Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-6: Mechanics of Materials — National Exams, December 2017. Closed-book, 3 hours;
any FIVE of the eight questions constitute a complete paper (all eight are answered below as a
complete study resource). Aid sheet permitted; geometric properties of
W-shape sections were supplied as an appendix (not required for any of the eight questions solved
here).
Applied torques: 150 kN·m clockwise at B; 100 kN·m counterclockwise at C;
40 kN·m counterclockwise at D. $G=80$ GPa, $\tau_y=280$ MPa.
Find. (a) τmax and its location/distribution; (b) rotation at D;
(c) the consequence of doubling every applied torque.
Stepped shaft (solid AB, hollow BC and CD) with the three applied torques.
Approach. Cut the shaft segment-by-segment starting from the free end D and
sum the applied torques on the free side of each cut to get the internal torque; convert to stress
with $\tau=Tr/J$ (max at the outer radius) and to twist with $\theta=TL/(JG)$,
summing segment twists for the total rotation at D.
Internal torques (counterclockwise positive, consistent with the "out of page"
labelling given):
$$T_{CD}=+40\text{ kN}\cdot\text{m}\ (\text{only the D torque acts beyond a cut in CD})$$
$$T_{BC}=+40+100=+140\text{ kN}\cdot\text{m}$$
$$T_{AB}=+40+100-150=-10\text{ kN}\cdot\text{m}$$
Shear stress in each segment (outer radius $r=60$ mm in all
segments):
$$\tau_{AB}=\frac{|{-10}|\times10^6(60)}{2.036\times10^7}=29.5\text{ MPa}$$
$$\tau_{BC}=\frac{140\times10^6(60)}{1.634\times10^7}=514.2\text{ MPa}$$
$$\tau_{CD}=\frac{40\times10^6(60)}{1.634\times10^7}=146.9\text{ MPa}$$
Maximum shear stress (part a).
$$\boxed{\tau_{max}=514.2\text{ MPa, in segment BC, at the outer radius (60 mm)}}$$
Torsional shear is linear in radius, measured from the shaft's own axis: on the hollow BC section
it runs linearly from $\tau=343\text{ MPa}$ at the 40 mm inner surface up to
$514.2$ MPa at the 60 mm outer surface (no material exists inside the 40 mm bore
to carry the near-axis portion of the profile that a solid shaft would show).
Rotation at the free end D (part b). Summing $\theta=TL/(JG)$ for
all three segments:
$$\theta_D = \frac{(-10\times10^6)(800)}{J_{solid}(80{,}000)}+\frac{(140\times10^6)(600)}{J_{hollow}(80{,}000)}+\frac{(40\times10^6)(1000)}{J_{hollow}(80{,}000)}$$
$$\boxed{\theta_D = 0.0900\text{ rad} = 5.15^\circ}$$
What happens if every torque is doubled (part c)? Doubling every applied
torque doubles every internal torque and, by direct proportion, doubles every elastic shear
stress: $\tau_{AB}\to58.9$ MPa (still well below yield), $\tau_{CD}\to293.8$
MPa (now exceeds the 280 MPa yield stress, unlike before), and
$\tau_{BC}\to1028.4$ MPa (far beyond yield). Segment BC is already at
514.2 MPa — beyond the 280 MPa shear yield stress — under the original loading
(see the callout below); doubling the torques would drive segment BC further into gross yielding
and would also push segment CD past yield for the first time, while AB alone would remain in the
elastic range.
Check: the elastic torsion formula $\tau=Tr/J$ gives
514.2 MPa in segment BC, which already exceeds the stated shear yield strength of 280 MPa.
This means segment BC has, in fact, already yielded under the given (un-doubled) torques —
the elastic-formula stress is reported as the requested "maximum shear stress" per the standard
undergraduate torsion formula (this course does not cover elastic-plastic torsion), but physically
the true stress there is capped near yield with the excess carried by plastic rotation. This is
flagged rather than silently corrected, since it changes the reading of part (c): BC is not newly
overstressed by doubling — it already was.
Quantity
Value
TAB, TBC, TCD
−10, 140, 40 kN·m
τAB, τBC, τCD
29.5, 514.2, 146.9 MPa
τmax
514.2 MPa in BC (already beyond the 280 MPa yield)
Rotation at D, θD
0.0900 rad = 5.15°
Doubled torques
BC further overstressed; CD now also exceeds yield (293.8 MPa); AB stays elastic (58.9 MPa)
