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CE 160 Notes: Virtu al Work Frame Example

The statically determinate frame from our internal force diagram example is made up of columns that are W8x48 (I = 184 in4) members and a beam that is a W10x22 (I = 118 in4). The modulus of elasticity of structural steel is 29,000 ksi, which yields the following bending stiffnesses:

EIABC = 5,336,000 k-in2 EIBDE = 3,422,000 k-in2

Find the horizontal displacement of point C using the method of virtual work.

Real Problem We found the Shear, Moment, and Axial force diagrams for this frame in a previous example earlier in the semester.

C

E

A

5 ft 6 ft

8 ft

D

8 k

1 k

10 ft

B

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Moment Diagram of Real or P-System (MP diagram)

(You should review your notes and verify this is the correct diagram)

Virtual Problem

Virtual System to measure C

C

B

D E

A

12 k-ft

18 k-ft

8 k-ft

10 k-ft

8 ft

6 ft

C

E

A

5 ft 6 ft

8 ft

D

1

10 ft

B

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Free Body Diagram of Virtual System (or Q-system)

Equilibrium ! ! ! ! ! ;! ! ! ! ! !! ! ! ! ! ! !!yields the following support reactions:

10 ft

C

B

5 ft 6 ft

8 ft

D

1

Ax

Ey

Ay

10 ft

C

B

5 ft 6 ft

8 ft

D

1

1

1.6364

E

A

1.6364

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Virtual Moment Diagram (MQ diagram)

(You should be able to verify this is the correct diagram)

C

B D

E

A

18 ft 8 ft

10 ft

13.091 ft

8 ft

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Use Table 4 in the text to evaluate the virtual work product integrals that come from the principle of virtual work:

! ! ! ! ! !!!1!"

! ! !!

!!" !! !!!

Integrate over three segments: AB, BC, and BDE

Segment AB

From Virtual Work Integral Table (see page 8):

!!

! ! ! ! ! !!

!! !" !!" !" !! !!" !" !!"

= !!!"!!! #$%&! ##= #()* +,,,# $%-. ! #

Segment BC

From Virtual Work Integral Table (see page 8):

!!! ! ! !

!

3! ! !!" ! !! !!" ! !!"

=170"***) #$%&! ##= /01 +02/ #$%-. ! #

C

B

D E

A

12 k-ft

18 k-ft

8 k-ft

10 k-ft

8 ft

6 ft

C

B D

E

A

18 ft 8 ft

10 ft

13.091 ft

8 ft

! "#$%&(&)# ! *#$%&(&)##

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Segment BDE -- Split into two segments at inflection point of MP diagram (point Z)

!" !!"

!!

!" !!" !! ! !

x = 3 ft

Segment BZ

From Virtual Work Integral Table (see page 8):

!6 ! ! ! ! ! ! ! ! !

!!

13!!"# !!"#$! !" !!" !" !k!!" ! !!" !

= 112"323#$%&! ##= )*!+1*2"32*) #$%-. ! #

B

E

12 k-ft

18 k-ft

8 ft

6 ft

B

E

18 ft 13.091 ft

8 ft

! "#$%&(&)# ! *#$%&(&)##

3 ft

3 ft

Z Z

D

B

18 ft 13.091 ft

Z

3 ft

B

18 k-ft

Z

3 ft

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Segment ZDE

From Virtual Work Integral Table (see page 8): Note that result is negative due to dissimilar curvatures

!!!

! ! ! ! ! ! ! !!6 !" !!"# !!" !" !! !!" ! !!" ! 6 !"

= -!**"(1(1 #$%&! ##= -*!!+!0,"(1 #$%-. ! #

#Total for segment BDE #

)*!+1*2"32*)#$ %-. ! - *!!+!0,"(1#$ %-. ! #4#2!,+,)2"/3 # Divide by EI of each segment

! ! !576!!!! !! !!" !

! !!!" !!!! !! !!" 2!

!"# !!"# !k!!" !

! !!!" !!!! !! -!" ! !!"# ,!"# !!" !! !!" !

! !!"" !!!! !! !!" !

! ! ! ! !!"#$ !in ! ! !!""#$ !!" ! ! !!"#!$ !!"

! ! ! ! !"# !!" Result is positive, so deflection is in the direction of the unit load to the right

12 k-ft

E

8 ft

13.091 ft

Z E

6 ft

Z

2 ft

D

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Product Integral Evaluation using Table 4 in text: