Question

A solid steel cylinder 500 mm long and 70 mm diameter is placed inside an aluminium cylinder having 75 mm inside diameter and 100 mm outside diameter. The aluminium cylinder is 0.16 mm. longer than the steel cylinder. An axial load of 500kN is applied to the bar and cylinder through rigid cover plates as shown in Fig.. Find the stresses developed in the steel cylinder and aluminium tube. Assume for steel, E = 220 GN/m² and for Al E = 70 GN/m²

Answer 1

Since the aluminium cylinder is 0.16 mm longer than the steel cylinder, the load required to compress this cylinder by 0.16 mm will be found as follows : 

E = stress/ strain = P.L/A.\(\delta\)L 

Or P = E.A.\(\delta\)L/L 

= 70 × 10⁹ × \(\pi\)/4 (0.1² - 0.075²) × 0.00016/0.50016 = 76944 N 

When the aluminium cylinder is compressed by its extra length 0.16 mm, the load then shared by both aluminium as well as steel cylinder will be, 

500000 – 76944 = 423056 N

Let e_s = strain in steel cylinder 

e_a = strain in aluminium cylinder 

\(\sigma\)_s = stress produced in steel cylinder 

\(\sigma\)_a = stress produced in aluminium cylinder 

E_s = 220 GN/m² 

E_a = 70 GN/m²

As both the cylinders are of the same length and are compressed by the same amount

e_s = e_a 

\(\sigma\)_s /E_s =\(\sigma\)_a /E_a 

or; \(\sigma\)s = E_s/E_a.\(\sigma\)_a= (220 × 10⁹ /70 × 10⁹ ). 

\(\sigma\)_a = (22/7). \(\sigma\)_a

Also P_s + P_a = P 

or; \(\sigma\)_s As + \(\sigma\)_a. A_a = 423056 (22/7). 

 As + \(\sigma\)a A_a = 423056 ...(i) 

A_s = \(\pi\)/4 (0.07² ) = 0.002199 m² 

A_a = \(\pi\)/4 (0.12 – 0.075² ) = 0.003436 m²

Putting the value of A_s and A_a in equation (i) we get

\(\sigma\)_a = 27.24 × 10⁶ N/m² = 27.24 MN/m² 

\(\sigma\)_S = 22/7 × 27.24 = 85.61 MN/m²

Stress in the aluminium cylinder due to load 76944 N

= 76944/ \(\pi\)/4 (0.12 - 0.0752) 

= 23.39 × 10⁹ N/m² = 22.39 MN/m²

Total stress in aluminium cylinder 

= 27.24 + 22.39 = 49.63 MN/m² 

and stress in steel cylinder = 85.61 MN/m²