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Structural Analysis
Chapter 4:
Analysis of statistically indeterminate structure
(Beam and Frame)
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Statistical indeterminate beams/frame
• Beam are statistically indeterminate generally because of their support system
– which have number untraceable (reaction to support) more than 3
– static equilibrium is not the solution
• Method to solve the problem:
– Slope deflection method
– Moment distribution methodDetermine reaction force, further shear strength and bending moment
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Slope deflection method
• The shear force and moment at the ends beam element are related to the end displacements rotations.
• Procedure
– Determine the fixed moment, moment result deflection and moment result deposit/support shift
– Determine slope at support
– Determine moment at support
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• Slope – deflection equation
θ1
θ2
M1 M2
δ1 δ2
L
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• Therefore bending moment diagram can be draw
• At any x cut, M=M1(1-x/L)-M2x/L
+veM1(1-x/L)
M1
M2
x
-ve
M2x/L
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• Relationship between curve and moment for beam
)3(6/)6/2/(
)2(2/)2/(
)1(/)/1(,
3
2
32
1
3
2
2
1
212
2
2
2
BAxLxMLxxMEIyor
ALxMLxxMdx
EIdyor
LxMLxMMdx
yEIdTherefore
Mdx
yEId
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• A and B can be defined using boundary method
When x= 0, y=0 placed in equation 3, B=0Also when x=0, dy.dx=θ1 in equation 2, A= -EIθ1
When x=L, y=δ, placed in equation (3)-EIδ = M1L
2/3 - M2L2/6 - EIθ1L equation (4)
Also when x = L ; dy/dx =θ2 place in equation (2)-EIθ2 = M1L/3 - M2L/2 - EIθ1 equation (5)
Rearrange equation (4) and (5)
M1=4EIθ1/L + 2EIθ2/L - 6EIδ/L2 equation (6)M2=4EIθ2/L + 2EIθ1/L - 6EIδ/L2 equation (7)
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Or it can be write (simplified equation)
L
L
EIM
L
EIM
)32(2
)32(2
122
211
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• Effect of Fixed end moment (FEM)(Page 4 of your lecturer notes)
– Defined as moment resultant at end to end outside tax incidence member that imposed when both its member is fixed
– When load is applied to the beam will produced FEM at both end
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• Include effect of the FEM
)32(2
)32(2
1212
2121
L
EIMM
L
EIMM
F
F
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Step of solution
• Write the slope for each end moment member
• Equilibrium equation obtainable to each joint when the total of joint member is zero (ΣM=0)
• Boundary condition placed at fixed end slope, where the displacement equal to zero
• After slope equation for value determined, all end moment defined with include slope value stated into equation
• Further member can be sorted for calculation reaction, shear strength and bending moment
• From here, shear force diagram and bending moment could be sketched
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Example 4.1
• Determine the moment value and shear force to each support and draw shear force diagram (SFD) and bending moment (BMD) or structure beam below. EI value is constant
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1. Finding the fixed end moment (FEM)
2. Check boundary condition
A = 0, Δ = 0 = 0, B = ?
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3. Slope deflection equation
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4. Equilibrium equation
5. End moment (From slope equation)
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6. Shear body diagram and bending moment Diagram
i. Reaction
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Exercise 4.1
• Determine the reaction at each support using slope deflection method and draw shear force diagram (SFD) and bending moment (BMD) for structure beam below. EI value is constant.
20 kN 20 kN
12.5 m 12.5 m 15 m 10 m
A B C
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Exercise 4.2
• Determine moment for beam below using slope deflection method. Beam also have a 40mm displacement at support B and 30mm at support C. (EI = 400kN.m2)
6 kN/m30 kN
B C
4 m 4 m 3 m 3 m
A D
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Moment distribution method
• Involve distribution moments to joint repetitively
• Accuracy depend on the number of repetition
• Value is dependant to
– fixed end moment (FEM)
– Factor bring side (Carry over factor-COF)
– Strength (Distribution factor-DF)
• Concept: determining element strength each structure which is depends on:
– Modulus flexibility/elasticity (E)
– Moment of inertia (I)
– Length of element/span (L)
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MBA
A B
θAMAB
AAB
BAAB
L
EIM
L
EIFEMM
4
)32(2
L
EIM AB
4 4EI/L I stiffness where EI/L is stiffness factor
θA can be define using slope deflection method
ABBA MM2
1 Where ½ is carry over factor (COF)
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K
M
K
KKK
MMMMM
KL
EM
KL
EM
KL
EM
AA
A
AADACAB
ADACABAA
AADA
AD
AACA
AC
AABA
AB
)(
)(
4
4
4
3
3
2
2
1
1
B C
D
A
MA
E3,I
3,L
3
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• Replace θA = MA/ΣK into moment at A (MA) equation:
AAD
AD
AAC
AC
AAB
AB
MK
KM
MK
KM
MK
KM
Expression Ki/ΣK known as distribution factor (DF), and ΣK<1, where:
Ki = Stiffness of member
ΣK = Total of stiffness for all member
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• Distribution factor (DF) at the end fixed
• Distribution factor (DF) at the end pinned
A
KAB
∞
A
KAB
0
0)(
)(
AB
ABAB
K
KDF
1)0(
)(
AB
ABAB
K
KDF
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Step of solution(Moment distribution method)
• Calculate distribution factor (DF)
• Fixed end Moment (FEM @MF)
• Moment distribution
– Table of distribution
• Finding the reaction
• Draw Shear Force Diagram & Bending Moment diagram
Joint A B C
Member AB BA BC CB
CF
DF
FEM
End moment
L
EIK
K
KDF
4,
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Example 4.3
• Draw BMD and SFD for the beam below
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1. Distribution factor (DF)
Joint Member K(4EI/L) ΣK DF(K/ΣK)
A AB 4EI/4.5 4EI/4.5+0 1
B
BA4EI/4.5
(4EI/4.5)+(8EI/3.5) =3.17EI
0.28
BC4E(2I)/3.5
0.72
C CB4E(2I)/3.5 8EI/3.5+∞
0
2. Fixed end moment (FEM)
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3. DistributionJoint A B C
Member AB BA BC CB
CF 0.5 0.5 0.5 0
DF 1 0.28 0.72 0
FEM -45 +45 -66.86 +66.86
Distr. +45 +6.12 +15.74 0
COF +3.06 +22.5 0 +7.87
Distr. -3.06 -6.3 -16.2 0
COF -3.15 -1.53 0 -8.1
Distr. +3.15 +0.43 +1.1 0
COF +0.22 +1.58 0 +0.55
Distr. -0.22 -0.44 -1.14 0
COF -0.22 -0.11 0 -0.57
Distr. +0.22 +0.03 +0.08 0
COF +0.02 +0.11 0 +0.04
Distr. -0.02 -0.03 -0.08 0
End moment 0 67.36 -67.36 +66.65
+45 – 66.86 = -21.8621.86 X 0.28= +6.12
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Exercise 4.3
• Draw SFD and BMD using moment distribution method
A
12.5m 12.5m 15m 10m
20kN 20kN
B C
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1. Distribution factor (DF)
Joint Member K(4EI/L) ΣK DF(K/ΣK)
A AB 4EI/25 4EI/25+∞ 0
B
BA4EI/25
8EI/25
0.5
BC4EI/25
0.5
C CB4EI/25 4EI/25+∞
0
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2. Fixed end moment (FEM)
kNmFEM
kNmFEM
kNmFEM
kNmFEM
CB
BC
BA
AB
7225
)10()15(20
4825
)10)(15(20
5.6225
)5.12()5.12(20
5.6225
)5.12)(5.12(20
2
2
2
2
2
2
2
2
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3. Distribution
Joint A B C
Member AB BA BC CB
CF 0 0.5 0.5 0
DF 0 0.5 0.5 0
FEM -62.5 +62.5 -48 +72
Distr. 0 -7.25 -7.25 0
COF -3.63 0 0 -3.63
Distr. 0 0 0 0
End moment -66.13 +55.25 -55.25 +68.37
+62.5 – 48 = +14.5-14.5 X 0.5
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4. Reaction
20kN
RA RB1
66.12 55.25
20kN
RB2 Rc
55.25 68.38
)(57.9;0
)(43.10
025.55)5.12(2012.66)(25
:0
1
kNRF
kNR
R
M
BY
A
A
B
)(47.7;0
)(53.12
0)(2538.68)15(2025.55
:0
2
kNRF
kNR
R
M
BY
C
C
B
Span A-B
Span B-C
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4. SFD and BMD20kN 20kN
+
-
+
-
10.44
9.56
7.47
12.53
AB
C
C
+ +- - -
66.12
=66.12+10.44(12.5)=64.31
55.26
68.38
AB
C
=55.25+7.47(15)=64.31
AB
RA=10.43 RB1=9.57 RB2=7.47 RC2=12.53
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Moment distribution method(Modify stiffness)
• To facilitate moment distribution process to be more precise
• Continues beam with either or both of them pin or roller
• Moment to pin or roller which is located at the end beam must null/empty
– No need to process bring side to support
– Modify should be done upper factor distribution
• Total moment at the end span , MAB = 0
AB
M
K1CK2
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• At B support
For span beam have end fixed and one more end are joint with pin, stiffness for span have extension pin stated is ¾ from original stiffness while calculation factor distribution.
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Example 4.5
• Determine end moment after moment distribution using the stiffness modifies method.
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1. Distribution factor (DF)
Joint Member K(4EI/L) ΣK DF(K/ΣK)
A AB(3/4(4EI/4.5))
=0.667EI0.667+0 1
B
BA(3/4)4EI/4.5 =0.0667EI
0.0667EI+2.286EI= 2.953
0.23
BC4E(2I)/3.5=2.286EI 0.77
C CB4E(2I)/3.5=2.286EI
2.286EI+∞0
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• Fixed end moment
• Moment distribution
Joint A B C
Member AB BA BC CB
CF 0.5 0 0.5 0
DF 1 0.23 0.77 0
FEM -45 +45 -66.86 +66.86
Distr. +45 +5.03 +16.83 0
COF 0 +22.5 0 +8.42
Distr. 0 -5.18 -17.32 0
COF 0 0 0 -8.6
Distr. 0 0 0 0
End moment 0 67.35 -67.35 +66.68
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Moment distribution method (rigid non sway Frame)
• In general, approach steps involved is the same as to beam. More practical by using modified stiffness.
• Rigid non-sway frame is the state where deformation / rigid framework movement will not cause joint or extension or join framework shift (Δ= 0).
• Case-study to rigid frames this make equivalent as in the case beam.
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• Shear force diagram
• Bending moment diagram
(+)
(+) (+)
(+)
(+)
(+)
(+)
(+)
(+)(+)
(+) (+)
(+)
(+)
(+)
(+) (+)
(+)
(+) (+)
(+)
(+)
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Example
Using moment distribution method, determine the end moment, reaction and draw shear force diagram and bending moment diagram for the frame below (EI constant)
100kN
20kN/m 4m
4m2m
A B
C
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1. Distribution factor (DF)
Joint Member K=(4EI/L) ΣK DF(K/ΣK)
A AB 4EI/6 4EI/6 +∞ 0
B
BA 4EI/6 4EI/6 + 4EI/4= 10EI/6
0.4
BC 4EI/4 0.6
C CB4EI/4 4EI/4+∞
0
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2) Fixed End Moment (FEM)
3) Moment distribution
kNmwl
FEM
kNmwl
FEM
kNmL
bPaFEM
kNmL
PabFEM
CB
BC
BA
AB
7.2612
420
12
7.2612
420
12
4.446
42100
9.886
42100
22
22
2
2
2
2
2
2
2
2
Joint A B C
Member AB BA BC CB
COF 0 0.5 0.5 0
DF 0 0.4 0.6 0
FEM -88.9 +44.4 -26.7 +26.7
Distr. 0 -7.1 -10.6 0
CO -3.6 0 0 -5.3
Distr. 0 0 0 0
End moment -92.5 +37.3 -37.3 +21.4
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Example 4.8
• Draw shear force and bending moment diagram to rigid frame structure as below.
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2
3
44
4
34
4
3
4
3EI
L
EIKKBC
stiffness equation for span with pinned end
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At joint B, the total of moment is 28.33-5.33+0 = +23
Therefore, Distribution value is
BA-23 X 0.25 = -5.75BC-23 x 0.5 = -11.5BD-23 x 0.25 = -5.75
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Moment distribution method (rigid frame with sway)
• Sway – framework experience some movement
• Cause moment lurch to which shift happen
• Moment due to sway (MS) with shift could be phrased as
Δ P2
P1
Δ
A
B C
D
2
.6
L
EI
2
.6
L
EI
Δ
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• If support end is pin or roller, then Moment due to sway (MS) becomes
• Sway to rigid frame will occur because of:
A
B C
D
0
32
DCS
CDS
M
L
EIM
Horizontal load Unsymmetrical vertical load
Unsymmetrical frame system (geometry or material (I))
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• Solution: 2 stage and combined (overlap principle)
Unequal support
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• First stage : No sway cause by constraint
• Second stage : sway framework due to Q counter direction to constraint discharge
Moment distribution can do as usual (FEM) Final moment consider as M1 (moment no sway)
horizontal to extension only (MS) Final moment consider as M2 (moment with sway)
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• Stage 1 + Stage 2 = Actual moment
Actual moment is , M = M1 + X.M2 where X is correction value If F case from no sway and Q is from the sway case:
F – X.Q = 0X = F/Q
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Example 4.9
• Draw shear force and bending moment diagram to rigid frame structure as below. The value of E is constant and assume EIΔ = 160
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• Calculation for F value
*F = 40 + 10.40 – 6.75 = 43.65 kN
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• Calculation for Q value
*Q = -(-22.24) – (-8.51) = 30.75 kN
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