Notes_09c Virtual Work for Frames

8/12/2019 Notes_09c Virtual Work for Frames

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Virtual Work for Frames

Applying the virtual work equations to a frame

structure is as simple as separating the frameinto a series of “beams” and summing the virtualwork for each section.

In addition, when evaluating the deformation ofa frame structure, you may have to considerboth bending and axial internal forcecomponents.


Compute the deflection deflection at point C on the frame

shown below. Include only the effects of bending in your

virtual work e uation no axial work .12 k

8 ft 8 ft

10 ft

A

B

C

D

E

E = 29,000 ksi I = 3,500 in4

A = 35 in2


The first step is to find the equation for moment ineach section of the frame due to the real loads real loads .

To do develop the moment expression we need thereaction a points A and E.

8 ft 8 ft

10 ft

A

B

C

D

E

A y E y

Ax

0 12y Y y F k E A

6y E k

0 12 (8 ) (16 )A Y M k ft E ft

6y A k

0x x F A


The next step is to find the equation for moment ineach section of the frame.

Consider section AB

0AB M 0M M

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

cu

MAB

y


Consider section BC

6BC M x 0 6M M k x

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

MBC

x

cu


Consider section DC

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

1

6DC

M x

10 6 ( )

cut DC M M k x

x1

MDC

CIVL 3121 Virtual Work for Frames 1/9




Consider section ED

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

0DE M

0cut DE M M

MDE

y


The next step is to find the equation for moment ineach section of the frame due to the virtual load virtual load .

To do develop the moment expression we need thereaction a points A and E.

8 ft 8 ft

10 ft

A

B

C

D

E

A y E y

Ax

0 1y Y y F E A

0.5y E

0 1(8 ) (16 )A Y M ft E ft

0.5y A

0x x F A


In this problem, the virtual moments are the realmoments divided by 12 (from superposition).

12 k

Section M m

10 ft

A

B

C

E

AB 0 0

BC 6x x/2

DC 6x1 x1/2

DE 0 0


The virtual work equations are:

B C C E

DC DC DE DE AB AB BC BC C

A B D D

M m M m M m M m dy dx dx dy

EI EI EI EI

Substituting the moment expressions into the virtualwork equation and integrating yields the following:

8 8

1 11

0 0

(6 )(6 )

2 2C

x x x x dx dx

EI EI

8 2

0

6x dx

EI

83

0

2x

EI

31,024 kft

EI

3 3 3

4

1,024 (1,728 / )

(29, 000 )(3,500 )

kft in ft

ksi in 0.017in


Compute the slope slope at point C on the frame shown below.Include only the effects of bending in your virtual workequation (no axial work).

12 k

8 ft 8 ft

10 ft

A

B

C

D

E

E = 29,000 psi I = 3,500 in4

A = 35 in2


Find the moments in the frame due to a virtual couple First, find the reaction in the frame to the virtual

couple

8 ft 8 ft

10 ft

A

B

C

D

E

A y E y

Ax

0y Y y F E A

116y E

0 1 (16 )A Y M ft E ft

116y A

0x x F A

1 ft





The next step is to find the equation for moment ineach section of the frame.

Consider section AB

0AB m 0M m

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

m AB

116

cu

y


Consider section BC

BC

x m 10M m x

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116

m BC

x

cu


Consider section DC

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116

1

16DC

x m

1

1160 ( )

cut DC M m x

x1

m DC


Consider section ED

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116

0DE m

0cut DE M m

m DE

y


The following table lists the moment expression due tothe real loading and the moment expression due to avirtual couple at point C

12 k

Section M m

10 ft

A

B

C

E

AB 0 0

BC 6x x/16

DC 6x1 -x1/16

DE 0 0


The virtual work equations are:

B C C D

DC DC DE DE AB AB BC BC C

A B D E

M m M m M m M m dy dx dx dy

EI EI EI EI

Substituting the moment expressions into the virtualwork equation and integrating yields the following:

8 8

1 11

0 0

(6 )(6 )

16 16C

x x x x dx dx

EI EI 0

The slope at point C is zero





Repeat the previous example and include the effects of

axial work. In order to compute the axial work, we need the axial

force in the real and virtual loading systems

12 k

8 ft 8 ft

10 ft

A

B

C

D

E

E = 29,000 psi I = 3,500 in4

A = 35 in2


Find the axial force in each section of the frame.

Consider section AB

6AB F k 0 6F F k

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

FAB

y

y


Consider section BC

0BC F 0F F

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

x

x

FBC


Consider section DC

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

0DC F

0x DC F F

x1

FDC


Consider section ED

6F k 0 6F F k

8 ft 8 ft

10 ft

A

B

C

D

E

6k 6k

FDE

y

y


In this problem, the virtual axial forces are the realaxial forces divided by 12 (from superposition).

12 k

Section N n

10 ft

A

B

C

E

AB -6k -0.5

BC 0 0

DC 0 0

DE -6k -0.5





The virtual work equations for axial forces are:

C nNL AE

virtual work equations yields the following:

720 k in

AE 2

720

(29,000 )(35 )

k in

ksi in 0.0007in

( 0.5)( 6 )(120 ) ( 0.5)( 6 )(120 )C

k in k in

AE AE


The displacement at point C due to bending moment

work and axial force work is:0.017C in 0.0007in 0.0177in

12 k

8 ft 8 ft

10 ft

A

B

C

D

E

E = 29,000 psi I = 3,500 in4

A = 35 in2

rom en ingmoment work

from axialforce work


Compute the axial forces in the frame due to thevirtual couple.

Recall we already have the frame reactions due to thevirtual couple

E = 29,000 psi I = 3,500 in4

A = 35 in2

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116


The next step is to find the axial force in each sectionof the frame.

Consider section AB

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116

116AB n

n AB

1160y AB F n

y


Consider section BC

0BC n 0F n

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116

x

x

n BC


Consider section DC

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116

x1 0DC n

0x DC F n

n DC





Consider section ED

1n 1

8 ft 8 ft

10 ft

A

B

C

D

E

1 ft

116

116

n AB

1616y DE

y


The real axial forces and the virtual axial forces due to

a unit virtual couple are:12 k

Section N n

10 ft

A

B

C

E

AB -6k -1/16

BC 0 0

DC 0 0

DE -6k 1/16


The virtual work equations for axial forces are:

C

nNL

AE

Substitutin the values for the axial forces into thevirtual work equations yields the following:

(6 )(120 ) (6 )(120 )

16 16C

k in k in

AE AE 0

The contribution to the slope at point C from the axialenergy is slope is zero.

The total slope at point C due to bending moment andaxial force work is zero.


In problems involving both bending and axialdeformation, be careful with the units.

Also, note that the contribution of the axial.

This is more or less typical of the relative size of thebending and the axial effects in frame-deflectionproblems.

Therefore, it is usually permissible to neglect theeffect of axial deformation in such cases.


The primary cause of deformation in beams and framesis due to bending strain bending strain .

However, in some structures additional deformation due

to axial axial and shear forces shear forces , torsion torsion , and perhapstemperature temperature may be important.

We have already discussed deformation due to bendingmoments and axial forces.

In this section, we will consider the effect of shear,torsion, and temperature on the deformation of linearelastic structures.


Virtual Strain Energy From Shear

Consider the following beam and a small element dx

w = w(x )

x

x

dx

V

V

dy

dx






The shearing deformation dy caused by the real loads isdy = dx, where is the shear strain.

V

V

dy

dx

Since we are assuming thematerial is linear and elastic, thenHooke’s law applies

The shear strain is related to theshear stress by = /G , where isthe shear stress and G is theshearing modulus of elasticity.



The shear stress may be calculated as = K(V/A)dx ,where K is a form factor that depends of the shape ofthe beam’s cross–sectional area A .

V

V

dy

dx

Combining these two expressionsgives dy = KV/(GA) dx .

The internal virtual work done bythe virtual shear force v acting onthe beam before it is deformed bythe real loads is dU i = v dy



Integrating the expression dU i = v dy over the entirebeam gives:

V

V

dy

dx

0

L

shear

vV U K dx

GA

Remember that v is the shear dueto the virtual load and V is the

shear due to the real loads.



Integrating the expression dU i = v dy over the entirebeam gives:

V

V

dy

dx

0

L

shear

vV U K dx

GA

The form factor K is based on thecross-sectional area:

K = 1.2 for rectangular sectionsK = 10/9 for circular sectionsK ≈ 1 for I-beams, where A is the area of the web

10 k

B C

0.5 k /ft


Compute the vertical deflection deflection androtation at point C on the frameshown.

Include the effects of bending and

both axial and shear forces in your

A

20 ft

10 ft

E = 29,000 ksi G = 12,000 ksi I = 1,000 in4

A = 25 in2

K = 1.2

.


Virtual Strain Energy From Torsion

For example, consider a circular cross-section where nowrapping of the section occurs.

For non-circular sections a more rigorous analysis isrequired.

dx

T T

d θ

c






This torque causes a shear strain: cd dx

dx

T T

d θ

c

For a linear-elastic response: G

Tc

J



The angle of twist:

T

dx

T T

d θ

c

c Gc GJ



If a virtual load is applied to the structure that causesan internal virtual torque t , then after applying the realloads, virtual strain ener will be:

dx

T T

d θ

c

t

tT dU t d dx

GJ



Integrating the virtual strain over the length of themember yields:

tTL

dx

T T

d θ

c

t GJ


Virtual Strain Energy From Temperature

Consider a structure member is subjected to atemperature difference across its depth.

For discussion, we will choose the most common case ofa beam having a neutral axis located at the mid-depth c of the beam

First compute the amount of rotation of a differentialelement dx of the beam caused by the thermalgradient.


T 1 dx


2

d

T m

T 1

T 2

δ x

δ x

M

1 2

2m

T T T

m x x





T 1 dx


T 2

d

T m

T 1

T 2

δ x

δ x

M

1 2

2m

T T T

x c



If a virtual load is applied to the structure that causesan internal virtual torque m , then after applying the realloads, virtual strain ener will be:

0

L m

temp

m T dU dx

c


Unless otherwise stated, in this course we will consideronly beam and frame deflections due to bending.

The additional deflections caused by the shear andaxia orce a ter t e e ections y on y a ew percentand are generally ignored for even “small” two- andthree-member frames of one-story height.

End of Virtual Work - Frames

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Notes_09c Virtual Work for Frames

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