Method of Consistent Deformation - Asad Iqbal · Structural Analysis By Aslam Kassimali Theory of...
Transcript of Method of Consistent Deformation - Asad Iqbal · Structural Analysis By Aslam Kassimali Theory of...
Moment-Distribution Method
Structural Analysis By
Aslam Kassimali
Theory of Structures-II
M Shahid MehmoodDepartment of Civil Engineering
Swedish College of Engineering & Technology, Wah Cantt
FRAMES
Analysis of Frames Without Sidesway
• The procedure for the analysis of frames without sidesway issimilar to that for the analysis of continuous beam.
• Unlike the continuous beams, more than two members may beconnected to a joint of a frame.
2
Example 1
• Determine the member end moments and reactions for the frameshown by the moment-distribution method.
3
A B
DC E
2 k/ft
40 k
10 ft
10 ft
I = 8
00 in
4
I = 8
00 in
4
I = 1,600 in4 I = 1,600 in4
30 ft 30 ft
E = 29,000 ksi
Solution
1.Distribution Factors
• Distribution Factors at Joint C,
4
571.0
30
1600
20
800
30
1600
429.0
30
1600
20
800
20
800
CD
CA
DF
DF
Checks 1571.0429.0 CDCA DFDF
• Distribution Factors at Joint D,
5
3.0
30
1600
4
3
30
1600
20
800
30
1600
4
3
4.0
30
1600
4
3
30
1600
20
800
30
1600
3.0
30
1600
4
3
30
1600
20
800
20
800
DE
DC
DB
DF
DF
DF
Checks 14.03.02 DCDEDB DFDFDF
• Distribution Factors at Joint E,
2.Fixed-End Moments (FEMs)
3.Moment Distribution
4.Final Moments
6
1EDDF
ft-k 150
ft-k 150
0
ft-k 100
ft-k 100
EDDC
DECD
DBBD
CA
AC
FEMFEM
FEMFEM
FEMFEM
FEM
FEM
7
Distribution Factors
1.Fixed-end Moments
ACMember Ends
2.Balance Joints
3.Carryover
4.Balance Joints
5.Carryover
7.Carryover
8.Balance Joints
9.Carryover
-10.7
10.Balance Joints
13.Final Moments +92.1 -115.9 +115.9 -186.4 -19.4 +205.6
CA CD DC DB DE ED BD
0.429 0.571 0.4 0.3 0.3 1
+100 -100 +150 -150 +150 -150
-21.4 -28.6 +150
-14.3 +75
-24.3 -18.2 -18.2
-12.2
6.Balance Joints +5.2 +7
+2.6 +3.5
-1.4 -1.1 -1.1
-0.7
+0.4+0.3
11. Carryover +0.2 +0.2
-0.1 -0.1 -0.112.Balance Joints
0
-9.1
-0.6
-9.7
Carryover
8
A
C
2 k/ft
40 k
92.1
115.9
115.9C
186.4D
B
D
9.7
19.4
2 k/ft
D E205.6
Analysis of Frames With Sidesway
Consider the rectangular frame shown in Figure.
A qualitative deflected shape of the frame for an arbitrary loadingis shown in the figure using an exaggerated scale.
9
A B
DC
w
P
EI = constant
Δ Δ
While the fixed joints A and B of the frame are completelyrestrained against rotation as well as translation, the joints C and Dare free to rotate and translate.
Since the members of the frame are assumed to be inextensibleand the deformations are assumed to be small, the joints C and Ddisplace by the same amount Δ, in the horizontal direction only.
10
A B
DC
w
P
Actual Frame
Δ Δ
The MD analysis of such a frame, with sidesway, is carried out intwo parts.
In the first part, the sidesway of the frame is prevented by addingan imaginary roller to the structure.
External loads are then applied to this frame, and MEM arecomputed by applying the MD process in the usual manner.
11
A B
DC
w
PFrame with sidesway prevented
With the MEM known, the restraining force (reaction) R thatdevelops at the imaginary support is evaluated by applying theequations of equilibrium.
In the second part of the analysis, the frame is subjected to theforce R, which is applied in the opposite direction, as shown in thenext slide.
12
A B
DC
w
P
R
Imaginary roller
Frame with sidesway prevented
The moments that develop at the member ends are determinedand superimposed on the moments computed in the first part toobtain the member end moments in the actual frame.
If M, MO, and MR denote, respectively, the MEM in the actualframe, the frame with sidesway prevented, and the framesubjected to R, then we can write
13
A B
D
C R
Δ Δ
Frame subjected to R
RO MMM
14
A B
DC
w
P
Δ Δ
A B
DC
w
P
R
A B
D
C R
Δ Δ
+
=
M Moments
MO Moments MR Moments
RO MMM
An important question that arises in the second part is “how todetermine the member end moments MR that develop when theframe undergoes sidesway under the action of R”.
The MDM cannot be used directly to compute the moments due tothe known lateral load R, we employ an indirect approach in whichthe frame is subjected to an arbitrary known joint translation Δ’caused by an unload load Q acting at the location and in thedirection of R.
From the known joint translation, Δ’, we determine the relativetranslation between the ends of each member, and we calculatethe member FEMs in the same manner as done previously in thecase of support settlements.
15
16
A B
D
C Q
Δ Δ
Frame subjected to an arbitrary Translation Δ’MQ Moments
The FEMs thus obtained are distributed by the MD process todetermine the MEMs MQ caused by the yet-unknown load Q.
Once the MEMs MQ have been determined, the magnitude of Qcan be evaluated by the application of equilibrium equations.
With the load Q and the corresponding moments MQ known, thedesired moments MR due to the lateral load R can now bedetermined easily by multiplying MQ by the ratio R/Q; that is
By substituting this Equation into the last Equation, w can expressthe MEMs in the actual frame as
17
QR MQ
RM
QO MQ
RMM
Example 2
• Determine the member end moments and reactions for the frameshown by the moment-distribution method.
18
A
B
DC
40 kN
7 m
EI = constant
4 m3 m
5 m
Solution
• Distribution Factors
At joint C
At joint D
19
A
B
DC
40 kN
7 m
EI = constant
4 m3 m
5 m
5.0
72
7 I
I
DFDF CDCA
583.0
57
5
417.0
57
7
II
I
DF
II
I
DF
DB
DC
Checks 1583.0417.0 DBDC DFDF
• Part 1: Sidesway Prevented
The sidesway of frame is prevented by adding an imaginary rollerat joint C.
Assuming that joint C and D of this frame are clamped againstrotation, we calculate the FEMs due to the external loads to be
20
A
B
DC
40 kN
Imaginary RollerFrame with sidesway prevented
0
m . kN 4.29
m . kN 2.39
DBBDCAAC
DC
CD
FEMFEMFEMFEM
FEM
FEM
R
The MD of these FEMs is then performed, as shown on the MDTable to determine the MEMs “MO” in the frame with sideswayprevented.
21
AC
-9.8
-12 -24 +23.9 -24 +24
CA CD DC DB
0.5 0.5 0.417 0.583
+39.2 -29.4
-19.6 -19.6
-9.8 +8.6
+4.1 +5.7
+2.1
-1.1 -1.1
-1.6
-0.6 +0.4 -0.6 +0.5
+0.3-0.2
+12
BD
+12.3 +17.1
+6.2
-3.1 -3.1
-1.6 +2.9
+0.7 +0.9
-0.2 +0.3
Member End Moments for Frame with Sidesway Prevented - MO
To evaluate the restraining force R that develops at the imaginaryroller support, we first calculate the shears at the lower ends of thecolumns AC and BD by considering the moment equilibrium of thefree bodies of the columns.
22
A
B
DC
5.14
12
7.2
12
7.2
24
5.14
24
7 m
5 m
Next, by considering the equilibrium of the horizontal forces actingon the entire frame, we determine the restraining force R to be
Restraining force acts to the right, indicating that if the roller wouldnot have been in place, the frame would have swayed to the left.
23
kNRRFx 06.2 02.714.5 0
A
B
DC
R = 2.06
5.14
7.2
• Part 2: Sidesway Permitted
Since the actual frame is not supported by a roller at joint C, weneutralize the effect of the restraining force by applying a lateralload R = 2.06 kN in the opposite direction to the frame.
MD method cannot be used directly to compute MEMs MR due to thelateral load R = 2.06 kN, we use an indirect approach in which theframe is subjected to an arbitrary known joint translation Δ’ causedby an unknown load Q acting at the location in the direction of R.
24
A
B
DC
R = 2.06 kN
Frame subjected to R = 2.06 kNMR Moments
Assuming that the joints C and D of the frame are clamped againstrotation as shown in figure on next slide, FEMs due to thetranslation Δ’ are given by
25
A
B
DCQ
Frame subjected to an ArbitraryTranslation Δ’MQ Moments
Δ’ Δ’
0
25
'6
5
'6
49
'6
7
'6
2
2
DCCD
DBBD
CAAC
FEMFEM
EIEIFEMFEM
EIEIFEMFEM
In which negative sign have been assign to the FEMs for thecolumns, because these moments must act in the clockwisedirection, as shown.
Instead of arbitrarily assuming a numerical value for Δ’ to computethe FEMs, it is usually more convenient to assume a numericalvalue for one of the FEMs, evaluate Δ’ from the expression of thatFEM, and use the value of Δ’ to compute the remaining FEMs.
26
A
B
DC
FEMs due to Known Translation Δ’
Δ’ Δ’
Thus, we arbitrarily assume the FEMAC to be -50 kN.m
by solving for Δ’, we obtain
by substituting this value of Δ’ into the expressions for FEMBD andFEMDB, we determine the consistent values of these moments tobe
These FEMs are then distributed by the usual MD process, todetermine the MEMs MQ caused by the yet unknown load Q.
27
kN.m 5049
'6
EIFEMFEM CAAC
EI
33.408'
kN.m98
25
33.4086
DBBD FEMFEM
28
AC
+12.5
-42.3 -34.5 +34.3 +45.4 -45.4
CA CD DC DB
0.5 0.5 0.417 0.583
+25 +25
+12.5 +28.6
-5.2 -7.3
-2.6
+1.3 +1.3
-5.2
+0.7 +1.1 +0.7 +1.5
-0.3-0.6
-71.8
BD
+40.9 +57.1
+20.5
-10.3 -10.3
-5.2 -3.7
+2.2 +3
-0.6 -0.4
Member End Moments Due to Known Translation Δ’- MQ
-50 -50 -98 -98
-0.3 -0.2 -0.3 -0.2
+0.1+0.1 +0.1 +0.2
To evaluate the magnitude of Q that corresponds to these MEMs,we first calculate shears at the lower ends of the columns byconsidering their moment equilibrium and then apply the equationof equilibrium in the horizontal direction to the entire frame
29 kNQQFx 41.34 044.2397.10 0
A
B
DC
10.97
42.3
23.44
71.8
23.44
45.4
10.97
34.5
7 m
5 m
A
B
DC
Q = 34.41
10.97
23.44
which indicates that the moments MQ computed are caused by alateral load Q = 34.41 kN.
Since the moments are linearly proportional to the magnitude ofthe load, the desired moment MR due to the lateral load R = 2.06kN must be equal to the moment MQ multiplied by the ratio R/Q =2.06/34.41.
• Actual Member End Moments
The actual MEMs , M, can now be determined by algebraicallysumming the MEMs MO and 2.06/34.41 times the MEMs MQ.
30
31
NS m . kN 7.78.7141.34
06.212
NS m . kN 3.214.4541.34
06.224
NS m . kN 3.214.4541.34
06.224
NS m . kN 263.3441.34
06.29.23
NS m . kN 1.265.3441.34
06.224
NS m . kN 5.143.4241.34
06.212
AM
AM
AM
AM
AM
AM
BD
DB
DC
CD
CA
AC
32
A
C
14.5
26.1
C21.3
D
B
D
7.7
21.3
40 kN
26.1
Actual Member End Moments (kN . m)
Example 3
• Determine the member end moments and reactions for the frameshown by the moment-distribution method.
33
A
B
DC30 k
20 ft12 ft
16 ft
EI = constant
12 ft
8 ft
Solution
• Distribution Factors
At joint C
34
A
B
DC30 k
20 ft12 ft
16 ft
EI = constant
12 ft
8 ft
5.0
202
20
I
I
DFDF CDCA
Solution
• Distribution Factors
At joint D
35
A
B
DC30 k
20 ft12 ft
16 ft
EI = constant
12 ft
8 ft
49.0
42.144
3
20
20
II
I
DFDC51.0
42.144
3
20
42.144
3
II
I
DFDB
MEMs due to an Arbitrary Sidesway Δ’
Since no external loads are applied to the members of the frame,the MEMs MO in the frame restrained against sidesway will bezero.
To determine the MEMs M due to the 30-k lateral load, we subjectthe frame to an arbitrary known horizontal translation Δ’ at joint C.
Figure on the next slide shows a qualitative deflected shape of theframe with all joints clamped against rotation and subjected to thehorizontal displacement Δ’ at joint C.
36
37
A
B
DC30 k
Δ’
4
3
5
5
4
3
3
2
13C’
D’
D1
3
213
D
D1
D’
Δ
32
13
5
4
3
'3
2
'4
3
FEMs due to an Arbitrary Translation Δ’
MEMs due to an Arbitrary Sidesway Δ’
Note that, since the frame members are assumed to beinextensible and deformations are assumed to be small, an end ofa member can translate only in a direction perpendicular to themember.
From this figure, we can see that the relative translation ΔAC
between the ends of members AC in the direction perpendicular tothe member can be expressed in terms of the joint translation Δ’ as
38
'25.1'4
5' CCAC
MEMs due to an Arbitrary Sidesway Δ’
Similarly, the relative translation for members CD and BD are givenby
The FEMs due to the relative translation are
39
'202.1'3
13'
'417.1'4
3'
3
2'1
DD
DD
BD
CD
2
2
2
42.14
'202.16
20
'417.16
20
'25.16
EIFEMFEM
EIFEMFEM
EIFEMFEM
DBBD
DCCD
CAAC
MEMs due to an Arbitrary Sidesway Δ’
in which the FEMs for members AC and BD are CCW (positive),whereas those for member CD are CW (negative).
If we arbitrarily assume that
then
and therefore
40
ft-k 10042.14
'202.162
EI
FEMFEM DBBD
2.2883'EI
ft-k 3.61
ft-k 1.54
DCCD
CAAC
FEMFEM
FEMFEM
The FEMs are distributed by the MD process to determine theMEMs MQ.
To determine the magnitude of the load Q that corresponds to theMEMs MQ we first calculate the shears at the ends of the girder CDby considering the moment equilibrium of the free body of thegirder as shown.
41
A
C
55.3
56.5
C55.2
D
B
D
55.2
56.4
11.17
11.17
5.58 5.58
5.58
5.585.58
5.58
8.32
8.32
Evaluation of Q
42
AC
+1.8
+55.3 +56.5 -56.4 -55.2 +55.2
CA CD DC DB
0.5 0.5 0.49 0.51
+3.6 +3.6
+1.8
+23.6 +24.6
+11.8
-5.9 -5.9
+2.4
-3 -0.6 -3
+1.5+0.3
0
BD
-19 -19.7
-9.5
+4.8 +4.8
+2.4
-1.2 -1.2
+0.3 +1.5
Member End Moments Due to Known Translation Δ’- MQ
+54.1 +54.1 +100 +100
+0.2 +0.8 +0.2
-0.1-0.4 -0.4 -0.1
1
-0.2
-61.3 -61.3
-0.2
-50
+0.1
-100
+0.1
MD TABLE
The girder shears (5.58 k) thus obtained are then applied to thefree bodies of the inclined members AC and BD.
Next, we apply the equations of moment equilibrium to membersAC and BD to calculate the horizontal forces at the lower ends ofthese members.
43
AB
DC
11.17
8.32
Evaluation of Q
The magnitude of Q can now be determined by considering theequilibrium of horizontal forces acting on the entire frame asshown below
44
AB
DCQ = 19.49 k
11.17
8.32
Evaluation of Q
k 49.19 032.817.11 0 QQFx
Actual MEMs
The actual MEMs, M, due to the 30-k lateral load can now beevaluated by multiplying the moments MQ computed in Table bythe ratio 30/Q=30/19.49:
45
NS 0
NS ft -k 852.5549.19
30
NS ft -k 852.5549.19
30
NS ft -k 8.864.5649.19
30
NS ft -k 875.5649.19
30
NS ft -k 1.853.5549.19
30
AM
AM
AM
AM
AM
AM
BD
DB
DC
CD
CA
AC
Member End Forces
46
A
C
85.1
87
C85
D
B
D
85
86.8
17.2
17.2
8.59 8.59
8.59
8.59
8.59
8.59
12.8
12.8
Evaluation of Q
12.8 12.8C
8.59
8.59
17.2
30 kD
8.59
12.8
8.59
Support Reactions
47
A
B
DC30 k
85.1 k-ft
17.2 k
8.59 k
8.59 k
12.8 k