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PROBLEM \( 3.112 \) A 50-mm-diameter cylinder is made of a brass for which the stress-strain diagram is as shown. Knowing that the angle of twist is \( 5^{\circ} \) in a \( 725-\mathrm{mm} \) length, determine by approximate means the magnitude \( T \) of torque applied to the shaft SOLUTION \( (a) \) \[ \begin{aligned} \varphi & =5^{\circ}=87.266 \times 10^{-3} \mathrm{rad} \\ c & =\frac{1}{2} d=0.025 \mathrm{~m} \quad L=0.725 \mathrm{~m} \\ \gamma_{\max } & =\frac{c \varphi}{L}=\frac{(0.025)\left(87.266 \times 10^{-3}\right)}{0.725}=0.00301 \end{aligned} \] Let \( \quad z=\frac{\gamma}{\gamma_{\max }}=\frac{\rho}{c} \) \( T=2 \pi \int_{0}^{c} \rho^{2} \tau d \rho=2 \pi c_{2}^{3} \int_{0}^{1} z^{2} \tau d z=2 \pi c_{2}^{3} I \quad \) where the integral \( I \) is given by \( I=\int_{1 / 3}^{1} z^{2} \tau d z \) Evaluate \( I \) using a method of numerical integration. If Simpson's rule is used, the integration formula is \[ I=\frac{\Delta z}{3} \sum w z^{2} \tau \] where \( w \) is a weighting factor. Using \( \Delta z=0.25 \), we get the values given in the table below. \[ \begin{array}{l} I=\frac{(0.25)\left(283.75 \times 10^{6}\right)}{3}=23.65 \times 10^{6} \mathrm{~Pa} \\ T=2 \pi c^{3} I=2 \pi(0.025)^{3}\left(23.65 \times 10^{6}\right)=2.32 \times 10^{3} \mathrm{~N} \cdot \mathrm{m} \quad T=2.32 \mathrm{kN} \cdot \mathrm{m} \end{array} \]

          PROBLEM \( 3.112 \)
A 50-mm-diameter cylinder is made of a brass for which the stress-strain diagram is as shown. Knowing that the angle of twist is \( 5^{\circ} \) in a \( 725-\mathrm{mm} \) length, determine by approximate means the magnitude \( T \) of torque applied to the shaft
SOLUTION
\( (a) \)
\[
\begin{aligned}
\varphi & =5^{\circ}=87.266 \times 10^{-3} \mathrm{rad} \\
c & =\frac{1}{2} d=0.025 \mathrm{~m} \quad L=0.725 \mathrm{~m} \\
\gamma_{\max } & =\frac{c \varphi}{L}=\frac{(0.025)\left(87.266 \times 10^{-3}\right)}{0.725}=0.00301
\end{aligned}
\]
Let \( \quad z=\frac{\gamma}{\gamma_{\max }}=\frac{\rho}{c} \)
\( T=2 \pi \int_{0}^{c} \rho^{2} \tau d \rho=2 \pi c_{2}^{3} \int_{0}^{1} z^{2} \tau d z=2 \pi c_{2}^{3} I \quad \) where the integral \( I \) is given by \( I=\int_{1 / 3}^{1} z^{2} \tau d z \)
Evaluate \( I \) using a method of numerical integration. If Simpson's rule is used, the integration formula is
\[
I=\frac{\Delta z}{3} \sum w z^{2} \tau
\]
where \( w \) is a weighting factor. Using \( \Delta z=0.25 \), we get the values given in the table below.
\[
\begin{array}{l}
I=\frac{(0.25)\left(283.75 \times 10^{6}\right)}{3}=23.65 \times 10^{6} \mathrm{~Pa} \\
T=2 \pi c^{3} I=2 \pi(0.025)^{3}\left(23.65 \times 10^{6}\right)=2.32 \times 10^{3} \mathrm{~N} \cdot \mathrm{m} \quad T=2.32 \mathrm{kN} \cdot \mathrm{m}
\end{array}
\]

PROBLEM 3.112
A 50-mm-diameter cylinder is made of a brass for which the stress-strain diagram is as shown. Knowing that the angle of twist is 5^∘ in a 725-mm length, determine by approximate means the magnitude T of torque applied to the shaft
SOLUTION
(a)

φ =5^∘=87.266 × 10^-3rad

c =(1)/(2) d=0.025 m L=0.725 m
γmax =(c φ)/(L)=((0.025)(87.266 × 10^-3))/(0.725)=0.00301

Let z=(γ)/(γmax)=(ρ)/(c)
T=2 π∫0^cρ^2τ d ρ=2 π c2^3∫0^1 z^2τ d z=2 π c2^3 I where the integral I is given by I=∫1 / 3^1 z^2τ d z
Evaluate I using a method of numerical integration. If Simpson's rule is used, the integration formula is

I=(Δ z)/(3)∑ w z^2τ

where w is a weighting factor. Using Δ z=0.25, we get the values given in the table below.

I=((0.25)(283.75 × 10^6))/(3)=23.65 × 10^6 Pa

T=2 π c^3 I=2 π(0.025)^3(23.65 × 10^6)=2.32 × 10^3 N·m T=2.32 kN·m

Added by Ibrahim A.

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