A steel specimen is tested in tension. The specimen is 1.0 in. wide by 0.25 in. thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 12.5 kips and fractured at 17.5 kips.

Determine the tensile stresses at yield and at fracture.

The tensile stresses at yield and at fracture for the steel specimen are 50 kips/in² and 70 kips/in², respectively. This calculation uses the specimen's cross-sectional area and the loads applied at yield and fracture.

Explanation:

The question involves calculating the tensile stresses at yield and at fracture for a steel specimen tested in tension. The given dimensions of the specimen are 1.0 in. wide and 0.25 in. thick, and the applied loads at yield and fracture are 12.5 kips and 17.5 kips, respectively.

To calculate the stress, the formula σ = F/A is used, where σ is the stress, F is the force, and A is the cross-sectional area. The area (A) of the specimen can be calculated as 1.0 in. × 0.25 in. = 0.25 in.². Therefore, the tensile stress at yield (σ⁰) is 12.5 kips / 0.25 in.² = 50 kips/in.², and the tensile stress at fracture (σₒ) is 17.5 kips / 0.25 in.² = 70 kips/in.².