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16-Civ-A1 Elementary Structural Analysis · December 2016

Question 3 of 8: Vertical deflection of the overhang tip on a continuous beam

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 3: Vertical deflection of the overhang tip on a continuous beam (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.

Supportspin at $B\,(x=2)$, roller at $D\,(x=10)$ — free overhang tips $A\,(x=0)$, $E\,(x=12)$
Loading$12\text{ kN/m}$ on each $2\text{ m}$ overhang; $6\text{ kN/m}$ on the $8\text{ m}$ interior span $B$–$D$
Stiffness$EI=2000\text{ kN}\cdot\text{m}^2$ (uniform)
Symmetrystructure and load symmetric about $C\,(x=6)$

Find. The vertical deflection $\delta_A$ of the left overhang tip.

126 kN/m12C (symm.)24 m4 m2ABCDEδ?
Q3 — symmetric two-overhang continuous beam; $\delta_A$ sought at the free tip.

Approach. The beam is determinate (pin + roller). Use the unit-load (virtual-work) method: $\delta_A=\dfrac{1}{EI}\int M\,m\,dx$, with $M$ the real moment and $m$ the moment from a unit vertical load placed at $A$.

  1. Real reactions. Total load $=12(2)+6(8)+12(2)=96\text{ kN}$; by symmetry $R_B=R_D=\boxed{48\text{ kN}}$.
  2. Virtual system. A unit downward load at $A$ gives, from $\sum M_B=0$, $r_D=-0.25$ and $r_B=+1.25$; the virtual moment $m(x)$ is the corresponding bending-moment diagram.
  3. Integrate. Evaluating $\displaystyle\int_0^{12} M\,m\,dx$ (done in closed form / numerically over all four segments) gives $-40\text{ kN}^2\text{m}^3$, hence $$\delta_A=\frac{-40}{2000}=-0.020\text{ m}.$$ The negative sign means the tip moves opposite to the assumed downward unit load.

Result: $\boxed{\delta_A = 20\text{ mm upward}}$. The heavy interior span hogs the beam over supports $B$ and $D$; the resulting rotation at $B$ lifts the lightly-loaded $2\text{ m}$ overhang more than its own load pushes it down. An independent stiffness-matrix (FEM) solution reproduces $+20.0\text{ mm}$ and $R_B=R_D=48\text{ kN}$.