Simply supported round beam with a point load, a concentrated couple and a triangular load (20 marks): 22-Mec-B1 Advanced Machine Design, Undated paper, Q4

Question 4: Simply supported round beam with a point load, a concentrated couple and a triangular load (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. A round steel beam on a pin at A and a roller at B, span $2a$, carrying three superimposed actions positioned as shown on the figure:

Given data — beam loading and section
Quantity	Symbol	Value
Reference length	$a$	1 m
Span, pin A to roller B	$2a$	2 m
Transverse point load, at $a/2$ from A	$P$	5 kN, downward
Concentrated couple, applied at $x=a$	$M_z$	9 kN$\cdot$m, counter-clockwise
Triangular load over the right half ($a$ to $2a$)	$w$	0 at $x=a$ rising to 9 kN/m at B
Beam diameter (solid round)	$d$	55 mm

Find. The reactions at A and B, the complete shear-force and bending-moment diagrams, and the maximum bending stress in the 55 mm round section.

Problem 4 — beam AB: point load P at a/2, counter-clockwise couple M_z at mid-span, and the triangular load rising to w at the roller.

Approach. Replace the triangular load by its resultant to write the two equilibrium equations for the reactions, then build the shear diagram by walking along the beam accumulating load, build the moment diagram as the running integral of shear while remembering that a concentrated couple produces a step in moment, and finally divide the largest moment by the section modulus.

Resultant of the triangular load. A load rising linearly from zero at $x=a$ to $w$ at $x=2a$ has resultant equal to the area of the triangle, acting through its centroid: $$W=\tfrac{1}{2}wa=\tfrac{1}{2}\left(9\right)\left(1\right)=4.5\ \text{kN}\ \text{at}\ \bar x=a+\tfrac{2}{3}a=1.667\ \text{m}$$ The centroid is two-thirds of the way along the loaded length from the zero-intensity end, not at its midpoint — the most frequent slip in this class of problem.
Reactions from equilibrium. The couple contributes its full magnitude to any moment equation regardless of where it is applied, and with the counter-clockwise sense shown by the dot-and-arc symbol it acts to reduce the required roller reaction. Taking moments about A, counter-clockwise positive: $$R_B\left(2a\right)-P\left(\tfrac{a}{2}\right)-W\bar x+M_z=0$$ $$R_B=\frac{5\left(0.5\right)+4.5\left(1.667\right)-9}{2}=\frac{2.5+7.5-9}{2}=\boxed{\ 0.50\ \text{kN}\uparrow\ }$$ Vertical equilibrium then gives $$R_A=P+W-R_B=5+4.5-0.5=\boxed{\ 9.00\ \text{kN}\uparrow\ }$$ Independent check — moments about B must also close: $-9\left(2\right)+5\left(1.5\right)+4.5\left(0.333\right)+9=-18+7.5+1.5+9=0$. The strongly lopsided result, with A carrying eighteen times the roller reaction, is a direct consequence of the couple: it is large enough (9 kN$\cdot$m against a load moment of 10 kN$\cdot$m about A) to very nearly unload the roller altogether. Had the couple been clockwise instead, $R_B$ would have come out at 9.5 kN and $R_A$ at zero — the sign of the couple is genuinely load-bearing information, so read the figure symbol carefully.
Shear-force diagram. Walking from A and accumulating the applied load, with sagging-positive sign convention:
- $0\lt x\lt 0.5$ m: $V=R_A=+9.0$ kN, constant.
- At $x=0.5$ m the 5 kN load drops the shear to $+4.0$ kN, which then holds constant to $x=1$ m. The couple at $x=1$ m does not affect shear.
- $1\lt x\lt 2$ m: the triangular load removes area parabolically, $V\left(x\right)=4.0-\tfrac{1}{2}\left(9\right)\left(x-1\right)^{2}$, so the diagram curves downward as a parabola.
- Shear crosses zero at $\left(x-1\right)=\sqrt{2\left(4.0\right)/9}=0.943$ m, i.e. $x=1.943$ m, and finishes at $V=4.0-4.5=-0.50$ kN at B, which is exactly $-R_B$ as it must be.
Bending-moment diagram. The moment is the running integral of the shear, plus a step wherever a couple is applied: $$M\left(x\right)=R_Ax-P\left\langle x-\tfrac{a}{2}\right\rangle-\frac{w\left\langle x-a\right\rangle^{3}}{6a}-M_z\left\langle x-a\right\rangle^{0}$$ Working through the stations: $M$ rises linearly at 9 kN$\cdot$m per metre to $M=4.50\ \text{kN}\cdot\text{m}$ under the point load, then more gently at 4 kN$\cdot$m per metre to $$M\left(a^{-}\right)=9\left(1\right)-5\left(0.5\right)=6.50\ \text{kN}\cdot\text{m}$$ just to the left of the couple. The counter-clockwise couple then drops the diagram vertically by its full 9 kN$\cdot$m, to $M\left(a^{+}\right)=-2.50\ \text{kN}\cdot\text{m}$. Over the triangular run the moment recovers as a cubic, reaching a local extremum of only $+0.014\ \text{kN}\cdot\text{m}$ where the shear crosses zero at $x=1.943$ m, and closes at $M=0$ at the roller — the arithmetic check that the whole diagram is right.
Locate the governing moment. Scanning the complete diagram, the largest magnitude is $$\boxed{\ \left|M\right|_{\max}=6.50\ \text{kN}\cdot\text{m}\ \text{immediately to the left of the couple at }x=1\ \text{m}\ }$$ This is the crucial reading of the problem: because a couple steps the moment diagram, the peak sits at a discontinuity, not at a point where the shear vanishes. Hunting only for $V=0$ would return the trivial 0.014 kN$\cdot$m extremum in the right-hand span and understate the design moment by a factor of nearly 500.
Section modulus of the round section. $$S=\frac{\pi d^{3}}{32}=\frac{\pi\left(55\right)^{3}}{32}=1.633\times10^{4}\ \text{mm}^{3}$$
Maximum bending stress. $$\sigma_{\max}=\frac{M_{\max}}{S}=\frac{32M_{\max}}{\pi d^{3}}=\frac{6.50\times10^{6}\ \text{N}\cdot\text{mm}}{1.633\times10^{4}\ \text{mm}^{3}}$$ $$\boxed{\ \sigma_{\max}=398\ \text{MPa}\ }$$ tensile on the bottom fibre and compressive on the top at that station, since the moment there is sagging. Transverse shear does not add to this figure: the bending stress peaks at the outer fibre where the transverse shear stress is zero, so 398 MPa is the full state of stress at the critical point.

Problem 4 — shear-force and bending-moment diagrams. The couple steps the moment down by its full 9 kN·m, so the governing 6.50 kN·m sits at the discontinuity rather than where the shear crosses zero.

Check: the stated 55 mm section is not adequate, and that is the answer. At 398 MPa the beam is above the yield strength of every ordinary structural or general-purpose steel — roughly 1.6 times $S_y$ for 250 MPa mild steel and still 13 % above a 350 MPa grade — so as dimensioned the beam would yield rather than merely deflect. The question asks only for the stress, and 398 MPa is the answer to what was asked; the engineering conclusion is that either a much higher-strength alloy steel (heat-treated, $S_y\gtrsim600$ MPa) or a larger section is required. Sizing for mild steel at a factor of safety of 1.5 gives an allowable of $250/1.5=167$ MPa and $$d\ge\left(\frac{32M_{\max}}{\pi\sigma_{\text{allow}}}\right)^{1/3}=\left(\frac{32\times6.50\times10^{6}}{\pi\times166.7}\right)^{1/3}=73.5\ \text{mm}\ \Rightarrow\ \text{use 75 mm round stock.}$$ Assumptions declared: the beam is prismatic and the couple is applied about the $z$ axis in the plane of bending; the supports are frictionless and apply no axial restraint; the 9 kN$\cdot$m couple is taken as counter-clockwise from the dot-and-arc symbol on the figure, which is the only sense that leaves both reactions non-zero.

Problem 4 — final results
Quantity	Symbol	Result
Triangular-load resultant and position	$W$, $\bar x$	4.50 kN at 1.667 m from A
Reaction at the pin A	$R_A$	9.00 kN upward
Reaction at the roller B	$R_B$	0.50 kN upward
Shear immediately right of A / right of P	$V$	+9.00 kN / +4.00 kN
Shear at the roller	$V\left(2a\right)$	$-0.50$ kN ($=-R_B$)
Zero-shear station in the triangular run	$x$	1.943 m
Moment just left / just right of the couple	$M$	$+6.50$ / $-2.50$ kN$\cdot$m (step of 9 kN$\cdot$m)
Governing bending moment	$\left\|M\right\|_{\max}$	6.50 kN$\cdot$m at $x=1$ m$^{-}$
Section modulus, $d=55$ mm	$S$	$1.633\times10^{4}$ mm$^3$
Maximum bending stress	$\sigma_{\max}$	398 MPa
Diameter for mild steel at $n=1.5$	$d_{\text{req}}$	73.5 mm, use 75 mm

Topic. Shear-force and bending-moment diagrams for a beam carrying a concentrated couple as well as point and distributed load, and the bending stress that follows.

Key relations. $dV/dx=-w\left(x\right)$ and $dM/dx=V$; a concentrated couple steps $M$ by its magnitude with no effect on $V$; triangular-load resultant $\tfrac12wa$ at two-thirds of the loaded length; $S=\pi d^{3}/32$ and $\sigma=M/S=32M/\pi d^{3}$.

Why this works. The two differential relations mean the diagrams can be drawn by inspection: constant load makes shear linear and moment parabolic, triangular load makes shear parabolic and moment cubic, and a point load steps shear while a couple steps moment. The couple is the one action that breaks the otherwise reliable habit of looking for the maximum moment where the shear crosses zero, because it inserts a jump in $M$ that can dwarf every smooth extremum on the beam.

Common pitfall. Three traps live in this one question: placing the triangular resultant at mid-length instead of the two-thirds centroid; omitting the couple from the moment equation on the grounds that it “has no lever arm”, when in fact it enters every moment equation at full value; and searching for the peak moment only where $V=0$, which here misses the true maximum entirely. Guessing the sense of the couple is a fourth — it decides whether $R_A$ is 9 kN or zero.

Source. Hibbeler, Mechanics of Materials, 10th ed., §6-1 to §6-2 (graphical construction of shear and moment diagrams, including the couple discontinuity) and §6-4 (the flexure formula). Budynas & Nisbett, Shigley’s Mechanical Engineering Design, 11th ed., §3-9 to §3-10 and Table A-18 for the round-section properties.

Question 4 of 6: Simply supported round beam with a point load, a concentrated couple and a triangular load (20 marks)

Question 4: Simply supported round beam with a point load, a concentrated couple and a triangular load (20 marks)