The following stressstrain curve was prepared based on a ten

The following stress-strain curve was prepared based on a tensile test of a specimen that had a circular cross-section. The gage diameter of the specimen was 0.875 inches and the gage length was 5 inches. The stress scale of the stress-strain diagram is given with the factor a = 20 ksi. Estimate:

(a) The modulus of elasticity.
(b) The ultimate strength.
(c) The yield strength (0.2% offset).
(d) The percent elongation at fracture

Solution

From graph,

a) Modulus of elasticity = slope of straight line upto elastic limit = 6.6a / .008 = (6.6X 20) /0.008 = 16500 ksi

b) Ultimate strength is a stress which corresponds to a point on a curve where slope is zero.

From observation, stress is 7.4a = 7.4 X 20 = 148 ksi

c) Yield strength is found by drawing a straight line with a slope of modulus of elasticity with an offset of 0.002 of strain which equals 6.8a = 6.8X 20 = 136 ksi

d) At the fracture point corresponding strain is 0.086.

Hence, % elongation is 8.6% approx.