160.12.1 Influence Lines - San Jose State University...shear, bending moment, axial force). Why Do...
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Constructing Influence Lines Steven Vukazich San Jose State University
Transcript of 160.12.1 Influence Lines - San Jose State University...shear, bending moment, axial force). Why Do...
Constructing Influence LinesStevenVukazich
SanJoseStateUniversity
In addition to supporting fixed gravity loads (Dead Load), structures must also support gravity loads that can vary in magnitude and position (Live Loads).
To design the components of a structure, it is important to understand how to place live loads to produce the maximum response for important design quantities (e.g. support reactions, internal shear, bending moment, axial force).
Why Do We Construct Influence Lines ?
The Influence Line for a response quantity is a tool to help place live loads to find the maximum response
Construction of Influence Lines
The overhanging beam shown has a fixed support at A, a roller support at C and an internal hinge at B. Construct influence lines for:
1. The roller support at C;2. The vertical reaction at the fixed support at A:3. The moment reaction at the fixed support at A
9 m 6 m 3 m
D
CB
A
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
Choose Reference Coordinate and Sign Convention
The overhanging beam shown has a fixed support at A, a roller support at C and an internal hinge at B. Construct influence lines for:
1. The roller support at C;2. The vertical reaction at the fixed support at A:3. The moment reaction at the fixed support at A
9 m 6 m
3 m
D
CB
A
x++
Sign Convention for Positive Support Reactions
Place Unit Load at x = 0 (Point A)
9 m 6 m
3 m
DB
x
Free-body Diagram
AxAy
Cy
1MA
4 Unknowns – 3 Equations of EquilibriumNeed to make a cut at the hinge at B
Place Unit Load at x = 0 (Point A)
9 m6 m 4 m
DB
x
Free-body Diagrams
AxAy
Cy
1MA
6 Unknowns – 6 Equations of Equilibrium
B
VB VB
FB FB
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
Cy = 0
FB = 0
VB = 0
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MA = 0
Ax = 0
Ay = 1
Place Unit Load at x = 5 m
5 m
6 m 3 m
DB
x
Free-body Diagrams
AxAy
Cy
1MA
6 Unknowns – 6 Equations of Equilibrium
B
VB VB
FB FB
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
Cy = 0
FB = 0
VB = 0
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MA = – 5 m
Ax = 0
Ay = 1
9 m
Place Unit Load at x = 9– m
5 m
6 m 3 m
DB
x
Free-body Diagrams
AxAy
Cy
1MA
6 Unknowns – 6 Equations of Equilibrium
B
VB VB
FB FB
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
Cy = 0
FB = 0
VB = 0
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MA = – 9 m
Ax = 0
Ay = 1
9 m
Place Unit Load at x = 12 m
3 m
6 m 3 m
DB
x
Free-body Diagrams
AxAy
Cy
1MA
6 Unknowns – 6 Equations of Equilibrium
B
VB VB
FB FB
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
Cy = 0.5
FB = 0
VB = 0.5
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MA = – 4.5 m
Ax = 0
Ay = 0.5
9 m
Place Unit Load at x = 15 m
3 m
6 m 3 m
DB
x
Free-body Diagrams
AxAy
Cy
1MA
6 Unknowns – 6 Equations of Equilibrium
B
VB VB
FB FB
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
Cy = 1
FB = 0
VB = 0
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MA = 0
Ax = 0
Ay = 0
9 m
Place Unit Load at x = 18 m
6 m 3 m
DB
x
Free-body Diagrams
AxAy
Cy
1MA
6 Unknowns – 6 Equations of Equilibrium
B
VB VB
FB FB
"𝑀$
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
Cy = 1.5
FB = 0
VB = – 0.5
"𝑀+
�
�
= 0+
"𝐹)
�
�
= 0+
"𝐹*
�
�
= 0+
MA = 4.5 m
Ax = 0
Ay = – 0.5
9 m
Cy 1.5
9 m 6 m 3 m
D
CB
A
0
x = Cy
0 05 m 0
9 – m 012 m 0.515 m 1.018 m 1.5
Plot the Influence Line for Cy
+
Ay
– 0.5
9 m 6 m 3 m
D
CB
A
0
x = Ay
0 15 m 1
9 – m 112 m 0.515 m 018 m – 0.5
Plot the Influence Line for Ay
+
1.0
MA
4.5 m
9 m 6 m 3 m
D
CB
A
0
x = MA
0 05 m – 5 m
9 – m – 9 m12 m – 4.5 m15 m 018 m 4.5 m
Plot the Influence Line for MA
+
– 9.0 m
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