Shear/Moment Diagrams for an Overhanging Beam with a Tip Moment: 04-BS-6, Undated paper, Q3

Question 3: Shear/Moment Diagrams for an Overhanging Beam with a Tip Moment

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Span A–B (pin at A, roller at B)	6 m
Overhang B–C (free end at C)	4 m
UDL $w$ (over A–B)	5 kN/m
Applied moment at C	60 kN·m, clockwise

Find. $V(x)$ and $M(x)$ over the full 10 m length; the SFD and BMD with $M_{max}^{+}$, $M_{max}^{-}$, and any inflection point identified.

Beam A-B-C: pin A, roller B (6 m), overhang to free end C (4 m); 5 kN/m over A-B; 60 kN.m CW at C.

Approach. Find reactions from global equilibrium (the applied end moment enters directly, independent of position), then integrate the load to get $V(x)$ and $M(x)$ in the two regions (loaded span, unloaded overhang).

Reactions. Taking moments about A (CCW+, clockwise applied moment enters as $-60$): $$\sum M_A=0:\ 6B_y-(5)(6)(3)-60=0 \Rightarrow B_y=25\text{ kN}$$ $$\sum F_y=0:\ A_y=5(6)-25=5\text{ kN}$$
Region 1, $0\le x\le6$ m (measured from A). $$V(x)=A_y-wx=5-5x\ \text{kN}$$ $$M(x)=A_y x-\tfrac{w}{2}x^2=5x-2.5x^2\ \text{kN}\!\cdot\!\text{m}$$ $V=0$ at $x=1$ m, giving the local maximum positive moment $M(1)=2.5$ kN·m. At $x=6^-$: $V=-25$ kN, $M(6)=-60$ kN·m; the roller reaction $B_y=+25$ kN then brings $V$ back to $0$ just past B.
Region 2, $6\le x\le10$ m (the overhang carries no load). $$V(x)=0,\qquad M(x)=-60\ \text{kN}\!\cdot\!\text{m (constant)}$$ This constant value is confirmed independently from the free body to the RIGHT of any cut in this region (just the 60 kN·m clockwise tip moment, giving internal $M=-60$ kN·m by equilibrium of that small free body) — consistent with the value carried through from Region 1 at $x=6$, so $M(x)$ has no jump at B (only $V$ jumps, from the roller force).
Inflection point. $M(x)=5x-2.5x^2=0\Rightarrow x=0$ or $x=2$ m — curvature reverses at $x=2$ m. $$\boxed{M_{max}^{+}=2.5\text{ kN}\cdot\text{m at }x=1\text{ m};\quad M_{max}^{-}=-60\text{ kN}\cdot\text{m from }x=6\text{ to }10\text{ m};\quad \text{inflection at }x=2\text{ m}}$$

Result	Value
$A_y$, $B_y$	5 kN, 25 kN
$V(x)$	$5-5x$ (0–6 m); $0$ (6–10 m)
$M(x)$	$5x-2.5x^2$ (0–6 m); $-60$ (6–10 m)
$M_{max}^{+}$ / $M_{max}^{-}$	$+2.5$ kN·m at $x=1$ m / $-60$ kN·m over the overhang
Inflection point	$x=2$ m

Topic: Shear-force and bending-moment diagrams for a beam with an overhang and an end couple.

Key relations: $dV/dx=-w(x)$; $dM/dx=V(x)$; a concentrated moment applied at a point produces a JUMP in $M(x)$ passing that point but no jump in $V(x)$ (the opposite of a point force).

Why this works: Because the applied couple sits exactly at the free tip, taking the free body from the cut to the tip contains only that couple — so $M(x)$ is a constant $-60$ kN·m throughout the whole overhang, with no gradient (matching $V=0$ there). The loaded span behaves as an ordinary simply-supported segment for $0\le x\le6$.

Common pitfall: Forgetting that a beam's inflection point (where curvature/concavity reverses, $M=0$) is not the same location as $M_{max}$; also, mishandling the sign of an applied end moment in the reaction equation.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 6 §6.1–6.2 (Shear and Moment Diagrams).

Question 3 of 8: Shear/Moment Diagrams for an Overhanging Beam with a Tip Moment

Question 3: Shear/Moment Diagrams for an Overhanging Beam with a Tip Moment