
Respuesta :
Answer:
a) 42.422 KN
b) 44.356 KN
Explanation:
Given data :
Diameter = 20 mm
yield strength = 350 MN/m^2
Torque ( T ) Â = 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 Â
                    = 63.662 MPa
Normal stress(  Гb + Гa )  = MY/ I  +  P/A
= [( 150 * 10^3) Â * 10 ] Â / Â [tex]\pi /32[/tex] * ( 20)^4 Â + 4P / [tex]\pi * 20^2[/tex]
= 190.9859 + 4P / [tex]\pi * 20^2[/tex] Â MPa
a) Using MSS theory
value of axial force = 42.422 KN
solution attached below
b) Using MDE Â theory
value of axial force = 44.356 KN
solution attached below
