To use hooke’s law to determine the value of an axial load applied to a rectangular shape and to determine its lateral expansion.

the element shown below is subjected to the axial compressive force p, which causes the shape to contract laterally in the x direction a distance of δx = 0.270 mm . let h = 144.5 mm , w = 82.50 mm , and l = 267.0 mm . assume the element is made from steel that has a modulus of elasticity of e=200 gpa and a value of poisson’s ratio of v=0.32.(figure 1)

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part a - value of the axial load

determine the value of the axial load p.

part b - lateral expansion in the y direction

determine the lateral expansion in the y direction due to the axial load.

part c - lateral expansion in the z direction

determine the lateral expansion in the z direction due to the axial load.

(a) 2411.039 kN

(b) 0.04676 mm

(c) 0.0267 mm

Explanation:

Given information

h=144.5 mm

w=82.5 mm

L=267.0 mm

v=0.32

The figure not given but I assumed the attached figure

Applying Hooke’s law

 where P is axial load, L is length, A is the area, is the stress along x-direction and E is the Young’s modulus of elasticity and making P the subject of the formula then

By substituting the given values

Therefore, the value of axial load is 2411.039 kN

(b)

Poison’s ratio, v is the ratio of lateral strain to longitudinal strain hence

Lateral strain in y-direction=-v* longitudinal strain

By substituting the given values then

Therefore, the lateral expansion in the y direction due to axial load is 0.04676 mm

(c)

In the z-direction, using the given poison’s ratio then

Lateral strain in z-direction=-v * Longitudinal strain

Substituting the information provided above

Therefore, the lateral expansion in the z-direction due to axial load is 0.0267 mm

Keywords: Stress, strain, Hooke's law, Poison's ratio

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can you make it more obvious. i can't see it properly