Question

A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter = 0.8 in. Yielding occurs at a load of 30,000 lb. The corresponding gage length = 1.8075 in, which is the 0.2 percent yield point. The maximum load of 56,050 lb is reached at a gage length = 2.35 in. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 2.5 in, determine the percent elongation. (e) If the specimen necked to an area=0.35 in^2, determine the percent reduction in area.

Answer 1

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Step-by-step explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%