steven vukazich san jose state university
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Constructing Influence Lines for Trusses
StevenVukazichSanJoseStateUniversity
1. Chooseareferencecoordinate;2. Chooseasignconventionforeachdiagram;3. Placeaunit,dimensionlessloadonthestructure;4. Useequilibriumanalysistofindtheresponse
quantity(e.g.supportreaction,internalforce)atthepositionoftheunit,dimensionless,load;
5. MoveunitloadtoanotherpositionandrepeatStep4;
6. Plotthevalueoftheresponsequantityversusthepositionoftheunit,dimensionless,load.
General procedure for the construction of influence lines
A
B C6 ft 6 ft
8 ft
DE
Truss Influence Line Example
6 ft
F G H
6 ft
The truss shown supports a bridge deck and has a pin support at E and a roller support at B. For a load moving across the bridge deck, construct influence lines for:1. The axial force in member GH;2. The axial force in member GD;3. The axial force in member CD;4. The axial force in member GC.
Bridge Deck
A B C6 ft 6 ft
8 ft
DE
6 ft
F G H
6 ft
Bridge Deck
Unit load moves along level of bridge deck
+Choose tension positive for axial force in truss members
Choose Reference Coordinate and Sign Convention
x
"𝑀$
�
�
= 0+
Unit Load at Point A
Ey = -0.333
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1
DE
6 ft
F G H
6 ft
"𝐹)
�
�
= 0+
By = 1.333
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1
DE
6 ft
F G H
"𝐹*
�
�
= 0+
Unit Load at Point A
We can use the method of sections to find member forces GH, GD, and CD.
Note that for the unit load at A, GC is a zero-force member
A
B C
6 ft 6 ft
8 ft
D
E
6 ft
F G H
6 ft
q
q1 1.333 0.333
Unit Load at Point A
FBD of the Section to the Right of Cut q-q
FGD
FGH
FCD
G
8 ft
D
E
6 ft
H
6 ft0.333
q
q"𝑀+
�
�
= 0+
FCD = -0.50
"𝑀,
�
�
= 0+
FGH = 0.25
FBD of the Section to the Right of Cut q-qFGH
FCD
G
8 ft
D
E
6 ft
H
6 ft0.333
q
q
FGD = 0.4167
"𝐹)
�
�
= 0+
35𝐹+,
3
45
45𝐹+,
"𝑀$
�
�
= 0+
Unit Load at Point B
Ey = 0
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1
DE
6 ft
F G H
6 ft
Unit Load at Point B
By = 1
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1
DE
6 ft
F G H
6 ft
"𝐹)
�
�
= 0+
Note that for the unit load at roller support B, no axial forces are developed in any of the the truss members.
A
B C
6 ft 6 ft
8 ft
D
E
6 ft
F G H
6 ft
q
q
1
Unit Load at Point B
1
"𝑀$
�
�
= 0+
Unit Load at Point C
Ey = 0.333
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1D
E
6 ft
F G H
6 ft
Unit Load at Point C
By = 0.667
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1D
E
6 ft
F G H
6 ft
"𝐹)
�
�
= 0+
We can use the method of sections to find member forces GH, GD, and CD.
We can use method of joints to find the member force GC.
A
B C
6 ft 6 ft
8 ft
D
E
6 ft
F G H
6 ft
q
q10.667 0.333
Unit Load at Point C
FBD of the Section to the Right of Cut q-q
FGD
FGH
FCD
G
8 ft
D
E
6 ft
H
6 ft0.333
q
q"𝑀+
�
�
= 0+
FCD = 0.50
"𝑀,
�
�
= 0+
FGH = -0.25
FBD of the Section to the Right of Cut q-qFGH
FCD
G
8 ft
D
E
6 ft
H
6 ft0.333
q
q
FGD = -0.4167
"𝐹)
�
�
= 0+
35𝐹+,
3
45
45𝐹+,
C
1
FBD of Joint C
FBC
FCG
FCG = 1
"𝐹)
�
�
= 0+
FCD
"𝑀$
�
�
= 0+
Unit Load at Point D
Ey = 0.667
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1D
E
6 ft
F G H
6 ft
Unit Load at Point D
By = 0.333
ExBy Ey
A
B C
6 ft 6 ft
8 ft
1D
E
6 ft
F G H
6 ft
"𝐹)
�
�
= 0+
A
B C
6 ft 6 ft
8 ft
DE
6 ft
F G H
6 ft
q
q10.333 0.667
Unit Load at Point D
We can use method of sections to find member forces GH, GD, and CD.
Note that for the unit load at D, GC is a zero-force member
FBD of the Section to the Right of Cut q-q
FGD
FGH
FCD
G
8 ft
D
E
6 ft
H
6 ft0.667
q
q"𝑀+
�
�
= 0+
FCD = 0.25
"𝑀,
�
�
= 0+
FGH = -0.5
1
FBD of the Section to the Right of Cut q-qFGH
FCD
G
8 ft
D
E
6 ft
H
6 ft
q
q
FGD = 0.4167
"𝐹)
�
�
= 0+
35𝐹+,
3
45
45𝐹+,
0.6671
FCD
0.25
x = FCD
0 (A) – 0.506 ft (B) 0
12 ft (C) 0.5018 (D) 0.2524 (E) 0
Plot the Influence Line for FCD
+0
0
0.50
– 0.50
A
B C DE
F G H
FGH
x = FGH
0 (A) 0.256 ft (B) 0
12 ft (C) – 0.2518 (D) – 0.524 (E) 0
Plot the Influence Line for FGH
+0
0
– 0.25
0.25
A
B C DE
F G H
– 0.5
FGD0.4167
x = FGD
0 (A) 0.41676 ft (B) 0
12 ft (C) – 0.416718 (D) 0.416724 (E) 0
Plot the Influence Line for FGD
+0
0
0.4167
– 0.4167
A
B C DE
F G H
FGC
0
x = FGC
0 (A) 06 ft (B) 0
12 ft (C) 118 (D) 024 (E) 0
Plot the Influence Line for FGC
+00
1
0
A
B C DE
F G H
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