Shear Force and Bending Moment Diagrams Questions and Answers

Answer: A

The beam is a cantilever, free at P and fixed at Q, and loaded with a UDL of 3 kN/m. The shear force at a section a distance x from the free end is \(w x\), which grows steadily towards the fixed end and is greatest at the fixed support:

The shear at the free end P is zero, and at the midpoint it is 7.5 kN and at 1 m from the fixed support it is 12 kN, both smaller than the 15 kN at Q.

Answer: C

The bending moment diagram shows the moment at the intermediate support B as 2.4 kN-m with a negative sign, since the moment there is hogging (the shaded region over B is marked negative). The fixing moment at the support is therefore -2.4 kN-m. The positive moments in the figure, 4 kN-m under the 3 kN load and 2 kN-m near the middle of BC, are sagging moments in the spans and are not the moment at the support. The options 1.4 and 1.8 kN-m have no basis in the figure (1.4 is a shear force value), and -2.8 kN-m is not shown.

Answer: C

For a cantilever carrying a uniformly distributed load over its whole length the bending moment diagram is parabolic. At a distance x from the free end the bending moment is \[ M = -\frac{wx^{2}}{2} \] which is a quadratic function of x, so the diagram is a parabola. A point load would give a linear diagram, and an elliptical shape does not arise from this loading.

Answer: A

The reaction at A is larger than that at B.

The triangular load has total magnitude \(\frac{1}{2}q_{0}l\), acting at its centroid, \(l/3\) from A, the heavy end. Taking moments about A and then summing vertical forces:

\[ \begin{aligned} R_{B}\,l &= \tfrac{1}{2}q_{0}l \cdot \tfrac{l}{3} \Rightarrow R_{B} = \tfrac{1}{6}q_{0}l \\ R_{A} &= \tfrac{1}{2}q_{0}l - \tfrac{1}{6}q_{0}l = \tfrac{1}{3}q_{0}l \end{aligned} \]

So \(R_{A} = 2R_{B}\). The reactions would be equal only for a load symmetric about mid-span. A simply supported beam has two unknown vertical reactions and two equilibrium equations, so the reactions are statically determinate.

Answer: A

The correct axial force diagram is (A): 3 kN in AB and zero in BC.

The 5 kN force has a 3:4:5 slope, so its components are 4 kN vertical and 3 kN horizontal, the horizontal part pointing towards A. The roller at C gives only a vertical reaction, so the whole horizontal component must be taken by the pin at A.

A section in AB therefore carries 3 kN of axial force (the force at B pushes towards A, which pushes back), while a section in BC has no horizontal force on either side and carries zero. Diagram (B) wrongly spreads the force over the whole beam, (C) splits it between the supports as if both resisted horizontally, and (D) puts it in the wrong span.

Answer: A

The shear force diagram is Option-I, a constant \(M_{o}/L\) of positive sign.

A couple has no net vertical force, so the support reactions form an equal and opposite pair. Taking moments about B for the counter-clockwise couple gives \(R_{A} = M_{o}/L\) upward and \(R_{B} = M_{o}/L\) downward. These are the only vertical forces, and a concentrated couple changes the bending moment, not the shear force, so the shear is \(M_{o}/L\) along the whole span. With the sign convention shown (upward force on the left face positive) it is positive.

Option II shows a jump in shear at the couple, which only a concentrated force can cause. Options III and IV show sloping shear, which needs a distributed load, and none acts on the beam. The jump at the couple belongs in the bending moment diagram instead.

Answer: A

The equivalent moment is \(M = Pa\).

Moving a force to a new point without changing its effect requires adding a couple equal to the force times the perpendicular distance between the old and new lines of action. P is vertical and acts at a horizontal distance a from the vertical leg, which is fixed at mid-span, so moving it to the beam gives Q = P plus

\[ M = P \times a \]

The height b does not count, because it is measured along the line of action of the vertical force and contributes no moment arm. That rules out Pb and P(a + b), and M = 0 would hold only if P acted directly above the attachment point.

Answer: B

The bending moment at R is Fl.

The roller at Q gives only a horizontal reaction H, along the member QR. Cut the frame just at R and take moments of the forces on the part Q-R about R:

• H acts along the line of QR, so it has no moment about R.
• The load F acts at the midpoint of QR, a distance l from R, giving a moment Fl.

So \(|M_{R}| = Fl\). (For completeness, moments about P give H = F/3, but its value does not affect the moment at R.)

Zero would need F to act at R, and 2Fl or 3Fl take lever arms of 2l or 3l, the length of QR or the frame's height, instead of the distance from F to R.

Answer: B

The point of contraflexure lies where the bending moment is zero or changes its sign.

Contraflexure is the point where the beam's curvature reverses, from sagging to hogging or the reverse. Since curvature is proportional to bending moment, this happens where the bending moment passes through zero and changes sign, as in overhanging and fixed beams.

A change in the sign of shear force, or zero shear force, marks the point of maximum bending moment, which is the opposite situation, so the shear force options and 'maximum bending moment' are wrong.

Answer: B

The maximum bending moment is \(\frac{wl^{2}}{9\sqrt{3}}\).

The total load is \(wl/2\), acting at \(2l/3\) from the zero end, so the reactions are \(R_{A} = wl/6\) at the zero end and \(R_{B} = wl/3\). The shear force is zero where \(wl/6 = wx^{2}/2l\), that is at \(x = l/\sqrt{3}\). There

\[ \begin{aligned} M &= \frac{wl}{6}x - \frac{wx^{3}}{6l} \\ &= \frac{wl^{2}}{6\sqrt{3}} - \frac{wl^{2}}{18\sqrt{3}} = \frac{wl^{2}}{9\sqrt{3}} \end{aligned} \]

\(wl^{2}/8\) is the result for a uniform load w over the whole span, which carries twice this total load. \(wl^{2}/4\) and \(wl^{3}/4\) match no standard case, and the last has the wrong dimensions.