Appreciate if someone can take the time to check my answers to the following question. If wrong, could you point me in the right direction.

A round bar, 800 mm long, CSA 15 mm^2 experiences an applied tensile load of 6 kN and stretches elastically by 3 mm.

a) Calculate the tensile stress in the bar.
My answer 400 MN/m^2

b) Calculate the strain on the rod.
My answer 0.00375 or 0.375%

c) Calculate the Modulus of Elasticity of the material.
My answer 106.67 GN/m^2

TIA

This should be the subject

The subject of your question seems to be related to mechanics of materials, specifically the concepts of stress, strain, and the modulus of elasticity. Let's go through your answers and verify if they are correct.

a) To calculate the tensile stress in the bar, you need to divide the applied tensile load by the cross-sectional area of the bar. The formula for stress is:

Stress = Force / Area

Given that the applied force is 6 kN (or 6000 N) and the cross-sectional area is 15 mm^2 (or 0.000015 m^2), we can calculate the stress as follows:

Stress = 6000 N / 0.000015 m^2 = 400 MN/m^2

Your answer of 400 MN/m^2 is correct.

b) To calculate the strain on the rod, you need to divide the elongation (change in length) by the original length of the bar. The formula for strain is:

Strain = ΔL / L

Given that the elongation is 3 mm (or 0.003 m) and the original length is 800 mm (or 0.8 m), we can calculate the strain as follows:

Strain = 0.003 m / 0.8 m = 0.00375 or 0.375%

Your answer of 0.00375 or 0.375% is correct.

c) To calculate the modulus of elasticity (also known as Young's modulus), you need to divide the stress by the strain. The formula for modulus of elasticity is:

Modulus of Elasticity = Stress / Strain

Given that the stress is 400 MN/m^2 and the strain is 0.00375, we can calculate the modulus of elasticity as follows:

Modulus of Elasticity = 400 MN/m^2 / 0.00375 = 106.67 GN/m^2

Your answer of 106.67 GN/m^2 is correct.

Overall, your answers are correct. Good job!