Answer:
a) 42.422 KN
b) 44.356 KN
Explanation:
Given data :
Diameter = 20 mm
yield strength = 350 MN/m^2
Torque ( T ) = 100 N.m
Bending moment = 150 N.m
Determine the value of the applied axial tensile force when yielding of rod occurs
first we will calculate the shear stress and normal stress
shear stress ( г ) = Tr / J = [( 100 * 10^3) * 10 ] / [tex]\pi /32[/tex] * ( 20)^4
= 63.662 MPa
Normal stress( Гb + Гa ) = MY/ I + P/A
= [( 150 * 10^3) * 10 ] / [tex]\pi /32[/tex] * ( 20)^4 + 4P / [tex]\pi * 20^2[/tex]
= 190.9859 + 4P / [tex]\pi * 20^2[/tex] MPa
a) Using MSS theory
value of axial force = 42.422 KN
solution attached below
b) Using MDE theory
value of axial force = 44.356 KN
solution attached below
