A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 83,051 N , and experien
ces an axial elongation of 0.18 mm. Assuming that the deformation is entirely elastic, calculate the elastic modulus of this steel in GPa. Answer Format: X (no decimal places) Unit: GPa
1 answer:
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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
<u>Determine the value of the applied axial tensile force when yielding of rod occurs </u>
first we will calculate the shear stress and normal stress
shear stress ( г ) = Tr / J = [( 100 * 10^3) * 10 ] / * ( 20)^4
= 63.662 MPa
Normal stress( Гb + Гa ) = MY/ I + P/A
= [( 150 * 10^3) * 10 ] / * ( 20)^4 + 4P /
= 190.9859 + 4P / MPa
<u>a) Using MSS theory </u>
value of axial force = 42.422 KN
solution attached below
<u>b) Using MDE theory </u>
value of axial force = 44.356 KN
solution attached below
