18 Virtual Work Frame F16 - San Jose State University Work Frame Example.pdf1 CE 160 Virtual Work Frame Example CE 160 Notes: Virtual Work Frame Example ... The modulus of elasticity

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CE 160 Notes: Virtual 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 δC

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 𝑀! = 0; 𝐹! = 0; 𝐹! = 0 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 ∙ 𝛿! = 1𝐸𝐼

𝑀!𝑀!

!

!𝑑𝑥

Integrate over three segments: AB, BC, and BDE

Segment AB

From Virtual Work Integral Table (see page 8):

13𝑀!𝑀!𝐿 =

13

10 ft 10 k-ft 10 ft

= 333.333 k-ft3 = 576,000 k-in3

Segment BC

From Virtual Work Integral Table (see page 8):

13𝑀!𝑀!𝐿 =

13

8 ft 8 k-ft 8 ft

=170.6667 k-ft3 = 294,912 k-in3

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

MQDiagram MPDiagram

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

18 𝑓𝑡𝑥 =

12 𝑓𝑡 5− 𝑥

x = 3 ft

Segment BZ

From Virtual Work Integral Table (see page 8):

16 𝑀! + 2𝑀! 𝑀!𝐿 =

16 13.091 ft +2 18 ft 18 k-ft 3 ft

= 441.818 k-ft3 = 763,461.8167 k-in3

B

E

12 k-ft

18 k-ft

8 ft

6 ft

B

E

18 ft 13.091 ft

8 ft

MQDiagram MPDiagram

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

−16𝑀!𝑀! 𝐿 + 𝑐 = −

16 13.091 ft 12 k-ft 8 ft+ 6 ft

= -366.5454 k-ft3 = -633,390.54 k-in3

Total for segment BDE

763,461.8167 k-in3 - 633,390.54 k-in3 = 130,071.28 Divide by EI of each segment

𝛿! =576,000 k-in3

5,336,000 k-in2 +294,912 k-in3

5,336,000 k-in2 +130,071.28 k-in3

3,422,000 k-in2

𝛿! = 0.1079 in+ 0.05527 in+ 0.03801 in

𝜹𝑪 = 𝟎.𝟐𝟎𝟏 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: