Question 2 of 8: Shear and Moment Functions, Overhang Beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2017 — 04-BS-6: Mechanics of Materials (3 hours, closed book, one hand-written aid sheet permitted). Any FIVE of the eight questions constitute a complete paper on the official exam; every question is answered below. A wide-flange (W-shape) section-property table is attached at the end of the official exam; every question below supplies its own built-up or standard section directly, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial members with initial gaps, shear/moment diagrams, deflection by direct integration, transformation of stress via Mohr's circle, combined axial+bending on eccentrically-loaded columns, torsion of stepped shafts, Euler column buckling, unsymmetric-section flexure of brittle materials).
Question 2: Shear and Moment Functions, Overhang Beam (20 marks)
Given. Pin at A (x=0), roller at B (x=6 m), free tip at the overhang end D (x=9 m). A 30 kN/m UDL acts over the main span [0,6] m, a 90 kN·m CCW couple acts at A, and a 50 kN point load acts downward at the tip D.
Quantity
Value
Main span, L1
6 m (A to B)
Overhang
3 m (B to D), total 9 m
UDL, w
30 kN/m over [0,6] m
Applied couple, M0
90 kN·m, CCW, at A
Tip load, P
50 kN down, at D (x=9 m)
Find. V(x), M(x) for 0≤x≤9 m and the corresponding SFD/BMD.
Overhang beam: pin at A, roller at B, UDL over the main span, CCW couple at A, point load at the free tip D.
Approach. Find the two reactions from statics (taking moments about A, with the CCW couple entering directly), then build V(x) and M(x) piecewise by direct integration of the load, checking both end conditions M(0+)=−M0 and M(9)=0.
Reactions. Taking moments about A (CCW+, couple enters as +90):
$$90+R_B(6)-30(6)(3)-50(9)=0\ \Rightarrow\ R_B=150\text{ kN}$$
and $R_A=30(6)+50-R_B=\boxed{80\text{ kN}}$, both upward.
V(x), 0≤x≤6. With $dV/dx=-w$,
$$V(x)=R_A-wx=80-30x\text{ kN}$$
so $V(0^+)=80$ kN and $V(6^-)=-100$ kN. At the roller, V jumps up by $R_B=150$ kN to $V(6^+)=50$ kN, constant to the tip since no load acts on the overhang itself.
M(x), 0≤x≤6. An applied CCW couple at a support drops M by M0 immediately to its right (the pin reaction carries no moment, but the separately-applied external couple still creates an internal jump), so
$$M(x)=R_Ax-\frac{wx^2}{2}-M_0=80x-15x^2-90\text{ kN}\!\cdot\!\text{m}$$
$$M(0^+)=\boxed{-90\text{ kN}\!\cdot\!\text{m}},\quad M(6)=\boxed{-150\text{ kN}\!\cdot\!\text{m}}$$
Local extremum and the overhang. $V=0$ at $x=R_A/w=2.667$ m, where
$$M(2.667)=\boxed{16.67\text{ kN}\!\cdot\!\text{m}}$$
a local (but not global) maximum. On the overhang, $V(x)=50$ kN constant and $M(x)=M(6)+50(x-6)$ for $6\le x\le 9$, giving $M(9)=\boxed{0}$ — the required free-end condition, confirming the reactions.
V(x) and M(x): V drops linearly from 80 to −100 kN over [0,6], jumps to 50 kN at the roller and stays constant to the tip; M jumps to −90 at A, rises to a local peak of 16.7 at x=2.67 m, falls to −150 at the roller, then climbs linearly back to 0 at the tip.
