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Method of Virtual Work Beam Deflection Example
Support MovementStevenVukazich
SanJoseStateUniversity
Beam Support Movement Deflection Example
The overhanging beam, from our previous example, has a fixed support at A, a roller support at C and an internal hinge at B. EIABC = 2,000,000 k-in2 and EICDE = 800,000 k-in2
For the support movements shown, find the following:
1. The vertical deflection at point E;2. The slope just to the left of the internal hinge at C;3. The slope just to the right of the internal hinge at C
8 ft 4 ft
D
C
BA E
8 ft8 ft
𝛿"
𝜃$%
𝜃$&1.0°
2.0 inches
Recall the General Form of the Principle of Virtual Work
𝑄𝛿( +*𝑅,
�
�
𝛿. = 0𝐹,𝑑𝐿(
�
�
Real Deformation
Virtual Loads
External Work due to Support Settlements
𝑄𝛿( +*𝑅,
�
�
𝛿. = 4 𝑀,
6
7
𝑀(
𝐸𝐼𝑑𝑥
General Form for Bending Deformation
Principle of Virtual Work for Bending Deformation
Internal work = 0 for this problem
Virtual support reactions
𝑄𝛿( +*𝑅,
�
�
𝛿. = 4 𝑀,
6
7
𝑀(
𝐸𝐼𝑑𝑥
For this problem, there is only support movement causing deformation, so the internal work term is zero.
In order to find the external work due to support movement, we need to find the support reaction for the virtual system.
Virtual load
Real support movementsReal deformation of interest
1
x
y
From an equilibrium analysis, find the support reactions for the virtual system: RQ
8 ft 4 ft
D
CBA
E
8 ft8 ft
Virtual System to Measure the Deflection at Point E
1y
8 ft 4 ft
CBA E
8 ft8 ft
Find the Support Reactions for the Virtual System
CAx
Ay Dy
MA VC VC
FC FC D
*𝑀$
�
�
= 0+
*𝐹<
�
�
= 0+
*𝐹=
�
�
= 0+
Dy = 1.5
FB = 0
VC = – 0.5
*𝑀>
�
�
= 0+
*𝐹<
�
�
= 0+
*𝐹=
�
�
= 0+
MA = 8 ft
Ax = 0
Ay = – 0.5
1
8 ft 4 ft
CB
A
E
8 ft8 ft
Support Reactions for the Virtual System
C1.5
8 ft 0.5
D
0.5
0.5
Evaluate the Virtual Work Expression
1 ⋅ 𝛿" +*𝑅,
�
�
𝛿. = 0
𝛿" + 8ft 0.01745312inft
− 1.5 2.0in = 0
Positive result, so deflection is in the same direction as the virtual unit load
𝛿" = 1.325indownward
1 ⋅ 𝛿" + 𝑀,>𝜃,> + 𝑅,S𝛿,S = 0
Need to convert 𝜃,>to radians
𝜃,> = 1°𝜋radians180°
= 0.017453radians
𝛿" + 1.6755in − 3.0in = 0
𝛿" = 1.325in
1
x
y
8 ft 4 ft
DCB
AE
8 ft8 ft
Virtual System to measure the Rotation Just to the Left of Point C
From an equilibrium analysis, find the support reactions for the virtual system: RQ
8 ft 4 ft
CB
A
E
8 ft8 ft
Support Reactions for the Virtual System
C
1
D
1
Evaluate the Virtual Work Expression
1 ⋅ 𝜃$% +*𝑅,
�
�
𝛿. = 0
𝜃$% + 1 0.017453 = 0
Negative result, so deflection is in the opposite direction as the virtual unit moment
1 ⋅ 𝜃$% + 𝑀,>𝜃,> + 𝑅,S𝛿,S = 0
Need to convert 𝜃,>to radians
𝜃,> = 1°𝜋radians180°
= 0.017453radians
𝜃$% = −0.017453rad = −1°
𝜃$% = 0.01743radians = −1°clockwise
Evaluate Product Integrals
𝜃$% =−32,256k−in2
2,000,000k−in2+
0800,000k−in2
1 ⋅ 𝜃$% =1
𝐸𝐼>]$4 𝑀,
6^_`
7𝑀(𝑑𝑥 +
1𝐸𝐼$S"
4 𝑀,
6`ab
7𝑀(𝑑𝑥
𝜃$% = −0.0161rad + 0= − 0.0161rad Negative result, so rotation is in the opposite direction of the virtual unit moment𝜃$% = 0.0161radiansclockwise
4 𝑀,
6^_`
7𝑀(𝑑𝑥 = −338 + 114k−ft2
12cin2
ft2= −32,256k−in2
4 𝑀,
6`ab
7𝑀(𝑑𝑥 = 0
1
x
y
8 ft 4 ft
DCB
AE
8 ft8 ft
Virtual System to Measure the Rotation Just to the Right of Point C
From an equilibrium analysis, find the support reactions for the virtual system: RQ
y
8 ft 4 ft
CBA E
8 ft8 ft
Find the Moment Diagram for the Virtual System
CAx
Ay Dy
MA VCVC
FC FC D
*𝑀$
�
�
= 0+
*𝐹<
�
�
= 0+
*𝐹=
�
�
= 0+
Dy = – 0.125 /ft
FB = 0
*𝑀>
�
�
= 0+
*𝐹<
�
�
= 0+
*𝐹=
�
�
= 0+
MA = – 2
Ax = 0
1
VC = 0.125 /ftAy = 0.125 /ft
8 ft 4 ft
CBAE
8 ft8 ft
Support Reactions for the Virtual System
C
2
D
10.125 /ft
0.125 /ft0.125 /ft0.125 /ft
Evaluate the Virtual Work Expression
1 ⋅ 𝜃$& +*𝑅,
�
�
𝛿. = 0
𝜃$& − 2 0.017453 + 0.125/ft 2.0inft12in
= 0
Negative result, so deflection is in the opposite direction as the virtual unit moment
1 ⋅ 𝜃$& + 𝑀,>𝜃,> + 𝑅,S𝛿,S = 0
Need to convert 𝜃,>to radians
𝜃,> = 1°𝜋radians180°
= 0.017453radians
𝜃$& − 0.034906rad + 0.020833rad = 0
𝜃$& = 0.01407radianscounter−clockwise
𝜃$& = 0.01407rad
Beam Support Movement Deflection Example
8 ft 4 ft
D
C
BA E
8 ft8 ft
1.0°
2.0 inches
0.01745radians = 1°
0.01407radians1.325inches