NivaarExam PrepOfficial exam papers ↗

16-Civ-A1 Elementary Structural Analysis · May 2014

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

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).

The paper directs candidates to answer Q1–Q4, then one of Q5/Q6 and one of Q7/Q8. For completeness all eight questions are worked here.

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, two point loads

24 kN32 kN2 m4 m6 m2(a): 24 kN at the free tip, pin at 2 m, 32 kN at 6 m, roller at 12 m.

Given. Free left tip at $x=0$ carrying 24 kN; pin B at $x=2$ m; 32 kN at $x=6$ m; roller C at $x=12$ m.

Find. Reactions and SFD/BMD extreme ordinates.

  1. Reactions ($\sum M_B=0$). $R_C(10) - 32(4) + 24(2) = 0 \Rightarrow R_C = \dfrac{128-48}{10} = \boxed{8\ \text{kN}\uparrow}$.
  2. Vertical equilibrium. $R_B = 24+32-8 = \boxed{48\ \text{kN}\uparrow}$.
  3. Shear. $-24$ kN over the overhang; jumps to $+24$ at B; falls to $-8$ at the 32 kN load; $R_C$ closes it. Extremes $\mathbf{+24}$ and $\mathbf{-24}$ kN.
  4. Bending moment. At the pin $M_B = -24(2) = \boxed{-48\ \text{kN}\cdot\text{m}}$ (hogging); at the 32 kN load $M = +R_C(6) = \boxed{+48\ \text{kN}\cdot\text{m}}$ (sagging); zero at tip and roller.
SFD 2(a) (kN)+24-24-8 BMD 2(a) (kN·m)-48+48

2(b) — Doubly-overhanging beam, full-length UDL

22 kN/m3 m11 m6 m2(b): UDL 22 kN/m over the whole 20 m; pin at 3 m, roller at 14 m.

Given. $w=22$ kN/m over the entire 20 m beam; pin A at $x=3$ m, roller B at $x=14$ m; 3 m and 6 m overhangs.

Find. Reactions and SFD/BMD extreme ordinates.

  1. Reactions ($\sum M_A=0$). $R_B(11) = 22(20)(10-3) \Rightarrow R_B = \dfrac{440(7)}{11} = \boxed{280\ \text{kN}\uparrow}$.
  2. Vertical equilibrium. $R_A = 22(20)-280 = \boxed{160\ \text{kN}\uparrow}$.
  3. Shear. $0\to-66$ over the left overhang; jumps to $+94$ at A; falls linearly through zero at $x=7.27$ m to $-148$ just left of B; jumps to $+132$ at B; back to $0$ at the right tip. Extremes $\mathbf{+132}$ / $\mathbf{-148}$ kN.
  4. Bending moment. $M_A = -22(3)^2/2 = \boxed{-99\ \text{kN}\cdot\text{m}}$; sagging peak at $x=7.27$ m, $M = \boxed{+101.8\ \text{kN}\cdot\text{m}}$; hogging over the roller $M_B = -22(6)^2/2 = \boxed{-396\ \text{kN}\cdot\text{m}}$; zero at both tips.
SFD 2(b) (kN)+94-148+132-66 BMD 2(b) (kN·m)-99+101.8-396

2(c) — Continuous cranked (Z) member

8 kN/mApin2 m8 m2(c): bottom member (free tip at A, roller at 2 m, UDL 8 kN/m to the corner at 10 m), 2 m riser, 2 m top arm to a pin.

Given. Continuous bent: bottom beam with free tip at $x=0$, roller at $x=2$ m, UDL $w=8$ kN/m from 0 to the corner at $x=10$ m; a 2 m riser up to a 2 m top arm ending at a pin.

Find. Reactions and the SFD/BMD ordinates of every member.

  1. Reactions. $\sum F_x=0 \Rightarrow$ pin horizontal $=0$. $\sum M_{\text{roller}}=0$: $6\,P_y - (8\times10)(5-2) = 0 \Rightarrow P_y = \boxed{40\ \text{kN}\uparrow}$; roller $= 80-40 = \boxed{40\ \text{kN}\uparrow}$.
  2. Bottom member BMD. $M = -16$ kN·m over the roller ($x=2$); rises to a sagging $\boxed{+20\ \text{kN}\cdot\text{m}}$ where $V=0$ at $x=5$ m; falls to $\boxed{-80\ \text{kN}\cdot\text{m}}$ at the corner. Shear $+24$ (just right of roller) to $-40$ kN (at the corner).
  3. Riser and top arm. The 40 kN pin reaction on the 2 m top arm gives $M = 40(2) = \boxed{80\ \text{kN}\cdot\text{m}}$ at the corner; with no transverse load the riser carries this $80$ kN·m constant to the corner, matching the bottom-member value there (moment is transferred through the continuous joint).
SFD 2(c) bottom member (kN)+24-40 BMD 2(c) bottom member (kN·m)-16+20-80
StructureReactions$M^{+}_{max}$$M^{-}_{max}$$V$ range
2(a)$R_B=48,\ R_C=8$ kN$+48$$-48$$+24/-24$
2(b)$R_A=160,\ R_B=280$ kN$+101.8$$-396$$+132/-148$
2(c)roller $=40$, pin $=40$ kN$+20$ (span); $+80$ (corner)$-80$$+24/-40$