Question 45° 200 lb B The bracket is supported by 1/2 -in.-diameter pins at A and B (the pin at B fits in the 450 slot in the bracket). Neglecting friction, determine the shear stresses in the pins, assuming single shear. 12 in. O A - 36 in.-

MWZ52N The Asker · Civil Engineering
Transcribed Image Text: 45° 200 lb B The bracket is supported by 1/2 -in.-diameter pins at A and B (the pin at B fits in the 450 slot in the bracket). Neglecting friction, determine the shear stresses in the pins, assuming single shear. 12 in. O A - 36 in.-
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Transcribed Image Text: 45° 200 lb B The bracket is supported by 1/2 -in.-diameter pins at A and B (the pin at B fits in the 450 slot in the bracket). Neglecting friction, determine the shear stresses in the pins, assuming single shear. 12 in. O A - 36 in.-
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Step 1Given:-Diameter of pin A^(¨) and pin B^(¨)=D=1//2-in.Required:- We have to calculate the Shear strasses at pin A and pin ' B '. 7Step 2Now apply Equilibrium Euation in order to get shear force at B^(˙)sumM_(B)=0; momoment about B^(¨)12 xxR_(Ax)-36(200)=0Resolve the forces horizontally & vertically.{:[:'sumf_(x)=0;=>R_(Ax)-E_(B)xx cos 45^(@)=0],[sumf_(y)=0;=>R_(Ay)+K_(B)xx sin 45^(@)-200=0]:}:'R_(Ax)=(36 xx200)/(12)=600.9 b.=>R_(Ax)=60006Now substivtute Rax in &quation (2), weget F_(B).{:[R_(Ax)-F_(B)cos 45^(@)=0],[600-F_(B)xx cos 45^(@)=0],[=>F_(B)=(600)/(cos 45^(@))=848.528*lb=F_(B)]:}:' Now substitute ' F_(B) ' value in euation(3).{:[:'quadR_(A ... See the full answer