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Constructing Beam Influence Lines Using the Muller-Breslau Principle Steven Vukazich San Jose State University
Constructing Beam Influence Lines Using the Muller-Breslau Principle
StevenVukazichSanJoseStateUniversity
Example Problem
x
24 ft 20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
18 ft
Q
A beam is pin supported at point A and roller supported at points B, D, and F. The beam contains internal hinges at points C and E.
Construct the influence line for the bending moment at point Q.
Muller-Breslau Principle to Find the Shape of the MQ Influence Line
24 ft 20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
18 ft
Q
1. Remove the ability of the beam to resist bending moment at point Q. This is the modified unstable structure;
2. Apply the response quantity, MQ, consistent with the chosen positive sign convention;
3. The rigid body motion of the modified structure is the shape of the MQ influence line.
MQMQ
Shape of MQ Influence Line
20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
MQ
0
18 ft
Q
6 ft
x
+
??
??
Use Shape of Influence Line to Place Unit Load
20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
MQ
0
18 ft
Q
6 ft
x
+
??
??
1
Place the unit load just to the left of Q and use statics to find the value of MQ
Place Unit Load at x = 18– ft (Just to the Left of Point Q)
18– ft F
BAx
Ay
Fy
9 Unknowns – 9 Equations of Equilibrium
DVC VC
FC
FE
Fy = 0
FE = 0
VE = 0
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
By = 0.75
Ax = 0
Ay = 0.25
QBy
20 ft
12 ft
8 ft
12 ft 8 ft18 ft
6 ft
x
1
VEFEDy
VE
Dy = 0
FC = 0
VC = 0
FBD of Segment QBC for Unit Load at x = 18– ft(Just to the Left of Point Q)
B
VC = 0FQ
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MQ = 4.5 ft
FQ = 0
VQ = – 0.75
Q
MQ
VQFC = 0
12 ft6 ft
0.75
Use Shape of Influence Line to Place Unit Load
20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
MQ
0
18 ft
Q
6 ft
x
+
?4.5 ft
??
1
Place the unit load just to the left of C and use statics to find the value of MQ
Place Unit Load at x = 36– ft (Just to the Left of Point C)
18– ft F
BAx
Ay
Fy
DVC VC
FC
FE
Fy = 0
FE = 0
VE = 0
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
By = 1.5
Ax = 0
Ay = – 0. 5
QBy
20 ft
12 ft
8 ft
12 ft 8 ft18 ft
6 ft
x
1
VEFEDy
VE
Dy = 0
FC = 0
VC = 0
FBD of Segment QBC for Unit Load at x = 36– ft(Just to the Left of Point C)
B
VC = 0FQ
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MQ = – 9.0 ft
FQ = 0
VQ = – 0.5
Q
MQ
VQFC = 0
12 ft6 ft
1.5
1
Use Shape of Influence Line to Place Unit Load
20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
MQ
0
18 ft
Q
6 ft
x
+
?4.5 ft
?
1
Place the unit load just to the left of E and use statics to find the value of MQ
– 9.0 ft
Place Unit Load at x = 56– ft (Just to the Left of Point E)
18– ft F
BAx
Ay
Fy
DVC VC
FC
FE
Fy = 0
FE = 0
VE = 0
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
By = – 2.25
Ax = 0
Ay = 0. 75
QBy
20 ft
12 ft
8 ft
12 ft 8 ft18 ft
6 ft
x
1
VEFEDy
VE
Dy = 2.5
FC = 0
VC = –1.5
FBD of Segment QBC for Unit Load at x = 56– ft(Just to the Left of Point C)
B
VC = 1.5FQ
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MQ = 13.5 ft
FQ = 0
VQ = 1.25
Q
MQ
VQFC = 0
12 ft6 ft
2.25
Use Shape of Influence Line to Place Unit Load
20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
MQ
0
18 ft
Q
6 ft
x
+4.5 ft
?
1
Place the unit load at G and use statics to find the value of MQ
13.5 ft
– 9.0 ft
Place Unit Load at x = 84 ft (At Point G)
18– ft F
BAx
Ay
Fy
DVC VC
FC
FE
Fy = 1.4
FE = 0
VE = – 0.4
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
By = 0.9
Ax = 0
Ay = – 0.3
QBy
20 ft
12 ft
8 ft
12 ft 8 ft18 ft
6 ft
x
1VEFEDy
VE
Dy = – 1.0
FC = 0
VC = 0.6
G
FBD of Segment QBC for Unit Load at x = 84 ft(At Point G)
B
VC = 0.6FQ
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MQ = – 5.4 ft
FQ = 0
VQ = – 0.3
Q
MQ
VQ
FC = 0
12 ft6 ft
0.9
MQ Influence Line
20 ft12 ft 8 ft12 ft 8 ftBA
CD
EF
G
MQ
0
18 ft
Q
6 ft
x
+
13.5 ft
4.5 ft
x = MQ
0 018– ft 4.5 ft18 + ft 4.5 ft24 ft 036– ft – 9.0 ft44 ft 056– ft 13.5 ft76 ft 084 ft – 5.4 ft
– 5.4 ft– 9.0 ft
