Question

PROBLEM \( 3.123 \) \( (a) \) Using \( \tau_{\text {all }}=70 \mathrm{MPa} \) and \( G=27 \mathrm{GPa} \), determine for each of the aluminum bars shown the largest torque \( \mathbf{T} \) that can be applied and the corresponding angle of twist at end \( B \). (b) SOLUTION \[ \tau_{\text {all }}=70 \times 10^{6} \mathrm{~Pa} \quad G=27 \times 10^{9} \mathrm{~Pa} \quad L=0.900 \mathrm{~m} \] (a) \( a=45 \mathrm{~mm} \quad b=15 \mathrm{~mm} \quad \frac{a}{b}=3.0 \quad \) From Table 3.1, \( c_{1}=0.267, \quad c_{2}=0.263 \) \[ \begin{aligned} \tau_{\max } & =\frac{T}{c_{1} a b^{2}} \quad T=c_{1} a b^{2} \tau_{\max } \\ T & =(0.267)(0.045)(0.015)^{2}\left(70 \times 10^{6}\right)=189.236 \mathrm{~N} \cdot \mathrm{m} \quad T=189.2 \mathrm{~N} \cdot \mathrm{m} \\ \varphi & =\frac{T L}{c_{2} a b^{3} G}=\frac{(189.236)(0.900)}{(0.263)(0.045)(0.015)^{3}\left(27 \times 10^{9}\right)}=157.921 \times 10^{-3} \mathrm{rad} \quad \varphi=9.05^{\circ} \end{aligned} \] (b) \( \quad a=25 \mathrm{~mm} \quad b=25 \mathrm{~mm}, \quad \frac{a}{b}=1.0 \quad \) From Table 3.1, \( c_{1}=0.208, \quad c_{2}=0.1406 \) \[ \begin{aligned} \tau_{\max }=\frac{T}{a b^{2}} \quad T & =c_{1} a b^{2} \tau_{\max } \\ & =(0.208)(0.025)(0.025)^{2}\left(70 \times 10^{6}\right) \\ & =227.5 \mathrm{~N} \cdot \mathrm{m} \quad T=228 \mathrm{~N} \cdot \mathrm{m} \\ \varphi=\frac{T L}{c_{2} a b^{3} G}= & \frac{(227.5)(0.900)}{(0.1406)(0.025)(0.025)^{3}\left(27 \times 10^{9}\right)}=138.075 \times 10^{-3} \mathrm{rad} \quad \varphi=7.91^{\circ} \end{aligned} \]

          PROBLEM \( 3.123 \)
\( (a) \)
Using \( \tau_{\text {all }}=70 \mathrm{MPa} \) and \( G=27 \mathrm{GPa} \), determine for each of the aluminum bars shown the largest torque \( \mathbf{T} \) that can be applied and the corresponding angle of twist at end \( B \).
(b) SOLUTION \[ \tau_{\text {all }}=70 \times 10^{6} \mathrm{~Pa} \quad G=27 \times 10^{9} \mathrm{~Pa} \quad L=0.900 \mathrm{~m} \]
(a) \( a=45 \mathrm{~mm} \quad b=15 \mathrm{~mm} \quad \frac{a}{b}=3.0 \quad \) From Table 3.1, \( c_{1}=0.267, \quad c_{2}=0.263 \)
\[
\begin{aligned}
\tau_{\max } & =\frac{T}{c_{1} a b^{2}} \quad T=c_{1} a b^{2} \tau_{\max } \\
T & =(0.267)(0.045)(0.015)^{2}\left(70 \times 10^{6}\right)=189.236 \mathrm{~N} \cdot \mathrm{m} \quad T=189.2 \mathrm{~N} \cdot \mathrm{m} \\
\varphi & =\frac{T L}{c_{2} a b^{3} G}=\frac{(189.236)(0.900)}{(0.263)(0.045)(0.015)^{3}\left(27 \times 10^{9}\right)}=157.921 \times 10^{-3} \mathrm{rad} \quad \varphi=9.05^{\circ}
\end{aligned}
\]
(b) \( \quad a=25 \mathrm{~mm} \quad b=25 \mathrm{~mm}, \quad \frac{a}{b}=1.0 \quad \) From Table 3.1, \( c_{1}=0.208, \quad c_{2}=0.1406 \)
\[
\begin{aligned}
\tau_{\max }=\frac{T}{a b^{2}} \quad T & =c_{1} a b^{2} \tau_{\max } \\
& =(0.208)(0.025)(0.025)^{2}\left(70 \times 10^{6}\right) \\
& =227.5 \mathrm{~N} \cdot \mathrm{m} \quad T=228 \mathrm{~N} \cdot \mathrm{m} \\
\varphi=\frac{T L}{c_{2} a b^{3} G}= & \frac{(227.5)(0.900)}{(0.1406)(0.025)(0.025)^{3}\left(27 \times 10^{9}\right)}=138.075 \times 10^{-3} \mathrm{rad} \quad \varphi=7.91^{\circ}
\end{aligned}
\]

PROBLEM 3.123
(a)
Using τall =70 MPa and G=27 GPa, determine for each of the aluminum bars shown the largest torque 𝐓 that can be applied and the corresponding angle of twist at end B.
(b) SOLUTION
τall =70 × 10^6 Pa G=27 × 10^9 Pa L=0.900 m

(a) a=45 mm b=15 mm (a)/(b)=3.0 From Table 3.1, c1=0.267, c2=0.263

τmax =(T)/(c1 a b^2) T=c1 a b^2τmax

T =(0.267)(0.045)(0.015)^2(70 × 10^6)=189.236 N·m T=189.2 N·m
φ =(T L)/(c2 a b^3 G)=((189.236)(0.900))/((0.263)(0.045)(0.015)^3(27 × 10^9))=157.921 × 10^-3rad φ=9.05^∘

(b) a=25 mm b=25 mm, (a)/(b)=1.0 From Table 3.1, c1=0.208, c2=0.1406

τmax=(T)/(a b^2) T =c1 a b^2τmax
=(0.208)(0.025)(0.025)^2(70 × 10^6)
=227.5 N·m T=228 N·m
φ=(T L)/(c2 a b^3 G)= ((227.5)(0.900))/((0.1406)(0.025)(0.025)^3(27 × 10^9))=138.075 × 10^-3rad φ=7.91^∘

Added by Ibrahim A.

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