theory of structures chapter 3 : moment distribution …
TRANSCRIPT
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by Saffuan Wan Ahmad
THEORY OF STRUCTURESCHAPTER 3 : MOMENT DISTRIBUTION
(FOR FRAME)PART 4
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 4 – Moment Distribution
• Aims– Determine the end moment for frame using Moment Distribution Method
• Expected Outcomes :– Able to do moment distribution for frame.
• 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
MDM for Frame without SIDE-SWAY• MDM- solving indeterminate structures is a process in which the
moment in the members are determined by successive approximation.• Does not result in moment diagram but it provides the magnitude and
sense of the internal moments at joint – to obtain the shear and bending moment.
• TERM USED– Fixed end moment (FEM)– Carry over factor– Stiffness or resistance to rotation of a member
Clockwise moments are considered positiveWhereas, counterclockwise is negative
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by Saffuan Wan Ahmad
MDM for Frame without SIDE-SWAYSummary• Stiffness for member: end is pinned
equal to:
• Stiffness for member: end is fixed equal to:
LEIK 3
LEIK 4
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by Saffuan Wan Ahmad
Frame WITHOUT Sidesway
Determine the final moment for frame ABCD shown below. Hence,Draw the SFD and BMD.EI is indicate in the figure.
100kN
Cable / Tie Rod
3m 3m
3m
A
B C
D
4EI
2EI
EI
EXAMPLE 1
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by Saffuan Wan Ahmad
Fixed End Moment (MF)
kNmMkNmL
PabM
MMMM
FCB
FBC
FDC
FCD
FBA
FAB
7575
0
2
2
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by Saffuan Wan Ahmad
Distribution Factor (DF)
JOINT MEMBER K K DF
A AB 0
BBA 0.8
BC 0.2
CCB 0.5
CD 0.5
D DC 0
3)4(4 EI
3)4(4 EI
6)2(4 EI
6)2(4 EI
34EI
34EI
3
16EI
3
4EI
320EI
38EI
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by Saffuan Wan Ahmad
Member AB BA BC CB CD DC
DF 0 0.8 0.2 0.5 0.5 0
FEM 0 0 -75 75 0 0
BalCO
030
600
15-18.75
-37.57.5
-37.50
0-18.75
BalCO
07.5
150
3.75-1.88
-3.751.88
-3.750
0-1.88
BalCO
00.75
1.50
0.37-0.47
-0.940.19
-0.940
0-0.47
Bal 0 0.38 0.1 -0.1 -0.1 0
End Moment 38.25 76.88 -76.88 42.28 -42.29 -21.10
Table Moment Distribution
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by Saffuan Wan Ahmad
Shear Force and Bending Moment Diagram
A
BC
D
_
_
+_
+
38.25
55.58
44.20
21.10 38.25
A
B C
D
76.85
90.50
21.00
42.23
SFD BMD
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by Saffuan Wan Ahmad
Frame WITHOUT Sidesway
Determine the final moment for frame ABCDF with overhanging member BEshown below. Draw the BMD.EI is indicate in the figure.
5 kN 15 kN 10 kN
5 kN/m
A D
EB C F
2m 2m 2m 2m3m
4m4EI
4EI 4EI
4EI
3EI
EXAMPLE 2
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by Saffuan Wan Ahmad
Fixed End Moment (MF)
kNmM
kNmM
kNmM
MM
kNmM
kNmM
kNmM
kNmM
FBE
FFC
FCF
FDC
FCD
FCB
FBC
FBA
FAB
10)2(5
2.7
8.4
0
75
5.7
67.6
67.6
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by Saffuan Wan Ahmad
Distribution Factor (DF)
JOINT MEMBER K K DF
B
BA 0.5
BC 0.5
BE 0
C
CB
CD
CF
4)4(4 EI
4)4(4 EI
0
4)4(4 EI
4)4(4 EI
5)3(3 EI
EI8
549EI
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by Saffuan Wan Ahmad
JOINT A B C D F E
Member AB BA BC BE CB CD CF DC FC EB
DF 0 0.5 0.5 0 0 1.0 0
FEM -6.67 6.67 -7.5 10.0 7.5 0 -4.8 0 7.2 0
BalCO
-4.59 -4.49 0 -7.2
BalCOBalCOBal
End Moment
0 0
Table Moment Distribution **Note:Any no. divide by infinity =0
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by Saffuan Wan Ahmad
5 kN 15 kN 10 kN
5 kN/m
A D
EB C F
2m 2m 2m 2m3m
4m4EI
4EI 4EI
4EI
3EI
9.13
1.78
11.7410 6.34
0.65
1.29
7.63
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by Saffuan Wan Ahmad
9.13
1.78
A
B C
D
FE
11.74
10
15
6.34
7.63
12
1.29
10
BMD
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by Saffuan Wan Ahmad
MDM for Frame with SIDE-SWAY• The frames that are non-symmetrical or subjected to non-symmetrical
loadings have a tendency to SIDE-SWAY• Application of this technique is illustrated as below.
R
P
Virtual Prop Force(No side-sway)
R’
Virtual Prop Force is REMOVED(side-sway)
∆ ∆
P
Horizontal displacement are EQUAL
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by Saffuan Wan Ahmad
Summary• Stiffness for member: end is pinned
equal to:
• Stiffness for member: end is fixed equal to:
LEIK 3
LEIK 4
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by Saffuan Wan Ahmad
Summary• Moment produced at each member
when one end of member is displaced relative to other and both ends are FIXED
• Moment produced at the near end of a member if the remote end is displaced relative to the near end with remote end held in position but allowed to rotate
26
LEIM
23
LEIM
L
L
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by Saffuan Wan Ahmad
A portal frame ABCD as shown in Figure below is subjected to point load of 100 kN atmember BC. EI is constant. Analyze using Moment Distribution Method.
3m
A
B
3m 3m 4m
C
D
100 kN
EXAMPLE 3
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SOLUTION EXAMPLE 3:
DA
B C
100 kN
DA
BC
100 kN
=
+∆
DA
BC
∆BC
∆CD
∆
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by Saffuan Wan Ahmad
Case 1:Fixed End Moment (MF): Non-sway Analysis
kNmMM
MMMMF
CBF
BC
FDC
FCD
FBA
FAB
75
0
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by Saffuan Wan Ahmad
Distribution Factor (DF)
JOINT MEMBER K K DF
A AB 0
BBA 0.67
BC 0.33
CCB 0.53
CD 0.47
D DC 1
34EI
34EI
64EI
64EI
53EI
53EI
3
4EI
53EI
EI2
1519EI
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by Saffuan Wan Ahmad
Member AB BA BC CB CD DC
DF 0 0.67 0.33 0.53 0.47 1
MF 0 0 -75 75 0 0
BalCO 25.2
50.3 24.7-19.9
-39.7512.4
-35.25-17.63
BalCO
06.7
13.3 6.6-3.3
-6.63.3
-5.8 17.63
BalCO
01.1
2.2 1.1-0.85
-1.70.55
-1.6
Bal 0 0.57 0.28 -0.29 -0.26
End Moments 33.0 66.37 -66.37 42.91 -42.91 0
Table Moment Distribution (Non-sway Analysis)
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by Saffuan Wan Ahmad
Horizontal Reactions
kNHH
M
B
B
A
12.33037.660.33)3(
,0
A
33.0
66.37
B
HA
HB
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66.37
HB HC
VB VC
42.91100
kNVVVF
kNVV
M
B
CB
y
C
C
B
9.530100
,01.46
091.42)3100(37.66)6(,0
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by Saffuan Wan Ahmad
kNHH
M
C
C
D
77.75091.42)41.46()3(
,0
- 42.91
D
C
HD
HC
46.1
VD
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by Saffuan Wan Ahmad
33.12 kN75.77 kNR
kNRR
HHRF
R
CB
H
65.42077.7512.33
00
find To
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by Saffuan Wan Ahmad
Case 2:Fixed End Moment (MF): Sway Analysis
DA
BC
∆ ∆
∆BC
∆CD3
434tan
BCBC
35
54sin
CDCD
BC
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5533
92
666
96
366
,
2
)35
(
2
2
)34
(
2
22
EIEILEIM
EIEILEIMM
EIEILEIMM
therefore
CDS
CBS
BCS
BAS
ABS
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by Saffuan Wan Ahmad
9:10:305
:9
2:9
6::
:,45
,,,
EIEIEIMMM
thereforeEIassume
DCCDS
CBBCS
BAABS
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Member AB BA BC CB CD DC
DF 0 0.67 0.33 0.53 0.47 1
MF 30 30 -10 -10 9 0
BalCO -6.7
-13.4 -6.60.27
0.53-3.3
0.470.24
BalCO
0-0.09
-0.18 -0.090.88
1.75-0.5
1.55
BalCO
0-0.3
-0.59 -0.290.14
0.27-0.15
0.23
Bal 0 -0.09 -0.05 0.08 0.07
Assume Sway Moment
22.91 15.74 -15.74 -11.32 11.32 0
Table Moment Distribution (Sway Analysis)
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by Saffuan Wan Ahmad
Horizontal Reactions
kNHH
M
B
B
A
88.12074.1591.22)3(
,0
A
22.91
15.74
B
HA
HB
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-15.74
HB HC
VB VC
-11.32100
kNVV
M
C
C
B
51.4032.1174.15)6(
,0
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by Saffuan Wan Ahmad
kNHH
M
C
C
D
79.90)451.4(32.11)3(
,0
11.32
D
C
HD
HC
4.51
VD
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12.88 kN9.79 kNR
kNRR
HHRF
R
CB
H
67.22079.988.12
00
find To
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Correction Factor and Final Moment
88.167.2265.42
'
RRASM
A B C DAssume sway moment
22.91 15.74 -15.74 -11.32 11.32 0
Actual sway moment (ASM)
43.07 29.59 -29.59 -21.28 21.28
(Non-sway moment)
33.0 66.37 -66.37 42.91 -42.91 0
Final Moments
76.07 95.96 -95.96 21.63 -21.63 0
**ASM x Assume sway moment
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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