Answer
$A_{\text {tube }}=\frac{\pi}{4}\left(d_{o}^{2}-d_{i}^{2}\right)=\frac{\pi}{4}\left(36^{2}-28^{2}\right)=402.12 \mathrm{mm}^{2}=402.12 \times 10^{-6} \mathrm{m}^{2}$
$A_{\mathrm{rod}}=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(25)^{2}=490.87 \mathrm{mm}^{2}=490.87 \times 10^{-6} \mathrm{m}^{2}$
$\delta_{\text {tube }}=\frac{P L}{E_{\text {tube }} A_{\text {tube }}}=\frac{P(0.250)}{\left(70 \times 10^{9}\right)\left(402.12 \times 10^{-6}\right)}=8.8815 \times 10^{-9} P$
$\delta_{\text {rod }}=-\frac{P L}{E_{\text {rod }} A_{\text {rod }}}=\frac{P(0.250)}{\left(105 \times 10^{6}\right)\left(490.87 \times 10^{-6}\right)}=-4.8505 \times 10^{-9} P$
$\delta^{*} =\left(\frac{1}{4} \text { turn }\right) \times 1.5 \mathrm{mm}=0.375 \mathrm{mm}=375 \times 10^{-6} \mathrm{m} $
$\delta_{\text {tube }} =\delta^{*}+\delta_{\text {rod }} \text { or } \delta_{\text {tube }}-\delta_{\text {rod }}=\delta^{*} $
$ 8.8815 \times 10^{-9} P+4.8505 \times 10^{-9} P =375 \times 10^{-6} $
$P =\frac{0.375 \times 10^{-3}}{(8.8815+4.8505)\left(10^{-9}\right)}=27.308 \times 10^{3} \mathrm{N} $
(a)
${\sigma_{\text {tube }}=\frac{P}{A_{\text {tube }}}} {=\frac{27.308 \times 10^{3}}{402.12 \times 10^{-6}}=67.9 \times 10^{6} \mathrm{Pa}} $
$$ \therefore {\sigma_{\text {tube }}=67.9 \mathrm{MPa}} $$
$ {\sigma_{\text {rod }}} {=-\frac{P}{A_{\text {rod }}}=-\frac{27.308 \times 10^{3}}{490.87 \times 10^{-6}}=-55.6 \times 10^{6} \mathrm{Pa}} $
$$\therefore {\sigma_{\text {rod }}=-55.6 \mathrm{MPa}}$$
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(b)
${\delta_{\text {that }}=\left(8.8815 \times 10^{-9}\right)\left(27.308 \times 10^{3}\right)=242.5 \times 10^{-6} \mathrm{m}} $
$$ \therefore {\delta_{\text {trube }}=0.243 \mathrm{mm}} $$
$ {\delta_{\text {rod }}=-\left(4.8505 \times 10^{-9}\right)\left(27.308 \times 10^{3}\right)=-132.5 \times 10^{-6} \mathrm{m}} $
$$\therefore {\delta_{\mathrm{rod}}=-0.1325 \mathrm{mm}} $$
Work Step by Step
$A_{\text {tube }}=\frac{\pi}{4}\left(d_{o}^{2}-d_{i}^{2}\right)=\frac{\pi}{4}\left(36^{2}-28^{2}\right)=402.12 \mathrm{mm}^{2}=402.12 \times 10^{-6} \mathrm{m}^{2}$
$A_{\mathrm{rod}}=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(25)^{2}=490.87 \mathrm{mm}^{2}=490.87 \times 10^{-6} \mathrm{m}^{2}$
$\delta_{\text {tube }}=\frac{P L}{E_{\text {tube }} A_{\text {tube }}}=\frac{P(0.250)}{\left(70 \times 10^{9}\right)\left(402.12 \times 10^{-6}\right)}=8.8815 \times 10^{-9} P$
$\delta_{\text {rod }}=-\frac{P L}{E_{\text {rod }} A_{\text {rod }}}=\frac{P(0.250)}{\left(105 \times 10^{6}\right)\left(490.87 \times 10^{-6}\right)}=-4.8505 \times 10^{-9} P$
$\delta^{*} =\left(\frac{1}{4} \text { turn }\right) \times 1.5 \mathrm{mm}=0.375 \mathrm{mm}=375 \times 10^{-6} \mathrm{m} $
$\delta_{\text {tube }} =\delta^{*}+\delta_{\text {rod }} \text { or } \delta_{\text {tube }}-\delta_{\text {rod }}=\delta^{*} $
$ 8.8815 \times 10^{-9} P+4.8505 \times 10^{-9} P =375 \times 10^{-6} $
$P =\frac{0.375 \times 10^{-3}}{(8.8815+4.8505)\left(10^{-9}\right)}=27.308 \times 10^{3} \mathrm{N} $
(a)
${\sigma_{\text {tube }}=\frac{P}{A_{\text {tube }}}} {=\frac{27.308 \times 10^{3}}{402.12 \times 10^{-6}}=67.9 \times 10^{6} \mathrm{Pa}} $
$$ \therefore {\sigma_{\text {tube }}=67.9 \mathrm{MPa}} $$
$ {\sigma_{\text {rod }}} {=-\frac{P}{A_{\text {rod }}}=-\frac{27.308 \times 10^{3}}{490.87 \times 10^{-6}}=-55.6 \times 10^{6} \mathrm{Pa}} $
$$\therefore {\sigma_{\text {rod }}=-55.6 \mathrm{MPa}}$$
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(b)
${\delta_{\text {that }}=\left(8.8815 \times 10^{-9}\right)\left(27.308 \times 10^{3}\right)=242.5 \times 10^{-6} \mathrm{m}} $
$$ \therefore {\delta_{\text {trube }}=0.243 \mathrm{mm}} $$
$ {\delta_{\text {rod }}=-\left(4.8505 \times 10^{-9}\right)\left(27.308 \times 10^{3}\right)=-132.5 \times 10^{-6} \mathrm{m}} $
$$\therefore {\delta_{\mathrm{rod}}=-0.1325 \mathrm{mm}} $$
