Question

A massless semicircular rod held fixed at end A is in the xy-plane as shown. A force P along the negative z direction is acting at point B on the rod. The unit vectors along x, y and z directions are denoted by i, j and k. Due to the applied force P, the cross-section of the rod at point D will be subjected to

• A. a twisting moment $PR(1-\cos\theta)\,\mathbf{i}$, a bending moment $PR\sin\theta\,\mathbf{j}$, and a shear force $-P\,\mathbf{k}$
• B. a twisting moment $PR(1-\sin\theta)\,\mathbf{i}$, a bending moment $PR\cos\theta\,\mathbf{j}$, and a shear force $P\,\mathbf{k}$
• C. a twisting moment $PR(\cos\theta-1)\,\mathbf{i}$, a bending moment $-PR\sin\theta\,\mathbf{j}$, and a shear force $-P\,\mathbf{k}$
• D. a twisting moment $PR\sin\theta\,\mathbf{i}$, a bending moment $PR(1-\cos\theta)\,\mathbf{j}$, and a shear force $P\,\mathbf{k}$

Hint:

While analyzing curved beams, always decompose the position vector into tangential and radial components to determine twisting and bending moments correctly.

Answer 1

The Correct Option is A

To determine the internal forces at point D, we express the position vector of point B relative to D along the semicircle. The force at B is $-P\mathbf{k}$ (negative z-direction).
The twisting moment is produced by the tangential component of the moment arm, while the bending moment arises from the radial component. The shear force is simply the applied force resolved at the section.
Evaluating the moment arm components along the semicircular arc gives the twisting moment term proportional to $R(1-\cos\theta)$ or $R(1-\sin\theta)$ depending on the coordinate choice. These correspond exactly to options (A) and (B).
Thus, both (A) and (B) represent correct possible expressions for the internal twisting moment and bending moment depending on the geometric definition of the angle $\theta$.