Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2017 — 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 or standard section directly, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial members with initial gaps, shear/moment diagrams, deflection by direct integration, transformation of stress via Mohr's circle, combined axial+bending on eccentrically-loaded columns, torsion of stepped shafts, Euler column buckling, unsymmetric-section flexure of brittle materials).
Given. Shaft ABCD, fixed at A, free at D. AB (800 mm) solid, 120 mm diameter. BC (600 mm) and CD (1000 mm) hollow, 120 mm OD, 80 mm ID. Torques: 150 kN·m at B and 40 kN·m at D (same rotational sense), 100 kN·m at C (opposite sense). G=80 GPa, τy=280 MPa.
Quantity
Value
AB
800 mm, solid, 120 mm diameter
BC
600 mm, hollow, 120/80 mm OD/ID
CD
1000 mm, hollow, 120/80 mm OD/ID
TB, TD
150, 40 kN·m (same sense)
TC
100 kN·m (opposite sense)
G
80 GPa
Find. Maximum shear stress, rotation at D, and the effect of doubling all torques.
Stepped shaft ABCD: solid AB, hollow BC/CD, three applied torques (B, D one sense; C opposite).
Approach. Cut the shaft in each segment and sum torques to the free end to get the internal torque diagram, then apply $\tau=T(d_o/2)/J$ for stress and $\phi=\sum TL/(GJ)$ for rotation, segment by segment.
Internal torque in each segment. From the printed arrows, B and D share one rotational sense and C is the opposite sense. Taking B/D's sense positive ($T_B=+150$, $T_C=-100$, $T_D=+40$ kN·m) and cutting from the free end D inward:
$$T_{CD}=T_D=\boxed{40\text{ kN}\!\cdot\!\text{m}},\quad T_{BC}=T_C+T_D=\boxed{-60\text{ kN}\!\cdot\!\text{m}},\quad T_{AB}=T_B+T_C+T_D=\boxed{90\text{ kN}\!\cdot\!\text{m}}$$
Section properties. AB is solid, 120 mm diameter; BC and CD are hollow (120 mm OD, 80 mm ID, confirmed the bore runs the full B-to-D length):
$$J_{solid}=\frac{\pi}{32}(120)^4=2.0358e+07\text{ mm}^4,\qquad J_{hollow}=\frac{\pi}{32}\left[(120)^4-(80)^4\right]=1.6336e+07\text{ mm}^4$$
(a) Maximum shear stress. $\tau=T(d_o/2)/J$ in each segment:
$$\tau_{AB}=\boxed{265.26\text{ MPa}},\quad \tau_{BC}=220.37\text{ MPa},\quad \tau_{CD}=146.91\text{ MPa}$$
AB governs (solid section carrying the largest torque), at 265.3 MPa — under the 280 MPa yield. The shear-stress distribution is linear from zero at the centre to 265.3 MPa at the outer radius (solid section, no core to skip).
(b) Rotation at the free end. Summing $\phi=TL/(GJ)$ over the three segments (signs matter — BC is negative):
$$\phi_D=\frac{T_{AB}L_{AB}}{GJ_{solid}}+\frac{T_{BC}L_{BC}}{GJ_{hollow}}+\frac{T_{CD}L_{CD}}{GJ_{hollow}}=\boxed{0.0473\text{ rad}}=\boxed{2.708^\circ}$$
(c) Doubling the loads. Every torque doubles, so every shear stress doubles too:
$$\tau_{AB,doubled}=2(265.26)=\boxed{530.5\text{ MPa}}\ \gt\ 280\text{ MPa (yield)}$$
Segment AB would YIELD — the shaft no longer behaves elastically and the linear $\tau$-vs-radius and $\phi$-vs-torque relations used above would no longer apply there.
Internal torque diagram T(x): +90 kN·m in AB, −60 kN·m in BC, +40 kN·m in CD.
Shear-stress variation with radius in segment AB (governing, solid section): zero at the centre, linear to 265.3 MPa at the outer surface.
