## Unpressurized Vessel

The below cross section is of an unpressurized tank vessel with a modulus of elasticity of 200x10^3 MPa, a Poisson’s ratio of 0.25, a coefficient of thermal expansion of 0.000012 m/m°C, and a yield strength of 350 MPa. If the internal pressure were to increase to P such that the wall stresses between Points A & B are below, what is the increased length (along the outer wall) between Points A & B?

##
__
__**Hint**

**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**

**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.

$$$\varepsilon_{x}=\frac{1}{200\cdot 10^3MPa}[24MPa-(0.25)(50MPa+0MPa)]$$$

$$$=\frac{1}{200\cdot 10^3MPa}[24MPa-12.5MPa]=\frac{11.5}{200\cdot 10^3}=5.75\cdot 10^{-5}$$$

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.

$$$5.75\cdot 10^{-5}=\frac{\Delta L}{1,250mm}$$$

$$$\Delta L =5.75\cdot 10^{-5}(1,250mm)=0.072\:mm$$$

0.072 mm