Problem 1 Determine the vertical displacement at joint H of the truss below using the method of virtual work. Each steel member has a cross-sectional area of 4.5 in², and E = 29(10³) ksi. Hint: For more efficient computation of internal axial forces, consider the concept of zero-force members and use the symmetry of the structure and the given loading. 12 ft- B 6 k -12 ft- H 8 k 12 ft- G D 6 k 12 ft- 9 ft
Problem 1 Determine the vertical displacement at joint H of the truss below using the method of virtual work. Each steel member has a cross-sectional area of 4.5 in², and E = 29(10³) ksi. Hint: For more efficient computation of internal axial forces, consider the concept of zero-force members and use the symmetry of the structure and the given loading. 12 ft- B 6 k -12 ft- H 8 k 12 ft- G D 6 k 12 ft- 9 ft
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question

Transcribed Image Text:Problem 1
Determine the vertical displacement at joint H of the truss below using the method of virtual work. Each steel
member has a cross-sectional area of 4.5 in², and E = 29(10³) ksi. Hint: For more efficient computation of internal
axial forces, consider the concept of zero-force members and use the symmetry of the structure and the given
loading.
A
J
—12 ft
B
6 k
12 ft
H
C
8 k
12 ft
G
D
6 k
- 12 ft
F
T
9 ft
E
![Table for:
M
M
Rectangle
M
Triangle
M
L
Triangle
M
L
Trapezoid
L
Triangle
-α** ·b·
L
Parabola
slope-0
L
slope-0
Parabola
slope = 0
L
Parabola
SMQdx
L
Parabola
L
M
M
M₁
M
M
slope - 0
www.
Q
Rectangle
L
LMQ
LMQ
2
LMQ
LMQ
(M,+M)
The values in the table represent the integration of the
product of the two shapes with a common length L.
Triangle
Triangle
Trapezoid Q
2LMQ
2LMQ
3
LMQ
LMQ
L
LMQ
2
LMQ
LMQ
6
MQ (L + a)
6
LO (M,+2M)
SLMQ
12
LMQ
QQ
IMQ
4
LMQ
12
L
LMQ
NE
LMQ
6
LMQ
3
(2M, + M)
LMQ
SLMQ
12
Ι.ΜΟ
LMQ
L
LM(Q+Q)
IM(Q+20)
LM (2Q+Q)
[Q(+b)
+Q₂(l+a)]
Q(2M,+ M)
1, +2M)]
+Q₂ (M₂+
IM (3Q+5Q)
12
[MsQ+3Q)
IM(Q+30)
(3Q+Q)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F95aeff90-eae1-41d1-a6d5-24c543733ee7%2F5fce1d3c-4d8d-4122-9c60-7cf5a1c5332a%2Fqc1id4n_processed.png&w=3840&q=75)
Transcribed Image Text:Table for:
M
M
Rectangle
M
Triangle
M
L
Triangle
M
L
Trapezoid
L
Triangle
-α** ·b·
L
Parabola
slope-0
L
slope-0
Parabola
slope = 0
L
Parabola
SMQdx
L
Parabola
L
M
M
M₁
M
M
slope - 0
www.
Q
Rectangle
L
LMQ
LMQ
2
LMQ
LMQ
(M,+M)
The values in the table represent the integration of the
product of the two shapes with a common length L.
Triangle
Triangle
Trapezoid Q
2LMQ
2LMQ
3
LMQ
LMQ
L
LMQ
2
LMQ
LMQ
6
MQ (L + a)
6
LO (M,+2M)
SLMQ
12
LMQ
QQ
IMQ
4
LMQ
12
L
LMQ
NE
LMQ
6
LMQ
3
(2M, + M)
LMQ
SLMQ
12
Ι.ΜΟ
LMQ
L
LM(Q+Q)
IM(Q+20)
LM (2Q+Q)
[Q(+b)
+Q₂(l+a)]
Q(2M,+ M)
1, +2M)]
+Q₂ (M₂+
IM (3Q+5Q)
12
[MsQ+3Q)
IM(Q+30)
(3Q+Q)
Expert Solution

Step 1
Given:
Area of the member, A = 4.5 in2
Modulus of elasticity, E = 29000 ksi
To find:
Vertical displacement at H
Step by step
Solved in 6 steps with 8 images

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