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
๐ฟ"
๐$%
๐$&1.0ยฐ
2.0 inches
Recall the General Form of the Principle of Virtual Work
๐๐ฟ( +*๐ ,
๏ฟฝ
๏ฟฝ
๐ฟ. = 0๐น,๐๐ฟ(
๏ฟฝ
๏ฟฝ
Real Deformation
Virtual Loads
External Work due to Support Settlements
๐๐ฟ( +*๐ ,
๏ฟฝ
๏ฟฝ
๐ฟ. = 4 ๐,
6
7
๐(
๐ธ๐ผ๐๐ฅ
General Form for Bending Deformation
Principle of Virtual Work for Bending Deformation
Internal work = 0 for this problem
Virtual support reactions
๐๐ฟ( +*๐ ,
๏ฟฝ
๏ฟฝ
๐ฟ. = 4 ๐,
6
7
๐(
๐ธ๐ผ๐๐ฅ
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
*๐$
๏ฟฝ
๏ฟฝ
= 0+
*๐น<
๏ฟฝ
๏ฟฝ
= 0+
*๐น=
๏ฟฝ
๏ฟฝ
= 0+
Dy = 1.5
FB = 0
VC = โ 0.5
*๐>
๏ฟฝ
๏ฟฝ
= 0+
*๐น<
๏ฟฝ
๏ฟฝ
= 0+
*๐น=
๏ฟฝ
๏ฟฝ
= 0+
MA = 8 ft
Ax = 0
Ay = โ 0.5
Evaluate the Virtual Work Expression
1 โ ๐ฟ" +*๐ ,
๏ฟฝ
๏ฟฝ
๐ฟ. = 0
๐ฟ" + 8ft 0.01745312inft
โ 1.5 2.0in = 0
Positive result, so deflection is in the same direction as the virtual unit load
๐ฟ" = 1.325indownward
1 โ ๐ฟ" + ๐,>๐,> + ๐ ,S๐ฟ,S = 0
Need to convert ๐,>to radians
๐,> = 1ยฐ๐radians180ยฐ
= 0.017453radians
๐ฟ" + 1.6755in โ 3.0in = 0
๐ฟ" = 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
Evaluate the Virtual Work Expression
1 โ ๐$% +*๐ ,
๏ฟฝ
๏ฟฝ
๐ฟ. = 0
๐$% + 1 0.017453 = 0
Negative result, so deflection is in the opposite direction as the virtual unit moment
1 โ ๐$% + ๐,>๐,> + ๐ ,S๐ฟ,S = 0
Need to convert ๐,>to radians
๐,> = 1ยฐ๐radians180ยฐ
= 0.017453radians
๐$% = โ0.017453rad = โ1ยฐ
๐$% = 0.01743radians = โ1ยฐclockwise
Evaluate Product Integrals
๐$% =โ32,256kโin2
2,000,000kโin2+
0800,000kโin2
1 โ ๐$% =1
๐ธ๐ผ>]$4 ๐,
6^_`
7๐(๐๐ฅ +
1๐ธ๐ผ$S"
4 ๐,
6`ab
7๐(๐๐ฅ
๐$% = โ0.0161rad + 0= โ 0.0161rad Negative result, so rotation is in the opposite direction of the virtual unit moment๐$% = 0.0161radiansclockwise
4 ๐,
6^_`
7๐(๐๐ฅ = โ338 + 114kโft2
12cin2
ft2= โ32,256kโin2
4 ๐,
6`ab
7๐(๐๐ฅ = 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
*๐$
๏ฟฝ
๏ฟฝ
= 0+
*๐น<
๏ฟฝ
๏ฟฝ
= 0+
*๐น=
๏ฟฝ
๏ฟฝ
= 0+
Dy = โ 0.125 /ft
FB = 0
*๐>
๏ฟฝ
๏ฟฝ
= 0+
*๐น<
๏ฟฝ
๏ฟฝ
= 0+
*๐น=
๏ฟฝ
๏ฟฝ
= 0+
MA = โ 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 โ ๐$& +*๐ ,
๏ฟฝ
๏ฟฝ
๐ฟ. = 0
๐$& โ 2 0.017453 + 0.125/ft 2.0inft12in
= 0
Negative result, so deflection is in the opposite direction as the virtual unit moment
1 โ ๐$& + ๐,>๐,> + ๐ ,S๐ฟ,S = 0
Need to convert ๐,>to radians
๐,> = 1ยฐ๐radians180ยฐ
= 0.017453radians
๐$& โ 0.034906rad + 0.020833rad = 0
๐$& = 0.01407radianscounterโclockwise
๐$& = 0.01407rad