Thin and Thick Cylinders Questions and Answers

Answer: A

The vessel yields at an internal pressure of about 3.55 MPa.

In a thin cylinder the principal stresses are the hoop stress, the axial stress (half of it) and a negligible radial stress. Tresca says yield occurs when the largest difference of principal stresses reaches the tensile yield strength; here that is hoop minus radial:

\[ \begin{aligned} \sigma_{h} &= \frac{pr}{t} = \frac{p(1800)}{20} = 90p \\ 90p &= 320 \Rightarrow p = 3.55\ \text{MPa} \end{aligned} \]

7.1 MPa uses the hoop-minus-axial difference (45p) as if the radial stress were not the minimum. 1.775 MPa is half the correct value, and 4.0 MPa does not follow from the hoop stress.

Answer: D

The hoop stress rises by about 2.02 %.

For a thin spherical shell the stress is the same in every direction in the wall:

\[ \sigma = \frac{pr}{2t} \]

With the pressure unchanged, the stress scales as r/t:

\[ \frac{\sigma_{new}}{\sigma} = \frac{1.01}{0.99} = 1.0202 \]

So the stress increases by 2.02 %. Adding the two 1 % changes gives 2 %, close but not exact; 1 % counts only one of the two changes; and 0 would need the changes to cancel, but a larger radius and a thinner wall both raise the stress.

Answer: A

The correct plot is (A): \(\sigma_{1}\) constant, \(\sigma_{2}\) rising linearly from zero and equal to \(\sigma_{1}\) at \(y = h/2\).

With \(y\) measured down from the liquid surface, the pressure at depth \(y\) is \(\rho g y\). The hoop (circumferential) stress of a thin cylinder is

\[ \sigma_{2} = \frac{pr}{t} = \frac{\rho g y\, r}{t} \]

which is zero at the surface and grows linearly with depth. Pressure on the vertical walls acts horizontally, so the only vertical load is the liquid's weight pressing on the base, \(\rho g h \cdot \pi r^{2}\). Hanging from the brim, every section of the wall carries that same load:

\[ \begin{aligned} \sigma_{1} &= \frac{\rho g h\,\pi r^{2}}{2\pi r t} \\ &= \frac{\rho g h\, r}{2t} \end{aligned} \]

which does not depend on \(y\). Setting \(\sigma_{2} = \sigma_{1}\) gives \(y = h/2\), and at the base \(\sigma_{2} = 2\sigma_{1}\), exactly as plot (A) shows. (B) swaps the two stresses. (C) and (D) make the axial stress grow with depth, which would need the wall load to change along its height.

Answer: D

The ratio is 2:1.

For a thin cylinder with closed ends, a force balance on a longitudinal section gives the hoop stress, and one on a transverse section gives the longitudinal stress:

\[ \begin{aligned} \sigma_{h} &= \frac{pd}{2t} \\ \sigma_{l} &= \frac{pd}{4t} \end{aligned} \]

so \(\sigma_{h}/\sigma_{l} = 2\). The hoop stress is the larger, which is why thin pipes split along their length. 1:2 inverts the ratio, and 1:1 would hold only for a thin sphere, where the stress is \(pd/4t\) in every direction.

Answer: A

The stresses are \(\sigma_{1} = \frac{pd}{2t}\); \(\sigma_{2} = \frac{pd}{4t}\).

\(\sigma_{1}\) is the hoop (circumferential) stress. Cutting a length \(\ell\) of the cylinder along a diameter, the pressure force \(p d \ell\) is resisted by two walls of area \(t\ell\): \(\sigma_{1} = \frac{pd}{2t}\).

\(\sigma_{2}\) is the longitudinal stress, set up because the ends are closed. The pressure on the end, \(p\frac{\pi d^{2}}{4}\), is carried by the wall ring of area about \(\pi d t\): \(\sigma_{2} = \frac{pd}{4t}\), half the hoop stress.

The option with \(\sigma_{1}\) and \(\sigma_{2}\) swapped assigns the hoop stress to the axial direction. \(\frac{pd}{t}\) forgets that two walls share the hoop load, and \(\sigma_{2} = 0\) would hold only for an open-ended cylinder, but this vessel is closed.

Answer: C

The maximum safe pressure is 1.87 MPa.

For a thin closed cylinder the wall stresses are the hoop stress, the longitudinal stress and, on the inside, a radial stress that is small enough to take as zero:

\[ \begin{aligned} \sigma_{h} &= \frac{pd}{2t} = \frac{300p}{4} = 75p \\ \sigma_{l} &= \frac{pd}{4t} = 37.5p, \quad \sigma_{r} \approx 0 \end{aligned} \]

The Tresca criterion uses the largest shear stress in three dimensions, half the difference between the largest and smallest principal stresses, here \(\sigma_{h}\) and \(\sigma_{r} = 0\):

\[ \begin{aligned} \tau_{max} &= \frac{75p - 0}{2} = 37.5p = 70 \\ \Rightarrow p &= 1.87\ \text{MPa} \end{aligned} \]

3.73 MPa is what the in-plane shear stress \((\sigma_{h} - \sigma_{l})/2 = 18.75p\) alone would allow; it ignores the larger out-of-plane shear. 7.46 and 5.60 MPa would put the absolute maximum shear stress above 70 MPa.

Answer: A

Thin cylindrical shells are designed on the basis of hoop stress.

For internal pressure p, diameter d and thickness t, the hoop stress is \(\frac{pd}{2t}\) and the longitudinal stress \(\frac{pd}{4t}\). The hoop stress is twice as large, so it governs the thickness needed.

Designing on the longitudinal stress, or on any mean of the two, would give a wall too thin to resist the larger hoop stress.

Answer: D

The ratio of hoop stress to longitudinal stress is 2 : 1.

\[ \begin{aligned} \sigma_{h} &= \frac{pd}{2t}, \quad \sigma_{l} = \frac{pd}{4t} \\ \frac{\sigma_{h}}{\sigma_{l}} &= 2 \end{aligned} \]

The hoop stress resists the pressure across a length of the cylinder, while the longitudinal stress resists it only over the circular end, which halves the ratio of force to wall area. 1:1 would hold only for a sphere, where both stresses equal \(\frac{pd}{4t}\).