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04-BS-6 · May 2016

Question 8 of 8: Shear and Bending-Moment Diagrams — Beam with an Overhang

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams May 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 W-shape section-property table is attached at the end of the official exam; it is a red herring here — every question below supplies its own built-up or standard section, and Question 2 cites the CISC-standard W610×125 properties directly.

Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (composite beams, deflection by integration, combined loading, transformation of stress, columns, torsion, shear/moment diagrams); CISC Handbook of Steel Construction (W-shape section properties).

Question 8: Shear and Bending-Moment Diagrams — Beam with an Overhang (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantityValue
Beamoverhang 0≤x≤4 m (free tip at x=0, roller at x=4 m), main span 4–10 m (pin at x=10 m)
Distributed loadw = 10 kN/m, over the full overhang (0≤x≤4 m)
Applied coupleM0 = 25 kN·m at the free tip (x=0), counterclockwise (by the arc's sweep in the printed figure)
SectionT-shape: flange 150×25 mm (top), web 30×200 mm
SteelFy = 350 MPa, τy = 75 MPa, E = 200 GPa
M0 = 25 kN·m w = 10 kN/m 4 m 6 m
Beam with overhang: UDL, applied couple, and support layout.
150 mm 25 mm 200 mm 30 mm Beam cross-section (T)
Beam cross-section (T-shape).

Find. Piecewise V(x) and M(x) over the whole beam, and the corresponding SFD and BMD with all critical values labeled.

Approach. Find the reactions from global equilibrium (the applied couple enters the moment equation directly), then build V(x), M(x) segment by segment using $dV/dx=-w(x)$, $dM/dx=V(x)$, with the standard rule that a counterclockwise applied couple produces an immediate DOWNWARD jump in M(x) just to its right (Macaulay's-method convention) while leaving V(x) unaffected.

  1. Reactions. Taking moments about x=0 (CCW+), with $R_R$ at x=4 m and $R_P$ at x=10 m, UDL resultant 40 kN at x=2 m, and the +25 kN·m applied couple: $$25 - 2(40) + 4R_R+10R_P=0\qquad R_R+R_P=40$$ $$\Rightarrow\quad R_R=57.5\ \text{kN (up)}\qquad R_P=-17.5\ \text{kN, i.e. 17.5 kN acting DOWNWARD}$$ (The overhang load and the applied couple together tend to lift the far end, so the pin must hold the beam down there — a perfectly valid pin reaction.)
  2. Segment 1 ($0\le x\le4$ m, overhang). Only the UDL contributes to V(x); the couple enters M(x) as an immediate jump right at x=0: $$V(x)=-wx = -10x\ \text{kN}\qquad M(x)=-M_0-\frac{wx^2}{2}=-25-5x^2\ \text{kN}\cdot\text{m}$$ $$V(0^+)=0\qquad V(4^-)=-40\ \text{kN}\qquad M(0^+)=-25\ \text{kN}\cdot\text{m}\qquad M(4)=-25-5(16)=-105\ \text{kN}\cdot\text{m}$$
  3. At the roller (x=4 m). V jumps by $+R_R$; M is continuous (no couple applied here): $$V(4^+)=-40+57.5=+17.5\ \text{kN}\qquad M(4)=-105\ \text{kN}\cdot\text{m}\ \text{(unchanged)}$$
  4. Segment 2 ($4\le x\le10$ m, main span). No load acts in this span, so V is constant and M is linear: $$V(x)=17.5\ \text{kN}\qquad M(x)=-105+17.5(x-4)\ \text{kN}\cdot\text{m}$$ $$M(10^-)=-105+17.5(6)=0\ \text{kN}\cdot\text{m}\quad\boxed{\text{closes to zero exactly at the pin, as required}}$$
x (m) V (kN) -40.0 17.5 x=4
Shear force diagram.
x (m) M (kN·m) M(0+)=-25 M(4)=-105
Bending moment diagram.
QuantityValue
Rroller (x=4 m)57.5 kN, up
Rpin (x=10 m)17.5 kN, down
V(x), 0≤x≤4−10x kN
V(x), 4≤x≤10+17.5 kN (constant)
M(x), 0≤x≤4−25−5x² kN·m
M(x), 4≤x≤10−105+17.5(x−4) kN·m
Critical M, at x=4 m−105 kN·m (numerically largest)
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