Question 1.14: A steel rod with a diameter of 10 mm passes centrally throug...

The area of steel rod is

\text { Area }=\frac{\pi}{4} \times 10^2=78.54  mm ^2

The area of copper tube is given by

\text { Area }=\frac{\pi}{4} \times\left(40^2-30^2\right)=549.78  mm ^2

When the nuts are tightened, they would compress the copper tube through the end plates. Thus, the copper tube will be in compression and consequently, the steel rod will be in tension. In the absence of any external load, compressive load on copper will be equal to tensile load on steel.
If stress on copper is \sigma_{ Cu } \text { and stress on steel is } \sigma_{ St } \text {, then }

\sigma_{ Cu } \times \frac{\pi}{4} \times\left(40^2-30^2\right)=\sigma_{ St } \times \frac{\pi}{4} \times(10)^2

or        \sigma_{ Cu }=\sigma_{ St } \times \frac{100}{700}

or        \sigma_{ St }=7 \sigma_{ Cu }          (1)

Now, shortening of copper tube is 0.4 mm. Therefore,

\text { Strain in copper tube }=\frac{0.4}{2000}=0.0002=∈_{ Cu }

Therefore, stress in copper tube is

\sigma_{ Cu }=∈_{ Cu } \times E_{ Cu }=0.0002 \times 100  kN / mm ^2=0.02  GPa =20  MPa

Putting this value in Eq. (1.33), we get

\sigma_{ St }=7 \times 20=140  MPa

So, compressive stress in copper tube and tensile stress in steel rod due to tightening of nut are 20 MPa and 140 MPa, respectively.

When temperature is raised, both copper and steel will expand; and if they are free to expand separately, they will expand by different amounts. However, as they are constrained and must have same length, one will have to contract a bit whereas the other will have to expand a bit; that is, copper (having higher value of ‘α’) will have to undergo compression of some amount whereas steel will have to elongate to maintain the same final length.

Therefore, we can express it mathematically in an equation which is as follows:

\alpha_{ St } \Delta T L_{ St }+\frac{\sigma_{ St }^{\prime}}{E_{ St }} \times L_{ St }=\alpha_{ Cu } \Delta T L_{ Cu }-\frac{\sigma_{ Cu }^{\prime}}{E_{ Cu }} \times L_{ Cu }           (2)

where \sigma_{ St }^{\prime} \text { and } \sigma_{ Cu }^{\prime}  are the stresses induced in steel and copper due to constrained expansion of the system. Now,

\begin{aligned} L_{ St } &=2+2 \times 0.02=2.04  m \\ L_{ Cu } &=2  m \\ \Delta t &=60^{\circ} C \end{aligned}

Substituting these values in Eq. (2), we get

12 \times 10^{-6} \times 60 \times 2.04+\frac{\sigma_{ St }^{\prime}}{210 \times 10^9} \times 2.04=17.5 \times 10^{-6} \times 60 \times 2-\frac{\sigma_{ Cu }^{\prime}}{100 \times 10^9} \times 2          (3)

Now, as thermal load is

\text { Thermal load }=\sigma_{ St }^{\prime} A_{ St }=\sigma_{ Cu }^{\prime} A_{ Cu }             (where A is area)

Therefore, we get

\sigma_{ St }^{\prime}=\sigma_{ Cu }^{\prime} \times \frac{A_{ Cu }}{A_{ St }}=\sigma_{ Cu }^{\prime} \times \frac{549.78}{78.54}=\sigma_{ Cu }^{\prime} \times 7            (4)

Combining Eqs. (3) and (4), we get

\sigma_{ St }^{\prime}=50.19  MPa \text { and } \sigma_{ Cu }^{\prime}=7.17  MPa

Combining these stresses arising due to change in temperature with those due to tightening, we get

\left(\sigma_{ Cu }\right)_{\text {overall }}=\underset{\left(\begin{array}{l} \text { Both the terms } \\ \text { are compressive } \end{array}\right)}{20+7.17}=\underset{\text { (Compressive) }}{27.17  MPa }

\left(\sigma_{ St }\right)_{\text {overall }}=\underset{\left(\begin{array}{l} \text { Both the terms } \\ \text { are tensile } \end{array}\right)}{140+50.19}=\underset{\text { (Compressive) }}{90.19  MPa }

So, overall stresses in copper and steel are 27.17 MPa (compressive) and 190.19 MPa (tensile), respectively.