Question 1 of 8: Shear-Force and Bending-Moment Diagrams
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).
Question 1: Shear-Force and Bending-Moment Diagrams (20 marks)
Given. A 6 m simply-supported beam (pin at A, roller at B) carries a triangularly distributed load rising linearly from 0 at A to 30 kN/m at B, plus an applied couple of 120 kN·m acting CLOCKWISE at A. The I-section is given in the figure.
Quantity
Value
Span, L
6 m
Distributed load, w(x)
triangular, 0 at A to 30 kN/m at B
Applied couple, M0
120 kN·m, clockwise, at A
Supports
pin at A (x=0), roller at B (x=6 m)
Find. V(x) and M(x) for 0 ≤ x ≤ 6 m, and the corresponding shear-force and bending-moment diagrams.
Loaded beam and I-section (100 × 300 mm, 15 mm flanges, 20 mm web).
Approach. Find the reactions from statics (treating the couple as a pure moment), integrate the distributed load to get V(x), then integrate V(x) to get M(x), fixing the two integration constants with the two known end conditions.
Reactions. The triangular load resultant is $W=\tfrac12(30)(6)=90$ kN acting at $x=\tfrac23(6)=4$ m from A. Taking moments about A (CCW+, so the clockwise 120 kN·m couple enters as $-120$):
$$R_B(6)-90(4)-120=0 \;\Rightarrow\; R_B=80\text{ kN}$$
and $R_A=90-80=10$ kN, both upward. Both reactions check against the independent moment sum about B.
V(x) by integration of the load. With $w(x)=5x$ kN/m (kN/m, $x$ in m) and $dV/dx=-w(x)$,
$$V(x)=R_A-\int_0^x w(s)\,ds = 10-2.5x^2\ \text{kN}$$
Check: $V(0^+)=+10$ kN and $V(6^-)=10-2.5(36)=-80$ kN, matching $-R_B$ exactly as required.
M(x) by integration of V(x). $dM/dx=V(x)$ gives
$$M(x)=M(0^+)+\int_0^x V(s)\,ds = M(0^+)+10x-\tfrac{5}{6}x^3$$
The clockwise couple at A produces a jump in the internal moment right at the support; $M(0^+)$ is fixed by the condition that the roller at B carries no applied moment, $M(6)=0$:
$$M(0^+)+60-180=0 \;\Rightarrow\; M(0^+)=120\text{ kN}\!\cdot\!\text{m}$$
so $\boxed{M(x)=120+10x-\tfrac{5}{6}x^{3}\ \text{kN}\!\cdot\!\text{m}}$.
Locate the critical points. $V(x)=0$ at $x=\sqrt{4}=2$ m, where M(x) is a local maximum:
$$M(2)=120+20-\tfrac{5}{6}(8)=\boxed{133.3\text{ kN}\!\cdot\!\text{m}}$$ this is the global maximum moment (it exceeds the $M(0^+)=120$ kN·m jump value and $M(6)=0$).
V(x) and M(x) diagrams. V runs linearly-squared from +10 kN at A to −80 kN at B, crossing zero at x=2 m; M jumps to 120 kN·m at A (the applied couple), rises to a peak of 133.3 kN·m at x=2 m, then falls to 0 at B.
