analysis of simple portal frame with sway
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DEPRATMENTOF CIVIL ENGINEERING
SHREE S’AD VIDHYA MANDAL INSTITUTE OF TECHNLOGY
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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
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ANALYSIS OF SIMPLEPORTAL FRAME WITH
SWAY
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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 .
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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.
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w KN/m
w
P
Ha
Hd
B
h
C
D
A
L1
L2
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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):
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12KN
10 KN2.4 KN/m
A4 m
BC
1.5 m
1.5 m
( I )
( I )
D
1.5 m
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(b) Slope – Deflection equation :
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(c) Equilibrium equation :
At joint B,
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(d) Final Moments :
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(d) Simply supported moments :
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0.30
4.810.20
15
4.8
9
9.15D
CBA
(e) Simply supported moments
B.M Diagram
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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.
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90 KN
AAC B
3 m 6 m9 m
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Fixed end moments (FEM):
Slope deflection equations:
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NETSETTLEMENT5mm
15mm20
mm
Slope deflection
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B.M. diagram
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120.03
180
59.90
A B C
+
+
-
B.M Diagram
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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
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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
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)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 :
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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 :
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(E) B.M DIAGRAM
mkNl
Wab
mkNwl
.24010
46100MBC,Span
.2408
8308
M AB,Span
moments. supportedSimply 22
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158.18
240
163.64
240
BA C