the shear tends to move the joint downwards. A torsional moment has no tendency to move the joint in any direction, and therefore no shear is shown at ends of torsional members. By inspection of Fig­ures 86 and 87 it is seen that joint B is acted upon by a downward force of 50,000 + 5,780 = + 55,780 pounds and joint C by a force of 50,000 + 4,590 = 54,590 pounds. These forces are 1.nalogous to the unbal­anced shear of previous problems. Since no reaction is provided under joints B and C, the joints will deflect downward until the shears in adjacent beams at each joint, due to deflection only, balance the above forces. The corrections in moments caused by this

B A

v + 5,7 eo*

F

+ 16,670 + 1,750 + 820 E + 100 +- 50 + 0 +19,390

deflection may be found in the- same way as the corrections due to sidesway. Joints B and C in turn are given an arbitrary deflec­tion multiplied by a correction factor. The moments, caused by these deflections, are computed and distributed as before. But, since torsion is present, moment distribu­tion in two directions must be made. The unknown correction factors are found from the conditions that shears caused by allow­ing joints B and C to deflect must be equal and opposite to the unbalanced shears or un­balanced downward forces due to loading.

-Kolding all joints rigid agains(~r-Qfation in any direction, let joint B deflect a dis-

c

E

'f +4,590i

/

;/0

FIGURE 87 - Shears Due to a Load of 500 lb. per Linear Foot on Beams FB and CE.

91

tance A . Then, by case IV in Figure 25(b), the moments at the ends of adjacent beams are

For this case it is convenient to let the value of

12 6E 1,000X

By the definition of deflections, a downward deflection of joint B is negative for beams AB and BF and positive for beam BC. The initial moments are then

MBA = 2 X

33 1 OQQX 102 '

+ 540X

MCB = - 2 x 33 1,ooox 152

- 240X

3 MBF = ~l,OOOX 252

+ 13X

The moment distribution for moments in the direction AD is shown in Table XXV. Since the torsional distribution factor for beams BF and CE is zero, no· moment is carried into these beams. The moment dis­tribution for moments in direction FB is shown in Table XXVI.

The shears at joints B and C,due to the moments of Tables XXV and XXVI, are shown in Figure 88. Torsional moments do not affect these shears.

Holding ::tll joints rigid and letting joint C deflect an amount

A = A'Y = 1.,g_ 1,000Y, 6E .

the initial end moments are, paying due attention to signs,

M 2 x 33

1 OOOY BC = 152 '

+ 240Y

M - 2 x 33 1 OOOY DC = 52 '

- 2,160Y TABLE XIV

A

- 90

9

2

+439X

Distribution of Moments in Direction AD.

8

-180

! - 18 I

4

+337X

F

0

0

I ~-2

v -240 -120 + 30

12 + 7

0 - 3 + I

f -240 + 60 - 60 t 15

6 + 2

2

0 -337X -231X

'f 0

92

Joint B Deflected

0

0

0

0

~ 0

-I

c I 2·

+ 45

+ 4

t231X

E

D . 'o /

+ 90

+ 23

+ 2

+115X

A

8 c ~=-116.31X ~=ti07.17X

Aar---r----.===::::;----r---r----;===::r---r--~o v ~ -77.60X -0.84X -37.87X -t37.87X +Q.IOX t69.20X

FIGURE 88 - Shears for. Joint B Def~ected.

MCE = MEC = 1 :0~3 l,OOOY

= 20Y

B ~=+107.28Y

The moments in direction AD are distri­buted in 'Th.ble XXVII, and moments in direc­tion FB are distributed in Table XXVIII. The corresponding shears are shown in Figure 89.

c ~=-474.30Y

~----------------------------------~------------~--~------~/0 v v v "' ;/,

+44.40Y +0.08Y +62.80Y -62.80Y -1.70Y

FIGURE 89 - Shears for Joint C Def~ected.

TABLE XXVI

Distribution of Moments in Direction FB. Joint B Def~ected

8 c %

A D -t-<--i0.45 0.30t->-l -1-<_,-0.20 o.s9r>-t ~ %

~ t ~ ~ ~ f f ~ ~ 0

+6

+I +7X

- 'f 0 l() +13 0

N -6 0 - 3 -4

t -IN

+I -I

-7X +lOX -3X

//~ ~/h J F v

+13 - 2

+I I X

93

0

0 +4 -I

t3X

' 0

0

._,

-IX

r v 0

-I

,___

- 0 N

d 0

t -IN -2

-2X

///~ >'////

E

==;::::=

-I X

0

+2

+2X

-72

-72

- 2

- 2

-148Y

+2

+2Y

TABLE XXVII

'q +240

0 - 96 +480 +240 - 48

-144 0 - 96 + 12 + 6 - 48

- 4 0 - 2 + 12 + 6 I

- 4 0 - 2 0 0 I

0 0 0 0 -296Y 0 +296Y +646Y

TABLE XXVIII

0

0

0

0 0

f.{

0

.l I + 36

+ 36

+

+- I

- 646Y

E

Distribution of Moments in Direction FB. Joint C Deflected

-2 -I 0 0 -IN 0 +I

- I

-2Y -IY +3Y -3Y +l6V -13V

F 'f E

' +20 /

-( - 2 -IY +ISY

94

+ 720

+ 18

+ 18

+

-1403V

0

+I

+13Y

Now for equilibrium, the summation of shears from Figures 87, 88, and 89 plus the concentrated loads of Figure 86 must equal zero at each joint. Therefore,

- 116.31X + 107.28Y - 55,780

+ 107.17X- 474.30Y - 54,590

As for the previous problem, it is con­venient to solve these equations in such a way that the effect of any unbalanced shear or force, at each joint, may be obtained sep­arately. This is done in Table XXIX for unbalanced downward forces of 1,000 pounds at each joint. Note that the equations have been multiplied through by -1.

TABLE :XXlX

Solution for Correction Factors

I 2 3 4

X y 1000 LB.AT 1000 LB.AT JOINT 8 JOINT C

116.31 - 107.28 1000 0

-107.17 +474.30 0 1000

I -0.9223 8.598 0

-I +4.4257 0 9.331

0 +3.5034 8.598 9.331 y 2.454 2.663

X -0.9223Y 8.598 0

X 10.861 2.456

In the following tables, flexural and torsional moments do not occur together in the same table. Thus Table XXX gives the correction in bending moments only, due to an unbalanced shear or load of 1,000 pounds qt joint B and no load at joint C. The sq,me correction for torsional moments is given in Table XXXI. These bending and torsional moments are found by performing the multiplications of Tables XXV, XXVI, XXVII, and XXVIII, using the values of X = 10.861 and Y = 2.454 in column 3 of Table XXIX. The same correction for a downward load of 1,000 pounds at joint C and no load at joint B is given in Tables XXXII and XXXIII and is found by perform­ing the multiplications of Tables 'XXV, XXVI, XXVII, and XXVIII, using the values of X = 2. 456 and Y = 2. 663 in column 4 of Table XXIX.

1\s found previously the unbalanced loads (shears) at joints B and C are + 55,780 pounds and + 54,590 pounds, respectively. The correction in moments due to these loads is now found simply by multiplying the final moments of Tables XXX and XXXl by the ratio 55,780/1,000 = 55.78 and the final moments of Tables XXXII and XXXIII by the ratio 54,590/1,000 = 54. 59. These computations are shown in lines 2 and 3 of Tables XXXIV and XXXV. In line 1 of these tables are the final moments of distribution in Table XXIV. The summation of lines 1, 2, and 3 yields the final moments caused by the loading shown in Figure 86.

TABLE XXX

A +4770

- 364 +4406

Correction in Bending Moments Due to +1,000 lb. at Joint B and No Load at Joint C

' 'f Yj

+3660 +108 -3660 +2510

- 726 2 + 726 +1585 -1585

+2934 +106 -2934 - 925 +28 + 925

F E y

t120 -II 2 +44 --

+118 +33

95

D 'f ~

+ 1250 -3440

-2190

TABLE XXXI

Correction in Torsional Moments Due to +1,000 lb. at Joint B and No Load at Joint C

A~~-----r~~----~----~----~~~~--~~0

A

'V +76 t 5 +81

+1080 - 394 + 686

-76 - 5 -81

F~---.

v ' 0

0 0

v -33 + -26

yt

+33 - 7 + 26

TABLE XXXII

0

Correction in Bending Moments Due to +1,000 lb. at Joint C and No Load at Joint B

v -q ' -q 'f 'l

+828 +25 -828 -568 - 2 +568 ~·

+282 -788 - 3 +·788 +1720 +43 -1720 ~3730

+ 40 +22 - 40 +1152 +41: -1152

F E ,. 'f

"' t27 - 2 - 3 +48 --+24 +46

TABLE XXXIII

Correction in Torsional Moments Due to +1,000 lb. at Joint C and No Load at Joint B

-3448

0

A ~~----~--~~--~------~----~--------~0 'V

+17 + 5

+22 +8

+I

96

A

A

.-14,04() ... 4,520 +I 200 - 8,320

TABLE XXXIV

Final Corrected Flexural Mo.ments Due to Loading Shown in Figure 86

/ +29,900 + 6,580 + 1,310 +37, 790

v + 19,420 + 1,840 + 2, 51 0 t 23,770

TABLE XXXV

c

/

Final Corrected Torsional Mo.ments Due to Loading Shown in Figure 86

v 'f ... 14,040 0 +4,160 -4,160 0 +15,380 - 4,520 0 - I ,450 + 1,450 0 - 3,0 10 - I 200 0 + 55 55 0 - 2 180 ... 8,320 0 +2,765 -2,765 0 +10, 190

F '!( 'f E 0 0 ,-0 0

_Q_ 0 0 0

D

-122,000 -188 000 -310,000

D

~15,380 +3,010, + 2 180 -10,190

Note the large reduction in torsional mo­ments (the difference between line 1 and

line 4 of Table XXXV) caused by the deflec­tion at joints B and C.

97

CHAPTER III

DESIGN DIAGRAMS

In order to expedite the design of rein­forced-concrete members, a collection of design diagrams is given in this chapter.

98

All diagrams are self-explanatory.

Figure 92 on page 104 may be used for any number of rows of steel.

I

u

'2.0.f'l ~ I\. I I

I\ II ~j ~'I

19.'' 0 \ 1\ ~00

1\1~ II

1\. 18.0 I

t'."oo " "' II 1\ I

jl""

17.0 "'1 ..... ' r'l~ ,oo

I' "'~P> 16.0 I\

~

1'\ l'_ I'""

15.0 ' \

I\

14. ~

~ P.

I

rF ~ -1-+-+-+-H,, '\ 1\ 0 +, -+--1*-+-HJ 1--H\+-1-t---H-1--+-+--H

~" 13.0 !""'

~~r--. ~~~~~~··~~~~~o ~ "o'

~..t---4---+\+-t-t-+-t~· 0 +- '-I-I\~ ~.-+-1-+---t'Ht-\+-+-+i-:--1 II++ \+-1-t-t--t---t-H 1-1-

"' 1-1- +-o l'o. \1 ~ \ I ' ii1 I~

12.0 1\;J 1\

3.0~~~+4~~4-~~~~-~~~~~~~,,~~~~~~~1\.~N,,~~~j~c,~~ "'- ~ ~ N ~

1.0 H-~~+4~~+-t-+-t-r-t--H-t-++-t-t-++!--t--H-t-t-+-+-t-++~~R, ~--- - - t- - H-- - -- -+v ~' ~e's f-o......lf+-~eg-+.....L.....f-+-~++--HH-+

o -c;-+-+-+-t-+-+--+ o, s '·or-r- -~ +--+-t- 1.1~ l 'l. o

G.A. =Gravity axis or section

p = Steel ratio ::: ~

M' d'

: Ne =- Mp+ N 1:'

fc. c M' :. b(f'L

fs ::-. nfe c~-""~)

n : E.s Ec

X- 0- 1448

FIGURE 90 - Bending and Direct Tension--Reinforcement on One Face of Section.

99

u

c.... 0

Ill 0)

:J

0

>

19D~1-+-~~++~~++~0:~~H-r+~~~~rr~~ ~~~~4-~+4~~~~~~+'~~~~-~~~~

IS.OJ-..4..~"1-1-1-.:l'llr.,.+-+-+ 1 ~-+-1~1 \r!-1--+-.11--!-+-+-+-~+-+-+-+-+-+-t ~+"~~"~~+*4\~~~4-H~~-~~~~~-+-~-+~

H-+-l:-+-++~H-~-+-1--t-l'w--t-l~ + ~~ ~~-+-~~ H--+-+r+-++-+~'~~--1--++-+- 0 T H-, ..... +-+--+-t-+-H

t1.0H-+-+1H-1 +++-~+-t-+-iH-+t-tL.H-~-+-t-+~rr~~ II' \ I\ 1\ I \ I I

~"\· I I

' " 1\! \I 0 1\ 16.0l-i-...p.,j"'.-#-++-+-H-+\:.;,_Hl-~HJ..r+~-!l-~-+-irr+-+-l

I\ N. I Yo 1 I I 0

I 1\~ 11 I

" 1\ 1\ 1\

" ,, I II If' "f-'_ II " l~, !I II

"-I\ 0~

~~~ I

1"'1 ·t- 11 I

'' 0~

ll - I\ : I \ 0 I I

13.01-+...p:j~++~

I! , l

1 I

i I' I I I

2.47·--'1'

rorce .>-

p: Steel ratio for one face of .section =--ft = If

f M M c. = c "bfll. e = N

d' fs = nfe [ (i~Ti) -t] f~ = nfc [ 1- ~ ( ~') J n - E.s -~

X- 0-1449

FIGURE 9~(a) - Bending and Direct Tension--Symmetrically Reinforced Section. d'/h = 0.05.

100

'2.0.0

19.0 ,., ll~

-~ ~a '" I~ ...... JO!>"' ~ I' ~~

~ r.v ~ 1\""4 -~ ~ ·,·

' J"7"" \ I\ 1\.

ll I

1\ \ .~ .... \

15.0

~ \ i \ '' 14 \ ,_

\ \ \ \

:\ \ , .. _ ~n~~~~?~~v~~\~~~+4'~~~~4-~~~~+4~~ 1.-oov

J?"'

-=-·~· \ 1\.:! \ \.i

'' \ '

'\ I \.

' \ \.

r \

\.'

'I\ 1\.

1~ rr 1\.

' 1\ I\

.,.. '" 11' " I{ l

~ l,'o·-~N·--~-~-'"~-H~·4'-'· .. l-l-·~--\.-l.:..+-l ..._, ,,... ~ \.' I\. ll. I{

~ ·~.1 ~ ! 1\. \-'~f'l\i\ ,. 1\ 1\ ~o~1,+4~~~~~m~~~~~-~+~·~~~~Y~~~~

y,, II.. ['. :'1 I '

~ ~~~~ ~ ~ ~ 1\ ' ' a.o wtP.~;~· .1~FI,c;;~,::..t.~~~-~~""',~~-'++.'rN"l.~"-+++-~HI.,+\.-lltrfr'l\-'~

- ~~~c ~ ~ \ ~ ~ ~~, ~1\'

---.

1.0 H-+-+-+-+-+~~+-+-+-+-++-+-+-.... +=~~~lr3...,;"""k-ii"".P"'-~"''.:N-"!IId!1~~J-I ""'IS

N= Tensile fo~

.cl'=ah ~ I

~- b •>: v fs e ofs 1

·~·-A- .;' I -")ooi if j<-

• A • I J;='.- -II\, -

i-At7J :At l 2

h ~~ A c.

: I d1 Z I

• r • j __ t='£~-.· ~ _y__ ....._ ~ f~} I I

->l fer<-

p= Steel rafio for one face of section= ~

bh

fc = c if,_ e= ~

f.= nfc [Q-~) -~ f~ ~ n f c [ 1- ~ ( ~') J

H-+-++-++-v1a 4~~ o1

0 .,UI u -+-+-+--~-+-f-·P',., Q 5 heet '2. of 4 X- 0-1449

FIGURE 9l(b) - Bending and Direct Tension--Symmetrically Reinforced Section. d' /h = 0.10.

101

u

'2.0.0 1-++-+--+-1-++-++1--++-++-lH-++-H-+-++--H-+-t-t--H N = Tensile force -----r--)oo-

~ rt--+-+--+-ir--t--t--t---t-t-t-t-· . -+-1-+-1-++-+-t-t--+-+-+--t--t--t-1 19.0 14--t-l-+-l-+-+-+--+-1-++-+-+-11-+++-+-1-t-++--H-+-t-+-t-t I e L

1<- b --~ v' d -:ah f · -~.!! <-

18.0 :.At h '2. ~ _ )

·~·-r-=-!=f-- h _1 s l f.: n:

l-+-+-+++---l-l--+-H-+++-++-+--+-1f-t-H-+t-+--l--+-1-t-H - t- . : -~ ].. - - - I"

H-+-+---1---4-t-t-++-t---t--1!-t-+++-t-- -++---t-+-t-+ ·- --t- -r--11.0 1-+-H-+-H-++--HH--H-+-t-t-t+-H-t-t+-H-t-H-t-1

1--1--++-+-+--t-- H- f- -\

[\I- - t-f- -v~ ~ r- r-;.; ++t--4f---+--.f.l\++-1--1-+++\t-H-t-t--i 16.0 o- 1-H\O

1-----P:+--!-+-t--+---'f'":__- t- t-f-l :_ -+-++--l-l--'--1--t-+-t--~-.!i r-t-\ t-l t- - --10. t- ---~- - - t-t- 'X r-t-

rt--+-+--+-ir--t--hl'_\"' : J t:t ~~-- t-t- ~ ·t-t;:~ ::o.oz - · • -

0 _c;--r-

IS.o r- li ,, 01 rf---t-~r+--f----f'"""""'oo;j-'-~ • t- ~t- --~- ··1-t-++-+-+. ~ 1-1- s- I-t-

1\ - i\

-f- --1\- f-t- t- -- I' -t-~"'~+\1 ++-+-+-tl-tt-\t-+~-t--'H---t-1 14.0 1-++-+-+\11-++-+~1-+H'-IH:-t-t-t--H+t-t--H-t-H-"H

'ol ' -- f- - --·- -1*-H-41-+1 \ ___ H-· - 1--\,---r-13.0 \ ~\ I..., I"

~~~g~4~l\+++1-rP~~+Iri-++"~Tt-t-++-t~H I 'Z.O t- ---- I\ !""' """ 1"•\.._.L-';,'-+--1-+-+1'\.~1-1+-\f-t--t--t---f+-1--1

'-"' '" \ _l 1*+--+---tt,l-+~1-\t-\t-·f'"'tr-.:i\-· __ t- f"K . -- - ~' --- -- - -- -lt-

1---f:-1 \+-++-l\-1\+++-+11 \r++-t-- r-'f'i - - -- \ ~ !\ 1\

~ I\ I~

Ac.~ : z k.h M ..1...... y: ~~z;. ........... \1 ......___. •• ~d' ~ • J __ lf. I

£.! ->I c I<· 5 I I

p= Sfee/ ratio ror one Face or sech'on = .it

bh

fc. :. M e = 1:1 c bh2 N

f s = nfe [G-!) _, J

f~ :. nfc ['-t c~·;J

2.0~+-+~~+-+~~~T~~~~~~~t-t-~~~~~l\rf~~-~rT~~~~ 'too I"- .... ~ ~ _l~tl.:

r.oi-++4-+-H-++--Hrr++~Hrt+~-rt++1,_r~~~~~~~~~~~~~ ~-~+-++-l~r++-l-r--+-r~i·~~h~~~++~H-~~-,~~~~~~~~~r~l,

t--)_t-_t-_t_:t-_:t-~_:t_:j-_·1~~!~ttr t/ r~.t-ot--t+-_+~-++-t-+-+~_t_;__t3+1 o+-t-+4-t-t_,-+-___,t-=t-f'4't-""'~"t~~J sheet 3 of 4 X- 0- 1449 4.oes..:,

FIGURE 9l(c) - Bending and Direc~ Tension--Symmetrically Reinforced Section. d'/h = 0.15.

102

u

t-+-+-+---+--+--+-1--+-+-+- ·+t-HH-++-H-!-H--++-l--+-1-~ rr+-+,~~+++4-+-H-4~~-+~--~--+-~++~ rr++,~~+++4~-H-· ... ····~+-+-+++-+-+-~~

19.0 H-+-++++1-+-+-+-H+H-++-+-4-+-~++-+-1-l-+--J.....j

18.0 H-H-++-+-H-+++++-1H-+++++1~+-I--+-1-l-l-l-l

11.0 H-++++-t-t+t-t+-t-1-1-+++--H-+-H-++-+-1-l-+--1--1 t-t-+-+++~-1-1--+-l··· .. ··+-+++-H-+--t-+--H-++-+-t-+-1

16.0 H-H-++-+-H-++-H+H-+++++-~++-+-1--f.l\1-l-1 H-++++-+-1-++++-+-+-1-++++-+--l--+·+-+++-+--+--\--+-l

,..

I\

lf2.:: 0." s 11.0 I 1\ I / .!"'-

p :: Stee I ratio f'or one race of section = fl.,*

f - C M e-M c- bJ1'L -N

cf fs = n fc L l-kh -t]

FIGURE 9~(d) - Bending and Direct Tension--Symmetrically Reinforced Section. d'/h = 0.20.

103

+If_ 1'1 :z. .3 .4

+9

+8~

+1~

+6~

+5

t +1-

t -z

-3

-8~

.3 .4

Values of k .s .6 .1 .8 .9

I

1!:

~

I!•: I

~ l ~§·

I

n,•· I

n.

I

._,

1.0

p =Steel ratio for first row of tensile steel=~

~ = t,- =I; r2 = ~ ; etc. e =.!;} n = fc-

A= snp (iD-E) = A section constant

B=6np(~D-1.5E+F)=A section constant

IN GENERAL THE EQUATION FOR k IS :::}c:k'+ n1k1 + n1 k+ n1 r:~o 0

TO FINO k TO FINO k

when N =Tensile force when N = Comp. force

n,•-3(t+~) n,=+~ ({--~)

n,• -6np f 0 +A n, :+6np ~ D +A

n1 •+6np fi" (D-E) -B n1•·6np ~(D-E)- B

Draw line from n, to n,,at Draw line from n, to no, ctf 1,. interse_;t. of n,,read value of k. intersect. of n,,read value ofk

+9

+t

L.I--J.+!-++t-t-t-t-ti-t-H-t+-+!J!I•:--;~,.-li~J.-· ~-TO FIND STRESSES FOR EITHER A TENSILE ~ OR COMPRESSIVE FORCE

.s .6

Values of k .8

~:

b,•-9 and b1 •-6B Draw line from b, to b,, at intersection of k,

read v9lue of b, (dotted curves)

or b •. ={+t.5k-k 2; then c= b:-A

and fc = ci\h; fe" fs2• etc. as in fig. above +=Tension;.-= Camp.

TO FIND k FO~ BENDING ONLY

n,=-3 n2 =-6npD n3•+6np (D-E) Draw line from n, to n3, at intersection of n, read value of k. Repeat, reversing the signs of n,, n, and n,. Use averaqe of the two values thus obtained. The stresses are found exactly as above.

t EXAMPLE r---.. 10!'····>1 o Given: ~- ~.;:J A :nember,width b•IO"~nd d_epth hxts" with ! At•I.S"··· ~ retnforcement on tens1on stde A,= l.&in~

1!! l: and a second row of reinforcement 1!: A2•o.1s;~ ? A2= 0.15in', subject to a moment M • !00,000 ! ~-T' • ..L in.lbs. and a direct load N • 10,000 lb. L '--- compression. !'l•IO;

e• ~~~~0~~0

•30.in.; t·*•2; ar' ~·0.1; o2 • ~·0.8

~ np- 11~::;-o.l; r2•0j:;=o.s; D•I+0.5=1.5

·t

-· -~

[ ~ E=O.IXI+0.8x0.5=0.5; F=O.I2 xi+O,B2 x0.5=0.33 A=6x0.1(~ xi.5-0.5) • +0.15; B=6x0.1 (~x 1.5-1.5x 0.5+0.33):+0.198_6

~ .9 1.0

n,= 3( 2-~) =.~.!'>; n,• 6><0.1><2 x1.5+0.15=+ 1.95

n1 • ·I>KO.Ix2 ( 1.5 -0.5) -0.198 • -1.398

b,• -9 (always); b3=-6x0.198•-1.19

Draw a straight line frorn n,•+4.5 to n,•-1.398 and ot the intersection of this line. with the curve n,•+l.95, (marked 0) read the value of k•0.~1 at top or bottom of sheet. Then draw a straight line from b,=-9 to b,• -1.19 and where this line inter­sects k• 0.31 read the value of b,• 0.9S on dotted curves for b1. ~arked El)

and

then c • 0_ 95~ 0_ 15 • 7.50

fc•i50x ~~~~ii' •IOOO lbs./in~

t, 10 x 1000 .fl:..Sil.1)• + 14,300 I bs./in.• I \ .37

f11• lOx 1000 (1 ~~78-0·- 4600 lbs./in.•

*For bending only k1 Is divided by ~.Since ~--.f·o.

h

t .s2 '-0

c .... 0

fll u ::s c; >

X--0-1450

FIGURE 92 - Bending and D:trect Stress in Reinforced Concrete Members.

104

c: ·a. u.. 0

VALUES OF k AND j FOR GIVEN VALUES OF pn AND p'n IN

RECTANGULAR BEAMS REINFORCED FOR COMPRESSION FOR RATIOS OF f FROM 0.05 TO 0.30

n

8 10 12.

n=t IS

A' p·= o-a A' n-l { Takinq into account the reduction of

(Approximate) p' = bds x --r;- compression area of concrete due d' to the presence of reinforcinq steel

, _ ( k -a) fsk _ M k fs - nfc \-k- fc = n (I- k) or fc- t)di X pnJ (l-k\

In the followinq diaqrams

k = V 'ln(p + p· ~)' + n1(p+p'}~- n (P+P'}

k1 (1-~)+ 2.p•n(k-i)(1- %) j = -------..:,.,..---_;,;,_

k2 + 2p'n (k-f)

Sheet I of 9

FIGURE 93laJ - Rectangular Beams Reinforced for Compression-­VaJ.ues of k and J for Given Values of pn and p'n. Ratios of d'/d from 0.5 to 0.30.

RECTANGULAR BEAMS REINFORCED FOR

COMPRESSION

lllO.IS ~----H--Jo~+++R~I-+--fr·fH~-1--Jff-+-~f--llr--7---ft--'1 1.1.1 ::J .J <(

>

<;heet '2. of 9

FIGURE 93(b) - Rectangular Beams Reinforced for Compression-­Values of k and j. d' /d = 0.05.

105

n-1 -n-

0.875 0.900 0.911 0.933

X-0-14!51

X· 0-14!51

c: ·a.

u. 0

~0.15 ... ::J .

..1

~

0.30

c: ·a. LL.. 0

Sheet 3 of 9

FIGURE 93(c) - Rectangular BeamB Reinforced for Co.mpression-­Values of k and j. d 1 /d = .10.

RECTANGULAR BEAMS REINFORCED FOR

COMPRESSION VALUES FOR j AND K

X-0-1451

~ 0.15 ...,-_---t-t.....,.H-; dr-I-+-H+-T+-f--IL-+..-K--+--+-+--I-+1'+---I--:-f--+--Jt---iof-~=:;_f---,L,c.-..f+---,~-7!--+T--+t :::> ...

..1 <(

>

0·%~.o~~~~~~~~~~~~~~-'~o.~zo~~~-o.Lzs~~._~o.~3o--'--'-o~.3~s---'--o~.4-0--d~' ---o~~

VALUES OF pn a :.15

Sheet 4- of 9 X-0-1451

FIGURE 93(d) - Rectangular Beams Reinforced for Co.mpression--Values of k and j. d 1 /d = .15.

106

0.30

o.w

c: ·a. u. 0

:30.15 => ..J ct ::>

0.10

-

RECTANGULAR BEAMS REINFORCED FOR

COMPRESSION

0 0: II·

0')

0·%~.0~~~~~~~~~~~~~~~~~~~~~-0~.30--L--L-0.3~5~-L--0~.4~0-~~.~-0~.45

0 =.ZO Sheet 5 of 9

FIGURE 93{e) - Rectangular Beams Reinforced for Co.mpression-­Val.ues of k and j. d' /d = .20.

RECTANGULAR BEAMS REINFORCED FOR

COMPRESSION

Sheet 6 of 9

FIGURE 93(f) - Rectangular Beams Reinforced for Compression-­Values of k and J. d' /d = .225.

107

X-D-14!SI

X- D-145;

0.""

0.1 ..

H-H-

0.10 O.IS 0.'20 0.15 0.~0 0.35 VALUE.S OF pn

Sheet 1 of 9

FIGURE 93(g) - Rect~ Beams Reinforced for Compression-­Values of k and J. d' /d = .25.

Sheet 8 of 9

FIGURE 93(h) - Rect~ Beams Reinforced for Compression-­Values of k and J. d' /d = .275.

108

0.40 d' .0.45

cr=.'l.S X-D-14el

X-D-1411

c: 0.

""" -

,,., : ·:··· '-:···- ·-t L\1 - ' \.' 0'1.0 o.os 0.10 0.15 0.20 0.'2.5

VALUES OF pn Sheet 9 of 9

0~ 0.35

FIGURE 93(i) - Rect~ Beams Reinforced for Co.mpression-­Values of k and j. d'/d = .30.

The clockwise moment shown causes diagonal tension and compression in direction a-a and Cl'-a' respectively. On opposite fClces direction of forces <:lre opposite .

• I. H--. ~::f-LC H--0::± ' + + H- H-H-

0.40 .i 0.45 d =.30

X-D-1451

H-1-+-~+++-l,-l--1--l--. -J....i±:J ~_tf+l . .LtfR= I rt-· r- , i+ tttm::ttl=l-ttiH-- =H-H- ~r-~tt- -4+·-t-~--t-+, +-. H. i-t-+-T_--;+, ._-H'-, .--t-+-+-f. HH-+-+--"rl, ·-++-1-H++-+-r.~--+-+-~++-1--H--+-HH++H gE§§~I!~~~~33~~~~,~-,-iff+Ef-i8~-r3r3~·~~~~ggE§~~ftE58B~!EfEE8~~if;£:§§~Iiff~~~liE£saaa~ 0

1.0 2.0 3.0 4.0 s.o 6.0 7.0 8.0 9.0 10.0

VCl{I.Aes of~

TORSIONAL SHEAR IN RECTANGULAR BEAMS X· C-145?

FIGURE 94 - Torsional Shear in Rectangular Beams.

109

900

:::::::::: ::;::::::: :::::::::--t:----- -- -t:;:::: -------t:::::::= :::::::::-::---

~ ----- .....--: 800

~ v ~ ~ ~ ~

/ v ~ ~ ~ v v

700 p= 0.005

/; ~~ ~-----p=OOI p ,o.oz

f/1 [7l I

z ----P :::10.03

0 600 --p=0.04

t/)

VI~ v 1.4 1.6 1.8 . '2.0 2.2 2.4 a: VALUES OF e;h u1

p- As 0..

~II - Ag ui As=- Total steel area CD -lSOO

i// Aq = bh

z 1 +Ge - fc=fa t+ch6e ~

v h

(./)

From joint committee report, (./)

LL.l section 861, page 73 radius ct. 400 t- of q('ration of transformed (f) stee areas is neq lected. IJ..J -l CD

N <I: -< A $300 I

I

0 I I -l ~- I

-l I • ! •• Q)

<( I I I I I I I I '·As. I

~(<t.. I t ..c - I -zoo I I

I I I

I I I I I

.... i •. : 1

I I

~--- b ---~ 100 fc·= zooo#/in~

fs = 16000#/ in.2

n = 15

00 0.'2. 0.4 0.6 0.8 1.0 1.'2. 1.4 VALUES OF e/h

SHEET I OF 4

FIGURE 95{a) - Approximate Values for Allowab~e Combined Axial and Bending Stress--Tied Co~umns. f

0' =-2000#/1n2.

110

X-0-14!5!

1100 -----~--~ -~ ~

~ ~ ~ -

l..---- ~ ~ ::::---

.,v:;:: ~ ~ ~ 1000 __....

%: v /

~ ~ ~ ~ /

900 ~ t'i ~ k;=0.005

//; ~ / .. p,O.OI "Y- ---P = 0.02

z h tj;, l7: Pi0.03 -~BOO

.. ~ -, --p=0.04

//; 'II 1.4 1.6 1.8 2.0 2.2 2.4 a: VA LUES OF e/h uJ 0.

///; I p = !: (f)

co As== Tota I steel area ~700

r111 Aq= bh z - 1+6€

~ fc = fo 1 + ch se

~II h

If)

From joint committee re-(/) w 0::600 port, section 861, pa~e 73 1-

rll radius of qyration o

If) transformed steer areas

UJ is neq!ected.

~soJ < N + I I

_l I

_l 1..-- I I • l ••

<( I Q) I I I I

I ,

I I I I ···As, I

I y ~--' ..c - - '(

400 I I , I I

, I , I , I , I ... ·'·. ~

1 I

k--- b --~ 300 fc'= '2.,500 #/in.'

fs = 16,000#/ in.2

n = 1?.

2000 0.'2. 0.4 0.6 0.8 1.0 1.2. 1.4 VALUES OF e;h

SHEET 20F4

FiaURE 95(b) - Approximate Values for Allowab~e Combine~ Axial and Bending Stress--Tied Co~umns. f 0 ' = 2500#/in •

111

x- o-1453

1300

!ZOO

z (j l/)1100 0:: Li:J a.. tri co .J

t:tooo ~ (/) (f)

LJ.1 a: r- 900 (/)

IJJ _l

i fi ~BOO~

roo~·

600

5000

~-~;: ~ l=~

~-

------~ ~ -

p=~ ~ v ~ ~ b::::: ~-e:::::: b:=== p =0.03-. ~ ~ __,_-

p =0.02,,

~ V- /' ~ ~ . .1 / ~ p=O.Ok / p :.0.005 ~ ~ ~

/

7; VI ~ t.4 1.6 1.8 2.0 2.'2. 2.4 VALUES OF e;h

Vjj w p= _&_

h v;~ v Ag As= Total steel qrea Aq= bh

!!J VII I+~ fc =(I +Ch 6E

h

From joint committee re-

~I. ~ port, section 861, page 73 . ! radius of qyration of

'I# transformed steel areas is neglected.

'I -< A N

I I I I x---- I

: •t • • I OJ I I I I I

I

I ' I

I ''As, I

<t-y I . ~ ..c. -:_ t---

\ -

I I I I I I I I I . I

... . i .. I

t I I

k---b ---~

fc• = 3,oooff:;in.2

f5 = 16,0QO#fin.1

n =-10 0.2 0.4 0.6 0.8 1.0 1.'2. l.4

VALUES OF e;h

SHEET 3 OF 4

FIGURE 95(c) - ApproxilDate Values for Allowable Combin~ Axial and Bending Stress--Tied Columns. f

0' = 3000/1/in •

112

X-D-1453

~

1400

1300

!

z -~1200

c:! uJ 0. ({) d) _rttoo z -,_;= V) (/) u.J· ct: 1000 . 1-lf)

uJ ...J co <(

3900 0 ..J .J <!

800

700

6000

::::::::: :::::::::: ~ ::::::::: ::.-------y ~ /

,__- = ~ ~ ~ ~ .;:;::::; --~ /: :;:: ~ -/ /

/ ~ ~ Z=o.oos

V/; ~ r--p 1=0.01

'~--P=0.02

h r-7-:z v-p~O.p3 --- p = O.D4

II! II 1.4 1.6 I.B 'l.O 2.'2. 2.4 VALUES OF o/h

p=..&

J/h I Aq As= Total steel area

II/ Aq= bh

~ I+~ fc= fa I+C ~

Ill Q From joint com.mittee re-port, section 861, paqe 13

'lt radius of qyration of transformed steel areas is neqlected

~ij < N A I

II I I I "-.,.-- I : • ! •• Q)

I ' I ' I I ·.

I I I I ··As, I

<L I y .c. - r--- -

I I . I I I . I I I I I

••.• i .• I y ____ I I

k---b --->1 fc' = 3,500 *lin.' fs = 16,00ott/i n~ n =to

0.'2. 0.4 0.6 0.8 l.O l.'l 1.4 VALUES OF e;h

SHEET 4 OF 4

FIGURE 95 (d) - Approximate Values f'or Allowable Combine~ Axial and Bending Stress--Tied Columns. fc' = 3500#/irr. •

113

X-D-1453

AMERICAN ~TEEL&WIRE COMPANY's STEEL WIRE GAUGE NO

0000

000

00

0

I

2

3

4

5

6

7

8

9

10

II

12

13

14

AREAS AND PERIMETERS PER FOOT OF WIDTH FOR VARIOUS SPACING OF WIRE

O::CJ) UJLJJ CENTER TO CENTER SPACING t-~ wu 2 .. 311 4" 611 a" 12

11

~2 c:x--z o- As lo As ~0 As ~0 As ~0 As ~0 As ~0

.394 .731 7.423 .487 4.949 .365 3.711 .244 2.474 .183 1.856 .122 1.237

.363 0 I 00 I • 0 ••• . . ... . . . . . .310 3.416 .206 2.278 . .... • 0 •• 0 ..... . .....

.331 .. . . . 0 •••• 0 •••• .. . . . .258 3.120 .172 2.080 • 0 ••• I o o o 0 ..... .. ... .. . 307 .443 5.777 .295 3.852 .221 2.889 .148 1.926 .II I 1.444 .074 .963

.283 .377 5.334 .252 3.556 .189 2.667 .126 1.778 .094 1.334 .063 .889

.263 .325 4.948 .216 3.299 .162 2.474 .108 1.649 .081 1.237 .054 .825

.244 .280 4.594 .187 3.062 .140 2.297 .093 1.531 .070 1.148 .047 .766

.225 .239 4.247 .159 2.831 .121 2.123 .080 1.416 .060 1.062 .040 .708

.207 .202 3.902 .135 2.601 .10 I 1.951 .067 1.301 .050 .975 .034 .650

.192 .174 3.619 .116 2.413 .087 1.810 .058 1.206 .043 .905 .029 .603

.177 .148 3.336 .098 2.224 .074 1.668 .049 I. I 12 .037 .834 .025 .556

.162 .124 3.054 .082 2.036 .062 1.527 .041 1.018 .031 .763 .021 .509

.148 .104 2.795 .069 1.864 .052 1.398 .035 .932 .026 .699 .017 .466

.135 .086 2.545 .057 1.696 .043 1.272 .029 .848 .021 .636 .014 .424

.121 .068 2.271 .046 1.514 .034 1.136 .023 .757 .017 .568 .0 II .379

.106 .052 1.989 .035 1.326 .026 .994 .017 .663 .013 .497 .009 .331

.092 .039 1.725 .026 1.150 .020 .862 .013 .575 .010 .431 .007 .287

.080 .030 1.508 .020 1.005 .015 .754 .012 .503 .008 .377 .005 251

SECT I 0 NAL AREAS AND PER I METERS OF AMERICAN WELDED WIRE FABRIC

16.._

As ~0

..... . . ...

. .... .. .. .

. .... . .... .055 .722

.047 .667

.041 .619

.035 .574

.030 .531

.025 .488

.022 .452

.018 .417

.015 .382

.013 .349

.0 II .318

.009 .284

.007 .249

.005 .216

.004 .188

X-D-1454

FIGURE 96 - Sectional Areas and Perimeters of Ama.rican Welded Wire Fabric.

114

•• AREAS AND PERIMETERS OF ROUND BARS PER FOOT •• tX 1- SPACING <! J.Ll :X: <( LLJ :E ~ 2" 3" 4" 5" 6" 7" 8'' 9" tO" 0 0: a: ex: w L&J 3= As ~0 As Lo As Lo As Lo As !o As l:o As Io As Lo As Lo a.

3.18 1104 1.18 .376 .66 7.08 .44 4JZ .33 3.54 .26 2.83 .22 2.36 .19 2.02 .17 1.77 .15 L51 .13 1.4'2 t,;z 1963 1.51 .66S 1.18 9.42 .79 6."28 .59 4.11 .41 3.77 .39 3.14 .34 2.69 .29 2.36 .26 2.09 .24 r.ss ·~ 3068 1.96 1.043 1.84 11.16 1.23 7.84 .92 5.88 .14 4.70 .61 3.92 .53 3.36 .46 2.94 .41 2.61 31 2.35 ~ .4418 2.36 1.502 2.65 14.16 1.77 9.44 l.33 7.08 1.06 5.66 .88 4.12 .76 4.05 .66 3.54 .59 3.15 .53 2.83 Ys 6013 2.15 2.044 3.61 16.50 2.41 11.00 1.80 8.1.5 1.44 6.60 1.20 5.50 1.03 4.11 .90 4.12 .so 3.61 .7Z 3.30 I 7854 3.14 2.670 4.71 18.84 3.l4 12.56 2.36 9.42 1.88 7.54 t.57 6.28 1.35 5.38 LIB 4.11 1.05 4.19 .94 3.17

• • AREAS AND PE.RIME.TERS OF ROUND BARS •• <!

a. \- NUMBER OF BARS <{ ~ :z:

u..l (!) 2 3 4- 5 6 7 8 9 tO 0 C! " Uj <! u.J

3: As Lo As Lo As Lo As Lo As Lo As 2:o As Io As ro As Lo a. Ye .1104 1.18 .376 .221'1.36 .33 3.54 .44 4.12 .55 5.90 .66 7.08 .71 8.26 .88 9.44 .99 10.61 l.tO 11.80 lA .1963 1.51 .668 .39 3.14 .59 4.11 .79 6.28 .98 1.85 1.18 9.4Z 1.31 10.99 1.51 12.56 1.17 14.13 1.96 15.10

~ .3068 1.96 1.043 .61 3.92 .92 5.88 1.23 7.84 1.53 9.80 1.84 11.16 2.15 13.12 2.45 15.68 2.76 11.64 3.07 19.60 34 /+418 236 1.502 .88 4.12 1.33 1.08 1.11 9.44 2.21 11.80 2.6~ 14.16 3.09 16.52 3.53 18.88 3.98 21.24 4.42 Z3.60

'Ya .6013 Z.15 Z.044 1.'2.0 5.50 1.80 8.1.5 2.41 11.00 3.01 13.15 3.61 16.50 4.11 19.25 4.81 22.00 5.41 24.15 6.01 21.50 I .7854 3.14 Z.GlO 1.51 6.1.8 2.36 9.42 3.14 IZ.56 3.93 15.10 4.11 18.84 5.50 21.98 6.28 25.12 7.07 28.26 1.85 31.40

II mf AREAS AND PERIMETERS OF SQUARE BARS PER FOOT • • LJJ

C' 1- SPACING <{ UJ ::r N IJ.l -~ tO '2." 3" 4" 5" 6" l" 8" gu 1011 - 0:: ffi u:i V) <! ~ As La As L:a As La As La As La As Eo As ~0 As Lo As La a. Yz .250 2.0 .850 1.50 1'2.0 1.00 8.0 .75 6.0 .60 4:80 .50 4.0 .43 3.43 .37 3.00 .33 2.61 .30 2.40 I 1.000 4.0 3.400 6.00 24.0 4.00 16.0 3.00 12.0 2.40 9.60 2.00 8.0 1.1f 6.86 1.50 6.00 1.33 5.33 1.20 4.80

IYa 1.266 4.5 4.303 7.60 27.0 5.06 18.0 3.80 13.5 3.04 10.80 2.53 9.0 2.17 7.71 1.90 6.15 1.69 6.00 1.52 5.40 I~ 1.562 5.0 5.313 9.38 30.0 6.25 20.0 4.69 15.0 3.75 12.0C 3.12 ro.o '2.68 8.57 2.34 7.50 2.08 6.61 1.87 6.00

II m AREAS AND PERlME.TERS OF SQUARE BARS •• .:( cr. ..... NUMBER OF BARS LL1 ~ :r:

N u.l (.!) 2. 3 4 5 6 7 8 9 10 - C! ~ Uj If) <! u.J As La As La As La As Lo As La As La As La As La As La 0. 3: Yz .250 2.0 .850 .50 4.0 .75 6.0 1.00 8.0 1.1.5 10.0 1.50 12.0 1.15 14.0 2.00 16.0 2.25 18.0 2.50 20.0 I 1.000 4.0 3.40C z.oc 8.0 3.00 12.0 4.00 16.0 5.00 20.0 6.00 24.0 l.OO 28.0 8.00 32.0 9.00 36.0 10.00 40.0

tYe f.'266 4.5 4.303 2.53 9.0 3.80 13.5 5.06 18.0 6.33 zz.o 1.60 27.0 8.86 3!.0 10.13 36.0 11.39 40.0 12.66 4&.0 t!4 1.562 5.0 5.313 3.12 10.0 4.69 15.0 6.25 20.0 7.81 zs.o 9.31 30.0 10.94 35.0 12.50 40.0 14.06 45.0 15.63 50.0

1-0-1455

- -FIGURE 97 - Sectional Areas, Perimeters and Weights of steel. Bars.

115

APPENDIX A

DERIVATION OF EQUATIONS

General Equations for Initial End Moments

In order to derive special equations for the beam constants for different condi tior.s of loading and methods of end supports, it is convenient to derive general expressions for the moments at the ends of a member. Then from these general equations any special equation may be derived for any of the beam consh.nts merely by introducing into the general equations the restricted conditions of the special case.

The general expressions for the end moments of a member will be derived from the principal propositions of the moment­area method. These propositions may be stated as follows:

1. When a member is subjected to flex­ure, the algebraic difference in the slope* of the elastic curve between any two points A and B, is equal to the algebraic sum of

M the area of the Eix diagram** for the por­

x tion of the member between A and B.

*For obtaining the difference in the slope between any two points, the same sign system as defined in Chapter II may be used. Rest3.ted in different terms, this sign system reads: If the tangent to any given point on the elastic curve has ro­tated in a clockwise direction from its original position, the angle through which it has rotated is called negative (- ). Con­versely, if the tangent has rot!lted in a counterclockwise direction, the angle is positive (+).

**For totaling the moment areas and mo­ments of moment areas, the sign system as defined in Chapter II cannot be used. The sign of a moment for this purpose must be defined with respect to the stress it causes in the member. Moments caus­ing tension on opposite sides of a member have opposite signs. Either moment may be called positive or negative, since only the numerical result is used.

117

2. When a member is subjected to flex­ure, the distance to any point A on the elastic curve, from a tangent drawn to the elastic curve at any other point B, is equal to the algebraic sum** of the static moment

M about A of the area of the EI x diagram

X between i\. and B. Measurements must be made normal to the original position of the member.

Referring to Figure 1, and propositions numbered 1 and 2 above, gives:

g A- 8B = (AAMAB- ABMBA

6 - L8 B

1 - Ao)E (1)

(AAXAMAB- ABXBMBA

1 - AOXO) E (2)

Equations (1) and (2) are general, except for the signs concerning the external load­ing. For most conditions of loading, how­ever, the signs of the end moments, as expressed by equations (5) and (6), may easily be determined by inspection. Thus, .if the loading shown in Figure 1 of Chapter I had been upward, the signs of equations ( 5) and (6) would have been reversed.

Since At:..._, AB, and A0 are defined to be M M

the areas of ~ diagrams r8.ther than Et X X

diagrams, equations (1) and (2) must be multiplied by the reciprocal of E as shown.

Solving equation ( 1) for MBA gives:

Substituting this value of MBA in equa­tion (2) and solving for MAB gives:

L- XB +A (X -X ) 9B A B A

(3)

Similarly, solving equation (1) for MAB and using that value in equation (2) gives

L AB(XB- Xp)

Ao(Xo- XA)

AB(XB- XA)

t] (4)

The last term of equations (3) and (4) is seen to contain quantities depending on the external loading and the shape of beam only. Therefore, if no deflection and ro­tation of the ends of the member takes place

(5)

and

(6)

It is recognized that equations (5) and (6) are the general expressions for the moments at the ends of a member of any shape, due to any loading, when both ends are held fixed in their original positions. These moments, called fixed end moments, are generally denoted by CAB and CBA·

118

It can be proved that the coefficient of 9B in equation (3) is equal to the coefficient of 9 A in equation ( 4·). The proof is as follows:

L- XB c::::::: XA

~A (XB - XA) ::::::- AB(XB - XA) (a)

Canceling the term (XB ~ XA), supposition (a) may be written

(b)

Referring to part (g) of Figure 1, the gener~l expression for A A is

(7)

and the general expression for XA is

LL-x ~ --u;- x dx

}L-X dx 0 Lix

(8)

Similarly, from part (h) of Figure 1,

L X AB = 1 - dx

0 Lix

and

L x2 t ~ dx

L X I LI dx 0 X

(9)

(10)

Substituting these values of AA, XA, and XB in supposition (b) gives

L 2 . I X dx

(L - O ~ )12_ dx-=::::

L x 0 Lix :::::-

-6 Lix dx

JL-x xdx

( )( LI ) }L-xdx 0 x

0 Llx L L- x I LI dx 0 X

Dividing out equal terms and rewriting, gives

L L x2 -=:::: LfO LXI dx = /0 - dx=

X Llx ::=-

L 10 L-xxdx

Lix

L x2 1- dx 0 LIX

L L/~ dx

0 Lix

Therefore, supposition (a) is proven and

L-X B

This relation may be used as a check on the correctness of the computations.

By observation of equations (3) and (4) it is seen that the sum of the coefficients of 9 A and 9-B in each equation is equal to

the respective coefficients of ~ . To make

this true is the reason for not dividing out L in the terms containing A •

Taking advantage of the above proof and observations and multiplyinf and dividing

the terms in parenthesis by ..£... , equations L . (3) and (4) may be written in the following convenient forms:

119

(12)

where

(13)

c2 L (L- XB)

=-Ic AA (XB- XA)

L XA =-

Ic AB(XB- XA) (14)

c3 L (L- XA)

Ic· AB(XB- XA) (15)

.[he reason for multiplying and dividing by

~ will be shown later.

Specific Equations for Beam Constants for Given Methods of End Supports

By use of equations (11) and (12), it is a. simple matter to derive equations for any g1ven method of end restraint. For the S'ike of illustration, the equations of case I. on .page 17 of Chapter I will now be derived. It 1s seen that in case I the beam is held fixed in its original position. Therefore, 9 A = 9B = A = 0. Then by equations (11), (12)., (5), and (6),

Ao(XB- Xo) MAB = CAB = ~~~--~

AA(XB- XA)

Ao(Xo- XA) AB(XB- XA)

(16)

(17)

For shapes of beams for which the co­efficients C1, C2, and c 3 are available, it is convenient to have CAB and CBA ex­pressed in terms of these coefficients. Re­writing equations (16) and (17), it is easily recognized by equations (13), (14), and (15) that

CAB = Ao[AA(X:B- XA)

- AA(X:- XA)J

Ao [cl- ~o (Cl + c2~f (18)

CBA = AofAB(::- XA)

-AB(::- XA)]

(19)

By definition, the stiffness factor at end A for case I in Figure 25( a) is the moment required at end A to rotate this end through an angle of one radian when end B is held fixed. That is, 9A = 1, 9B = A = 0, and, since the rotation is due to the end moment only, CAB= CBA = 0. Therefore, by equa-tion (11)

(20)

Similarly, for obtaining the stiffness at end B, 9B = 1 and 9 A = A = CAB = CBA = 0, and by equation (12)

Eic KBA = L (C3)

Eic L (L- XA)

L ~ AB(XB - XA)

120

E(L- XA)

AB(XB- XA) (21)

By definition, the carry-over factor for M

moments from A to B is BA. MBA is MAB

caused by MAB only. The conditions are that 9B = A = 0 ·and 9A. has any value. Then by equations (11) and (12)

Eic L (C2QA)

Eic-L (C1eA)

(22)

The conditions for carrying- over mo­ments from B to A are that eA = A = 0 and 9B has any value. Then by equations (11) and (12)

c2 AB(L- XB)

c3 AA(L- XA) (23)

For beams of constant section the areas AA and AB are triangles with the length

L of the beam ~s their base and 1.. as their I

height. Therefore,

L AA = AB = 2f

The respective dist~nces from their centroids to end A are

and

Using these values of AA, .AB, XA, and XB in equations (13), (14), and (15) gives

C 1 = 4; C 2 = 2; and C 3 = 4

Now~ from equations (18), (19), (20), (21), (22J, and (23)

Xo I * CAB = A6(4- 6L) L

Xo I * A(-2+6-. )-o. L L

4EI L

2 r AB = rBA = 4

1 2

Deriv-ation of the beam constants for other·cases in Figures 25(a) and (b) is made in a similar manner arid will not be shown.

Equations for Plotting Curves of the Slope­Deflection Coeffictents c1, c2, and c3

Inserting the general expressions for AA, AB, XA, and XB of equations (7), (8), (9), and (10) into equation (13) gives

(24)

It is necessary to eliminate Ic and L from this equation in order to make the ex­pression for C1 applicable to any beam of any length. This may be done by letting

Ix = 0 Ic x = zL and dx = Ldz

*Since I is constant, it would not be neces­sary to multiply by I if it were omitted when computing A0 • X0 is not ·affected by I wh~n constant. Theref?re, X0 is also the distance to the centroid of tlie simple beam moment diagram.

121

When making these substitutions in equation (24), it must be remembered that the limits of integration are now from 0 to 1, since 1 is the value of z when x = L. Further, when making the above substitu­tions, it will be apparent why equations (3)

and ( 4) were multiplied and divided by 1c

L as shown in equations (11) and (12). If this had not been done, it would not have been possible to eliminate L and· Ic in equation (24).

Making the above suggested substitutions in the equation for C 1 and reducing gives

1 z2 -6 -w-·dZ

~1

(25)

If, to the denominator of equation (25) the two quantities

1 z2 1 z . 2 'o "T dz { "J dz

and [ 'o1 '¢2

dz r are added and subtracted, the equation for C 1 changes to

Let

1 z2 = A (27) L -dz

0 0

~~dz B (28)

'o1 ~dz D (29)

Then

A (30) A.B- n2

and it may be shown that

D (31) AB- n2

and

B (32) AB- n2

In this form the work of computing numerical values for given beams is sometimes simpli­fied a great deal.

Specific values of c1, c2, and C3 will now be derived for a member of two constant sections, as shown in Figure 98.

:C------- ---X: ZL-------------1

FIGURE 98 - Beam w1 th Two Constant Sections.

For this beam Ic = bt: ; for the heavier part

· bd3 m3 3 of the beam Ix = 12 = m Ic. By the

substitution on page 121 the moment of inertia at any point x is Ix = ¢Ic. Therefore, for

the portion aL S ¢ = 1, and for the portion (1 - a)L, 0 = m .

'Then, by equation (27),

a a A = I z2 dz + - 1- I z2 dz

0 m3 1

122

by equation (28)

B = la(1 - z)2 dz + ~3 11

(1 - z)2 dz 0 m a

3 3(a - a2 + .!-.) (m3 - 1) + 1

3 3m3

~nd by equation (29)

a 1 D = I z(1 ... z) dz + 1

3. I z(1 - a) dz

0 m a

a2(3 - 2a) (m3 - 1) + 1 6m3

A family of curves may now be plotted for c1, c2, and c3 with a and m as vari-ables. These curves are shown in Figures 30, 31, and 32.

Equations for Plotting Curves for Initial End Moments

The general expressions for A0 and X0 are

(33)

{34)

Using these expressions in the equation for CAB as shown for case I on page 17, Fig-ure 25(a) gives

(35)

Now, by reducing and resorting to the same substitutions for Ix, x, and dx as were used when deriving the coefficients Cv C2, and C3 as shown on page 121 gives

This is a general expression for CAB due to any loading and shape of beam. In order to proceed further, a definite loading and shape of beam must be assumed. For the purpose of illustration, assume a uniformly distributed load of w pounds per linear foot applied to the beam shown in Figure 98. The reaction of this load on a simply supported

beam is w2L , and the moment at any point

x is

wL x _ wx2 2 2

wL2 2 -2-z

wL2 --z 2

For values of z from 0 to a, 0 = 1, and

for values of z from a to 1, 0 =~. Using m

these values of Mx and 0 in equation (36) gives

w~2 [cl ~a (z - z2) dz

+ ~ 1\z -z2) dz m3 a

c1 + c2 1 2 J m 3 ~ ( z - z ) z dz

123

Using the values of C1 and C2 as found on page 122 for this beam, a set of curves may be plotted for CAB in terms of wL2. These curves are shown in Figure 33. Deri­vation for other initial end moments is simi­lar and will not be shown.

Equations for Plotting Curves for Torsional Stress and Stiffness

Plotting of the curves for torsional stress and stiffness is based on Saint Ve­nant's theory. According to this theory the relation between the torsional moment Mt and the rotation 't per unit length of a rectangular beam is -

+ 0.0045~ (37)

where

b is the short side and h is the long side in a rectangular beam

TJ - E - 2(1 + ~)

~ is Poisson's ratio

E is Young's modulus of elasticity

By definition, the torsional stiffness K is the moment required to twist the free end of a beam through an angle of one radian, that is 'tL = 1. Therefore,

K = b3h TJ ( 'T:L) _l [1 - 192b (tan h .!!h. L 3 "5h 2b

J b3h E

+ o.oo45) = ~ L 2(1 + ~) (38)

Values of ~ are plotted in Figure 49.

Saint Venant's theory further states that the maximum unit shear stress Vt at the middle of the long side is approximately

Substituting for ~ from equation (37) g~es -

[ 8 nh] M 1- ~sec h 2b

v - _._t_ " t - b2h .l [1 - 192b (tanh nh + 0.0045)]

3 n5h 2b

(40)

The maximum unit stress vt' at the middle of the short side is

v' t

8 ( nh 1 nh -ryrb 2 tan h 2b - 2 tan h 3 2b n 3

+ ... ) (41)

Substituting for the value of 1: from equation (37) gives

(42)

Values of P in equation (40) and ~' in equation (42) are plotted in Figure 94.

Check on Moment Distribution

That moments and shears be in static equilibrium is a necess'1.ry condition, but it is not sufficient as a check on the correctness of the computations. If deformations due to shears 1-nd direct stresses are disregarded, as previously assumed, the accuracy of the moment distribution is finally checked if it can also be shown that the deformationS-S3.t­isfy the conditions of continuity. Continuity requires that the angular deformation at the ends of all members meeting at a joint be equal and that the linear deformations do not imply any change in lengths of members.

124

In cases where no linear movement or definite, known movements ll have taken place, a simple criterion of continuity be­tween any two members meeting at a joint may be derived as follows: Rearranging equations (11) and (12) gives

MAB - [cAB - E~c (Cl + C2) t] Eic L (C19 A+ C29B) = Mab (43)

and

MBA- [cBA- E~c (C2 + C3) ~] Eic

= T (C29 A + C39B) = Mba (44)

It should be noted carefully that Mab and Mba' as defined by equations (43) and (44), respectively, are the algebraic differences between the final moments after distribution and the initial end moments. The initial end moments include the effect of any external loading plus the effect of any settlement of supports or known deflections ll and are subtracted algebraically from the final moments.

Solving for 9B in equation (44) gives

LMba C2 9B = ----9.~

C3Eic c3 -(45)

Substituting this value of 9B in equation (43) and reducing, gives

(46)

Solving for 9A in equation (43) gives

g - Mab L - c2 9 (47) A - c1Eic c1 B

Substituting this value of 9A in equation (44) and reducing, gives

2 c2 Eic ( c2 )

Mba- c1

Mab = c3T 1 - c1c

3 9B

(48)

Using the stiffness factors and carry­over factors of case I in Figure 25(a), equa­tions (46) and (48) may be written as follows:

KAB (1 - r ABrBA) e A

(49)

KBA (1 - r ABrBA) 9B

(50)

Now let equations {49) and (50) refer to the right span AR and the left span LA, respectively, in the following manner:

L

~-Joint A

Join-t··.)"

R

Joint·····~

Then by the conditions of continuity, 9 A =

9B, and therefore

125

MAR- rRAMRA = KA.R ( 1 - r ARrRA)

MAL- rLAMLA KAL 1 -r ALrLA

(51)

For members of constant section, equa­tion (51) reduces to

KAR. (52)

The criterion of continuity as expressed by equations (51) and (52) may, of course, be applied to any two members meeting at a joint.

In cases where 6 is not a predetermined quantity, its effect cannot be included in the initial end moments and a simple criterion of continuity cannot be derived. The check for continuity must be made by solving for A as well as 9 A and 9B· Having only two ~qlfations available (equations 11 and 12), 1t 1s necessary to start from points, such as fixed ends, where either 9A. or 9B is known.

APPENDIX B

DERIVATION OF EQUATIONS FOR PLOTTING BENDING AND DIRECT­

STRESS·DIAGRAMS FOR REINFORCED-CONCRETE MEMBERS

Assumptions

This study is based on the conventional assumptions in reinforced-concrete design. Reduction of compressive concrete area due to compressive reinforcement is disregarded.

Nomenclature

The derivations herein are taken from notes prepared previously for other work. To avoid errors, the nomenclature as used in the notes is unchanged and applies to these derivations only.

l\.

B

b

c

a section constant.

a section constant.

constants of the cubical equation for C.

width of section in com­pression.

f b h2 c 1V1

e eccentricity.

h

k

M

maximum unit concrete stress.

Young's modulus of elas­ticity of concrete in compression.

Young's modulus of elas­ticity of steel.

total depth of section.

fraction of h in com­pression.

bending moment.about a point half way between faces of section.

bending moment about center of gravity of section.

126

N = direct tensile or com­pressive force.

n

p

X

constants of the cubical equation for k.

steel ratio for fir~t row of tensile steel = b; .

value of cubic equation for approximate value of k.

A 2

k x = correction to nl + n2

be added or subtracted from k when value of cubic equation is x.

Other notations are defined where they first appear.

Case I - Steel in One Face of Section Direct Tension and Bending

With notations shown in Figure 90 and the conditions

2! moments = 0 2! forces = 0

the following equations are obtained. Mo­ments are taken about a point d/2 from the steel.

N = f bd[.!!E (1 - k) +! (l - !)] £ c 2 k 22 3 e

(53) and

N (54)

Equating equations (53) and (54) and solving for d/ e gives

d e

(55)

Taking moments again about a point d/2 from the steel

M' = 1 2[ 1-k) Ne = 2 fcbd np (-k-

+k(---) 1 k ~ 2 3

{56)

It should be observed that the moment MF is about center of gravity of section. This moment must be transferred to the point d/2 from the steel, such that

e = (MJ ±distance from center of

gravity to ~ ) = ~ (57)

Solving equation {56) for fc gives

2 M' np( 1 - k) +k(l- !) bd2

k 2 3

(58)

and hence

d ~ 1 - k k] - = np(--)-- C e k 2

(59)

Equation (59) is plotted in Figure 90. By use of this diagram and equations (58) and {60), steel and concrete stresses may be evaluated.

1 - k) fs = nfc (-k- (60)

127

Case Ii- Symmetrical Steel in Two Opposite Faces of Section -Direct Tension and Bending

In· this case the cover distance of the steel ·d' has to be introduced. Four dia­grams have been plotted with different values

d' of d' expressed as !i ratio a =h. For ·

other values of a interpolate between two diagrams, since practically a straight-line variation exists.

From notations shown in Figures 91(a) to 91(d), inclusive, and the conditions

l moments = 0 ~ forces = 0

the following equations are obtained:

fc

fs

f' s

h e

c

C M bh2

nfc (1 ~ a - 1)

nfc (1 - ~)

6n:e (1 - 2k~ - 3k2

12np ( ~ - a 2) + ~ k2 - k3

h e

k 1 - + np(2- -) 2 ·. k

(61)

(62)

(63)

(64)

(65)

Note the difference between equations (61)" and (58). Equations (64) and {65) are the basis of Figures 91(a) through 91(d), in­clusive.

Case III - General Case

The general diagram in Figure 92 covers bending with or without direct compressi.on or tension. It is not as simple of application as the special diagrams, but it is useful for problems where special diagrams are not available.

From the conditions of equilibrium

and by inspection of Figure 99(b), two equa­tions for N, the direct stress, are obtained.

For bending and direct tension

N = fcbh ~{.!£- [ (1 - au - k)(~- au)ru

1 + (1 - a2t - k)(2 - a2t)r2t

and

1 + (k - alc)(2 - alc)rlc

+ (k - a2c)(~ - a2c)r2c]

+! (1- !)l 2 2 3j

N = fcbh (-F [(1 - alt- k)rlt

(66)

- (k - a2clr2c J- ~} (67)

for bending and direct compression

N = fcbh ~ ~ [(1 - au-k)(!- au)ru

and

N

. 1 + (1 - a2t - k)(2 - a2t)r2t

+ (k - a2c)(~ - a2c)r2c]

+! (1- !)} 2 2 3

(68)

(69)

128

Equating the two expressions for N 111

each case and solving for k gives cubic equations that for both direct tension and compression may be written

The constants n1, n2, and n3, as shown later, are easily computed for a given case, and then k is obtained from Figure 92. For a given beam section it is convenient first to compute two constants, A and B, that properly may be called section constants. The expressions for A and B, obtained when solving for k from equations (66) and (70) are given below in equations (71) and (72), and later simplified.

A [ 1 1

6np (2 - au)ru + (2 - a2t)r2t

- (1 - a )rl - cl - a2 ) r2c1(71) 2 · lc c 2 c ~J

B

1 - r1cC2 - alc)alc

- r2c(~ - a2c)a2c] (72)

r, with a subscript referring to a row of steel, is the ratio of steel in that row, to the steel in the first row of tensile steel. There­fore r 1t is equal to 1. By using notations

·shown in F.igure 92, the depth, as a fraction of h, to the compressive steel ale and a2c may be written (1 - a4) and (1 - a3), respec­tively. Substituting this in equations (71) and (72) and letting

D = r1 + r2 + ... etc. (summation of r ratios for each row of steel)

E a1r1 + a2r2 + ... etc. (summation of ar values for each row of steel)

F a12r1 + a22r2 + ... etc. (summa­tion of a 2r values for each row of steel)

the equations are simplified to 1 .

A 6np (-D-E) (73) · 2 .

B 6np ( i D - 1. 5E + F) (74)

These are extremely simple equations obtained by adding up r ratios, products of ar and a 2r for each row of steel and then multiplying by 6rip.

Using the substitutions for ale' a2c' etc.,· resorted to above, the constants of equation (70) are: .

For bending and direct tension:

n1 = - 3(i + T{ (75)

- 6np ~ D +A. h

(76)

n3 = + 6np E (D - E) - B (77)

For bending and direct compression:

n1 = + 3(~ -l) h 2

+ 6np ED+ A

(78)

(79)

~ = - 6np E (D - E) - B ( 80)

The arithmetical work of de'riving the above constants is not shown but is a direct result of equating the two expressions for N in equations (66) to (69).

As pure bending is approached, e/h in-. creases without limit. It is seen that as e/h becomes very large, the effect of the additive. quantities, not factors of e/h in equations (75), (76), and (77) is insignificant, and may be dropped. The constants then may be written

nl = -3~ (81) h

n2 e

- 6np h D (82)

n3 + 6np E (D - E) (83)

129

Since the constants increase without limit, as e/h increases without limit, it would seem that Figure 92 could not be used for pure bending. However, with some in­accuracy, the scales for n1 and n3, as they appear in Figure 92, may be considered to run from 0 to + 100, + 1,000, or any higher number. (The scale of n2 must be changed correspondingly.) e/h in equations (81), (82), and (83) is, therefore, merely a scale fac­tor and may be set equal to 1. The con­stants for pure' bending then are

(84)

- 6np D (85)

n3 = + 6np (D - E) (86)

The end number of the scale for n1 and n3, used in constructing Figure 92, is 10. If a higher number had been chosen, the dia­gram would have been more nearly symmet­rical, but more difficult to read. The higher the end number of the scale, the closer the diagram would be to symmetry. The con­stants for pure bending are therefore based on a symmetrical diagram. Had equations (78), (79), and (80) been used in deriving the constants for pure bending, they would have appeared with opposite signs. This may be taken advantage of by first obtaining k, using the signs as shown in equations (84), (85), and (86), then repeating, using opposite signs. The average of the two k values thus obtained is correct, subject, of course, to the accuracy of the diagram. The effect of this procedure is to counteract the effect of non­symmetry of the diagram. As e/h increases, conditions of pure bending are approached rapidly. For values of e/h larger than 10, k changes very little. In most design prob­lems, the value of the constants n1 and n3 will lie between 0 and 10. Whenever any of the values exceed 10, extend the scales or consider the side scale to run from 0 to 100, and change the scale of n2 by the S'3.me ratio. The inaccuracy in k will be small.

At first glance, Figure 92 may appear complicated, but on closer examination, it is seen to be extremely simple. Starting from the tensile side of a section, the constants are chiefly summations of r, ar, and a2r, one term for each row of steel, without re­gard to the location .of the neutral axis. This is true no matter how many rows of steel there are. Consequently, this method gives a

direct solution for k, hence the stresses. In tabular form, the computations are further simplified as will be shown in the examples.

Having found k from Figure 92, the :con­crete stresses may be found quickly by using the same diagram, as shown by the following derivation:

Rewriting equation (66)

+ (K - ale) ( ! -ale) r le + (k - a2c) ( ! -r2e]

+! ( i -* >} (87)

The bracketed {} quantity in equation (87) is equal to the reciprocal of C as de­fined in the nomenclature.

Therefore:

... etc] (88)

Simplifying equation (88) gives:

3 2 6 k - 1. 5k + ( C + A)k - B = 0 (89)

In general, equation (89) may be written as follows:

k3 + b1 k2 + b2k + b3 = 0 (90)

Since k already has been found, the only unknown quantity in equation (90) is b2, and b2 may be obtained from Figure 92. For greater accuracy in reading the dia­gram, the constants of equation (90) have been multiplied by 6 so that

b1 - 1. 5 x 6 = - 9 (always) (91)

b3 - B X 6 = - 6B (92)

Multiplying b2 by 6 is accomplished by changing the scale in the diagram so that b2

130

is read fr·om the dotted curves, or b2 may be computed from:

(93)

Having found b2, C = b2 ~ A (94)

The concrete and steel stresses are then obtained from equations shown in Figure 92.

The accuracy obtained from Figure 92 is -usually very good. If greater accuracy is wanted, apply a correction to k derived as · follows:

For a given case, compute n1, n2, n3, then read k from the diagram and insert in equation (70). Since k, as read from the diagram is necessarily approximate, equation (70) will not equal zero, but to some value x.

Therefore:

X (95)

Now correct this value of k by an amount A such that

+ n3 = 0 (96)

E~and equation (96), subtract equation (95) from it. Disregarding squares and higher powers, and solving for A gives

A = 2

: (approximately) (97) n1 + n2

It is easily seen from the signs of the const~nts, in each case, whether A 'must be· added to or subtracted from k.

Effect of Location of Steel

If more than one row of steel is used in either face it is not correct, theoretically,­to consider the steel concentrated at the center of gravity of the steel. In the majority of cases the error involved is small and may be disregarded. The present discussion is included to show what errors may be ex­pected.

'

I I I

I I I I I I I

~Y------ -- - '""""""""'~.....,"'"""''""-"~

y

,NetAtral Qxis

(a) Rectangular beam1bendinq only,~r bending ctnd direct stress.

,..~~Steel stress=nfc[1-k0 '* -t] _\:. _____ _

- .1' • - -+-----L-, - ---· --1--'i- [ 1- Q l ~A tt r2t : · ... Steel stress= nfc -k-· _lt -1

I I I I

h

/Steel stress= nfc [t- ~lc] I

(b) Rectanqt..tlar beam with severa I rows of steel.

face

30" All be~rs are 1"+

(c) T· beam, bending and direct tension.

FIGURE 99 - Effect of Location of Stee~ in Beams.

131

For pure bending, the center of gravity may be used to find k, but it cannot be used to find the resisting moment. As an illustra­tion, consider the beam section shown in Figure 99(a). For bending, k depends on the properties of the section only and may be found by using the transformed section. Thus,

or

ny(A1 + A.2) + nA1 c1 - nA2c2

= 1. kbh2 2

.! kbh2 2

(98)

Since C1 and C2 are the distances from the center of gravity of A1 and A2, respec­tively, C 1 A1 = G2.A2. Equation (98) is therefore:

(99)

showing that to find k, the steel may be considered concentrated at the center of gravity. The resisting moment, however, would be in error. Let the steel stress at the center of gravity be fsc' and consider the steel concentrated at this point. Then, by taking moments about the neutral axis, the resisting moment of the steel is

(100)

Y + c1 The stress in A 1 is fsc Y , and

the stress in A2 is fsc Y - C2 . Taking y

moments again about the neutral axis, using actual distances to the steel area, the actual resisting moment is

Y + c1 Ms = fsc A1(y + C1) y

Y- c2 + fsc y A2(y - C2) (101)

This equation may be reduced to . 1 ( 2 Ms = Me of g + fsc y C1 A1

(102)

132

(C12A1 + c2 2A.2) is the moment of inertia

of the two steel areas about its center of gravity. This quantity is neglected. if the resisting moment is obtained by consider­ing the steel concentrated at the centroid. The error is small whenever (C1 + C2) is small compared to the depth of the beam .

For bending combined with direct stress, the above illustration for the resisting mo­ment holds. The expression for k however, is not valid since k no longer depends on the section properties alone but also on the direct stress.

Examples

~~~E~l~~~~~tr~~~~~~~~~~g ~nj jg~c!. ~I}_S!_o_g.

Given:

M 50,000 ft. -lb.

N 20,000 lb. tension

b 12 in.

h 20 in.

n 10

Reinforced with 2.4 sq. in. of steel in top and bottom.

Cover to center of steel, d ', is 2 inches.

Required:

Concrete and steel stresses.

Solution:

This problem will be solved twice: First, by use of Figure 91; second, by use of Figure 92.

Solution by Figure 91:

e ~ = 30 inches

h e

np

d' h

0.67

10 2.4

X 12 X 20

a = 0.1

0.1

Enter Figure 91 with h/e = 0.67. Read C and k at intersection with the curve np = 0.1; C = 5.3, k = 0.23. Then from equations on the diagram:

f = 5. 3 50,000 X 12 C 12 X 202

660 lb. per

sq. in.

fs 660 x 10 ( 1 - 0·1 - 1) 0.23

·19 ,200 lb. per sq. in.

f~ 660 x 10 (1 - o~2~ ) 3, 700 lb. per sq. in.

Solution by Figure 92:

e

h

a1

a2

np

r1

E

F

A

=

1.5

0.1

0.9

0.1

r2 2.4 1 = 2.4

0.1x1+0.9x1 = 1.0

0.12 X 1 + 0. 92 X 1 0. 82

0. 6 (~ X 2 - 1. 0) = 0

B = 0.6(1 - 1. 5 + 0.82) = 0.192

- 3(i + 1. 5) = - 6

- 0. 6 X 1. 5 X 2 + 0 = - 1. 8

n3 = 0. 6 x 1. 5(2 - 1. 0) - 0.192

133

+ 0. 708

- 9

- 6 X 0.192 = - 1.15

Hold a straightedge from n1 = - 6 to n3 = + 0. 708. Read k = 0. 23 at intersection of n2 = - 1. 8. Then move the straightedge to b1 =- 9 and b3 =- 1.15. Read b2 = 1.14.

Then,

6 c = 1.14 = 5.27

f I s

5. 27 50,000 x 12 = 660 lb. per 12 x 202 sq. in.

fs1 = 660 x 1D ( 1 - 0· 1 - 1) 0.23

= 19,200 lb. per sq. in.

( 1-09 ) fs2 = 660 x 10 0.23 - 1

= - 3, 700 lb. per sq. in. (- indicating comp.)

Correct values are:

k = 0.228

660 lb. per sq. in.

19,500 lb. per sq. in.

f~ = 3,700 lb. per sq. in.

Given:

Same beam as used in Example 1 but without direct stress.

Required:

Concrete and steel stresses.

Solution:

This problem also will be solved twice: First, by use of Figure 91; second, by use of Figure 92 .

.Solution by Figure 91:

Enter the diagram with h/e =0. Read k = 0. 29 and C = 5. 9 at intersection with the curve np = 0.1.

·Then:

f = 5.9 50,000 X 12 C 12 X 202

740 lb. per sq. in.

f' s

740 X 10 ( 1 - 0· 1 - 1) 0.29

15,500 lb. per sq. in.

7 40 X 10 ( 1 - _QJ_ ) 0.29

4,850 lb. per sq. in.

Solution by Figure 92:

Since this is a case of pure bending, . the constants are obtained from equations (84), (85), and (86).

- 3

- 0.6 X 2 - 1.2

ns = + 0. 6(2 - 1) = + 0. 6

The section constants A and B remain the same, Since the section is not changed Therefore, b3 = - 1.15 as before. b1 = - 9 alw'iys. Hold straightedge from n1 =- 3.0 to. n3 = + 0. 6. Read k = 0. 30 at intersection of n2 = ~ 1. 2. Now repeat, using n1 = + 3.0, . n3 =- 0.6, and n2 = + 1.2. Read k = 0.28. The average value is the correct k = 0.29. · Move straightedge to b1 =- 9, b3 = - 1.15, and read b2 = 1~0 at.intersection of k = 0.29.

134

Then:

c 6 6 r-:-o =

B X 50,000 X 12 12 X 202

750 lb. per sq. in.

f8

fs1 = 750 x 10 ( 1 - 0· 1 - 1) 0.29

= 15,800 lb. per sq. in.

f I f 750 10 ( 1 - 0· 9 1) s s2 = x 0.29 -

= - 4,900 lb. per sq. in.

Correct v1.lues are:

k = 0.29

fc = 740 lb. per sq. in.

15,500 lb. per sq. in.

fs2 . = f~ = 4,850 lb. per sq. in.

~~_gl£1~ _g_.: 1:.b~~m_::_b!2.n.§~g_ ~n.£ Qi!e_£:t tension. ----

Given:

or

A T-beam as shown in Figure 99(c).

M = 80,000 ft. -lb.

Mp = 84,200 ft.-lb.

N = 20,000 lb. tension

n = 10

Required:

. Concrete and steel stresses.

Solution:

This problem will be solved twice, by use of Figure 92: First, on the assumptior:

that the steel in eacn face is concentrated at its center of gravity, and then by using the actual location of the steel.

First solution- Steel concentrated at its centroid. Distance to centroid of tensile steel is 7. 88 inches from face of beam.

a1

a2

e

e 1i

np

r1

r2

D

E

F

73~8 = 0.263

27 0.9 - = 30

80 2000 X 12 20,000

48 inches

48 30 = 1. 6

8 X 0. 785 X 10 30 X 30

0.07

8 = 1 8

6 0.75 8

1 + 0. 75 = 1. 75

0.263 X 1 + 0.9 X 0.75 = 0.938

0.2632 X 1 + 0.92 X 0. 75 = 0. 676

A 6x0.07{~x1.75-0.938)

- 0.0265

B 1 0.42 (2 X 1. 75 - 1. 5 X 0.938

+ 0.676) = + 0.061

- 3 {0. 5 + 1. 6) = - 6. 3

- 0.42 X 1.6 X 1.75

- 0.0265 = .... 1.20

+ 0.42 X 1.6 X {1.75

- 0.938) - 0.061 = 0.49

135

- 9

b3 - 6 X 0.061 = - 0.366

From Figure 92, k = 0. 21, b2 = 0. 55.

Then:

6 c = 0. 55 + 0.0265 10·4

10.4 80,000 x 12 = 370 lb. per 30 x 302 sq. in.

370 X 10 ( 1 - 0· 1 - 1) 0.21

12,200 lb. per sq. in.

370 X 10 ( 1 - O. 9 - 1) .21

- 1,940 lb. per sq. in.

Second solution - Steel in its actual loca­tion

np = 3 X 0. 785 X 10 _ O 026 30 A 30 - .

TABLE XXXVI

Tabulation of Computations for Constants, S~etrical Beam

Steel a2r row a r ar

1 0.1 1.0 0.1 0.01

2 0.2 1.0 0.2 0.04

3 0.6 0.67 0.4 0.24'

4 0.9 2.0 1.8 1.62

Summation D = 4.67 E = 2.5 F = 1.91

By equations (73) and (74) the section con­stants are

A

B

6 X 0.026 (2.33 - 2. 5) = - 0.026

0.156 (2.33- 1.5 X 2.5 + 1.91)

0.077

By equations (75), (76), and (77) the con­sh.nts for k are

- 3 (0.5 + 1.6) = - 6.3, as before.

n2 - 6 X 0.026 X 1.6 X 4.67

- 0.026 = ..;. 1.19

+ 6 X 0.026 X 1.6 (4.67- 2.5)

- 0.077 = + 0.473

By equations (91) and (92),

- 6 X 0.077

- 0.462

Then from Figure 92, k = 0. 2 and b2 = 0.64.

t<--- b=241-->! I I

_"1-- -- ---7)-1'7\ W I 11

\..!.) 1 4-1 ° Bars I I

-+-2-1" c Bars I I I

--!-2-1" c Bars =o I.D II

· .:¥-- 2-1" c Bars I I

-t-e--<3--e+·-+-3-1" c Bars .....__ _____ y_

FIGURE 100 - Unsymmetrical Beam w1 th Several Rows of Steel--Bendins and Direct Compression.

C 6 = 9 = 0.64 + 0.026

9 x 80,000 x 12 = 320 lb. per 30 x 302 sq. in.

320 X 10 ( 1 - 0·1 - 1) 0.20

11,200 lb. per sq. in.

320 X 10 ( 1 - O. 9 - 1) 0.2

- 1,600 lb. per sq. in.

136

Given:

A beam section as shown in Figure 100 and subjected to a moment. M = 600,000 ft. -lb. and 9. direct force, N = 60,000 lb. comp. n = 10.

e = 600,000 60,000

10ft.

e 10 _ 2 h 5 -

6np = 6 x 10 x4 60 X 24

0.167

. e 6np h = 0.167 x 2 = 0.334

TABLE XXXVII

Tabulation of Computations for Constants, Uns1JID!letr1cal Beam

Steel a2r row a r ar

1 0.1 1.0 0.1 0.01

2 0.3 0.5 0.15 0.045

3 0.5 0.5 0.25 0.125.

4 0.7 0.5 0.35 0.245

5 0.9 0.75 0.675 0.608

Summation D = 3.25 E = 1.526 F = 1.033

Now by equations (73), (74), (78), (79), and (80)

A 1 0.167 (2 X 3. 25 - 1. 525) = 0. 0167

B 1 0.167 (2 X 3. 25 - 1. 5 X 1. 525

+ 1.033) = 0.0617

n1 +3(2.0-0.5) = +4.5

+ 0.334 X 3.25 + 0.0167

+ 1.10

- 0. 334 (3. 25 - 1. 525)

. 0.0617 = - 0.637

and by equations (91) and (92)

- 6 X 0.0617

.;. 0.37

Now by Figure 92,

k = 0.27 and b2 = 0.55

and by computations using equations given on Figure 92

6 c = 0.55- 0.0167 = 11. 25

11.25 600,000 X 12 24 x 6o2

9 340 ( 1 - 0· 1 - 1) ' 0.27

934 lb. per sq. in.

t 21,700 lb. per sq. in.

9 340 ( 1 "!' o. 3 - 1)

' o~27

14,850 lb. per sq. in.

: fs3

. 9 340 ( 1 - 0· 5 - 1) ' 0.27

7,940 lb. per sq. in.

9,340 ( 1 - 0· 7 - 1) 0.27 .

1,025 lb. per sq. in.

9 340 ( 1 - 0· 9 - 1)' ' 0.27

- 5, 880 lb. per sq .. in.

Exact values are:

fc = 910 lb. per sq. in.

21,400 lb. per sq. in.

14,700 lb. per sq. in.

7,850 lb. per sq. in.

1,020 lb. per sq. in.

fs5 = - 5,700 lb. per sq. in.

Verification of the accuracy of the values may be done by checking to see that z M = 0 and EN= 0.

137 ~~U.S. GOVERNMENT PRINTING OFFICE 1975 ·678-G97{317 Ret. 8