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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VukazichFall2016
1 CE160VirtualWorkFrameExample
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:
β’ EIABC = 5,336,000 k-in2 β’ 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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2 CE160VirtualWorkFrameExample
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 Ξ΄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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3 CE160VirtualWorkFrameExample
Free Body Diagram of Virtual System (or Q-system)
Equilibrium π! = 0; πΉ! = 0; πΉ! = 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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4 CE160VirtualWorkFrameExample
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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5 CE160VirtualWorkFrameExample
Use Table 4 in the text to evaluate the virtual work product integrals that come from the principle of virtual work:
1 β πΏ! = 1πΈπΌ
π!π!
!
!ππ₯
Integrate over three segments: AB, BC, and BDE
Segment AB
From Virtual Work Integral Table (see page 8):
13π!π!πΏ =
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):
13π!π!πΏ =
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 ππ‘π₯ =
12 ππ‘ 5β π₯
x = 3 ft
Segment BZ
From Virtual Work Integral Table (see page 8):
16 π! + 2π! π!πΏ =
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
β16π!π! πΏ + π = β
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
πΏ! =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
πΏ! = 0.1079 in+ 0.05527 in+ 0.03801 in
πΉπͺ = π.πππ in Result is positive, so deflection is in the direction of the unit load β to the right
12 k-ft
E
8 ft
13.091 ft
Z E
6 ft
Z
2 ft
D
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9 CE160VirtualWorkFrameExample
Product Integral Evaluation using Table 4 in text:
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