Shear and Moment Functions, Diagrams and Inflection Point: 04-BS-6, December 2013, Q6

Question 6: Shear and Moment Functions, Diagrams and Inflection Point (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 = 4 m, M₀ = 1 kN·m CCW at A, UDL w = 2 kN/m over the last 3 m.

Given.

Quantity	Value
Span L	4 m
Applied couple M₀ (CCW, at A)	1 kN·m
UDL w	2 kN/m over x ∈ [1, 4] m

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

Approach. Find the reactions, then cut the beam at a general x in each of the two load regions (before and after the UDL begins) and sum forces/moments of everything to the left of the cut.

Reactions. UDL resultant $=2(3)=6$ kN at its centroid $x=1+1.5=2.5$ m. Summing moments about A (CCW+): $$R_B(4) - 6(2.5) + 1 = 0 \Rightarrow R_B=3.5\ \text{kN}, \qquad R_A = 6-3.5=2.5\ \text{kN}$$
Region 1 (0 ≤ x ≤ 1 m, before the UDL). $$V(x)=R_A=2.5\ \text{kN (constant)}, \qquad M(x)=R_A x - M_0 = 2.5x-1\quad[\text{kN, kN}\cdot\text{m; }x\text{ in m}]$$
Region 2 (1 ≤ x ≤ 4 m, UDL active). $$V(x)=R_A-w(x-1)=2.5-2(x-1)$$ $$M(x)=R_A x - M_0 - \frac{w(x-1)^2}{2} = 2.5x-1-(x-1)^2$$ Check: $M(4)=2.5(4)-1-9=0$ (roller) and $V(4)=2.5-6=-3.5=-R_B$, both as required.
Maximum moments, from V(x)=0. In region 2, $V=0$ at $x=1+1.25=2.25$ m, giving the maximum positive (sagging) moment: $$M(2.25)=2.5(2.25)-1-(1.25)^2=\boxed{+3.06\ \text{kN}\cdot\text{m}}$$ The maximum negative (hogging) moment occurs right at the left support, where the applied couple produces a jump from the (zero) moment just outside the span to $$M(0)=\boxed{-1.00\ \text{kN}\cdot\text{m}}$$
Inflection point. M(x) is continuous throughout the interior of the span (the applied couple sits at the boundary, x=0, not mid-span, so there is no internal jump). Setting the region-1 expression to zero: $2.5x-1=0 \Rightarrow \boxed{x=0.40\ \text{m}}$, where the curvature genuinely reverses from hogging to sagging.

Shear force and bending moment diagrams, with the maximum/minimum moments and the inflection point labelled.

Final Results.

Quantity	Value
R_A, R_B	2.5 kN, 3.5 kN
V(x), M(x)	region 1: V=2.5, M=2.5x−1; region 2: V=2.5−2(x−1), M=2.5x−1−(x−1)²
Maximum positive moment	+3.06 kN·m at x = 2.25 m
Maximum negative moment	−1.00 kN·m at x = 0
Inflection point	x = 0.40 m

Topic: Direct-integration shear/moment functions for a beam with a support couple and a partial UDL.

Key relations: $V(x)=dM/dx$; a concentrated couple applied exactly at a support produces a step in M(x) right at that boundary but does not, by itself, create an inflection point inside the span (only an interior discontinuity would).

Why this works: Because the loading changes character only once (the UDL switches on at x=1 m), the whole beam needs only two moment expressions, matched for continuity at x=1 m and anchored by the M=0 conditions at both simple supports.

Common pitfall: Reporting the moment jump at the support itself as an "inflection point" — an inflection point is a genuine sign change of a CONTINUOUS M(x) inside the span, which here is the separate root at x=0.40 m.

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

Question 6 of 8: Shear and Moment Functions, Diagrams and Inflection Point

Question 6: Shear and Moment Functions, Diagrams and Inflection Point (20 marks)