Module3 rajesh sir

Structural Analysis - II

Approximate Methods

Dr. Rajesh K. N.Assistant professor in Civil EngineeringAssistant professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

Module IIIModule III

Approximate Methods of Analysis of Multi-storey Frames

• Analysis for vertical loads - Substitute frames-Loading conditions for maximum positive and negative bending moments in beams and maximum bending moment in columns

• Analysis for lateral loads - Portal method–Cantilever method–yFactor method.


2

Why approximate analysis?

• Rapid check on computer aided analysis

y pp x y

• Preliminary dimensioning before exact analysis

Advantage? • Fasterg

Disadvantage? • Results are approximateg pp

• Approximate methods are particularly useful for multi-storey frames taller than 3 storeys.


Approximate analysis for Vertical LoadsApproximate analysis for Vertical Loads

SUBSTITUTE FRAME METHOD

• Analyse only a part of the frame – substitute frame

SUBSTITUTE FRAME METHOD

Analyse only a part of the frame substitute frame

• Carry out a two-cycle moment distribution


Substitute frame

Actual frame


• Analysis done for:y

• Beam span moments• Beam support moments• Column moments

• Liveload positioning for the worst condition

• For the same frame, liveload positions for maximum span t t t d l t moments, support moments and column moments may

be different

• For maximum moments at different points, liveload positions may de different


positions may de different

LL positions for maximum positive span moment at Bp p p

BInfluence

D d l d Li l d

line for MB

Dead loads Live loads


LL positions for maximum negative support moment at A pos t o s o ax u egat e suppo t o e t at

AInfluence

Live loads

line for MA

Dead loads


Dead loads

LL positions for maximum column moment M1 at C pos t o s o ax u co u o e t M1 at C

C

M1

Live loadsLive loads

Dead loads


LL positions for maximum column moment M2 at D pos t o s o ax u co u o e t M2 at

D

M2

D

Live loadsLive loads

Dead loads


Problem 1: Total dead load is 12 kN/m. Total live load is 20 ob e : N/kN/m. Analyse the frame for midspan positive moment on BC.

4 m

6 m 6 m 6 m

B C DA

4m

4 m


11

12+20 kN/m

B D

12 kN/m 12 kN/m

B C DA 6 m 6 m 6 m

Fi d d t2 212 6 36AB

wlFEM kNm− − ×= = = − 36BAFEM kNm=

Fixed end moments

12 12AB

232 6 96FEM kN− × 96FEM kNm=9612BCFEM kNm= = −

36CD DCFEM FEM kNm− = =

96CBFEM kNm=


36CD DCFEM FEM kNm

Distribution factors

1 4 6 0.254 6 4 4 4 4AB DC

K EIDF DFK K K EI EI EI

= = = =1 2 3 4 6 4 4 4 4AB DCK K K EI EI EI+ + + +

4 6K EI1

1 2 3 4

4 6 0.24 4 4 6 4 4 4 6BA

K EIDFK K K K EI EI EI EI

= = =+ + + + + +

0.2BC CD CB BADF DF DF DF= = = =


A B C D

0.2 0.2* * * * * * FEM

0.2 0.20.25 DFs0.25

* * * * * *

* * * * CO

Dist

* *

* * Final Moments

Dist


A B C D

0.2 0.2-36 36 -96 96 -36 36 FEM

0.2 0.20.25 DFs0.25

9 12 12 -12 -12 -9

6 4.5 -6 6 -4.5 -6 CO

Dist

2.25 0.3 0.3 -0.3 -0.3 -2.25

-18.75 52.8 -89.7 89.7 52.8 18.75 Final Moments

Dist


B89.7kNm 89.7kNm32kN mA B89.7kNm 32kN m

23 32 6×

Midspan positive moment on BC,

3 32 689.7 32 3 54.32 2EM kNm×

= − − × + × =


Problem 2: Analyse the frame for beam negative moment at B. M t f i ti f b i 1 5 ti th t f l T t l d d Moment of inertia of beams is 1.5 times that of columns. Total dead load is 14 kN/m and total live load is 9 kN/m.

6 m 4m4 m

3.5

m.5

m

BA

m3

DC

3.5

m3.

5 m


17

3

14+9 kN/m 14+9 kN/mI

B D6 m

14 kN/mI

B C DA 6 m 4 m 4 m

I1.5I 1.5I 1.5I

Fi d d t2 223 6 69AB

wlFEM kNm− − ×= = = − 69BAFEM kNm=

Fixed end moments

12 12AB

223 4 30.67BC CBFEM FEM kNm×− = = = 30.67

12BC CBFEM FEM kNm

214 4 36CD DCFEM FEM kNm×− = = =


3612CD DCFEM FEM kNm

Distribution factors

( )( )

1 4 61.5 0.3044 6 4 3 5 4 3 5AB

K E IDFK K K E EI EI

= = =( )1 2 3 4 6 4 3.5 4 3.51.5AB K K K E EI EII+ + + +

1 5 6K I1

1 2 3 4

1.5 6 0.2091.5 6 3.5 3.5 1.5 4BA

K IDFK K K K I I I I

= = =+ + + + + +

1

1 2 3 4

1.5 4 0.3131.5 6 3.5 3.5 1.5 4BC

K IDFK K K K I I I I

= = =+ + + + + +1 2 3 4

0.284, 0.284, 0.396CB CD DCDF DF DF= = =


A B C D

0.209 0.313 0.284 0.2840.304 DFs0.3960.209 0.313* * * * ** * * *

FEM

Dist

0. 8 0. 8 0.396

* *

* *CODist

* * Final Moments


A B C D

0.209 0.313-69 69 -30.67 30.67 -18.67 FEM

0.284 0.2840.304 DFs0.396

20.98 -8.01 -12 -3.41

10.49 -1.71 CO

Dist

-1.84 -2.75

69.64 -47.13 Final Moments

Dist

B 69 64kNm47 13kNm B 69.64kNm47.13kNm

Max. beam negative moment at B 69.64 kNm=


Approximate analysis for Horizontal LoadsApproximate analysis for Horizontal Loads

1 P t l th d1. Portal method

2. Cantilever method

3. Factor method


22

PORTAL METHODPORTAL METHOD

Assumptions

1. The points of contraflexure in all the members lie at their

midpoints.

2. Horizontal shear taken by each interior column is double y

that taken by each exterior column.

Horizontal forces are assumed to act only at the joints.


CA B C DAP1

P 2P 2P PP

P

2P

2P

2P

2P

P

P

FE GHP2

Q 2Q 2Q Q

Q 2Q 2Q Q

J K LI


24

P1 2 2P P P P P= + + + 1

6PP⇒ =


2 2P P Q Q Q Q+ = + + + 1 2P PQ +⇒ =1 2 2 2P P Q Q Q Q+ = + + + 1 2

6Q⇒ =


Problem 3: Analyse the frame using portal method.ob e 3: y g p

B C DAm

120 kN

7 m 3.5 m 5 m

FE G H

3.5

m

180 kN

m

H

3.5

m

J K LI J K LI


27

Horizontal shears:

1, 2 2For the top storey P P P P P= + + +120 206

P kN⇒ = =

1 2,6

P PFor the bottom storey Q +=

120 180 506

kN+= =,

6y Q

6


120kN 3 5 mA

Moments:35kNm

m

3.5 mMoments:

35kN

20kN

1.75

m 35kNm 10kN

20kN

35kNm 35kNmB

35kNm 35kNm

40kN10kN70kNm

10kN


29

Beam moments:

B C DA

Beam moments:35

35

35

3535

3535 35 35

FE G H122.5

122.5

122.5 122.5

122.5 122.5

J K LI


30

Column moments:

B C DA 7035 kNm70

35 kNm

Column moments:

FE G H35 87.5 87.5175 70 3570 175

J K LI87.5 87.5175 175


31

Beam and Column moments:

35B D35 35 35

7035 7035 35 35

35 87 5 87.517570 3570

175

122.5 122.5

122 5

122.5

122 535 87.5 122.5 122.5 122.5

87 5 87 5175 17587.5 87.5175 175


32

Home work

B CA40 kN B CA

5 m

40 kN

5 m 7.5 m

FED3.

580 kN

m5

m

H IG


33

CANTILEVER METHODCANTILEVER METHOD

• Frame considered as a vertical cantilever

Assumptions

1. The points of contraflexure in all the members lie at

their midpoints.

2. The direct stresses (axial stresses) in the columns are

directly proportional to their distance from the directly proportional to their distance from the

centroidal vertical axis of the frame.


P1

y1y2 y3

y4

P2

A1 A2A3 A4

Area of cross ti

Centroidal vertical axis of the frame

section

1 1 2 2 3 3 4 4Ad A d Ad A dyA A A A+ + +

=+ + +

To locate centroidal vertical axis of the frame,

Dept. of CE, GCE Kannur Dr.RajeshKN35

1 2 3 4A A A A+ + +

V1 V2 V3 V4

xMyI

σ =

MMI

is constant at a given height (of the ‘vertical cantilever’).

1 2 3 4

1 2 3 4y y y yσ σ σ σ

= = = 31 2 431 2 4

1 2 3 4

V AV A V A V Ay y y y

⇒ = = = ( )1


1 2 3 4y y y y1 2 3 4y y y y

1Ph m1

m2

H1 H2 H3 H4

2 B

V1 V2 V3 V4

1 1 1 2 2 3 3 4 42BhM P Vm V m V m V m⇒ = + − −∑ ( )2

( ) ( ) 1 2 3 4, , , , .1 2From and V V V V can be found( ) ( ) 1 2 3 4, , , , .1 2From and V V V V can be found


1P

H1 H2 H3 H4

1 1 2 3 4P H H H H= + + +


Problem 4: Analyse the frame using cantilever method, if all ob e : y g ,the columns have the same area of cross section.

B C DAm

120 kN

7 m 3.5 m 5 m

FE G H

3.5

m

180 kN

m

H

3.5

m

J K LI J K LI


39


1 1 1 10 7 10.5 15.5A A A AyA A A A

× + × + × + ×=

+ + +


33 8.254

m= =1 1 1 1A A A A+ + + 4

120

8.25 1.25 2.257.25

180

Also, 31 2 431 2 4V AV A V A V A= = = 1 2 3 4V V V V

⇒ = = =


40

Also,8.25 1.25 2.25 7.25

= = =8.25 1.25 2.25 7.25

⇒

1 25 2 25 7 25V V V1 1 12 3 4

1.25 2.25 7.25, ,8.25 8.25 8.25

V V VV V V= = =

1P

2h m1

m2

OH1 H2 H3 H4

2 O

V1 V2 V3 V4

1 1 1 2 2 3 3 4 42OhM P Vm V m V m V m⇒ = + − −∑,For the top storey

1 2 3 43.5120 15.5 8.5 5 02

V V V V⇒ × = × + × − × − ×


41

3 5 ⎛ ⎞ ⎛ ⎞1 11

3.5 1.25 2.25120 15.5 8.5 52 8.25 8.25

V VV ⎛ ⎞ ⎛ ⎞⇒ × = × + × − ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 13.615V kN⇒ =

21.25 13.615 2.063 ,

8.252 25 13 615

V kN×= =

32.25 13.615 3.713 ,

8.257 25 13 615

V kN×= =

×4

7.25 13.615 11.9658.25

V kN×= =

: 13.615 2.063 3.713 11.965 0Check + − − =


H1 H2 H3 H4

V1 V2 V3 V4

O

V1 V2 V3 4

3 53 5⎛ ⎞

,For the bottom storey

1 2 3 43.53.5120 180 15.5 8.5 5 03.522OM V V V V⎛ ⎞⇒ × + × = × + × − × − ×+⎜ ⎟

⎝ ⎠∑


3 5 1 25 2 253 5 V V⎛ ⎞ ⎛ ⎞⎛ ⎞ 1 11

3.5 1.25 2.253.5120 180 15.5 8.5 53.52 8.25 8.252

V VV ⎛ ⎞ ⎛ ⎞⎛ ⎞⇒ × + × = × + × − ×+⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 61.267V kN⇒ =

1 25 61 267×2

1.25 61.267 9.283 ,8.25

2.25 61.267 16 709

V kN

V kN

×= =

×3 16.709 ,

8.257.25 61.267 53 841

V kN

V kN

= =

×= =4 53.841

8.25V kN= =

: 61 267 9 283 16 709 53 841 0Check + − − =


: 61.267 9.283 16.709 53.841 0Check + − − =

120kNMoments:

47.652kNmA120kN 3.5 m

Moments: A

1.75

m 47.652kNm 13.615kN

27.3kN

k13.615kN

Dept. of CE, GCE Kannur Dr.RajeshKN45

Beam and Column moments:

29.9B D47.6 27.4 29.9

7547.6 57.3B D

47.6 27.4 29.9

47 6 119 2 74.8187.857.3 29.975

143.2

166.8 96

96

104.7

104 747.6 119.2 74.8187.8 143.2166.8 96 104.7

119 2 74 8187 8 143 2119.2 74.8187.8 143.2


46

BA25 kNHome workBA25 kN

6 m

5 m

DC50 kN

3.5

3.5

m

k FE55 kN

4.5

m

HG


47

FACTOR METHODFACTOR METHOD

• More accurate than Portal and Cantilever methodsMore accurate than Portal and Cantilever methods

• Specially useful when moments of inertia of various

b d ffmembers are different.

Basis:

• At any joint the total moment is shared by all the members

in proportion to their stiffnesses

• Half the moment gets carried over to the far endHalf the moment gets carried over to the far end


Gi d d l f t Girder and column factors:

• Relative stiffness of a member Ik =

Girder factor at a joint

Relative stiffness of a member kL

=

Girder factor at a joint

,k of all meeting at the jointg

columns= ∑

,g

k of all members meeting at the joint=∑

Column factor at a joint

,1

,k of all meeting at the joint

c gk of all members meeting at the joint

beams= = −∑∑


, f g j∑

Moment factor for a member

,mC c k for a column= ,,

m

m

fG g k for a beam=

m

m

where c c half of column facor of far endand g g half of girder facor of far end

= += +m

storeC sum of column moment factors for a y→∑G sum of beam moment factors for a joint→∑


Problem 5: Analyse the frame using factor method.

H IG40 kN

ob e 5: y g

H IG

5 m

40 kN

5 m 7.5 m

FED3.

580 kN

m5

m

B CA


Total column moment above DEF 40 3.5 140kNm= × =

Total column moment above ABC 40 8 5 80 5 740kNm= × + × =Total column moment above ABC 40 8.5 80 5 740kNm× + ×


1 2 3 4 5 6 7 8 9 10 11 12JOINT MEMBER k=I/L Ʃk FACTOR c/2 5+6 MOMENT Total Col DFB BeamJOINT MEMBER k I/L Ʃk FACTOR c/2,

g/2 from far end

5+6 MOMENT FACTOR

Total Col.

Mom, MT

Col. Mom, MC =MT×

C/ƩC

DFB

=G/ƩGBeam Mom

= MC ×DFBCol Beam Col Beam c

=Ʃk(beg

=Ʃk(ccm gm C=

cm ×G = gm × C/ƩC(

ams)/Ʃk

(olumns)/Ʃk

k k

DDA 0.2

0.6860.29 0.5 0.79 0.158 740 99.4

DE 0.2 0.71 0.3 1.01 0.202 1 122.6DG 0 286 0 29 0 3 0 59 0 169 140 23 2DG 0.286 0.29 0.3 0.59 0.169 140 23.2

E

ED 0.2

0.819

0.59 0.36 0.95 0.19 0.59 83.25EH 0.286 0.41 0.27 0.68 0.194 140 26.6

EF 0.133 0.59 0.4 0.99 0.132 0.41 57.85EB 0.2 0.41 0.5 0.91 0.182 740 114.5

FE 0 133 0 79 0 3 1 09 0 145 1 133 03F

FE 0.1330.772

0.79 0.3 1.09 0.145 1 133.03FI 0.286 0.21 0.16 0.37 0.106 140 14.6FC 0.2 0.21 0.5 0.71 0.142 740 89.3

G GD 0.286 0.486 0.59 0.15 0.74 0.212 140 29.1GH 0.2 0.41 0.23 0.64 0.128 1 29.1

HHG 0.2

0.7720.46 0.21 0.67 0.134 0.559 16.5

HE 0.286 0.54 0.21 0.75 0.215 140 29.5HI 0.133 0.46 0.34 0.8 0.106 0.441 13.0

I IH 0.133 0.419 0.68 0.23 0.91 0.121 1 16.58IF 0 286 0 32 0 11 0 43 0 123 140 16 58IF 0.286 0.32 0.11 0.43 0.123 140 16.58

A AD 0.2 X 0.2 1 0 0.15 1.15 0.23 740 144.7B BE 0.2 X 0.2 1 0 0.21 1.21 0.242 740 152.2C CF 0.2 X 0.2 1 0 0.11 1.11 0.222 740 139.6

0 169 0 194 0 106 0 212 0 215 0 123 1 019F h C∑Dept. of CE, GCE Kannur Dr.RajeshKN

, 0.169 0.194 0.106 0.212 0.215 0.123 1.019For the top storey C = + + + + + =∑, 0.158 0.182 0.142 0.23 0.242 0.222 1.176For the bottom storey C = + + + + + =∑

Home work

B CA120 kN

6 m 6 m

4 m

60 kN FED60 kN6

m

H IG


54

SummarySummary

Approximate Methods of Analysis of Multi-storey Frames

• Analysis for vertical loads - Substitute frames-Loading conditions for maximum positive and negative bending moments in beams and maximum bending moment in columns

• Analysis for lateral loads - Portal method–Cantilever method–yFactor method.
