Axial Stress
Consider an air cylinder’s pressure gauge reads 3,000 kPa. If the cylinder is made of a 5 mm steel rolled plate, with an internal diameter of 600 mm, what is the axial stress in MPa?
Expand Hint
The cylinder can be considered thin-walled if:
$$$t\leq \frac{d_i}{20}$$$
where
$$t$$
is the wall thickness, and
$$d_i$$
is the inner diameter.
Hint 2
Axial stress:
$$$\sigma_a=\frac{P_ir}{2t}$$$
where
$$P_i$$
is the internal pressure,
$$r$$
is the mean radius, and
$$t$$
is the wall thickness.
The cylinder can be considered thin-walled if
$$t\leq \frac{d_i}{20}$$
where
$$t$$
is the wall thickness, and
$$d_i$$
is the inner diameter.
$$$t\leq \frac{600mm}{20}=30\:mm$$$
Since
$$5\:mm< 30\:mm$$
, the cylinder is thin-walled.
The axial stress formula for thin-walled vessels is:
$$$\sigma_a=\frac{P_ir}{2t}$$$
where
$$P_i$$
is the internal pressure,
$$r$$
is the mean radius, and
$$t$$
is the wall thickness.
$$$r=\frac{r_{inner}+r_{outer}}{2}=\frac{(600mm/2)+[(600mm/2)+5mm]}{2}$$$
$$$=\frac{300mm+305mm}{2}=302.5mm$$$
Thus, the axial stress is:
$$$\sigma_a=\frac{(3MPa)(302.5mm)}{2(5mm)}=\frac{907.5MPa}{10}=90.75\:MPa$$$
90.75 MPa
Time Analysis
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