Flexural Analysis of Beams - s3.amazonaws.com This Powerpoint presentation was prepared by Dr. Terry...

Flexural Analysis of Beams

1

Acknowledgement

This Powerpoint presentation was prepared by Dr. Terry Weigel, University of Louisville.

SI units are added by Dr. Tarek Ragab, University at Buffalo (SUNY)

2

Introduction

3

Gradual application of load on a simply supported beam until it fails

Three behavior stages:

(a) Concrete uncracked (b)Concrete cracked – elastic stresses (c) Concrete cracked – Ultimate strength

Uncracked Concrete Stage

4

Tensile stresses in concrete are less than modulus of rupture

Entire beam cross-section resists bending

Compression on one side and tension on the other

The effect of reinforcing steel is negligible on the beam properties

Uncracked Concrete Stage

5

Concrete Cracked – Elastic Stresses

6

Cracks develop on tension side of beam

Moment at which cracks begin – cracking moment (Mcr)

Cracks spread toward the neutral axis

Neutral axis moves upward (for positive moment)

Concrete Cracked – Elastic Stresses

7

Concrete Cracked – Elastic Stresses

8

Cracked concrete cannot resist tension

Tensile stress must be resisted by steel

This stage continues until concrete stresses are about one-half concrete compression strength and steel stress is less than yield

Stress varies linearly with strain

Concrete Cracked – Elastic Stresses

9

This is the stage occurring under service-load conditions

Stresses are computed using the transformed area method

Service or working loads are considerably higher than the cracking load

Concrete Cracked – Ultimate Strength Stage

10

Tensile cracks and neutral axis move upward (for positive moment)

Stresses are no longer linearly related to strain

Reinforcing bars yield

Concrete Cracked – Ultimate Strength Stage

11

Moment-Curvature Relationship

12

Cracking Moment

13

Effect of reinforcement is negligible until concrete is cracked

Stress in the beam may be calculated using:

g

Myf

I

ACI Section 9.5.2.3 – cracking moment is calculated using ACI equation 9-9:

r g

cr

t

f IM

y

Cracking Moment

14

Cracking stress – ACI Equation 9-10:

'0.62r cf f

yt is the distance from the centroidal axis to the extreme tension fiber

is a parameter to account for lightweight concrete:

= 1 for normal weight concrete = 0.85 for sand-lightweight concrete

= 0.75 for all-lightweight concrete

Example 2.1

15

For the beam shown in the figure, compute the bending stresses for a moment of 32 KN.m. Use normal weight concrete, a concrete strength of 28MPa and a modulus of rupture of 3.3MPa and determine the cracking moment.

Example 2.1

16

Example 2.1

17

(a) Calculate the bending stresses assuming the section uncracked

33 4

4

1 1300 450 2.278 9 mm

12 12

32 6 . (225 )3.15

2.278 9 mm

g

g

I bh mm mm E

E N mm mmMyf MPa

I E

Example 2.1

18

(b) Calculate the cracking moment

43.3 2.278 9 mm

225

33.41 6 . 33.41 .

r g

cr r

MPa Ef IM f S

y mm

E N mm KN m

Elastic Stresses - Cracked Concrete

19

All concrete in the tensile zone is cracked and is neglected

Perfect bond between the tension steel and concrete – strains in two materials are equal

Stresses are not equal because of differences in moduli

Elastic Stresses - Cracked Concrete

20

Ratio of the steel to concrete stress is given by the modular ratio

s

c

En

E

Area of tension steel (As) is equivalent to an equivalent concrete area of nAs

Transformed area

Elastic Stresses - Cracked Concrete

21

Figure 2.6

Example 2.2

22

Calculate the bending stresses in the shown beam using the transformed area method. Use a concrete strength of 21MPa, n = 9 and M = 90KN.m.

Example 2.2

23

Example 2.2

24

(a) Locate the neutral axis by summing moments of areas about the neutral axis

2

2

mm300 mm 9 1935 mm 425 mm

2

150 7.4 6 17,415

164

xmm x mm x

x E x

x mm

171 mm – Consider this value in next calculations

Example 2.2

25

(b) Compute the moment of inertia of the transformed area

3 22

4

1 164300 164 (300)(164)( ) (17,415) 425 164

12 2

1.627 9 mm

crI

E

where Icr is the cracked, transformed moment of inertia

Example 2.2

26

(c) Compute the bending stresses

4

4

(90 6 . ) 1648.84

1.67 9 mm

90 6 . 425 164( )9

1.67 9 mm

126.6

c

cr

s

cr

E N mm mmMxf MPa

I E

E N mmM d xf n

I E

MPa

Example 2.3

27

Determine the allowable bending moment that may be applied to the beam of Example 2.2 if the allowable stresses are 8MPa for concrete in compression and 140MPa for reinforcing steel in tension.

Example 2.3

28

4

4

8 1.627 9 mm79 6 .

164 mm

79KN.m

140 1.627 9 mm97 6 .

( ) 9 425 164

97 .

c crc

s crt

MPa Ef IM E N mm

x

MPa Ef IM E N mm

n d x

KN m

The beam capacity is controlled by the concrete stress and is 79KN.m

Example 2.4

29

Calculate the bending stresses in the notched beam shown using the transformed area method. Use n = 8 and M = 150KN.m.

Example 2.4

30

Example 2.4

31

(a) Locate the neutral axis (NA) by summing moments of areas about the neutral axis – assume the NA below the notch

2

2

mm450 mm 150 150 mm 75

2

8 3,276 mm 575mm mm

225 22,500 1.6875 6 15.0696 6 26,208

235

xmm x mm mm x mm

x

x x E E x

x mm

Note that the NA is below the depth of the slot, as assumed. If x had been less than 150 mm, the calculated value of x would not have been valid.

Example 2.4

32

(b) Compute the moment of inertia of the transformed area

3 3

22 4

1 1150 235 2 150 85

3 3

8 3,276mm 575 235 4.358 9 mm

crI

E

Example 2.4

33

(c) Compute the bending stresses

4

4

150 6 . 2358.1

4.358 9 mm

150 6 . 575 235( )8

4.358 9 mm

93.6

c

cr

s

cr

E N mm mmMxf MPa

I E

E N mmM d xf n

I E

MPa

Doubly-Reinforced Beams

34

Compression steel

Smaller beams

Reduction of long-term deflection

Stress in compression bars doubles over time

Support for stirrups

Example 2.5

35

Calculate the bending stresses in the beam using the transformed area method. Use n = 10 and M = 165KN.m.

Example 2.5

36 As time goes by, stress in compression steel is assumed to be doubled; 2nAs

Example 2.5

37

(a) Locate the neutral axis by summing moments of areas about the neutral axis

2

2

2

350 1,290 mm 20 1 502

10 2,580 mm 450

175 24,510 1.2255 6 11.61 6 25,800

139

xx x

x

x x E E x

x mm

10.38E6

Example 2.5

38

(b) Compute the moment of inertia of the transformed area

3 22

22 4

1350 139 20 -1 1,290 mm 89 mm

3

10 2,580 mm 311 mm 3 9 mm

crI

E

Example 2.5

39

(c) Compute the bending stresses

4

'

4

4

165 6 . 1397.65

3 9 mm

165 6 . 139-50( ')2 20 98

3 9 mm

165 6 . 450 139( )10 171

3 9 mm

c

cr

s

cr

s

cr

E N mm mmMxf MPa

I E

E N mmM x df n MPa

I E

E N mmM d xf n MPa

I E

Ultimate Moment

40

Tensile bars are stressed to yield before concrete crushes

b1 depends on concrete strength

Constant concrete compressive stress – 85% of ultimate stress

Rectangular stress block extends b1 times c

Whitney

stress

block

Example 2.6

41

Calculate the nominal flexural strength of the beam if the yield stress of the steel is 420MPa and the strength of the concrete is 21MPa.

Example 2.6

42

Example 2.6

43

(a) Calculate the tensile force (T) and the compressive force (C)

2

'

1935 420 812,700

0.85 0.85 21 350 6,247.5

s y

c

T A f mm MPa N

C f ab MPa a a

(b) Determine a by equating T and C

812,700 6,247.5

130

N a

a mm

Example 2.6

44

(c) Compute the moment arm and the moment

130525 460

2 2

( ) 812,700 460 373.8 6 .2

373.8 .

n

ad mm

aM T d N mm E N mm

KN m

Example 2.7

45

Calculate the nominal flexural strength of the beam if the yield stress of the steel is 420MPa and the strength of the concrete is 21MPa.

Example 2.7

46

Example 2.7

47

(a) Calculate the tensile force (T) and the compressive force (C); determine the area of concrete necessary to equilibrate T

2

'

2

'

2

2580 mm 420 1083.6

0.85

1083.6 360,705

0.85 0.85 21

(150)(150) ( 150)(450) 60,705

235

s y

c c

c

c

T A f MPa KN

C f A

C T

T EA mm

f MPa

a mm

a mm

Example 2.7

48

(c) Locate the NA and determine Mn.

Defining y as the distance from the top of the section to the centroid of the compression area,

8522500 75 38205 150

2149

60705

525 149 376

1083600 376 407 .n

y mm

d y mm

M N mm KN m