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(𝐸𝐼

𝑑𝑥

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(𝐸𝐼

𝑑𝑥

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$% = −0.0161rad + 0= − 0.0161radNegative result, so rotation is in the opposite direction of the virtual unit moment𝜃$% = 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