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16-Civ-A1 Elementary Structural Analysis · December 2014

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

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

Notes on this paper

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

Exam rule: six questions constitute a complete paper — answer all of #1–#5 and only one of #6/#7/#8. All eight are solved here as a complete study resource.

Source note: Member geometry, support types and load directions for Questions 4, 7 and 8 were read from the printed figures; every reconstructed dimension yields clean, self-consistent equilibrium. Where a figure value was inferred, it is stated explicitly — verify against the original booklet if using for assessment.

Question 2: Reactions and shear/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

20 kN8 kN/m2 m8 m4 m
Pin at 2 m, roller at 10 m; 20 kN at the free left end; 8 kN/m over the 8 m interior span; 4 m unloaded right cantilever.

Given. $P=20$ kN at $x=0$; $w=8$ kN/m over $2\le x\le10$; pin $A$ at $x=2$, roller $B$ at $x=10$. Find. reactions and SFD/BMD extremes.

  1. Reactions from equilibrium. Taking moments about $A$ (UDL resultant $64$ kN at $x=6$): $\sum M_A = 20(2) - 64(4) + 8R_B = 0$ gives $R_B = 27$ kN. Then $\sum F_y$: $R_A = 84-27 = \boxed{57\text{ kN}}$, $R_B=\boxed{27\text{ kN}}$ (both up).
  2. Shear. $V=-20$ over the left overhang; it jumps to $+37$ just right of $A$, falls linearly under the UDL to $-27$ just left of $B$, and returns to $0$ over the cantilever. Zero shear at $x=2+37/8=6.625$ m.
  3. Bending moment. $M_A = -20(2) = -40$ kN·m (hogging over the support). Peak sagging at $x=6.625$: $M = -20(6.625)+57(4.625)-4(4.625)^2 = \boxed{+45.56\text{ kN}\cdot\text{m}}$. $M=0$ at $B$ and along the cantilever.
-4045.56
Bending-moment diagram (kN·m): −40 at A, +45.56 mid-span, zero at B and over the cantilever.
QuantityValue
$R_A,\;R_B$57 kN, 27 kN (up)
Max +V / max −V+37 / −27 kN
Max +M / max −M+45.56 / −40 kN·m

2(b) — Inclined-and-horizontal bent

10 kN/m10 kN/m4.5 m5.5 m2 m6 mP (pin)K
Pin at the foot of a 7.5 m incline (4.5 m × 6 m, a 3-4-5 slope) carrying 10 kN/m perpendicular; a horizontal beam with 10 kN/m over 5.5 m to a roller and a 2 m cantilever.

Given. Pin $P=(0,0)$, knee $K=(4.5,6)$, roller at $(10,6)$, tip $(12,6)$. Perpendicular UDL 10 kN/m on the incline (total 75 kN, components $(60,-45)$ at the incline midpoint); vertical UDL 10 kN/m over the first 5.5 m of the beam (55 kN). Find. reactions and diagram extremes.

  1. Horizontal equilibrium. The incline load contributes $+60$ kN horizontally, so $\sum F_x$ gives the pin horizontal $P_x = \boxed{60\text{ kN}\;(\leftarrow)}$.
  2. Moment about the pin. $\sum M_P: 2.25(-45)-3(60)+7.25(-55)+10R = 0 \Rightarrow R = \boxed{68\text{ kN}\;(\uparrow)}$ at the roller.
  3. Vertical equilibrium. $P_y = (45+55)-68 = \boxed{32\text{ kN}\;(\uparrow)}$. (The pin resultant is $\sqrt{60^2+32^2}=68$ kN.)
  4. Bending moment. Along the incline, measuring $s$ from the pin, $M(s) = -67.2\,s + 5\,s^2$; at the knee $M_K = -222.75$ kN·m, and the extreme is $\boxed{-225.8\text{ kN}\cdot\text{m}}$ at $s=6.72$ m. On the beam the moment falls from $222.75$ kN·m at the knee to $0$ at the roller, then $0$ along the cantilever. The whole frame is in single-sense (hogging, tension on the outer face) bending.
QuantityValue
Pin reaction$P_x=60$ kN ($\leftarrow$), $P_y=32$ kN ($\uparrow$)
Roller reaction68 kN ($\uparrow$)
Max bending moment−225.8 kN·m (on incline near knee)

2(c) — Compound beam with internal hinge

hinge4 kN/m2 kN/m2020 kN4 m4 m4 m4 m
Built-in left end, roller at 12 m, internal hinge at 8 m; 4 kN/m over 0–4 m, 20 kN at 4 m, 2 kN/m over 12–16 m, 20 kN at the tip.

Given. Fixed end at $x=0$, roller at $x=12$, hinge at $x=8$; loads as shown. Four reaction unknowns ($V_0,M_0,R_{12}$ plus $H_0=0$) with three equilibrium equations and one hinge condition — determinate. Find. reactions and diagram extremes.

  1. Hinge condition (right side). $\sum M_{\text{hinge}}^{\,right}: 4R_{12} - 8(6) - 20(8) = 0 \Rightarrow R_{12} = \boxed{52\text{ kN}}$.
  2. Global equilibrium. $\sum F_y: V_0 = 64-52 = \boxed{12\text{ kN}}$. $\sum M_0$ gives the fixing moment $M_0 = \boxed{80\text{ kN}\cdot\text{m}}$ (internal moment at the wall is $+80$, sagging).
  3. Diagram. Peak sagging $+98$ kN·m at $x=3$ m (zero shear); $M=0$ at the hinge; maximum hogging $\boxed{-96\text{ kN}\cdot\text{m}}$ at the roller; $M=0$ at the free tip. Shear runs $+12 \to -24$ across the first span, jumps to $+28$ at the roller, back to $0$ at the tip.
809896-96
Bending-moment diagram (kN·m): +80 at the wall, +98 at x=3, 0 at the hinge, −96 at the roller, 0 at the tip.
QuantityValue
Reactions$V_0=12$ kN, $M_0=80$ kN·m, $R_{12}=52$ kN
Max +M / max −M+98 / −96 kN·m
Max +V / max −V+28 / −24 kN