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16-Civ-A1 Elementary Structural Analysis · December 2016

Question 2 of 8: Reactions, shear 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 (Ch. 2), method of joints/sections (Ch. 3), shear & moment diagrams (Ch. 4), virtual-work deflections (Ch. 8–9), influence lines (Ch. 6), moment distribution / slope–deflection (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged and two-hinged frames. Sign convention: sagging bending moment positive; upward reactions positive; member tension positive (T), compression negative (C).

The paper directs candidates to answer Q1–Q5 and then one of Q6/Q7/Q8. For completeness all eight questions are worked here. Because every structure is defined entirely by its figure, each drawing has been read from the printed figure and redrawn to scale below.

Question 2: Reactions, shear 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) — simply-supported beam with overhang

Given. Pin at $A$ ($x=0$), roller at $B$ ($x=12\text{ m}$); a $60\text{ kN}$ point load at $x=6\text{ m}$; a $6\text{ kN/m}$ UDL over $x\in[6,16]\text{ m}$; free end at $x=16\text{ m}$.

Find. $R_A,R_B$ and the complete shear and moment diagrams with extreme ordinates.

6 kN/m60 kN6 m6 m4 mAB
2(a) — loading and dimensions.

Approach. Take moments about $A$ for $R_B$, then $\sum F_y$ for $R_A$; section the beam to build $V(x)$ and $M(x)$.

  1. Reactions. $\sum M_A=0:\;R_B(12)=60(6)+\big(6\times10\big)(11)=360+660=1020$, so $R_B=\boxed{85\text{ kN}}$. Then $R_A=60+60-85=\boxed{35\text{ kN}}$.
  2. Shear. $V=+35$ from $A$ to the load; drops $60$ to $-25$ at $x=6$; the UDL then removes $6\text{ kN/m}$ so $V=-25-6(6)=-61$ just left of $B$; the reaction lifts it to $-61+85=+24$, and the overhang UDL brings it to $0$ at the free end. Extremes $V_{\max}=+35,\;V_{\min}=-61\text{ kN}$.
  3. Moment. $M$ rises linearly to $M(6)=35(6)=+210\text{ kN}\cdot\text{m}$ (the peak, since $V$ is negative beyond the load), then $M(12)=35(12)-60(6)-6\tfrac{6^2}{2}=-48\text{ kN}\cdot\text{m}$ over the roller; on the overhang $M=-6\tfrac{(16-x)^2}{2}$ returns to $0$. Peak sagging $\boxed{+210\text{ kN}\cdot\text{m}}$ at the load, peak hogging $\boxed{-48\text{ kN}\cdot\text{m}}$ at $B$.
Shear (kN)+35−61+24Bending Moment (kN·m)+210−48
2(a) — shear (kN) and bending-moment (kN·m) diagrams; sagging plotted below the axis.

2(b) — propped cantilever with an internal hinge

Given. Fixed wall at $x=0$; internal hinge at $x=4\text{ m}$; roller at $x=10\text{ m}$; free end at $x=14\text{ m}$; $6\text{ kN/m}$ over the whole $14\text{ m}$.

Find. Wall reactions, roller reaction, and the $V,M$ diagrams.

6 kN/m4 m6 m4 mhinge
2(b) — the beam has an internal hinge 4 m from the wall.

Approach. The hinge splits the beam. Analyse the right segment (hinge–free end) first — it is a simple overhang on the roller — then carry the hinge force into the left cantilever.

  1. Right segment ($x=4$ to $14$, roller at $10$): $\sum M_{\text{hinge}}=0:\;R_B(6)=\big(6\times10\big)(5)$, so $R_B=\boxed{50\text{ kN}}$; the hinge carries $V_h=60-50=10\text{ kN}$.
  2. Left cantilever ($x=0$ to $4$): $\sum F_y:\;R_A=6(4)+10=\boxed{34\text{ kN}}$; $\sum M_A:\;M_A=6(4)(2)+10(4)=\boxed{88\text{ kN}\cdot\text{m}}$ (hogging).
  3. Diagrams. $M(x)=-88+34x-3x^2$ gives $M(0)=-88$ (max hogging), $M(4)=0$ at the hinge (as required), and $M(10)=-48\text{ kN}\cdot\text{m}$ over the roller. $V=34-6x=0$ at $x=5.67\text{ m}$ where $M_{\max}=+8.3\text{ kN}\cdot\text{m}$. Shear runs $+34\to-26$, then $+24\to0$.
Shear (kN)+34−26+24Bending Moment (kN·m)−88+8.3−48M=0
2(b) — shear and moment; the moment passes exactly through zero at the hinge.

2(c) — inclined (gable) member, UDL on the horizontal projection

Given. Pin at $\text{(1)}=(0,0)$, apex $=(12,9)\text{ m}$, roller at $\text{(3)}=(15,5)\text{ m}$ (horizontal-surface roller ⇒ vertical reaction). A $10\text{ kN/m}$ UDL acts on the horizontal projection of the rising member, $x\in[0,12]$ (total $120\text{ kN}$). The descending member is unloaded.

Find. Reactions and the bending-moment diagram along the bent axis.

10 kN/m12 m3 m9 m4 m①apex③
2(c) — pitched member; the load is a vertical UDL over the 12 m horizontal projection.

Approach. The pin carries no horizontal reaction (loads are vertical, roller reaction vertical); take moments about the pin for the roller, then track $M$ as the moment of the left forces about each section.

  1. Reactions. $\sum M_{(1)}=0:\;V_3(15)=120(6)\Rightarrow V_3=\boxed{48\text{ kN}}$; $\sum F_y:\;V_1=120-48=\boxed{72\text{ kN}}$; $\sum F_x:\;H_1=0$.
  2. Moment on the rising member (horizontal coordinate $x$): $M(x)=V_1x-10x\cdot\tfrac{x}{2}=72x-5x^2$. Setting $\dfrac{dM}{dx}=72-10x=0$ gives $x=7.2\text{ m}$ and $M_{\max}=\boxed{259.2\text{ kN}\cdot\text{m}}$ (sagging).
  3. Apex and descending member. $M_{\text{apex}}=72(12)-5(12)^2=\boxed{144\text{ kN}\cdot\text{m}}$; check from the right, $V_3(15-12)=48(3)=144$ ✓. The unloaded member carries a straight line $144\to0$ down to the roller.