Answer:
(a) 2411.039 kN
(b) 0.04676 mm
(c) 0.0267 mm
Explanation:
Given information
[tex]\sigma_x= 0.270 mm[/tex]
h=144.5 mm
w=82.5 mm
L=267.0 mm
[tex]E=200Gpa= 200\times 10^{3} N/mm^{2}[/tex]
v=0.32
The figure not given but I assumed the attached figure
Applying Hooke’s law
[tex]\sigma_x=\frac {PL}{AE}[/tex] where P is axial load, L is length, A is the area, [tex]\sigma_x[/tex] is the stress along x-direction and E is the Young’s modulus of elasticity and making P the subject of the formula then
[tex]P=\frac {\sigma_x AE}{L}=\frac {\sigma_x (h\times w) E}{L}[/tex]
By substituting the given values
[tex]\frac {(0.270)(144.5\times 82.5)(200\times 10^{3})}{267}= 2411039 N= 2411.039 kN[/tex]
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
[tex]\frac {\triangle h}{h}=-v\frac {\sigma_x}{L}[/tex]
By substituting the given values then
[tex]\frac {\triangle h}{144.5}=-0.32(\frac {-.270}{267})[/tex]
[tex]\triangle h= 0.0467596 mm\approx 0.04676 mm[/tex]
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
[tex]\frac {\triangle w}{w}=-v\frac {\sigma_x}{L}[/tex]
Substituting the information provided above
[tex]\frac {\triangle w}{82.5}=-0.32(\frac {-0.270}{267})[/tex]
[tex]\triangle w= 0.0266966 mm\approx 0.0267 mm[/tex]
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: https://brainly.com/question/14288907
