Question

Maximum shear stress in a thin cylindrical shell subjected to internal pressure \( p \) is

• A. \( \dfrac{pd}{t} \)
• B. \( \dfrac{pd}{2t} \)
• C. \( \dfrac{pd}{4t} \)
• D. \( \dfrac{pd}{8t} \)

Hint:

In thin cylinders, maximum shear stress occurs midway between hoop and longitudinal stresses.

Answer 1

The Correct Option is C

Step 1: Identifying principal stresses.
In a thin cylindrical shell subjected to internal pressure, the two principal stresses are:
Hoop stress: \[ \sigma_h = \frac{pd}{2t} \] Longitudinal stress: \[ \sigma_l = \frac{pd}{4t} \]
Step 2: Formula for maximum shear stress.
Maximum shear stress is given by: \[ \tau_{\max} = \frac{\sigma_h - \sigma_l}{2} \]
Step 3: Substituting values.
\[ \tau_{\max} = \frac{\frac{pd}{2t} - \frac{pd}{4t}}{2} \] \[ \tau_{\max} = \frac{pd}{4t} \]
Step 4: Conclusion.
The maximum shear stress in the cylindrical shell is \( \dfrac{pd}{4t} \).

Answer 2

In a thin cylindrical shell under internal pressure, the three principal stresses at a point on the wall are the hoop stress \( \sigma_h = \dfrac{pd}{2t} \), the longitudinal stress \( \sigma_l = \dfrac{pd}{4t} \), and the radial stress, which is approximately zero for a thin wall (negligible compared to the other two). The true maximum shear stress is half the difference between the largest and smallest of these three principal stresses, not just the two in-plane ones. Testing each option:

1. \( \dfrac{pd}{t} \): This is twice the hoop stress itself and does not correspond to any half-difference of principal stresses in this problem, so it is too large to be a shear stress here.
2. \( \dfrac{pd}{2t} \): This equals the hoop stress itself, a normal stress, not a shear stress; the maximum shear stress must be half the difference between two principal stresses, not a principal stress value on its own.
3. \( \dfrac{pd}{4t} \): Taking the largest principal stress as the hoop stress \( \dfrac{pd}{2t} \) and the smallest as the (near-zero) radial stress, the true maximum shear stress is \( \dfrac{1}{2}\left(\dfrac{pd}{2t} - 0\right) = \dfrac{pd}{4t} \). This correctly accounts for all three principal stresses, not just the two in-plane ones.
4. \( \dfrac{pd}{8t} \): This is the shear stress obtained only from the difference between the hoop and longitudinal stresses, \( \dfrac{1}{2}\left(\dfrac{pd}{2t}-\dfrac{pd}{4t}\right) \), which is the maximum in-plane shear stress on the plane containing those two directions, but not the true (absolute) maximum shear stress once the near-zero radial stress is included in the comparison.

Since the true maximum shear stress must be measured between the largest and smallest of all three principal stresses, and the radial stress is effectively zero, the correct value is \( \dfrac{pd}{4t} \).

Therefore, the correct answer is \( \dfrac{pd}{4t} \).