Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2016 — 04-BS-6: Mechanics of Materials (3 hours, closed book, one hand-written aid sheet permitted). Any FIVE of the eight questions constitute a complete paper on the official exam; every question is answered below. A wide-flange (W-shape) section-property table is attached at the end of the official exam; every question below supplies its own built-up section, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (shear/moment diagrams, deflection by integration, combined axial+bending, transformation of stress via Mohr's circle, torsion of stepped shafts, axial members and pin connections, Euler column buckling).
Given. A solid stepped shaft fixed at A, free at D, with three segments: AB (140 mm dia, 1600 mm), BC (100 mm dia, 2000 mm), CD (60 mm dia, 700 mm). A concentrated 140 kN·m torque acts at B (clockwise, opposite sense to the others); a distributed torque of 20 kN·m/m acts along the full BC segment; a concentrated 6 kN·m torque acts at the free end D (same sense as the distributed load). $G=80$ GPa, $\tau_y=250$ MPa.
Quantity
Value
AB
140 mm dia, 1600 mm long
BC
100 mm dia, 2000 mm long, 20 kN·m/m distributed
CD
60 mm dia, 700 mm long
Concentrated at B
140 kN·m, clockwise
Concentrated at D
6 kN·m, counter-clockwise
Find. The maximum shear stress in the shaft and the rotation at the free end D.
Stepped shaft with applied torques and the resulting internal-torque diagram T(x) (positive = the distributed-load / 6 kN·m sense).
Approach. Adopt the distributed-load's sense as positive, sum torques on the free-end side of a cut in each segment to get the internal torque (constant in AB and CD, LINEARLY VARYING in BC because the distributed torque is spread along it), then apply $\tau=Tr/J$ per segment and sum $\int T\,dx/(GJ)$ for the total rotation.
Internal torque, segment CD. Only the 6 kN·m end torque acts beyond any cut in CD:
$$T_{CD}=+6\text{ kN}\!\cdot\!\text{m (constant)}$$
Internal torque, segment BC. A cut at distance s from B carries the remaining distributed torque over $(2000-s)$ mm plus the 6 kN·m at D:
$$T_{BC}(s)=20\Big(2-\tfrac{s}{1000}\Big)+6\ \text{kN}\!\cdot\!\text{m}\ \Rightarrow\ T_{BC}(0)=\boxed{46}, \quad T_{BC}(2000)=\boxed{6}\ \text{kN}\!\cdot\!\text{m}$$
so the torque diagram is a straight line from 46 kN·m at B down to 6 kN·m at C.
Internal torque, segment AB. Beyond any cut in AB: the $-140$ kN·m at B (opposite sense), the full 40 kN·m of distributed torque, and the 6 kN·m at D:
$$T_{AB}=-140+40+6=\boxed{-94\text{ kN}\!\cdot\!\text{m}}\ (94\text{ kN}\!\cdot\!\text{m, opposite sense})$$
Shear stress in each segment. $\tau=T(d/2)/J$ with $J=\pi d^4/32$; because BC's torque varies, both of its ends must be checked:
Location
T
d
τ
AB
94 kN·m
140 mm
174.47 MPa
BC at B
46 kN·m
100 mm
234.28 MPa
BC at C
6 kN·m
100 mm
30.56 MPa
CD
6 kN·m
60 mm
141.47 MPa
The much smaller diameter of BC more than offsets its lower torque, so $$\boxed{\tau_{max}=234.28\text{ MPa, at the B end of segment BC}}$$
still under the 250 MPa shear yield. At that location τ varies LINEARLY with radius, from zero at the shaft centre to 234.28 MPa at the outer surface (r=50 mm).
Rotation at the free end D. Summing $TL/(GJ)$ for AB and CD, and $\int_0^{2000}T_{BC}(s)\,ds/(GJ_{BC})$ for BC (evaluated symbolically):
$$\phi_D=\phi_{AB}+\phi_{BC}+\phi_{CD}=0.057623\text{ rad}=\boxed{3.302^{\circ}}$$
