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16-Civ-A1 Elementary Structural Analysis · December 2017

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

Paper format. National Exams, 16-Civ-A1 Elementary Structural Analysis, December 2017, 3 hours, closed book (approved Sharp or Casio calculator only). A complete paper is six questions: candidates answer all of Q1–Q5 and exactly one of Q6, Q7 or Q8. All eight questions are worked below — Q6, Q7 and Q8 each in full — so the set is a complete study resource. Marks: Q1 [6], Q2–Q5 [18 each], Q6–Q8 [22 each].

Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy & stability (Ch. 2), method of joints/sections (Ch. 3), shear & moment diagrams (Ch. 4), deflections by virtual work (Ch. 8–9), influence lines (Ch. 6), slope–deflection and moment distribution (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — compound (Gerber) beams, internal hinges and condition equations, indeterminate frames. Sign convention: sagging bending moment positive (tension on the underside); truss forces tension (T) positive, compression (C) negative.

Reading the exam figures. Every support, member and load below was read directly from the printed figures. The Q4(b) truss carries a stray “3 m” label that is the perpendicular spacing between the two parallel chords (confirmed below), and the Q8 frame carries a stray “4.8 m” dimension that is not a load — it is the offset from the beam down to the lower dimension line (the columns are 4 m, stated on both outer columns).

Question 2 — Reactions and shear / bending-moment diagrams [18]

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.

Part (a) — Overhanging beam

6 kN/m12 kN2 m6 m2 mA
Beam: pin at 2 m, roller at 8 m; UDL 6 kN/m over the left 8 m; 12 kN at the right tip (10 m).

Given. Span layout below (origin at the left tip):

PinRollerUDLTip load
$x=2$ m$x=8$ m$6$ kN/m over $[0,8]$$12$ kN ↓ at $x=10$

Find. $R_{\text{pin}},R_{\text{roller}}$; the shear and moment diagrams with peak ordinates.

Approach. Two vertical reactions from global equilibrium, then march the shear from the left and integrate for the moment.

  1. Resultants. UDL $=6(8)=48$ kN at $x=4$; tip $12$ kN at $x=10$.
  2. Moments about the pin ($x=2$). $R_{r}(6)=48(2)+12(8)=192\Rightarrow \boxed{R_{\text{roller}}=32\text{ kN}\uparrow}$.
  3. Vertical equilibrium. $R_{\text{pin}}=48+12-32=\boxed{28\text{ kN}\uparrow}$ (horizontal reaction $=0$).
  4. Key moments. At the pin the $2$ m cantilever gives $M=-6(2)(1)=-12$ kN·m (hogging); at the roller $M=-12(2)=-24$ kN·m (hogging). Shear is zero at $x=28/6=4.67$ m where $M_{\max}=+9.33$ kN·m (sagging).
-12+16-20+12-12Shear V (kN)x
Shear force diagram, Q2(a). Ordinates $-12,\,+16,\,-20,\,+12$ kN.
-12+9.33-24Moment M (kN·m)x
Bending-moment diagram, Q2(a). Max sag $+9.33$; hogging $-12$ (pin) and $-24$ (roller).
QuantityValue
$R_{\text{pin}}$ / $R_{\text{roller}}$$28$ / $32$ kN (up)
Max sagging moment$+9.33$ kN·m at $x=4.67$ m
Max hogging moment$-24$ kN·m (at the roller)
Shear extremes$+16$ / $-20$ kN

Part (b) — L-shaped frame

4 kN/m20 kN2 m10 m10 mABC
Horizontal beam A(0)–B(2, roller)–C(12, corner) with UDL 4 kN/m over [2,12] and a 20 kN tip load at A; vertical column C–D (10 m) with a pin at D.

Given. $20$ kN↓ at the free tip A ($x=0$); roller at B ($x=2$); UDL $4$ kN/m over $[2,12]$; the beam turns down at the corner C ($x=12$) into a $10$ m column to a pin at D.

Find. Reactions $R_B,\,D_x,\,D_y$; the beam and column diagrams.

Approach. The pin at D and the roller at B give three reactions; no horizontal load makes $D_x=0$, so the vertical column carries axial force only and its bending moment is zero.

  1. Horizontal equilibrium. No applied horizontal force ⇒ $\boxed{D_x=0}$; the column C–D is a pure strut ($M\equiv0$ along it).
  2. Moments about D ($x=12$). UDL resultant $=4(10)=40$ kN at $x=7$. $R_B(10)=20(12)+40(5)=440\Rightarrow\boxed{R_B=44\text{ kN}\uparrow}$.
  3. Vertical equilibrium. $D_y=20+40-44=\boxed{16\text{ kN}\uparrow}$.
  4. Beam moments. At B, $M=-20(2)=-40$ kN·m (hogging from the tip cantilever). Beam shear vanishes at $x=8$ where $M_{\max}=+32$ kN·m (sagging). At the corner C, $M=0$ (consistent with the moment-free column below).
-20+24-16Beam shear V (kN)x
Beam shear, Q2(b): $-20$ kN on the overhang, $+24$ kN just right of B, falling to $-16$ kN at C.
-40+320Beam moment M (kN·m)x
Beam bending moment, Q2(b): $-40$ at B, $+32$ sag at $x=8$, $0$ at the corner C. Column C–D: $M=0$ (axial 16 kN).
QuantityValue
$R_B$ (roller)$44$ kN ↑
Pin at D $(D_x,D_y)$$(0,\;16)$ kN
Beam: max sag / max hog$+32$ / $-40$ kN·m
Column C–Daxial $16$ kN, $M=0$, $V=0$

Part (c) — Trapezoidal frame with an internal hinge

4.8 kN/m24 kN24 kN5 m16 m5 m12 mABCD
A(0,0) pin, B(5,12), internal hinge at C(21,12), D(26,0) pin; UDL 4.8 kN/m over the 16 m beam B–C, with 24 kN at B and at C.

Given. A pin at A$(0,0)$ and a pin at D$(26,0)$; inclined legs A–B and C–D (each $13$ m, $5$:$12$); horizontal top beam B$(5,12)$–C$(21,12)$, $16$ m, carrying UDL $4.8$ kN/m ($=76.8$ kN) plus $24$ kN↓ at each end. An internal hinge sits at C.

Find. The four reaction components and the frame diagrams.

Approach. With the hinge at C and no load along leg C–D, that leg is a two-force member: the reaction at D acts along D→C. That gives the fourth equation ($D=0$ was replaced by the hinge). Solve equilibrium; the analysis then shows leg A–B is also two-force and beam B–C behaves as a simply supported span.

  1. Two-force leg C–D. Reaction at D directed along D$(26,0)\to$C$(21,12)$, i.e. proportional to $(-5,12)$. Write $D=(-5t,12t)$.
  2. Moments about A. $-24(5)-24(21)-76.8(13)+ (12t)(26)=0\Rightarrow312t=1622.4\Rightarrow t=5.2$, so $D=(-26,\,62.4)$ kN and the leg carries $13t=\boxed{67.6\text{ kN (C)}}$.
  3. Global equilibrium. $A_x=-D_x=\boxed{26\text{ kN}}$, $A_y=124.8-62.4=\boxed{62.4\text{ kN}}$. (Reaction at A also lies along A→B, so leg A–B is two-force too, $67.6$ kN.)
  4. Beam B–C. With both legs axial-only, the beam is a simply supported $16$ m span under $4.8$ kN/m: end shears $\pm38.4$ kN, $M_B=M_C=0$, and $$M_{\max}=\frac{wL^2}{8}=\frac{4.8(16)^2}{8}=\boxed{153.6\text{ kN}\cdot\text{m}}\ (\text{sag, midspan}).$$
0+153.60Beam B–C moment (kN·m)x
Bending moment, Q2(c). Legs A–B and C–D are axial only ($M=0$); beam B–C is a simple-span parabola peaking $+153.6$ kN·m at midspan.
QuantityValue
Reaction at A $(A_x,A_y)$$(26,\;62.4)$ kN
Reaction at D $(D_x,D_y)$$(-26,\;62.4)$ kN
Legs A–B, C–D (axial)$67.6$ kN compression, $M=0$
Beam B–C max moment$+153.6$ kN·m (midspan sag)
Check. Both legs turn out to be two-force members and the beam simply supported — a consequence of the loads landing exactly on the joints and the single hinge at C. The independent finite-element model reproduces $A=(26,62.4)$, $D=(-26,62.4)$ and $M_{\text{mid}}=153.6$ kN·m exactly.