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## Q. 4.22

A hollow steel tube is assembled at 25°C with fixed ends as shown in Fig. 4.70(a). At this temperature, there is no stress in the tube. The length and cross–sectional area of the tube are 200 mm and 300 mm2 respectively. During operating conditions, the temperature of the tube increases to 250°C. It is observed that at this temperature, the fixed ends are separated by 0.15 mm as shown in Fig. 4.70 (b). The modulus of elasticity and coefficient of thermal expansion of steel are 207 000 N/mm² and $10.8 \times 10^{-6}$ per °C respectively. Calculate the force acting on the tube and the resultant stress.

## Verified Solution

$\text { Given } \Delta T=(250-25)^{\circ} C \quad A=300 mm ^{2}$.

l = 200 mm E = 207 000 N/mm².

$\alpha=10.8 \times 10^{-6} \text { per }{ }^{\circ} C$.

joint separation = 0.15 mm
Step I Calculation of expansion of tube
From Eq. (4.68),

$\delta=\alpha l \Delta T$              (4.68).

$\delta=\alpha l \Delta T=\left(10.8 \times 10^{-6}\right)(200)(250-25)=0.486 mm$.

Step II Calculation of net compression of tube
When the tube is free to expand, its length will increase by 0.486 mm. However, the fixed ends are separated by 0.15 mm only. Therefore,
Net compression of tube = 0.486–0.15 = 0.336 mm.

Step III Calculation of force

$\delta=\frac{P L}{A E} \quad \text { or } \quad 0.336=\frac{P(200)}{(300)(207000)}$.

$\therefore \quad P=104328 N$.

Step IV Calculation of resultant stress

$\sigma=\frac{P}{A}=\frac{104328}{300}=347.76 N / mm ^{2}$.