Shear and Moment Diagrams for a Simply-Supported Beam: 04-BS-6, December 2019, Q2

Question 2: Shear and Moment Diagrams for a Simply-Supported Beam (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.

Given. Simply-supported steel beam, span $L=6\text{ m}$ (simple supports at $x=0$ and $x=6\text{ m}$ — the printed figure shows two support symbols, not a wall). UDL $w=3\text{ kN/m}$ over the last $2\text{ m}$ of the span ($4\le x\le6\text{ m}$). I-section: $d=220\text{ mm}$, $b_f=100\text{ mm}$, all thicknesses $=10\text{ mm}$ (used in Q3, not needed for $V,M$).

Find. $V(x)$, $M(x)$ over the full span, the diagrams, and the locations of maximum positive/negative moment and any inflection point.

Simply-supported 6 m beam, UDL 3 kN/m over the last 2 m of the span.

Solved below as the simply-supported beam the figure actually shows.

Approach. Find the two reactions from statics on the whole span, then write $V(x)$ and $M(x)$ piecewise by direct integration of the load (no superposition), splitting at $x=4\text{ m}$ where the UDL begins.

Reactions. UDL resultant $=3\times2=6\text{ kN}$ acting at its centroid $x=5\text{ m}$. $\sum M_{x=0}=0$: $R_6(6)-6(5)=0\Rightarrow R_6=5\text{ kN}$. $\sum F_y=0$: $R_0=6-5=\boxed{1\text{ kN}}$, $R_6=\boxed{5\text{ kN}}$.
$V(x)$, $0\le x\le4$. No load in this stretch: $V(x)=R_0=1\text{ kN}$ (constant).
$V(x)$, $4\le x\le6$. $V(x)=R_0-w(x-4)=1-3(x-4)\text{ kN}$; at $x=6^-$, $V=1-6=-5\text{ kN}=-R_6$, checks.
$M(x)$, $0\le x\le4$. $M(x)=R_0\,x=x\text{ kN}\cdot\text{m}$ (linear, $M(4)=4\text{ kN}\cdot\text{m}$).
$M(x)$, $4\le x\le6$. $M(x)=R_0x-\dfrac{w}{2}(x-4)^2=x-1.5(x-4)^2\text{ kN}\cdot\text{m}$; $M(6)=6-1.5(4)=0$, checks (simple support).
Maximum moment. $V=0$ at $1-3(x-4)=0\Rightarrow x=4.333\text{ m}$: $M(4.333)=4.333-1.5(0.333)^2=\boxed{4.17\text{ kN}\cdot\text{m}}$ (positive/sagging). Since $V\ge0$ for $x\le4.333$ and $M(0)=M(6)=0$ at the two simple supports, $M(x)\ge0$ over the ENTIRE span — there is no negative-moment region and no inflection point for this support/load configuration.

Final results
Quantity	Value
$R_0$ (left)	1 kN
$R_6$ (right)	5 kN
$M_{max}$ (at $x=4.333\text{ m}$)	+4.17 kN·m (sagging)
Negative moment / inflection point	none — $M(x)\ge0$ throughout

Topic: Direct-integration shear/moment diagrams, partial UDL

Key relations: $dV/dx=-w$, $dM/dx=V$; reactions from global statics; $M$ is continuous and piecewise-quadratic under a UDL, piecewise-linear where $V$ is constant.

Why this works: Because the load only acts on part of the span, $V(x)$ and $M(x)$ must be written piecewise, matched at $x=4\text{ m}$; this is exactly what the "no superposition" instruction is testing — write $V,M$ from the free-body of a cut section directly, not by adding a UDL-only and point-load-only textbook case.

Common pitfall: Misreading the support (wall vs. simple support) without checking the figure; and assuming every simply-supported span must have a sagging-then-hogging moment shape — a load confined to one end of the span can leave $M(x)$ single-signed with no inflection point.

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

Question 2 of 8: Shear and Moment Diagrams for a Simply-Supported Beam

Question 2: Shear and Moment Diagrams for a Simply-Supported Beam (20 marks)