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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

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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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: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
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
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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