A cylindrical specimen of a brass alloy 10.4 mm (0.4094 in.) in diameter and 119

Question

A cylindrical specimen of a brass alloy 10.4 mm (0.4094 in.) in diameter and 119.9 mm (4.720 in.) long is pulled in tension with a force of 12700 N (2855 lbf); the force is subsequently released. The tensile stress-strain behavior for this alloy is shown in the graph below.
(a) Compute the final length (mm) of the specimen at this time.
(b) Compute the final specimen length (mm) when the load is increased to 30200 N (6789 lbf) and then released.

A cylindrical specimen of a brass alloy 10.4 mm (0.4094 in.) in diameter and 119.9 mm (4.720 in.) long is pulled in tension with a force of 12700 N (2855 lbf); the force is subsequently released. The tensile stress-strain behavior for this alloy is shown in the graph below. (a) Compute the final length (mm) of the specimen at this time. (b) Compute the final specimen length (mm) when the load is increased to 30200 N (6789 lbf) and then released.

Explanation / Answer

The slope of the elastic region in graph gives the young's modulus of the material

so slope = 250 Mpa / (0.003) {see the graph where line parallel to elastic region is drawn, and compare}

Young's modulus = 83 Gpa

so final length of specimen can be obtained by stress/strain = young's modulus

stress = 12700/(pi * D2/4) = 149.5 Mpa

strain = delta L / L

which gives (delta L )= 0.215 mm , so final length = 119.9 + 0.215 = 120.11mm

b) When a load of 30200N is applied the stress is 355 Mpa,

so there will be permanent deformation with strain 0.075 ( from graph)

therefore deltaL = 0.075 *119.9 = 8.9925mm

Final length = 128.8925mm

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