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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