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16-Civ-A1 Elementary Structural Analysis · May 2013

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

National Exams, May 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; tension member forces positive (T), compression negative (C).

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.

Given. Three planar structures (all statically determinate): (a) a two-span beam with an overhang and one internal hinge; (b) a propped cantilever with an internal hinge and an overhang; (c) an unequal-leg portal frame. Loads and dimensions as drawn.

Find. All support reactions and the complete SFD/BMD for each, with the extreme ordinates labelled.

Approach. Each structure is determinate. Where an internal hinge is present, isolate the "suspended" segment first (its moment at the hinge is zero), find the hinge reaction, then work back to the main span; the frame is solved by overall equilibrium (three equations, three reaction unknowns).

(a) Two-span beam with overhang and internal hinge

2 kN/m9.6 kNhingeABC2 m8 m2 m8 m
(a) Beam: 9.6 kN at the free tip, UDL 2 kN/m over the right 8 m, internal hinge 2 m right of support B.
  1. Suspended right segment (hinge–C). The 8 m length carries the UDL $W=2(8)=16$ kN at its midpoint. Taking moments about the hinge, $C_y(8)-16(4)=0\Rightarrow C_y=8.0$ kN; vertical equilibrium gives the hinge shear $V_h=16-8=8.0$ kN delivered down onto the left part.
  2. Left part (tip–hinge) reactions. Downward loads are 9.6 kN at the tip ($x=0$) and the 8 kN hinge force at $x=12$; supports A ($x=2$) and B ($x=10$). Moment about A: $8B_y=9.6(2)+8(10)-... $ ⇒ solving, ==**$B_y=7.6$ kN**==. Then $A_y=9.6+8-7.6=$ ==**$10.0$ kN**==. (Check: $A_y+B_y+C_y=25.6=9.6+16$ ✓.)
  3. Shear. Starting from the tip: $-9.6$ (0–2 m) $\to$ $+0.4$ after A $\to$ $+8.0$ after B $\to$ falls linearly under the UDL to $-8.0$ at C, closed by $C_y$. Zero shear (peak moment) occurs 4 m into the UDL, at $x=16$ m.
  4. Bending moment. $M_A=-9.6(2)=-19.2$ (hog); $M_B=-19.2+0.4(8)=-16.0$ (hog); $M=0$ at the hinge (as required); rising to a sagging peak $M_{max}=8(4)-2\tfrac{4^2}{2}=$ ==**$+16.0$ kN·m at $x=16$ m**==; back to $0$ at C.
Shear-force diagram (a)-9.6+8.0-8.0kN
(a) SFD: $V_{max}=+8.0$, $V_{min}=-9.6$ kN.
Bending-moment diagram (a)-19.2-16.0+16.0kN·m
(a) BMD (sagging +): $M_{max}=+16.0$, $M_{min}=-19.2$ kN·m.

(b) Propped cantilever with internal hinge and overhang

2 kN/mhingewallB2 m4 m2 m
(b) Cantilever fixed at the wall, internal hinge 2 m out, roller at 6 m, free end at 8 m; UDL 2 kN/m over the hinge–tip length (6 m).
  1. Suspended part (hinge–free end), roller at 6 m. The UDL runs the full 6 m from the hinge ($x=2$) to the free end ($x=8$): $W=2(6)=12$ kN at $x=5$. Moments about the roller ($x=6$): $V_h(2-6)+12(1)=0\Rightarrow V_h=3.0$ kN up on this part. Then $B_y=12-3=$ ==**$9.0$ kN**==.
  2. Cantilever stub (wall–hinge). It carries only the 3 kN the hinge hands back (downward), 2 m out. Wall reactions: vertical ==**$3.0$ kN**== up and fixing moment $M_w=3(2)=$ ==**$6.0$ kN·m (hogging)**==.
  3. Shear. $+3.0$ over 0–2 m; falls under the UDL to $-5.0$ just left of the roller; jumps by $+9$ to $+4.0$; falls to $0$ at the free end.
  4. Bending moment. $M_w=-6.0$ (hog) $\to 0$ at the hinge $\to$ sagging peak $M=3(1.5)-2\tfrac{1.5^2}{2}=$ ==**$+2.25$ kN·m at $x=3.5$ m**== $\to -4.0$ at the roller $\to 0$ at the tip.
Shear-force diagram (b)+3.0-5.0+4.0kN
(b) SFD: $V_{max}=+4.0$, $V_{min}=-5.0$ kN.
Bending-moment diagram (b)-6.0+2.25-4.0kN·m
(b) BMD: $M_{max}=+2.25$, $M_{min}=-6.0$ kN·m.

(c) Unequal-leg portal frame

2 kN/m20 kNA (roller)TLTRD (pin)2 m10 m2 m2 m
(c) Portal frame: roller at A (left leg 2 m), pinned at D (right leg 4 m); UDL 2 kN/m over the 10 m beam; 20 kN horizontal at mid-height of the right leg.
  1. Reactions. Unknowns $A_y$ (roller, vertical), $D_x,D_y$ (pin). Horizontal: $D_x+20=0\Rightarrow D_x=-20$ kN (i.e. ==**20 kN acting left**==). Moments about D: $-10A_y+20(5)-20(2)=0\Rightarrow$ ==**$A_y=6.0$ kN ↑**==. Vertical: $D_y=20-6=$ ==**$14.0$ kN ↑**==.
  2. Left column & beam. With the roller giving no horizontal force, the left column carries no moment; the beam behaves like a simple span with end forces $6$ (left) and $14$ (right) plus UDL. $V(x)=6-2x$ (zero at $x=3$ m); $M(x)=6x-x^2\Rightarrow$ sagging peak ==**$+9.0$ kN·m at 3 m**==, and $M=-40$ kN·m at the top-right corner (hogging).
  3. Right column. Below the 20 kN load the net horizontal is $D_x=-20$, giving a shear of 20 kN and a moment growing to $20(2)=$ ==**$-40$ kN·m at the load level**==; above the load the applied 20 kN cancels $D_x$, so the moment is constant at $-40$ up to the corner (matching the beam). Axial in the right column $=14$ kN (compression).
QuantityValue
Reaction $A_y$ (roller)6.0 kN ↑
Reaction at D (pin)$D_x=20$ kN ←, $D_y=14$ kN ↑
Beam max sagging moment+9.0 kN·m at 3 m
Corner / right-column moment−40 kN·m (hogging)
BMD of the beam TL–TR (c)+9.0-40kN·m
(c) Beam bending moment; the −40 kN·m corner value is carried down the right leg to the load, then held constant to the pin.