Hooke’s Law

#573 / Mechanics of Materials Medium

The shown cross section is of an unpressurized tank vessel with a modulus of elasticity of 100x10^3 MPa, a Poisson’s ratio of 0.3, a coefficient of thermal expansion of 0.000012 m/m°C, and a yield strength of 300 MPa. If the internal pressure were to increase to P such that the wall stresses between Points A & B are below, what is the original length (along the outer wall) between Points A & B in meters if the measured change in length is 0.1 mm?

Expand Hint
For longitudinal strain without a temperature rise:
$$$\varepsilon_{x}=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)]$$$
where $$E$$ is the modulus of elasticity, $$\nu$$ is Poisson’s ratio, and $$\sigma$$ is the normal stress.
Hint 2
Engineering strain:
$$$\varepsilon=\frac{\Delta L}{L_o}$$$
where $$\Delta L$$ is the change in length of member, and $$L_o$$ is the original member length.
For longitudinal strain without a temperature rise:
$$$\varepsilon_{x}=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)]$$$
where $$E$$ is the modulus of elasticity, $$\nu$$ is Poisson’s ratio, and $$\sigma$$ is the normal stress.
$$$\varepsilon_{x}=\frac{1}{100\cdot 10^3MPa}[23MPa-(0.3)(55MPa+0MPa)]$$$
$$$=\frac{1}{100\cdot 10^3MPa}[23MPa-16.5MPa]=\frac{6.5}{100\cdot 10^3}=6.5\cdot 10^{-5}$$$
For engineering strain:
$$$\varepsilon=\frac{\Delta L}{L_o}$$$
where $$\Delta L$$ is the change in length of member, and $$L_o$$ is the original member length.
$$$L_o=\frac{0.1mm}{6.5\cdot 10^{-5}}=1,538mm=1.5\:m$$$
1.5 m


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