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16-Civ-A1 Elementary Structural Analysis · May 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, method of joints/sections, virtual-work deflections, moment distribution / slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams and three-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.

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) — beam with an internal hinge

Given. Pin at $A$ ($x=0$), internal hinge at $x=6\text{ m}$, rollers at $B$ ($x=8\text{ m}$) and $C$ ($x=16\text{ m}$); a $30\text{ kN}$ point load at $x=4\text{ m}$ and a $4\text{ kN/m}$ UDL over $B\!-\!C$ ($8\text{ m}$).

Find. $V_A,V_B,V_C$ and the SFD/BMD with extreme ordinates.

4 kN/m30 kN4 m2 m2 m8 m
2(a) Loading, supports and internal hinge.

Approach. The hinge supplies the extra equation: taking the left free body up to the hinge, its bending moment is zero.

  1. Hinge condition (left part $A\!-\!$hinge). $\sum M_{\text{hinge}}=0:\;V_A(6)-30(6-4)=0 \Rightarrow V_A=\boxed{10\text{ kN}}$.
  2. Global equilibrium. $\sum F_y:\;V_A+V_B+V_C=30+4(8)=62$. $\sum M_A:\;8V_B+16V_C=30(4)+32(12)=504$. Solving with $V_A=10$: $V_B=\boxed{41\text{ kN}}$, $V_C=\boxed{11\text{ kN}}$.
  3. Shear. $+10$ from $A$ to the load, $-20$ from the load through the hinge to $B$, jumps to $+21$ at $B$ then falls linearly under the UDL to $-11$ at $C^-$, and closes to $0$. Maximum $+21\text{ kN}$, minimum $-20\text{ kN}$.
  4. Moment. $M=+40\text{ kN}\cdot\text{m}$ (sagging) under the $30\text{ kN}$ load, back to $0$ at the hinge, $-40\text{ kN}\cdot\text{m}$ (hogging) over support $B$, then a sagging peak $+15.1\text{ kN}\cdot\text{m}$ in span $B\!-\!C$ before closing at $C$.
+10−20+21−112(a) Shear (kN)
2(a) Shear force diagram.
+40hinge 0−40+15.12(a) Bending moment (kN·m)
2(a) Bending moment diagram (sagging +).

2(b) — overhanging beam

Given. Free left end at $x=0$, pin $A$ at $x=2\text{ m}$, roller $C$ at $x=12\text{ m}$; a $12\text{ kN/m}$ UDL over $0\!-\!7\text{ m}$ (ending at the point load) and a $44\text{ kN}$ load at $x=7\text{ m}$.

Find. $V_A,V_C$ and the SFD/BMD.

12 kN/m44 kN2 m5 m5 m
2(b) Overhanging beam under partial UDL and a point load.
  1. Reactions. UDL resultant $=12(7)=84\text{ kN}$ at $x=3.5\text{ m}$. $\sum M_A:\;V_C(10)=84(1.5)+44(5)=346 \Rightarrow V_C=\boxed{34.6\text{ kN}}$. $\sum F_y:\;V_A=84+44-34.6=\boxed{93.4\text{ kN}}$.
  2. Shear. Falls to $-24\text{ kN}$ at $A^-$ (UDL on the overhang), jumps to $+69.4$, decreases to $+9.4$ at the load, drops to $-34.6$, constant to $C$. Maximum $+69.4\text{ kN}$, minimum $-34.6\text{ kN}$.
  3. Moment. $-24\text{ kN}\cdot\text{m}$ (hogging) over support $A$; rises to the sagging maximum $M=93.4(5)-84(3.5)=\boxed{173\text{ kN}\cdot\text{m}}$ under the $44\text{ kN}$ load, then closes to zero at $C$.
−24+69.4+9.4−34.62(b) Shear (kN)
2(b) Shear force diagram.
−24+1732(b) Bending moment (kN·m)
2(b) Bending moment diagram.

2(c) — inclined member

Given. A straight member on a $3\!:\!4$ slope from a pin at the top $(0,12)$ to a roller at the foot $(16,0)$; the roller reaction is perpendicular to the member. Vertical loads $5\text{ kN}$ at $x=6\text{ m}$ and $7\text{ kN}$ at $x=10\text{ m}$ (horizontal distances from the top).

Find. Reactions and the bending-moment / shear diagrams of the member.

5 kN7 kNpin at top, roller ⊥ to a 3:4 member at foot
2(c) Inclined member; foot roller normal to the axis.
  1. Roller reaction (moment about the pin). With $\hat n=\tfrac{1}{5}(3,4)$ at the foot $(16,0)$, $\sum M_A=0:\;-5(6)-7(10)+20R=0\Rightarrow R=\boxed{5\text{ kN}}$ (components $R_x=3,\;R_y=4\text{ kN}$).
  2. Pin reaction. $\sum F_x:\;H_A=-3\text{ kN}\;(\boxed{3\text{ kN}\leftarrow})$; $\sum F_y:\;V_A=5+7-4=\boxed{8\text{ kN}\uparrow}$.
  3. Bending moment. Taking moments of the left-side forces about each section: $M=0$ at the pin, $-34.5\text{ kN}\cdot\text{m}$ at the $5\text{ kN}$ point, $\boxed{-37.5\text{ kN}\cdot\text{m}}$ (maximum magnitude) at the $7\text{ kN}$ point, and $0$ at the roller. The member sags in single curvature throughout.
  4. Transverse shear (component $\perp$ axis): $+4.6\text{ kN}$ from pin to the $5\text{ kN}$ load, $+0.6\text{ kN}$ to the $7\text{ kN}$ load, $-5.0\text{ kN}$ to the roller.
−34.5−37.52(c) Bending moment along member (kN·m)
2(c) Bending moment diagram (plotted along the member axis).
CaseReactionsMax +MMax −M
2(a)$V_A=10,\;V_B=41,\;V_C=11$ kN$+40$ kN·m$-40$ kN·m
2(b)$V_A=93.4,\;V_C=34.6$ kN$+173$ kN·m$-24$ kN·m
2(c)$H_A=3\;(\leftarrow),\;V_A=8\;(\uparrow),\;R=5\;(\perp)$ kN$|M|_{\max}=37.5$ kN·m