Question 2 of 8: Shear & Moment Diagrams for an Overhanging W460×82 Beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-6: Mechanics of Materials — National Exams, December 2017. Closed-book, 3 hours;
any FIVE of the eight questions constitute a complete paper (all eight are answered below as a
complete study resource). Aid sheet permitted; geometric properties of
W-shape sections were supplied as an appendix (not required for any of the eight questions solved
here).
Pin at A (x=0), roller at B (x=6 m); overhang B→C = 4 m
Find.$V(x)$ and $M(x)$ over the full 10 m beam, with
the diagrams labelled.
Loading diagram with the resulting V(x) and M(x) diagrams for the overhanging beam.
Approach. Find the two reactions from statics (sum of moments taking the
applied 500 kN·m moment as a direct contribution, since a couple's moment about any point
equals itself), then integrate the load intensity segment-by-segment using
$dV/dx=-w(x)$, $dM/dx=V(x)$, cross-checked with an independent free body
from the free end.
Reactions. Taking moments about A (counterclockwise positive; the applied
500 kN·m clockwise couple enters as −500):
$$\sum F_y=0:\ R_A+R_B-20(6)-40=0$$
$$\sum M_A=0:\ R_B(6)-20(6)(3)-40(6)-500=0$$
Solving simultaneously:
$$\boxed{R_A=-23.33\text{ kN} \quad (\text{i.e. }23.33\text{ kN downward}),\qquad R_B=183.33\text{ kN (up)}}$$
Segment A–B, $0\le x\le 6$ m (origin at A, sagging-positive
convention). With only $R_A$ acting up to this point:
$$V(x)=R_A-wx=-23.33-20x\ \text{kN}$$
$$M(x)=R_A x-\frac{w x^2}{2}=-23.33x-10x^2\ \text{kN}\cdot\text{m}$$
At $x=6^-$: $V=-143.33$ kN, $M=-500$ kN·m.
Jump at B ($x=6$ m). The roller reaction (+183.33 kN) and the
40 kN point load (−40 kN) both act here:
$$V(6^+)=-143.33+183.33-40=0\ \text{kN}$$
Because the net shear becomes exactly zero, $M(x)$ stops changing for the rest of the
span.
Segment B–C, $6\le x\le 10$ m. No further transverse loads
act until the tip, so
$$V(x)=0,\qquad M(x)=-500\ \text{kN}\cdot\text{m (constant)}$$
Check at the free end C. The applied 500 kN·m clockwise couple is
itself the last event on the beam; adding it to the internal moment just to its left gives
$$M(10^-) + (+500) = -500+500=0$$
confirming $M=0$ at the true free end, exactly as a free end requires — an
independent cross-check (via a right-side free body through C, not shown) reproduces the same
$V(x)$, $M(x)$ for every $x$, confirming the reactions.
Location
V (kN)
M (kN·m)
x = 0 (A)
−23.33
0
x = 6− (B, left)
−143.33
−500 (global minimum)
x = 6+–10 (B–C)
0
−500 (constant)
x = 10 (C, just before the applied couple)
0
−500 → 0 (after couple)
No true inflection point exists: $M(x)$ is zero only at $x=0$ and is
negative (hogging) everywhere else on the span, so the curvature never reverses sign.
