DEPRATMENTOF CIVIL ENGINEERING

SHREE S’AD VIDHYA MANDAL INSTITUTE OF TECHNLOGY

FACULTY:- HOD RUCHI GUPTASUBJECT:- STRUCTURAL ANALYSIS -2GROUP NO:-7

ENROLLMENT NO. NAME

130450106033 Patel Jignasha Kanaiyalal

130450106035 Patel Margi Mauleshbhai

130450106044 Shah Ishani Milankumar

130450106045 Shah Richaben Umeshbhai

130450106047 Taira Naznin Iqbal

ANALYSIS OF SIMPLEPORTAL FRAME WITH

SWAY

PORTAL FRAMES WITH SIDE SWAY Causes of side sway :

1 Unsymmetrical loading (eccentric loading)2 Unsymmetrical out-line of portal frame 3 Different end conditions of the columns of the

portal frame .4 Non-uniform sections (M.I.) of the members of the

frame.5 Horizontal loading on the columns of the frame.6 settlement of the supports of the frame .

In case of portal frame with side sway , the joint translations become additional unknown quantities .

Some additional conditions will , therefore , be required for analysing the frame .

The additional conditions of equilibrium are obtained form the consideration of the shear force exerted on the structure by the external loading .

The horizontal shear exerted by a member is equal to the algebric sum of the moments at the ends divided by the length of the member .

All the end moments are assumed clockwise in calculating the horizontal reactions.

w KN/m

w

P

Ha

Hd

B

h

C

D

A

L1

L2

Determine support moments using slope deflection method for the frame shown in figure. Also draw bending moment diagram.

Solution:

Example 1 :

(a)Fixed end moments (FEM):

12KN

10 KN2.4 KN/m

A4 m

BC

1.5 m

1.5 m

( I )

( I )

D

1.5 m

(b) Slope – Deflection equation :

(c) Equilibrium equation :

At joint B,

(d) Final Moments :

(d) Simply supported moments :

0.30

4.810.20

15

4.8

9

9.15D

CBA

(e) Simply supported moments

B.M Diagram

EXAMPLE 2 :

A beam AB of uniform section of span 9m and constant EI=3.6×104 Nm² is partially fixed at ends when the beam carries a point load 90 kN at distance 3m from the left end A. The following displecements were observed.

i) rotation at A =0.001 rad (clockwise) and settlement at A=20mm

ii) rotation at B=0.0075rad (anticlockwise) and settlement at B=15mm.

Analyse using slope deflection method.

90 KN

AAC B

3 m 6 m9 m

Fixed end moments (FEM):

Slope deflection equations:

NETSETTLEMENT5mm

15mm20

mm

Slope deflection

B.M. diagram

120.03

180

59.90

A B C

+

+

-

B.M Diagram

100 KN30 KN/m

A B C8 m 6 m 4 m

( I ) ( 2I )

Using slope deflection method analyse the continuous beam shown in figure.Draw the bending moment diagram.

EXAMPLE : 3

mkNl

WbaCBM

mkNl

WabBCM

mkNwlBAM

mkNwlABM

f

f

f

f

.14410

64100

.9610

46100

.16012

.16012

83012

2

2

2

2

2

2

2

2

2

22

)2.(..........5.0160

0022160

322

)1.......(25.0160

008

2160

322

B

B

ABfBA

B

B

BAfAB

EIlEI

llEIBAMM

EI

EIll

EIABMM

(A) FIXED END MOMENTS

(b) Slope Deflection equations

)4(..........8.04.0144

322

)3.........(4.08.096

0210

)2(296

322

CB

BCfCB

CB

CB

CBfBC

EIEIll

EICBMM

EIEI

IEll

EIBCMM

iseanticlockwEi

clockwiseEI

BEIEIM

AEIEIEIEIEI

MMBM

AMM

C

CB

CB

cB

CBB

BCBA

CB

BCBA

.....63.183

.....272.7

2unknownsequationsMode.calculatorby (B) and (A)equation Solving

).....(1448.04.0,0

).......(644.03.10)4.08.096()5.0160(

0).......(0

).......(0

B

(C) EQUILIBRIUM EQUATIONS :

090.14691.2144

63.1838.0272.74.0144

8.04.0144.64.163

45.7382.596

63.1834.0272.78.096

4.08.096.64.163

272.75.0160

.18.158

272.725.0160

25.0160

EIEI

EIEI

EIEIMmkN

EIEI

EIEI

EIEIMmkN

EIEIM

mkNEI

EI

EIM

CBCB

CBBC

BA

BAB

(D) FINAL MOMENTS :

(E) B.M DIAGRAM

mkNl

Wab

mkNwl

.24010

46100MBC,Span

.2408

8308

M AB,Span

moments. supportedSimply 22

158.18

240

163.64

240

BA C