Homework SourseA 38 inch long torsion bar is made from a 025 inch diameter
A 38 inch long torsion bar is made from a 0.25 inch diameter C1050 steel bar, as rolled. What is the maximum twisting angle in degrees to which the bar can be subjected before failure?
Solution
From the mechanical properties of the cold drawn AISI 1050 carbon steel
Yield strength 580 Mpa
Shear modulus (G) = 80 GPa = 80 x 109 Pa
Now, From maximum shear stress theory
Maximum shear stress () = 0.5 x Yield strength= 290 MPa = 290 x 106 Pa
Given,
Length of the bar (L) = 38 in = 0.965 m
Diameter (d) = 0.25 in = 0.00635 m
From Torsion Formula
/r = G /L .......................(1)
or = L/Gr
Where, is angle of twist, r = d/2 = 0.003175 m
Substituting the values to equation (1)
= 1.102 radian = 63.14o
| Yield strength 580 Mpa Shear modulus (G) = 80 GPa = 80 x 109 Pa Now, From maximum shear stress theory Maximum shear stress () = 0.5 x Yield strength= 290 MPa = 290 x 106 Pa Given, Length of the bar (L) = 38 in = 0.965 m Diameter (d) = 0.25 in = 0.00635 m From Torsion Formula /r = G /L .......................(1) or = L/Gr Where, is angle of twist, r = d/2 = 0.003175 m Substituting the values to equation (1) = 1.102 radian = 63.14o |
