Question

A semi-circular bar of radius R m, in a vertical plane, is fixed at the end G, as shown in the figure. A horizontal load of magnitude P kN is applied at the end H. The magnitude of the axial force, shear force, and bending moment at point Q for \(\theta = 45^\circ\), respectively, are 
 

• A. \(\frac{P}{\sqrt{2}}\) kN, \(\frac{P}{\sqrt{2}}\) kN, and \(\frac{PR}{\sqrt{2}}\) kNm
• B. \(\frac{P}{\sqrt{2}}\) kN, \(\frac{P}{\sqrt{2}}\) kN, and 0 kNm
• C. 0 kN, \(\frac{P}{\sqrt{2}}\) kN, and \(\frac{PR}{\sqrt{2}}\) kNm
• D. \(\frac{P}{\sqrt{2}}\) kN, 0 kN, and \(\frac{PR}{\sqrt{2}}\) kNm

Hint:

For a semi-circular bar under a horizontal load, the axial force and shear force can be found by resolving the load components, while the bending moment is calculated using the moment arm and the load.

Answer 1

The Correct Option is A

In this problem, we are asked to find the axial force, shear force, and bending moment at point Q for a semi-circular bar subjected to a horizontal load \(P\) at point H. The axial force, shear force, and bending moment can be determined by resolving the forces in the direction of the applied load and using equilibrium equations. - Axial Force: The axial force at point Q is the force component in the direction of the bar, given by the horizontal component of the applied load. The axial force is therefore \( \frac{P}{\sqrt{2}} \). - Shear Force: The shear force at point Q is the vertical component of the applied load, also given by \( \frac{P}{\sqrt{2}} \). - Bending Moment: The bending moment at point Q is due to the horizontal load acting at point H. The moment at Q is calculated as \( \frac{PR}{\sqrt{2}} \), where \(R\) is the radius of the semi-circular bar. Thus, the correct magnitudes of the axial force, shear force, and bending moment at point Q are \( \frac{P}{\sqrt{2}} \) kN, \( \frac{P}{\sqrt{2}} \) kN, and \( \frac{PR}{\sqrt{2}} \) kNm, which corresponds to option (A).