A wood structure is assembled to support the force applied.All pieces of wood have a width of 5 in. The normal failurestress of the wood is o fail = 6000 psi. The shear failure stressof the wood is T fail = 2500 psi. The factor of safety againstfailure is 5.Hint:1. Analyze the whole structure first to find the normalforces in members AB and ACbB0₁ᎾᎥzFAF010₁CValues for the figure are given in the following table. Notethe figure may not be to scate. Values for the figure are given in the following table. Notethe figure may not be to scale.VariableF0₁a. Determine normal force in member AC, NAC.b. Determine normal force in member AB, NAB.c. Determine the shear force on the bottom supportmember 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.Round your final answers to 3 significant digits/figures.NAC =NAB =V =allowTallowb =t =Value5000 lb15 degrees=lbininlblbpsipsi

Find outDetermine the normal force in member AC, Determine the normal force in member AB, Determine…

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. 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 o 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

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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