Problem 2 Determine the rotation and vertical displacement at the end A of the beam below using the method of virtual work. E = 29(10³) ksi, I = 170 in*. A 2 k/ft -10 ft- B -10 ft 10 k -10 ft
Problem 2 Determine the rotation and vertical displacement at the end A of the beam below using the method of virtual work. E = 29(10³) ksi, I = 170 in*. A 2 k/ft -10 ft- B -10 ft 10 k -10 ft
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![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)
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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)
Transcribed Image Text:Problem 2
Determine the rotation and vertical displacement at the end A of the beam below using the method of virtual work.
E = 29(10³) ksi, I = 170 inª.
A
2 k/ft
-10 ft-
B
-10 ft
10 k
-10 ft
с
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