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Question : A 12-in-long strip of steel is 1/8 in thick and 1 in wide, a...

A 12-in-long strip of steel is 18\frac{1}{8} in thick and 1 in wide, as shown in Fig. 3–28. If the allowable shear stress is 11 500 psi and the shear modulus is 11.5(106)11.5\left(10^{6}\right) psi, find the torque corresponding to the allowable shear stress and the angle of twist, in degrees,(a) using Eq. (3–47) and (b) using Eqs. (3–40) and (3–41).

τ=Gθ1c=3TLc2\tau=G \theta_{1} c=\frac{3 T}{L c^{2}}    (3–47)

τmax⁡=Tαbc2≈Tbc2(3+1.8b/c)\tau_{\max }=\frac{T}{\alpha b c^{2}} \approx \frac{T}{b c^{2}}\left(3+\frac{1.8}{b / c}\right)    (3–40)

θ=Tlβbc3G\theta=\frac{T l}{\beta b c^{3} G}    (3–41)

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(a) The length of the median line is 1 in. From Eq. (3–47),

 

T=Lc2τ3=(1)(1/8)2115003=59.90lbf⋅inT=\frac{L c^{2} \tau}{3}=\frac{(1)(1 / 8)^{2} 11500}{3}=59.90 lbf \cdot in

 

θ=θ1l=τlGc=11500(12)11.5(106)(1/8)=0.0960rad=5.5∘\theta=\theta_{1} l=\frac{\tau l}{G c}=\frac{11500(12)}{11.5\left(10^{6}\right)(1 / 8)}=0.0960 rad =5.5^{\circ}

 

A torsional spring rate ktk_{t} can be expressed as T/θT / \theta :

 

kt=59.90/0.0960=624lbf⋅ in /radk_{t}=59.90 / 0.0960=624 lbf \cdot \text { in } / rad

 

(b) From Eq. (3–40),

 

T=τmax⁡bc23+1.8/(b/c)=11500(1)(0.125)23+1.8/(1/0.125)=55.72lbf⋅ in T=\frac{\tau_{\max } b c^{2}}{3+1.8 /(b / c)}=\frac{11500(1)(0.125)^{2}}{3+1.8 /(1 / 0.125)}=55.72 lbf \cdot \text { in }

 

From Eq. (3–41), with b/c=1/0.125=8b / c=1 / 0.125=8

 

θ=Tlβbc3G=55.72(12)0.307(1)0.1253(11.5)106=0.0970rad=5.6∘\theta=\frac{T l}{\beta b c^{3} G}=\frac{55.72(12)}{0.307(1) 0.125^{3}(11.5) 10^{6}}=0.0970 rad =5.6^{\circ}

 

kt=55.72/0.0970=574lbf⋅in/radk_{t}=55.72 / 0.0970=574 lbf \cdot in / rad

 

The cross section is not thin, where b should be greater than c by at least a factor of 10. In estimating the torque, Eq. (3–47) provided a value of 7.5 percent higher than Eq. (3–40), and 8.5 percent higher than the table on page 116.

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