Question

A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 84998 N , and experiences an axial elongation of 0.18 mm . Assuming that the deformation is entirely elastic, calculate the elastic modulus of this steel in GPa.Answer Format: X (no decimal places)

Answer 1

227 Gpa

Step-by-step explanation:

∆L = PL/AE

E = PL/A∆L

E is Elastic Modulus

L is length

A is Area

L = 300 mm = 300* 10^-3

A = (25 * 10 ^-3)^2

P = 84998N

∆L = 0.18mm = 0.18*10^-3

E = 84998*300*10^-3/((25*10^-3)^2*0.18*10^-3

E = 226661333333.3Pa

= 226.7 * 10^9Pa

10^9Pa = 1 GPA

E = 226.7 Gpa

E = 227 no decimal

Answer 2

Elastic Modulus = 227 GPa

Step-by-step explanation:

Given,

Load = 84998 N

Length of bar = 300 mm = 0.3 m

Elongation = 0.18 mm = 0.00018 m

Cross sectional Area of the bar = (25mm × 25mm) = 0.025 × 0.025 = 0.000625 m²

From Hooke's law, the stress experienced by a material is proportional to the strain experienced by the same body, as long as the elastic limit isn't exceeded.

Stress ∝ strain

The coefficient of proportionality is the elastic modulus, E.

Stress = E × (Strain)

Stress = (Load)/(Cross sectional Area)

Stress = (84998 ÷ 0.000625) = 135,996,800 N/m²

Strain = (Change in length)/(Original length)

Strain = (ΔL/L) = 0.00018 ÷ 0.3 = 0.0006

E = (Stress/Strain)

E = 135,996,800 ÷ 0.0006 = 226,661,333,333.3 Pa = (2.267 × 10¹¹) Pa

1 GPa = 10⁹ Pa

(2.267 × 10¹¹) = 2.267 × 10² × 10⁹ = 226.7 GPa = 227 GPa to the nearest GPa. (No decimal place)

Hope this Helps!!!