Answer:
(a) 2411.039 kN
(b) 0.04676 mm
(c) 0.0267 mm
Explanation:
<u>Given information</u>
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
Learn more: brainly.com/question/14288907