Shear and Moment Functions and Diagrams: 04-BS-6, May 2014, Q2

Question 2: Shear and Moment Functions and Diagrams (20 marks)

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.

Simply supported beam: L = 10 m; 20 kN at x = 4 m; UDL 6 kN/m over x ∈ [4, 10] m.

Given.

Quantity	Value
Span L	10 m, pin at A, roller at B
Concentrated load	20 kN, downward, at x = 4 m
UDL w	6 kN/m over x ∈ [4, 10] m

Find. V(x) and M(x) as explicit functions over the whole span; the shear and moment diagrams; the maximum positive moment; any inflection point.

Approach. Find the reactions from statics, then cut the beam at a general x in each of the two load regions and sum forces/moments of everything to the left of the cut.

Reactions. UDL resultant $=6(6)=36$ kN at its centroid $x=4+3=7$ m. Summing moments about A (CCW+): $$R_B(10) - 20(4) - 36(7) = 0 \ \Rightarrow\ R_B=33.2\ \text{kN}, \qquad R_A = 20+36-33.2=22.8\ \text{kN}$$
Region 1 (0 ≤ x ≤ 4 m). $$V(x) = R_A = 22.8\ \text{kN (constant)}, \qquad M(x) = R_A x = 22.8x \quad [\text{kN, kN}\cdot\text{m; }x\text{ in m}]$$
Region 2 (4 ≤ x ≤ 10 m). $$V(x) = R_A - 20 - 6(x-4) = 2.8 - 6(x-4)$$ $$M(x) = R_A x - 20(x-4) - \frac{6(x-4)^2}{2} = 22.8x - 20(x-4) - 3(x-4)^2$$ Check: $M(10) = 22.8(10)-20(6)-3(36) = 228-120-108=0$ (roller) and $V(10)=2.8-36=-33.2=-R_B$, both as required.
Maximum moment, from V(x)=0. In region 2, $V=0$ at $x=4+2.8/6=4.467$ m, giving the maximum sagging moment: $$M(4.467) = 22.8(4.467)-20(0.467)-3(0.467)^2 = \boxed{+91.9\ \text{kN}\cdot\text{m}}$$
Inflection point. $M(x)=22.8x\ge0$ throughout region 1 (zero only at the pin itself), and $M(x)$ in region 2 stays positive all the way to the roller (both supports carry only downward loads, so the beam sags everywhere). $$\boxed{\text{No interior inflection point exists} - M(x)\ge 0 \text{ over the whole span}}$$

Shear force and bending moment diagrams, with the maximum moment labelled; M(x) never changes sign, so there is no inflection point.

Final Results.

Quantity	Value
R_A, R_B	22.8 kN, 33.2 kN
V(x), M(x)	region 1: V=22.8, M=22.8x; region 2: V=2.8−6(x−4), M=22.8x−20(x−4)−3(x−4)²
Maximum positive moment	+91.9 kN·m at x = 4.467 m
Inflection point	none — M(x) ≥ 0 for the entire span

Topic: Direct free-body derivation of shear/moment functions for a beam with a point load and a partial UDL.

Key relations: $V(x)=dM/dx$; the moment function must be split at every load discontinuity, matched for continuity, and anchored by $M=0$ at both simple supports.

Why this works: A downward-only load set on a simply supported beam always sags (never hogs) between the supports, so M(x) is single-signed and there is no genuine curvature reversal — a correct diagram must show that rather than force an inflection point where none physically exists.

Common pitfall: Reporting an inflection point just because the question asks for one to be labelled "if any" — always test whether M(x) actually changes sign inside the span before claiming one.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 6 (shear and moment diagrams).

Question 2 of 8: Shear and Moment Functions and Diagrams

Question 2: Shear and Moment Functions and Diagrams (20 marks)