18 Virtual Work Frame F16 - San Jose State University Work Frame Example.pdf1 CE 160 Virtual Work...
Transcript of 18 Virtual Work Frame F16 - San Jose State University Work Frame Example.pdf1 CE 160 Virtual Work...
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:
• 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
VukazichFall2016
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 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
VukazichFall2016
3 CE160VirtualWorkFrameExample
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
VukazichFall2016
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
VukazichFall2016
5 CE160VirtualWorkFrameExample
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
VukazichFall2016
6 CE160VirtualWorkFrameExample
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
VukazichFall2016
7 CE160VirtualWorkFrameExample
VukazichFall2016
8 CE160VirtualWorkFrameExample
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
VukazichFall2016
9 CE160VirtualWorkFrameExample
Product Integral Evaluation using Table 4 in text: