A) What is the axial force along AB?FAR =unitsB) What must be the minimum diameter of the pin at B?(Remember it is double shear.)Pin diameter at B =units: Assume P=200 lbThere is a pin at B with double shear.The shear stress at failure for the pin material is 16 ksiFactor of safety= 2.545°

Given:P=200 lbFoS=2.5shear stress at failure for the pin material= 16 ksiPart a:Axial force along ABApplying lami's theoremPsin 135=FABsin 90FAB=Psin 135×sin 90FAB=2000.707×1FAB=282.88 lb ,Axial force Part B:Allowable shear stress=max. shear stress FoSallowable shear stress,ς=16 2.5=6.4 ksiallowable shear stress=shear Forceshear areashear force at pin B is FABς=FAB2×Apin( taking Area twice because this is double shear )Apin=π4×dp26.4=282.882×π4×dp2dp2=282.88×46.4×2×πdp=282.88×46.4×2×πdp=5.3045 inch or 13.47 cm

Answered: A) What is the axial force along AB?…

Engineering

Mechanical Engineering

A) What is the axial force along AB? FAR = units B) What must be the minimum diameter of the pin at B? (Remember it is double shear.)

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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