Method of Virtual Work Beam Deflection Example - San Jose ...

Method of Virtual Work Beam Deflection Example

StevenVukazichSanJoseStateUniversity

Summary of Procedure for Finding Bending Deformation Using Virtual Work

We want to find the deflection at point A and the slope at point B due to the applied loads

PMw

x

y

𝜃"

Modulus of Elasticity = EMoment of Inertia = I

AB

ab

L

𝛿$

Step 1 – Remove all loads and apply a virtual force (or moment) to measure the deformation at the point of interest

Qx

y

Aa

L

From an equilibrium analysis, find the internal bending moment function for the virtual system: MQ(x)

Convenient to set Q = 1

PMw

x

y

L

Step 2 – Replace all of the loads on the structure and perform the real analysis

From an equilibrium analysis, find the internal bending moment function for the real system: MP(x)

𝑄𝛿$ = ' 𝑀)

*

+

𝑀,

𝐸𝐼𝑑𝑥

If the bending stiffness, EI, is constant:

𝑄𝛿$ =1𝐸𝐼' 𝑀)

*

+𝑀,𝑑𝑥

Table in textbook appendix is provided to help evaluate product integrals of this type

Step 3 – Evaluate the virtual work product integrals and solve for the deformation of interest

Appendix Table.2

Table to Evaluate Virtual Work Product Integrals

' 𝑀)

*

+𝑀,𝑑𝑥

Table is as useful tool to evaluate product integrals of the form:

Beam Deflection Example

The overhanging beam shown 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 loads 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 ftD

C

BA

E

8 ft8 ft

6 k19 k

𝛿2

𝜃34

𝜃35

Step 1 – Remove all loads and apply a virtual force (or moment) to measure the deformation at the point of interest

1

x

y

From an equilibrium analysis, find the internal bending moment function for the virtual system: MQ(x)

8 ft 4 ft

D

CBA

E

8 ft8 ft

Find the Deflection at Point E

1y

8 ft 4 ft

CBA E

8 ft8 ft

Find the Moment Diagram for the Virtual System

CAx

AyDy

MA VCVC

FCFC D

6𝑀3

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 0+

Dy = 1.5

FB = 0

VC = – 0.5

6𝑀$

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 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

Moment Diagram for the Virtual System

1.5

8 ft

0.5

1

8 ft 4 ft

DCBA E

8 ft8 ft

+

+VQ

MQ

– 0.5

– 4 ft

8 ft

1.0

0

Step 2 – Replace all of the loads on the structure and perform the real analysis

From an equilibrium analysis, find the internal bending moment function for the real system: MP(x)

8 ft 4 ft

D

CBA

E

8 ft8 ft

6 k19 k

6 ky

8 ft 4 ft

CBA E

8 ft8 ft

Find the Moment Diagram for the Real System

CAx

AyDy

MA VCVC

FCFC D

6𝑀3

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 0+

Dy = 9 k

FB = 0

VC = – 3 k

6𝑀$

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 0+

MA = – 104 k-ft

Ax = 0

Ay = 16 k

19 k

8 ft 4 ft

CBA

E

8 ft8 ft

Support Reactions for the Real System

C9 k

104 k-ft 3 k

D

3 k

16 k

6 k19 k

Moment Diagram for the Real System

9 k16 k

6 k

8 ft 4 ft

DCBA E

8 ft8 ft

+

+VP

MP

16 k

– 24 k-ft– 104 k-ft

0

19 k104 k-ft

6 k

– 3 k

24 k-ft

1 ⋅ 𝛿2 =1𝐸𝐼' 𝑀)

*

+𝑀,𝑑𝑥

Use Table to evaluate product integrals

Step 3 – Evaluate the virtual work product integrals and solve for the deformation of interest

+MQ

– 4 ft

8 ft

0

+

8 ft 4 ft8 ft8 ft

DCBA E

EIABC = 2,000,000 k-in2 EICDE = 800,000 k-in2

x

X

104𝑥

=248 − 𝑥 832 = 128𝑥 𝑥 = 6.5ft

MP

– 24 k-ft– 104 k-ft

024 k-ft

8 – 0.5x = 4.75 ft

MQ

– 4 ft

8 ft

MP– 24 k-ft

– 104 k-ft

24 k-ft

1.5 ft

4 ft8 ft

8 ft

DCXA E6.5 ft

Evaluate Product Integrals

9.5 ftX

B

00 0

00 0 0 0

C

4.75 ft4.75 ft

Appendix Table.2

Table to Evaluate Virtual Work Product Integrals

' 𝑀)

*

+𝑀,𝑑𝑥

Table is as useful tool to evaluate product integrals of the form:AX

XC CE

Evaluate Product Integrals

M3 for𝑐 ≤ 𝑎:

13−

𝑎 − 𝑐 O

6𝑎𝑑𝑀P𝑀Q𝐿

16𝑀P𝑀Q 𝐿 + 𝑐1

6𝑀P + 2𝑀O 𝑀Q𝐿

164.75 + 2 8 −104 6.5

– 2337.83 k-ft3

164.75 24 9.5 + 8

332.5 k-ft3 384 k-ft3

13−4 −24 12

MQ

– 4 ft

8 ft

MP– 24 k-ft

– 104 k-ft

24 k-ft

1.5 ft 4 ft8 ft8 ft D

CXA E6.5 ft 9.5 ftX

B

00 0

00 0 0 0

C

4.75 ft4.75 ft

M2 M1

L L

M1

cd

ba

dcM3

M3

L

𝑐 = 𝑎 = 8ft

12 ft

Evaluate Product Integrals

– 2337.83 k-ft3

332.5 k-ft3

384 k-ft3

1 ⋅ 𝛿2 =1𝐸𝐼' 𝑀)

*

+𝑀,𝑑𝑥

MQ

– 4 ft

8 ft

0

8 ft 4 ft8 ft8 ft

DCBA E

6.5 ft

XMP

– 24 k-ft– 104 k-ft

024 k-ft

EIABC = 2,000,000 k-in2 EICDE = 800,000 k-in2

Segment XC

Segment CDE

' 𝑀)

*UVW

+𝑀,𝑑𝑥 = −2337.83 + 332.5k−ft3

12Qin3

ft3= −3,465,216.0k−in3

' 𝑀)

*W\]

+𝑀,𝑑𝑥 = 384k−ft3

12Qin3

ft3= 663,552k−in3

Segment AX

Evaluate Product Integrals

𝛿2 =−3,465,216.0k−in3

2,000,000k−in2+663,552k−in3

800,000k−in2

' 𝑀)

*W\]

+𝑀,𝑑𝑥 = 384k−ft3

12Qin3

ft3= 663,552k−in3

1 ⋅ 𝛿2 =1

𝐸𝐼$"3' 𝑀)

*UVW

+𝑀,𝑑𝑥 +

1𝐸𝐼3^2

' 𝑀)

*W\]

+𝑀,𝑑𝑥

𝛿2 = −1.733in + 0.8294in= − 0.903in Negative result, so deflection is in the opposite direction of the virtual unit load𝛿2 = 0.903inupward

' 𝑀)

*UVW

+𝑀,𝑑𝑥 = −2337.83 + 332.5k−ft3

12Qin3

ft3= −3,465,216.0k−in3

Step 1 – Remove all loads and apply a virtual force (or moment) to measure the deformation at the point of interest

1

x

y

From an equilibrium analysis, find the internal bending moment function for the virtual system: MQ(x)

8 ft 4 ft

DCB

AE

8 ft8 ft

Find the Rotation Just to the Left of Point C

y

8 ft 4 ft

CBA E

8 ft8 ft

Find the Moment Diagram for the Virtual System

CAx

AyDy

MA VC VC

FCFC D

6𝑀3

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 0+

Dy = 0

FB = 0

VC = 0

6𝑀$

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 0+

MA = 1

Ax = 0

Ay = 0

1

8 ft 4 ft

CB

A

E

8 ft8 ft

Support Reactions for the Virtual System

C

1

D

1

Moment Diagram for the Virtual System

1

8 ft 4 ft

DCBA E

8 ft8 ft

+

+VQ

MQ

0

1

0

0

1

Moment Diagram for the Real System

9 k16 k

6 k

8 ft 4 ft

DCBA E

8 ft8 ft

+

+VP

MP

16 k

– 24 k-ft– 104 k-ft

0

19 k104 k-ft

6 k

– 3 k

24 k-ft

1 ⋅ 𝜃34 =1𝐸𝐼' 𝑀)

*

+𝑀,𝑑𝑥

Use Table to evaluate product integrals

+

+

8 ft 4 ft8 ft8 ft

DCBA E

Evaluate the Virtual Work Product Integrals

MQ

1

0

MP

– 24 k-ft– 104 k-ft

024 k-ft

6.5 ft

EIABC = 2,000,000 k-in2 EICDE = 800,000 k-in2

Appendix Table.2

Table to Evaluate Virtual Work Product Integrals

' 𝑀)

*

+𝑀,𝑑𝑥

Table is as useful tool to evaluate product integrals of the form:

AX

XC

MQ1

1.5 ft 8 ft

CXA 6.5 ft

Evaluate Product Integrals

9.5 ftX

B

0

M1

L L

M1

M3

L

121 −104 6.5

– 338 k-ft2

121 24 9.5

1

114 k-ft2 0

M3

MP

– 104 k-ft

24 k-ft00 0

12𝑀P𝑀Q𝐿

1 1

– 24 k-ft

4 ft8 ft D

E

0 0

0 0

C

M3

12 ft

12𝑀P𝑀Q𝐿

Evaluate Product Integrals

– 338 k-ft2

114 k-ft2

0

1 ⋅ 𝜃34 =1𝐸𝐼' 𝑀)

*

+𝑀,𝑑𝑥8 ft 4 ft8 ft8 ft

DCBA E

6.5 ft

XMP

– 24 k-ft– 104 k-ft

024 k-ft

EIABC = 2,000,000 k-in2 EICDE = 800,000 k-in2

Segment XC

Segment CDE

' 𝑀)

*UVW

+𝑀,𝑑𝑥 = −338 + 114k−ft2

12Oin2

ft2= −32,256k−in2

' 𝑀)

*W\]

+𝑀,𝑑𝑥 = 0

Segment AXMQ

1

0

Evaluate Product Integrals

𝜃34 =−32,256k−in2

2,000,000k−in2+

0800,000k−in2

1 ⋅ 𝜃34 =1

𝐸𝐼$"3' 𝑀)

*UVW

+𝑀,𝑑𝑥 +

1𝐸𝐼3^2

' 𝑀)

*W\]

+𝑀,𝑑𝑥

𝜃34 = −0.0161rad + 0= − 0.0161rad Negative result, so rotation is in the opposite direction of the virtual unit moment𝜃34 = 0.0161radiansclockwise

' 𝑀)

*UVW

+𝑀,𝑑𝑥 = −338 + 114k−ft2

12Oin2

ft2= −32,256k−in2

' 𝑀)

*W\]

+𝑀,𝑑𝑥 = 0

Step 1 – Remove all loads and apply a virtual force (or moment) to measure the deformation at the point of interest

1

x

y

From an equilibrium analysis, find the internal bending moment function for the virtual system: MQ(x)

8 ft 4 ft

DCB

AE

8 ft8 ft

Find the Rotation Just to the Right of Point C

y

8 ft 4 ft

CBA E

8 ft8 ft

Find the Moment Diagram for the Virtual System

CAx

AyDy

MA VCVC

FCFC D

6𝑀3

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 0+

Dy = – 0.125 /ft

FB = 0

6𝑀$

�

�

= 0+

6𝐹;

�

�

= 0+

6𝐹<

�

�

= 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

Moment Diagram for the Virtual System

2

8 ft 4 ft

DCBA E

8 ft8 ft

+

+VQ

MQ

0

– 2

0

0

1

0.125 /ft0.125 /ft

– 1

0.125 /ft

Moment Diagram for the Real System

9 k16 k

6 k

8 ft 4 ft

DCBA E

8 ft8 ft

+

+VP

MP

16 k

– 24 k-ft– 104 k-ft

0

19 k104 k-ft

6 k

– 3 k

24 k-ft

1 ⋅ 𝜃35 =1𝐸𝐼' 𝑀)

*

+𝑀,𝑑𝑥

Use Table to evaluate product integrals

+

+

8 ft 4 ft8 ft8 ft

DCBA E

Evaluate the Virtual Work Product Integrals

MQ

– 2

0

– 1– 1.1875

MP

– 24 k-ft– 104 k-ft

024 k-ft

6.5 ft

EIABC = 2,000,000 k-in2 EICDE = 800,000 k-in2

Appendix Table.2

Table to Evaluate Virtual Work Product Integrals

' 𝑀)

*

+𝑀,𝑑𝑥

Table is as useful tool to evaluate product integrals of the form:AX

XC

CD

MQ– 2

– 24 k-ft

1.5 ft

8 ft

8 ft

DCXA E6.5 ft

Evaluate Product Integrals Using the Table

9.5 ftX

B

0

0

0 0

C

M2 M1

L L

M1M3

M3

4 ftL

16𝑀P + 2𝑀O 𝑀Q𝐿

16−1.1875 + 2 −2 −104 6.5

584.458 k-ft2

16−1.1875 24 9.5 + 8

– 1.1875

0 0– 1

– 83.125 k-ft2

16𝑀P𝑀Q𝐿

16−1 −24 8

32 k-ft2

M1

0

D

M3

MP

– 104 k-ft

24 k-ft00 0

– 1.1875

16𝑀P𝑀Q 𝐿 + 𝑐

cb

1 ⋅ 𝜃35 =1𝐸𝐼' 𝑀)

*

+𝑀,𝑑𝑥

8 ft 4 ft8 ft8 ft

DCBA E

Evaluate the Virtual Work Product Integrals

MQ

– 2

0

– 1– 1.1875

MP

– 24 k-ft– 104 k-ft

024 k-ft

6.5 ft

EIABC = 2,000,000 k-in2 EICDE = 800,000 k-in2

584.458 k-ft2

– 83.125 k-ft2

0

Segment XC

Segment CD

' 𝑀)

*UVW

+𝑀,𝑑𝑥 = 584.458 − 83.125k−ft2

12Oin2

ft2= 72,191.95k−in2

' 𝑀)

*W\]

+𝑀,𝑑𝑥 = 32k−ft2

12Oin2

ft2= 4608k−in2

Segment AX

32 k-ft2

Segment DE

Evaluate Product Integrals

𝜃35 =72,191.95k−in2

2,000,000k−in2+

4608k−in2

800,000k−in2

1 ⋅ 𝜃35 =1

𝐸𝐼$"3' 𝑀)

*UVW

+𝑀,𝑑𝑥 +

1𝐸𝐼3^2

' 𝑀)

*W\]

+𝑀,𝑑𝑥

𝜃35 = 0.0361 + 0.00576rad=0.0419rad Positive result, so rotation is in the same direction of the virtual unit moment

𝜃35 = 0.0419radianscounter−clockwise

' 𝑀)

*W\]

+𝑀,𝑑𝑥 = 32k−ft2

12Oin2

ft2= 4608k−in2

' 𝑀)

*UVW

+𝑀,𝑑𝑥 = 584.458 − 83.125k−ft2

12Oin2

ft2= 72,191.95k−in2

Beam Deflection Example Results

The overhanging beam shown 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 loads 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 ftD

C

BA

E

8 ft8 ft

6 k19 k

0.0161radians

0.0419radians

0.903inches