Answer:
a) 116.67 MPa
b) 368.85 GPa
c) -1.0122 *10^{-4}
Explanation:
Given data:
length of bar is 1.50 m
total elongation in bar = 1.22 mm
axial load = 35.0 kN
cross section dimension = 6 by 50 mm
proportional limit = 300 MPa
a) AXIAL STRESS
Stress =force/area
[tex]stress = \frac{35 \times 10^3}{6\times 50} = 116.67 MPa[/tex]
b) modulus of elasticity
[tex]\sigma_y = \epsilon_y \times E[/tex]
where E = modulus of elasticity
[tex]\epsilon = \frac{\delta L}{l_o}[/tex]
[tex]\epsilon = \frac{1.22}{1.5 \times 10^3} = 0.8133 \times 10^{-3}[/tex]
[tex]E= \frac{300}{0.8133 \times 10^{-3}}[/tex]
E = 36.8.85 GPa
c) [tex]\epsilon_y = \frac{-\mu \sigma}{E}[/tex]
where [tex]\mu[/tex] is poisson's ration= 0.32
[tex] \epsilon_y = -0.32 \times \frac{116.67}{368.85 \times 10^{-3}}[/tex]
[tex]\epsilon_y = - 1.0122 \times 10^{-4}[/tex]
