Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 · 98-Mar-B1 Advanced Machine Design.
Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; candidates answer any three of the four Part II problems (3–6). For study value, complete worked solutions to all six problems are provided below.
Check — paper content and assumptions.
This paper, although listed under Applied Thermodynamics and Heat Transfer, is headed “98-Mar-B1” with printed title “Advanced Machine Design”; it is a general mechanical machine-design paper — green-design/yield-theory short answers, Acme power-screw torque, beam reactions/SFD/BMD/bending stress, disk-clutch sizing, shaft fatigue with deflections, and notched-bar fatigue — with zero thermodynamics or heat-transfer content, and it is solved as the exam actually printed.
Engineering assumptions stated per the exam rubric: (1) “ton” is read as the US short ton (2000 lb), consistent with the inch/ft·min/hp unit set of Problem 2. (2) In Problem 3 the concentrated couple Mz is taken counter-clockwise (out of the page, per the dot symbol in the figure); a clockwise reading would give RA = 0, RB = 9.5 kN. (3) In Problem 5 the fluctuating torque is assumed transmitted over the 18 in from the drive end to the load; no stress concentration is used as the problem directs.
Given. Pin at A ($x=0$) and roller at B ($x=2a=2$ m). Downward load $P=5$ kN at $x=a/2=0.5$ m; CCW couple $M_z=9$ kN·m at $x=a=1$ m; triangular load rising from $0$ at $x=a$ to $w=9$ kN/m at B. Round section $d=55$ mm, $a=1$ m.
Problem 3: loading of beam AB — point load $P$ at $a/2$, couple $M_z$ at $a$, triangular load peaking at B.
Find. Support reactions $R_A,R_B$; the shear and bending-moment diagrams; and the maximum bending stress in the 55 mm round section.
Approach. Replace the triangular load by its resultant, apply the two equilibrium equations for the reactions, build $V(x)$ and $M(x)$ across the three load events (note the step in $M$ at the applied couple), then take $\sigma_{max}=M_{max}\,c/I$ with $I=\pi d^4/64$.
Resultant of the triangular load. Its magnitude and line of action are
$$W=\tfrac12 w a=\tfrac12(9)(1)=4.5\ \text{kN at } \bar x = a+\tfrac23 a = 1.667\ \text{m}.$$
Reactions. Taking moments about A (counter-clockwise positive, with $M_z$ CCW),
$$\sum M_A=0:\ R_B(2)-P(0.5)-W(1.667)+M_z=0$$
$$R_B=\frac{5(0.5)+4.5(1.667)-9}{2}=\boxed{0.5\ \text{kN}\ (\uparrow)},\qquad R_A=P+W-R_B=\boxed{9.0\ \text{kN}\ (\uparrow)}.$$
Shear diagram. Starting from A: $V=+9.0$ kN until the load $P$ drops it to $+4.0$ kN at $x=0.5$; it holds $+4.0$ kN to $x=1$ (the couple adds no shear), then the triangular load bleeds it down as $V(x)=4-\tfrac{9}{2}(x-1)^2$, reaching $-0.5$ kN at B, where $R_B$ closes it to zero. $V=0$ at $x=1.943$ m.
Bending-moment diagram. $M$ rises from $0$ at A to $M=9(0.5)=4.5$ at the load point, then to
$$M(1^-)=R_A(1)-P(0.5)=9-2.5=6.5\ \text{kN}\!\cdot\!\text{m}.$$
The applied CCW couple makes $M$ step down by $9$ kN·m at $x=1$, to $M(1^+)=-2.5$ kN·m; over the triangular span $M$ recovers to $0$ at B (confirming equilibrium). The largest magnitude therefore occurs just left of the couple:
$$\boxed{M_{max}=6.5\ \text{kN}\!\cdot\!\text{m}\ \text{at } x=1\ \text{m}.}$$
Section properties and maximum bending stress. For the round section,
$$I=\frac{\pi d^4}{64}=\frac{\pi(0.055)^4}{64}=4.49\times10^{-7}\ \text{m}^4,\quad c=\frac{d}{2}=0.0275\ \text{m}.$$
$$\sigma_{max}=\frac{M_{max}\,c}{I}=\frac{6500(0.0275)}{4.49\times10^{-7}}=\boxed{398\ \text{MPa}}.$$
This exceeds the yield of ordinary structural steel, so a larger diameter (or higher-grade steel) would be required in practice — a useful design observation from the stress result.
