A wood structure is assembled to support the force applied. All pieces of wood have a width of 5 in. The normal failure stress of the wood is fail = 6000 psi. The shear failure stress of the wood is T fail = 2500 psi. The factor of safety against failure is 5. Hint: 1. Analyze the whole structure first to find the normal forces in members AB and AC
A wood structure is assembled to support the force applied. All pieces of wood have a width of 5 in. The normal failure stress of the wood is fail = 6000 psi. The shear failure stress of the wood is T fail = 2500 psi. The factor of safety against failure is 5. Hint: 1. Analyze the whole structure first to find the normal forces in members AB and AC
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Answer the two last questions
Transcribed Image Text:13:23
Hint:
A wood structure is assembled to support the force applied.
All pieces of wood have a width of 5 in. The normal failure
stress of the wood is o fail = 6000 psi. The shear failure stress
of the wood is T fail = 2500 psi. The factor of safety against
failure is 5.
wamap.org
1. Analyze the whole structure first to find the normal
forces in members AB and AC
b
B
F
0₁
C
F
*
Ort
01
Values for the figure are given in the following table. Note
the figure may not be to scale
Transcribed Image Text:Values for the figure are given in the following table. Note
the figure may not be to scale.
Variable
F
0₁
c. Determine the shear force on the bottom support
member needed to find length b, V.
d. Determine the allowable normal stress, allow.
e. Determine the allowable shear stress, Tallow.
f. Determine minimum length, b.
g. Determine minimum thickness, t.
a. Determine normal force in member AC, NAC.
b. Determine normal force in member AB, NAB.
Round your final answers to 3 significant digits/figures.
NAC
Value
5000 lb
15 degrees
NAB= 9659
V = 9330
=9659
Gallow
b =
Tallow =
t =
1200
500
in
in
✓olb
Olb
Olb
opsi
opsi
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