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16-Civ-A1 Elementary Structural Analysis · December 2013

Question 2 of 8: Reactions, shear-force and bending-moment diagrams

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

Notes on this paper

National Exams, December 2013 — 98-Civ-A1 Elementary Structural Analysis. Three-hour, closed-book examination (approved Sharp/Casio calculator only). Format: six questions constitute a complete paper — answer all of #1–#5 and any one of #6, #7 or #8. All eight questions are solved below for completeness.

Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy, method of joints/sections, virtual-work deflections, moment distribution, slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges and compound structures. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C).

Question 2: Reactions, shear-force and bending-moment diagrams (18 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.

2(a) — Overhanging beam with UDL and two point loads

10 kN 25 kN 4 kN/m A B (pin) C (roller) 3 m 6 m 6 m
Beam 2(a): 10 kN at the free tip, 4 kN/m over the full 15 m, 25 kN at x = 9 m; pin at B (x = 3 m), roller at C (x = 15 m).

Given. Left overhang 3 m; pin support B at x = 3 m; roller C at x = 15 m; 10 kN downward at the free tip (x = 0); UDL w = 4 kN/m over the whole 15 m; 25 kN downward at x = 9 m.

Find. Reactions and the SFD/BMD with max positive and negative ordinates.

  1. Reactions by moments about the pin B. $\sum M_B = 0$: $+10(3) - 60(4.5) - 25(6) + R_C(12) = 0$ (the 60 kN UDL resultant acts at x = 7.5 m, i.e. 4.5 m right of B). Hence $R_C = \dfrac{270+150-30}{12} = \boxed{32.5\ \text{kN}\uparrow}$.
  2. Vertical equilibrium. $R_B = (10+60+25) - 32.5 = \boxed{62.5\ \text{kN}\uparrow}$.
  3. Shear. From the tip: $V$ falls under the UDL to $-22$ kN just left of B, jumps to $-22+62.5 = \mathbf{+40.5}$ kN at B, falls to $+16.5$ kN just left of the 25 kN load, drops to $-8.5$ kN, then falls to $-32.5$ kN just left of C, where $R_C$ closes it to zero. Shear is zero only across the point loads — the maximum sagging moment sits at the 25 kN load.
  4. Bending moment. At the pin, $M_B = -\bigl[10(3)+4(3)(1.5)\bigr] = \boxed{-48\ \text{kN}\cdot\text{m}}$ (hogging). At the 25 kN load, $M = -10(9)-4(9)(4.5)+62.5(6) = \boxed{+123\ \text{kN}\cdot\text{m}}$ (sagging). $M$ returns to zero at the free tip and at roller C.
−48 +123 0 0
BMD 2(a): hogging −48 kN·m over the pin, sagging peak +123 kN·m at the 25 kN load, zero at both ends.

2(b) — Compound beam with an internal hinge

4 kN/m 24 kN A hinge B C 4 m 2 m 6 m
Beam 2(b): roller A (x = 0), internal hinge at x = 4 m, pin B (x = 6 m), roller C (x = 12 m); 4 kN/m over 0–6 m; 24 kN at x = 10 m.

Given. Roller A at x = 0; internal hinge at x = 4 m; pin B at x = 6 m; roller C at x = 12 m; UDL 4 kN/m over 0–6 m (24 kN, resultant at x = 3 m); 24 kN downward at x = 10 m.

Find. Reactions and the SFD/BMD with extreme ordinates.

  1. Use the hinge to find $R_A$. The moment is zero at the hinge. Taking the left segment (0–4 m) about the hinge: $R_A(4) - (4\times4)(2) = 0 \Rightarrow R_A = \boxed{8\ \text{kN}\uparrow}$.
  2. Whole-beam moments about pin B. $\sum M_B = 0$: $R_A(-6) + 24(-3)_{\text{(UDL)}} + 24(-4)_{\text{(pt)}} + R_C(6) = 0$ (arms measured right-positive from B) $\Rightarrow R_C = \boxed{12\ \text{kN}\uparrow}$.
  3. Vertical equilibrium. $R_B = 48 - 8 - 12 = \boxed{28\ \text{kN}\uparrow}$.
  4. Shear. $+8$ at A, falling under the UDL through zero at x = 2 m to $-16$ kN just left of B; jumps to $+12$ kN at B; constant to the 24 kN load, then $-12$ kN to C. Extremes $\mathbf{+12}$ and $\mathbf{-16}$ kN.
  5. Bending moment. Sagging $+8$ kN·m at x = 2 m; zero at the hinge (as required); hogging $\boxed{-24\ \text{kN}\cdot\text{m}}$ over pin B; sagging $\boxed{+24\ \text{kN}\cdot\text{m}}$ under the 24 kN load; zero at C.

2(c) — Cranked beam on an inclined roller and a pin

33.8 kN/m A B (pin) 3 m 12 m 2 m 5 m
Beam 2(c): bent member A(3,0)–B(15,5) with short overhangs; an inclined roller at A (reaction ⟂ to the 5:12 incline) and a pin at B; 33.8 kN/m applied vertically.

Given. A cranked beam rising 5 m over a 12 m horizontal run (a 5–12–13 incline). The left foot A carries a roller whose surface is parallel to the incline, so its reaction acts perpendicular to the 5:12 member (unit normal $(-5,12)/13$); the upper node B is a pin. A downward UDL of 33.8 kN/m acts over the 12 m horizontal projection of the loaded span (total $W = 33.8\times12 = 405.6$ kN at mid-span, x = 9 m).

Find. The three reaction components.

  1. Moment about pin B to isolate the inclined-roller reaction. Its line of action passes at perpendicular distance 13 m from B (the full member length), so $\sum M_B = 0$ gives $-13\,R_A + 6W = 0$, i.e. $R_A = \dfrac{6(405.6)}{13} = \boxed{187.2\ \text{kN}}$ directed along $(-5,12)/13$.
  2. Resolve $R_A$. $R_{Ax} = 187.2(-5/13) = -72.0$ kN, $R_{Ay} = 187.2(12/13) = +172.8$ kN.
  3. Global equilibrium for the pin. $\sum F_x = 0 \Rightarrow B_x = \boxed{72.0\ \text{kN}\rightarrow}$; $\sum F_y = 0 \Rightarrow B_y = 405.6 - 172.8 = \boxed{232.8\ \text{kN}\uparrow}$.

The solution takes the drawn support at A as an inclined roller (reaction normal to the 5:12 leg) and applies the 33.8 kN/m over the 12 m loaded projection between the supports. If instead the UDL is spread over the full 17 m or the roller reaction is vertical, the numbers change; the method — one moment equation to isolate the single-direction reaction, then two force equations — is unchanged.

StructureReactions / key ordinates
2(a)$R_B=62.5$, $R_C=32.5$ kN; $M_{max}^{+}=+123$, $M_{max}^{-}=-48$ kN·m; $V:\,+40.5/-32.5$ kN
2(b)$R_A=8$, $R_B=28$, $R_C=12$ kN; $M_{max}^{+}=+24$, $M_{max}^{-}=-24$ kN·m; $V:\,+12/-16$ kN
2(c)$R_A=187.2$ kN ⟂ incline; $B_x=72.0$, $B_y=232.8$ kN