Question

Example: Since the safe shear stress value for the pipe in the figure is \( 120 \mathrm{MPa} \), the safety the torsional moment that can be applied within the limits and corresponding to this moment find the minimum shear stress. \[ T_{\max }=\frac{\mathrm{T} \cdot \mathrm{c}}{\mathrm{J}} \longrightarrow T=\frac{J \tau_{\max }}{c} \] a: radius The polar moment of inertia of the hollow section is:

          Example:
Since the safe shear stress value for the pipe in the figure is \( 120 \mathrm{MPa} \), the safety the torsional moment that can be applied within the limits and corresponding to this moment find the minimum shear stress.
\[
T_{\max }=\frac{\mathrm{T} \cdot \mathrm{c}}{\mathrm{J}} \longrightarrow T=\frac{J \tau_{\max }}{c}
\]
a: radius
The polar moment of inertia of the hollow section is:

Example:
Since the safe shear stress value for the pipe in the figure is 120 MPa, the safety the torsional moment that can be applied within the limits and corresponding to this moment find the minimum shear stress.

Tmax=(T·c)/(J)⟶ T=(J τmax)/(c)

a: radius
The polar moment of inertia of the hollow section is:

Added by Dilvin S.

Mechanics of Materials

R. C. Hibbeler 9th Edition

Chapter 8

Instant Answer

Solved by Expert Abid Hussain

Step 1

First, we need to find the polar moment of inertia (J) for the hollow section. The formula for the polar moment of inertia for a hollow circular section is: \[ J = \frac{\pi}{2} (R^4 - r^4) \] where R is the outer radius and r is the inner radius. Show more…

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