Tensile Load change in mm?

Question

A steel tube 30 mm outside diameter and 25 mm internal diameter and length 500 mm is subjected to a tensile load of 39 KN, determine the change in length in mm to 3 decimal places when the load was applied. For the material E=210 GPa

Answer 1

Given:

Outside diameter of the steel tube, d_O = 30 mm = 0.03 m

Inside diameter of the steel rod, d_i = 25 mm = 0.025 m

Length of the steel tube, l = 500 mm = 0.5 m

Tensile load, P = 39 kN

Young's modulus , E = 210 GPa =210×10⁶ kPa

Calculate the area of the rod by using the following formula:

A = (π/4)( ( d_O ) ² -( d_i ) ² )

    =(π/4) ( ( 0.03 ) ² - ( 0.025 ) ²)

    =0.000216 m²

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Calculate the change in length of the rod by using the following formula:

δl = Pl/AE

    =(39)(0.5)/(0.000216)(210×10⁶ )

    =4.3014×10^-4 m

    =0.43014 mm

Therefore, the change in length of the rod is 0.43014 mm