Lecture 25 Deformation Example Problem

Lecture 25
Example Problem

Deformation

a 100 inch long bar with a 10 kip load applied axially in tension

Given:
the 100 inch 1/2 inch diameter steel bar loaded in axial tension with the 10k load (P) as shown.
Determine:
the elongation of the bar.
Solution:
Knowing that the Modulus of Elasticity = stress/strain, and that strain is delta L/L, it is possible to solve for delta L. The Modulus of Elasticity of normal grades of steel is 29,000 ksi. Now, write out the full equation for the Modulus of Elasticity.

E = stress/strain

29,000 ksi = (P/A) / (delta L/L)

Because we are solving for delta L, rearrange the equation to isolate delta L.

delta L = (P)(L) / (A)(E)

Finally, substitute all the known values to solve for delta L;

delta L = (10k)(100in) / 0.196 in²) (29,000 ksi)

delta L = 0.176 in

Lecture 25 Example Problem

Lecture 25
Example Problem