A steel rod with a diameter of 10 mm passes centrally through a tube made of copper (Figure 1.35). The outer diameter of copper tube is 40 mm and inner diameter is 30 mm. The length of the tube is 2 m. The tube is closed at both ends with 20 mm thick steel plates which are fastened by nuts. The

Question

A steel rod with a diameter of 10 mm passes centrally through a tube made of copper (Figure 1.35). The outer diameter of copper tube is 40 mm and inner diameter is 30 mm. The length of the tube is 2 m. The tube is closed at both ends with 20 mm thick steel plates which are fastened by nuts. The nuts are tightened until the copper tube is shortened by 0.4 mm. Find out the stresses developed in steel rod and copper tube. Also find out the stresses in the two components (rod and tube) when temperature of the assembly is raised by [latex] 60^{\circ} C . E_{S t}=210 GPa , E_{C u}=100 GPa , \alpha_{S t}=12 imes 10^{-6} /{ }^{\circ} C , \alpha_{C u}=17.5 imes 10^{-6} /{ }^{\circ} C [/latex]

Accepted Answer

The area of steel rod is

[latex] ext { Area }=\frac{\pi}{4} imes 10^2=78.54 mm ^2 [/latex]

The area of copper tube is given by

[latex] ext { Area }=\frac{\pi}{4} imes\left(40^2-30^2 ight)=549.78 mm ^2 [/latex]

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 [latex] \sigma_{ Cu } ext { and stress on steel is } \sigma_{ St } ext {, then } [/latex]

[latex] \sigma_{ Cu } imes \frac{\pi}{4} imes\left(40^2-30^2 ight)=\sigma_{ St } imes \frac{\pi}{4} imes(10)^2 [/latex]

or [latex] \sigma_{ Cu }=\sigma_{ St } imes \frac{100}{700} [/latex]

or [latex] \sigma_{ St }=7 \sigma_{ Cu } [/latex] (1)

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

[latex] ext { Strain in copper tube }=\frac{0.4}{2000}=0.0002=∈_{ Cu } [/latex]

Therefore, stress in copper tube is

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

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

[latex] \sigma_{ St }=7 imes 20=140 MPa [/latex]

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:

[latex] \alpha_{ St } \Delta T L_{ St }+\frac{\sigma_{ St }^{\prime}}{E_{ St }} imes L_{ St }=\alpha_{ Cu } \Delta T L_{ Cu }-\frac{\sigma_{ Cu }^{\prime}}{E_{ Cu }} imes L_{ Cu } [/latex] (2)

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

[latex] \begin{aligned} L_{ St } &=2+2 imes 0.02=2.04 m \ L_{ Cu } &=2 m \ \Delta t &=60^{\circ} C \end{aligned} [/latex]

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

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

Now, as thermal load is

[latex] ext { Thermal load }=\sigma_{ St }^{\prime} A_{ St }=\sigma_{ Cu }^{\prime} A_{ Cu } [/latex] (where A is area)

Therefore, we get

[latex] \sigma_{ St }^{\prime}=\sigma_{ Cu }^{\prime} imes \frac{A_{ Cu }}{A_{ St }}=\sigma_{ Cu }^{\prime} imes \frac{549.78}{78.54}=\sigma_{ Cu }^{\prime} imes 7 [/latex] (4)

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

[latex] \sigma_{ St }^{\prime}=50.19 MPa ext { and } \sigma_{ Cu }^{\prime}=7.17 MPa [/latex]

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

[latex] \left(\sigma_{ Cu } ight)_{ ext {overall }}=\underset{\left(\begin{array}{l} ext { Both the terms } \ ext { are compressive } \end{array} ight)}{20+7.17}=\underset{ ext { (Compressive) }}{27.17 MPa } [/latex]

[latex] \left(\sigma_{ St } ight)_{ ext {overall }}=\underset{\left(\begin{array}{l} ext { Both the terms } \ ext { are tensile } \end{array} ight)}{140+50.19}=\underset{ ext { (Compressive) }}{90.19 MPa } [/latex]

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

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

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