theory of structures chapter 3 : slope deflection …
TRANSCRIPT
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by Saffuan Wan Ahmad
THEORY OF STRUCTURESCHAPTER 3 : SLOPE DEFLECTION
(FOR FRAME)PART 2
bySaffuan Wan Ahmad
Faculty of Civil Engineering & Earth [email protected]
For updated version, please click on http://ocw.ump.edu.my
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by Saffuan Wan Ahmad
Chapter 3 : Part 2 – Slope Deflection
• Aims– Determine the end moment for frame using Slope Deflection Method.
• Expected Outcomes :– Able to identify the frame – with or without side sway.– Able to determine end moment at critical points.
• References– Mechanics of Materials, R.C. Hibbeler, 7th Edition, Prentice Hall – Structural Analysis, Hibbeler, 7th Edition, Prentice Hall– Structural Analysis, SI Edition by Aslam Kassimali,Cengage Learning– Structural Analysis, Coates, Coatie and Kong – Structural Analysis - A Classical and Matrix Approach, Jack C. McCormac
and James K. Nelson, Jr., 4th Edition, John Wiley
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by Saffuan Wan Ahmad
FRAME
WITH SIDE SWAY
Displace to the side when the body or the loading acting on it is
nonsymmetric
WITHOUT SIDE SWAY
Properly restrainedSymmetric with
respect both loading and geometry
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by Saffuan Wan Ahmad
DEPEND ON CONFIGURATION
SUPPORT SYSTEMLOADING SYSTEM
MEMBER SIZES
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by Saffuan Wan Ahmad
FRAME WITH SIDESWAY
FRAME WITHOUT SIDESWAY
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by Saffuan Wan Ahmad
ANALYSIS - FRAME WITHOUT SIDESWAY.
By using SDM, determine the bending moment at critical points. Assume EI is constant.
EXAMPLE 1
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by Saffuan Wan Ahmad
SOLUTION
FIXED END MOMENT
0 FSR
FRS
FQP
FPQ MMMM
kNm .75338
)6(458
kNm 75.338
)6(458
PLM
PLM
FRQ
FQR
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by Saffuan Wan Ahmad
SLOPE DEFLECTION EQUATION 0 SP
FPQ
QPPQ M
LEI
LEI
LEIM
2
624
3Q
PQ
EIM
32 Q
QP
EIM
FQP
PQQP M
LEI
LEI
LEI
M
2
624
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by Saffuan Wan Ahmad
75.3333
2 RQ
QREIEI
M
75.3333
2 QR
RQ
EIEIM
FRQ
QRRQ M
LEI
LEI
LEIM
2
624
FQR
RQQR M
LEI
LEI
LEI
M
2
624
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by Saffuan Wan Ahmad
32 R
RSEIM
3R
SREIM
FRS
SRRS M
LEI
LEI
LEIM
2
624
FSR
RSSR M
LEI
LEI
LEIM
2
624
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by Saffuan Wan Ahmad
EQUILIBRIUM AT JOINT
0QM
0RM
1
2
75.3333
40
RQ
QRQP
EIEIMM
75.3333
4
0
QR
RSRQ
EIEI
MM
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by Saffuan Wan Ahmad
75.3375.33
34
31
31
34
R
Q
EIEI
BY USING CALCULATOR
75.3375.33
R
Q
EIEI
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by Saffuan Wan Ahmad
SUBSTITUTING INTO SDE
kNmMkNmM
kNmMkNmM
kNmMkNmM
SR
RS
RQ
QR
QP
PQ
25.115.22
5.225.22
5.2225.11
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by Saffuan Wan Ahmad
Determine end moment at critical point.
10kN/m
A
B C
D
3m 3m
6m
EI
2EI 2EI
FOOD OF MIND
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by Saffuan Wan Ahmad
SOLUTION
FIXED END MOMENT
0 FDC
FCD
FBA
FAB MMMM
kNmMM FCB
FBC 30
SLOPE DEFLECTION EQUATION 0 DA
FABBAAB M
LLEIM
32)2(2
32 B
ABEIM
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by Saffuan Wan Ahmad
FBAABBA M
LLEIM
32)2(2
34 B
ABEIM
FBCCBBC M
LLEIM
322
3033
2 CB
BCEIEIM
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by Saffuan Wan Ahmad
FCBBCCB M
LLEIM
322
FCDDCCD M
LLEIM
32)2(2
34 C
CDEIM
3033
2 BC
CBEIEIM
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by Saffuan Wan Ahmad
FDCCDDC M
LLEIM
32)2(2
32 C
DCEIM
EQUILIBRIUM AT JOINT
00
00
CDCB
C
BCBA
B
MMM
MMM
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by Saffuan Wan Ahmad
3030
231
312
C
B
EIEI
Solving by using Calculator
1818
C
B
EIEI
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by Saffuan Wan Ahmad
kNmMkNmMkNmMkNmM
kNmMkNmM
DC
CD
CB
BC
BA
AB
12242424
2412
SUBSTITUTING INTO SDE
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by Saffuan Wan Ahmad
ANALYSIS OF FRAME WITH SIDE SWAY.
DEPEND ON CONFIGURATION
SUPPORT SYSTEMLOADING SYSTEM
MEMBER SIZES
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by Saffuan Wan Ahmad
FUNDAMENTAL ASSUMPTIONS1. That axial deformation is ignored
2. That transverse end displacement do not affect the member lengthB B’ C C’
EI
EI
3EI
B B’ C C’
EI
EI
3EI
B’BA A’ C C’
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by Saffuan Wan Ahmad
DETERMINE END MOMENT AT CRITICAL POINT. ASSUME EI IS CONSTANT.
200 kN
6m
4m
5m
C
D
A
B
EXAMPLE 2
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by Saffuan Wan Ahmad
200 kN
6m
4m
5m
C
D
A
B
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by Saffuan Wan Ahmad
SOLUTION
0 FDC
FCD
FCB
FBC
FBA
FAB MMMMMM
Fixed End Moment
83
2
EIEIM B
AB
SLOPE DEFLECTION EQUATION
83
EIEIM BBA
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by Saffuan Wan Ahmad
52
54 CB
BCEIEIM
52
55 BC
CBEIEIM
63
EIEIM C
DC
632
EIEIM C
CD
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by Saffuan Wan Ahmad
0BM
Equilibrium at joint
0 BCBA MM
05
25
48
3
CB
BEIEIEIEI
08
35
25
9
EIEIEI CB 1
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by Saffuan Wan Ahmad
0CM
0 CDCB MM
063
25
25
4
EIEIEIEI CBC
0615
225
2
EIEIEI CB 2
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by Saffuan Wan Ahmad
0H
0200 DA HH
4
0)4(0
BAABA
ABAAB
B
MMH
HMMM
Consider member AB
A
B
AH
BH BAM
ABM
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by Saffuan Wan Ahmad
6
0)6(0
CDDCD
DDCCD
C
MMH
HMMM
Consider member CD
DH
CH CDM
DCM
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by Saffuan Wan Ahmad
48004466
20064
CDDCABBA
CDDCABBA
MMMM
MMMM
Insert the value……
48006
353
129
EIEIEI C
B 3
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by Saffuan Wan Ahmad
MATRIX FORM
480000
635
3129
61
1522
52
83
52
59
EIEIEI
C
B
Solving by using calculator
86.125066.7578.243
EIEIEI
C
B
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by Saffuan Wan Ahmad
Substituting into SDE
kNmM
kNmM
kNmM
kNmM
kNmM
kNmM
DC
CD
CB
BC
BA
AB
26.1836
88.1250366.75
04.1586
88.12503
)66.75(2
04.1585
)78.243(25
)66.75(4
29.2255
)66.75(25
)78.243(4
29.2258
)88.1250(378.243
16.3478
)88.1250(32
78.243
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by Saffuan Wan Ahmad
THANKS
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Author Information
Mohd Arif Bin SulaimanMohd Faizal Bin Md. Jaafar
Mohammad Amirulkhairi Bin ZubirRokiah Binti Othman
Norhaiza Binti GhazaliShariza Binti Mat Aris