Question

A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel (Es = 200 GPa and αs = 11.7 × 10-6/°C) and portion BC is made of brass (Eb = 105 GPa and αb = 20.9 × 10-6/°C). Knowing that the rod is initially unstressed, determine the compressive force induced in ABC when there is a temperature rise of 55°C.

Answer 1

Complete question:

A rod consisting of two cylindrical portions AB  and BC is restrained at both ends. Portion AB is made of steel (Es = 200 GPa and αs = 11.7 × 10-6/°C) and portion BC is made of brass (Eb = 105 GPa and αb = 20.9 × 10-6/°C). (diameter of AB is 30mm, Length of AB is 250 mm) and (diameter of BC is 50mm, Length of BC is 300 mm).

Knowing that the rod is initially unstressed, determine the compressive force induced in ABC when there is a temperature rise of 55°C

Answer:

The compressive force induced in ABC is 156.902 kN

Explanation:

Area of portion AB:

$A_{AB} = \frac{\pi }{4}d^2_{AB} = \frac{\pi }{4} (30)^2 = 706.95 mm^2$= 7.0695 X 10⁻⁴ m²

Area of portion BC:

$A_{BC} = \frac{\pi }{4}d^2_{BC} = \frac{\pi }{4} (50)^2 = 1963.75 mm^2$ = 1.96373 X 10⁻³ m²

Free thermal expansion:

$\delta_T = L_{AB}\alpha_s(\delta T) +L_{BC}\alpha_b(\delta T)$ = (0.25)(11.7 X 10⁻⁶)(55) +  (0.3)(20.9 X 10⁻⁶)(55)

= 505.725 X 10⁻⁶m

Shortening due to induced compressive force P:

$\delta_p = \frac{PL}{E_SA_{AB}} + \frac{PL}{E_bA_{BC}}$

$\delta_p = \frac{0.25P}{(200 X10^9)(7.0695 X10^{-4})} + \frac{0.3P}{(105 X10^9)(1.96373 X10^{-3})}$

    = 1.7682 X 10⁻⁹P + 1.455 10⁻⁹P

    = 3.2232 X 10⁻⁹P

For zero net deflection, $\delta _p = \delta _T$

∴ 3.2232 X 10⁻⁹P = 505.725 X 10⁻⁶

$P = \frac{505.725 X 10^{-6}}{3.2232 X 10^{-9}}$

P = 156.902 X 10³ N = 156.902 kN

Therefore, the compressive force induced in ABC is 156.902 kN