Question

Two gage marks are placed exactly 320 mm apart on a 12-mm diameter solid metal rod. When an axial tension force of 15 kN is applied to the rod, the distance between the gage marks is precisely measured as 320.5250 mm. Determine the modulus of elasticity of metal in Units of GPa.

Answer 1

Answer:

E = 80.381 GPa

Explanation:

given,

Distance between two gauges, L = 320 mm

diameter,d = 12 mm

Load, P = 15 x 10³ N

distance between gauge marks = 320.5280 mm

Now, elongation of rod is equal to

δ = 320.5280 - 320 = 0.5280 mm

Using elongation formula

$\delta = \dfrac{PL}{AE}$

$E = \dfrac{PL}{A\delta}$

$E = \dfrac{15\times 10^3\times 320}{\dfrac{\pi}{4}\times d^2\times 0.5280}$

$E = \dfrac{15\times 10^3\times 320}{\dfrac{\pi}{4}\times 12^2\times 0.5280}$

E = 80381.28 MPa

E = 80.381 x 10³ MPa

E = 80.381 GPa

Hence, Modulus of elasticity of the metal is equal to E = 80.381 GPa