Shear Force And Bending Moment in Strength Of Materials : Practice MCQ Questions & Answers with Explanations

Concept: Net weight of the UDL = wL = W RA + RB = wL Due to symmetry, ${RA} = \frac{{wL}}{2};{RB} = \frac{{wL}}{2}$ Taking a moment about B $M = {RA}x - \frac{{w{x^2}}}{{2}} = \frac{{wL}}{2}x - \frac{{w{x^2}}}{2}$ For maximum B.M, $\frac{{dM}}{{dx}} = 0\;i.e.\;\frac{{wL}}{2} - \frac{{w.2x}}{2} = 0$ $x = \frac{L}{2}$ ${M{max}} = \frac{{wL}}{2} \times \frac{L}{2} - \frac{{w{L^2}}}{8} = \frac{{w{L^2}}}{8}$

Explanation: Different type of support: Fixed support: The fixed support does not allow the rotation and translation motion to the structural members. It offers vertical, horizontal, and moment reactions. Pinned or hinge support: It resists the horizontal and vertical loads but is unable to resist the moment.

Explanation: Shear Force: Shear force at any section is the summation of all transverse loading, either to the left or to the right of any section. Bending Moment: Bending moment at any section is a resultant algebraic sum of moments caused by transverse forces either from the left or right of the section. Relationship between bending moment, shear force, and loading: The rate of change of shear force at any section is equal to load intensity at that section.

Explanation: Bending moment Diagram: Bending moment diagrams are simply plots of the bending moment (on the y-axis) versus the position of various points along the beam (on the x-axis). For UDL(Uniformly Distributed Load) Relationship between Shear force, Bending moment and Loading rate $-(\frac {ds}{dx}) =w$ i.e. negative slope of Shear force diagram at any section will be equal to the load intensity at that section $\frac {dM}{dx} = ~S$ i.e.