Vertical deflections of a non-prismatic beam: 16-Civ-A1 Elementary Structural Analysis, December 2018, Q3

Question 3: Vertical deflections of a non-prismatic 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. Beam $A(0)\!-\!B(6)\!-\!C(12)\!-\!D(13.5\text{ m})$: pin at $A$, roller at $C$, free tip at $D$. Stiffness $A\!-\!B=EI_0$ and $B\!-\!C\!-\!D=2EI_0$ with $EI_0=18\,000\text{kN}\cdot\text{m}^2$; a single $24\text{ kN}\downarrow$ at the overhang tip $D$. Find. The vertical deflections $\delta_B$ and $\delta_D$.

Question 3 — simple span $A\!-\!C$ with a $1.5\text{ m}$ loaded overhang (heavier line $=2EI_0$).

Reactions (determinate). $\sum M_{A}=0:\;R_{C}(12)=24(13.5)\Rightarrow \boxed{R_C=27\text{ kN}}$; then $\boxed{R_A=24-27=-3\text{ kN}}$ — the pin at $A$ pulls down (uplift), because the overhang load lifts the back span.
Real moment diagram. $M(x)=-3x$ on $[0,12]$ (so $M_B=-18,\;M_C=-36\text{kN}\cdot\text{m}$, all hogging) and $M=24x-324$ on the overhang, returning to $0$ at $D$.
Unit-load (virtual-work) integrals, split at every $EI$ break. $\delta=\displaystyle\int \frac{M\,m}{EI(x)}\,dx$. For $\delta_B$ place a unit load at $B$: $\int_0^6\frac{(-3x)(0.5x)}{EI_0}+\int_6^{12}\frac{(-3x)(6-0.5x)}{2EI_0}=\frac{-108-108}{EI_0}$.
Deflection at $B$. $\displaystyle \delta_B=\frac{-216}{18\,000}=-0.0120\text{ m}=\boxed{12\text{ mm }\uparrow}$ — the negative sign means the back span rises, consistent with the uplift reaction.
Deflection at $D$. With a unit load at the tip, $m=-0.125x$ on $[0,12]$ and $m=x-13.5$ on the overhang; the three integrals give $\dfrac{27+94.5+13.5}{EI_0}=\dfrac{135}{18\,000}=0.0075\text{ m}=\boxed{7.5\text{ mm }\downarrow}$.

Quantity	Value	Sense
$R_A$	$-3\text{ kN}$	downward (uplift)
$R_C$	$27\text{ kN}$	upward
$\delta_B$	$12\text{ mm}$	upward
$\delta_D$	$7.5\text{ mm}$	downward

Topic. Deflection of a non-prismatic determinate beam by the unit-load (virtual-work) method.

Key relations. $\delta=\int Mm/EI(x)\,dx$, split wherever $EI$ or the $M$/$m$ expression changes; reactions first from $\sum F=\sum M=0$.

Why this works. Betti–Maxwell virtual work equates $1\cdot\delta$ to $\int Mm/EI$. Because the $B\!-\!D$ portion is twice as stiff, its contribution is halved; keeping the $M$–$m$ product intact while switching only the flexibility handles the step change cleanly. An overhang load hogs the interior span, which is why $B$ moves up.

Common pitfall. Using one $EI$ for the whole beam, mistaking the pinned $A$ for a fixed support (it is a pin — the beam is determinate), or dropping the sign that reveals the uplift.

Source. Hibbeler 10e, Ch. 8–9; Kassimali 6e, Ch. 7.

Question 3 of 8: Vertical deflections of a non-prismatic beam

Question 3: Vertical deflections of a non-prismatic beam (18 marks)