A welded steel tube is 40 in long, has a $\frac{1}{8}$ -in wall thickness, and a 2.5-in by 3.6-in rectangular cross section as shown in Fig. 3–26. Assume an allowable shear stress of 11 500 psi and a shear modulus of $11.5(10^{6})$ psi.
(a) Estimate the allowable torque T.
(b) Estimate the angle of twist due to the torque.

Question

A welded steel tube is 40 in long, has a [latex]\frac{1}{8}[/latex] -in wall thickness, and a 2.5-in by 3.6-in rectangular cross section as shown in Fig. 3–26. Assume an allowable shear stress of 11 500 psi and a shear modulus of [latex]11.5(10^{6})[/latex] psi.
(a) Estimate the allowable torque T.
(b) Estimate the angle of twist due to the torque.

Accepted Answer

(a) Within the section median line, the area enclosed is

[latex]A_{m} = (2.5 − 0.125)(3.6 − 0.125) = 8.253 in^{2}[/latex]

and the length of the median perimeter is

[latex]L_{m} = 2[(2.5 − 0.125) + (3.6 − 0.125)] = 11.70 in[/latex]

From Eq. (3–45) the torque T is

[latex]τ =\frac{T}{2A_{m}t}[/latex] (3–45)

[latex]T = 2A_{m}tτ = 2(8.253)0.125(11 500) = 23 730 lbf · in[/latex]

(b) The angle of twist θ from Eq. (3–46) is

[latex]θ = θ_{1}l =\frac{T L_{m}}{4GA^{2}_{m}t}l =\frac {23 730(11.70)}{4(11.5 × 10^{6})(8.253^{2})(0.125)}(40) = 0.0284 rad =1.62◦[/latex]

Question 3.10: A welded steel tube is 40 in long, has a 1/8 -in wall thickn...

Related Answered Questions