Question

PROBLEM $ 3.120 $ A torque $ \mathbf{T} $ applied to a solid rod made of an elastoplastic material is increased until the rod is fully plastic and them removed. (a) Show that the distribution of residual shearing stresses is as represented in the figure. (b) Determine the magnitude of the torque due to the stresses acting on the portion of the rod located within a circle of radius $ c_{0} $. SOLUTION Loading Unloading Residual $ (a) $ After loading: \[ \rho_{Y}=0, \quad T_{\text {load }}=\frac{4}{3} T_{Y}=\frac{4}{3} \frac{\pi}{2} c^{3} \tau_{Y}=\frac{2 \pi}{3} c^{3} \tau_{Y} \] Unloading: Residual: \[ \begin{array}{l} \tau^{\prime}=\frac{T c}{J}=\frac{2 T}{\pi c^{3}}=\frac{2\left(T_{\text {load }}\right)}{\pi c^{3}}=\frac{4}{3} \tau_{Y} \quad \text { at } \rho=c \\ \tau^{\prime}=\frac{4}{3} \tau_{Y} \frac{\rho}{c} \\ \tau_{\mathrm{res}}=\tau_{Y}-\frac{4}{3} \tau_{Y} \frac{\rho}{c}=\tau_{Y}\left(1-\frac{4 \rho}{3 c}\right) \\ 0=1-\frac{4 c_{0}}{3 c} \quad \therefore \quad c_{0}=\frac{3}{4} c \\ c_{0}=0.150 c \\ \end{array} \] To find $ c_{0} $ set, $ \quad \tau_{\text {res }}=0 $ and $ \rho=c_{0} $ $ (b) $ \[ \begin{array}{l} T_{0}=2 \pi \int_{0}^{c_{0}} \rho^{2} \tau d \rho=2 \pi \int_{0}^{(3 / 4) c} \rho^{2} \tau_{Y}\left(1-\frac{4}{3} \frac{\rho}{c}\right) d \rho \\ =\left.2 \pi \tau_{Y}\left(\frac{\rho^{3}}{3}-\frac{4}{3} \frac{\rho^{4}}{4 c}\right)\right|_{0} ^{(34) c}=2 \pi \tau_{Y} c^{3}\left\{\frac{1}{3}\left(\frac{3}{4}\right)^{3}-\left(\frac{4}{3}\right) \frac{1}{4}\left(\frac{3}{4}\right)^{4}\right\} \\ =2 \pi \tau_{Y} c^{3}\left\{\frac{9}{64}-\frac{27}{256}\right\}=\frac{9 \pi}{128} \tau_{Y} c^{3}=0.2209 \tau_{Y} c^{3} \\ T_{0}=0.221 \tau_{Y} c^{3} \\ \end{array} \]

          PROBLEM \( 3.120 \)
A torque \( \mathbf{T} \) applied to a solid rod made of an elastoplastic material is increased until the rod is fully plastic and them removed. (a) Show that the distribution of residual shearing stresses is as represented in the figure. (b) Determine the magnitude of the torque due to the stresses acting on the portion of the rod located within a circle of radius \( c_{0} \).
SOLUTION
Loading
Unloading
Residual
\( (a) \)
After loading:
\[
\rho_{Y}=0, \quad T_{\text {load }}=\frac{4}{3} T_{Y}=\frac{4}{3} \frac{\pi}{2} c^{3} \tau_{Y}=\frac{2 \pi}{3} c^{3} \tau_{Y}
\]
Unloading:
Residual:
\[
\begin{array}{l}
\tau^{\prime}=\frac{T c}{J}=\frac{2 T}{\pi c^{3}}=\frac{2\left(T_{\text {load }}\right)}{\pi c^{3}}=\frac{4}{3} \tau_{Y} \quad \text { at } \rho=c \\
\tau^{\prime}=\frac{4}{3} \tau_{Y} \frac{\rho}{c} \\
\tau_{\mathrm{res}}=\tau_{Y}-\frac{4}{3} \tau_{Y} \frac{\rho}{c}=\tau_{Y}\left(1-\frac{4 \rho}{3 c}\right) \\
0=1-\frac{4 c_{0}}{3 c} \quad \therefore \quad c_{0}=\frac{3}{4} c \\
c_{0}=0.150 c \\
\end{array}
\]
To find \( c_{0} \) set, \( \quad \tau_{\text {res }}=0 \) and \( \rho=c_{0} \)
\( (b) \)
\[
\begin{array}{l}
T_{0}=2 \pi \int_{0}^{c_{0}} \rho^{2} \tau d \rho=2 \pi \int_{0}^{(3 / 4) c} \rho^{2} \tau_{Y}\left(1-\frac{4}{3} \frac{\rho}{c}\right) d \rho \\
=\left.2 \pi \tau_{Y}\left(\frac{\rho^{3}}{3}-\frac{4}{3} \frac{\rho^{4}}{4 c}\right)\right|_{0} ^{(34) c}=2 \pi \tau_{Y} c^{3}\left\{\frac{1}{3}\left(\frac{3}{4}\right)^{3}-\left(\frac{4}{3}\right) \frac{1}{4}\left(\frac{3}{4}\right)^{4}\right\} \\
=2 \pi \tau_{Y} c^{3}\left\{\frac{9}{64}-\frac{27}{256}\right\}=\frac{9 \pi}{128} \tau_{Y} c^{3}=0.2209 \tau_{Y} c^{3} \\
T_{0}=0.221 \tau_{Y} c^{3} \\
\end{array}
\]

$PROBLEM 3.120 A torque 𝐓 applied to a solid rod made of an elastoplastic material is increased until the rod is fully plastic and them removed. (a) Show that the distribution of residual shearing stresses is as represented in the figure. (b) Determine the magnitude of the torque due to the stresses acting on the portion of the rod located within a circle of radius c0. SOLUTION Loading Unloading Residual (a) After loading: ρY=0, Tload =(4)/(3) TY=(4)/(3)(π)/(2) c^3τY=(2 π)/(3) c^3τY Unloading: Residual: τ^'=(T c)/(J)=(2 T)/(π c^3)=(2(Tload ))/(π c^3)=(4)/(3)τY at ρ=c τ^'=(4)/(3)τY(ρ)/(c) τres=τY-(4)/(3)τY(ρ)/(c)=τY(1-(4 ρ)/(3 c)) 0=1-(4 c0)/(3 c) ∴ c0=(3)/(4) c c0=0.150 c To find c0 set, τres =0 and ρ=c0 (b) T0=2 π∫0^c0ρ^2τ d ρ=2 π∫0^(3 / 4) cρ^2τY(1-(4)/(3)(ρ)/(c)) d ρ =.2 πτY((ρ^3)/(3)-(4)/(3)(ρ^4)/(4 c))|0 ^(34) c=2 πτY c^3{(1)/(3)((3)/(4))^3-((4)/(3)) (1)/(4)((3)/(4))^4} =2 πτY c^3{(9)/(64)-(27)/(256)}=(9 π)/(128)τY c^3=0.2209 τY c^3 T0=0.221 τY c^3$

Added by Ibrahim A.

Question

Please give Ace some feedback