Question 3 of 8: Vertical deflection at B of a non-prismatic beam

Question 3: Vertical deflection at B of a non-prismatic beam (16 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. Simple span A–D, 9 m, supports at A (pin) and D (roller); 81 kN at B (x=3 m), 72 kN at C (x=6 m); flexural rigidity $$EI$$ on A–B, $$3EI$$ on B–C, $$EI$$ on C–D, with $$EI=3.0\times10^4\text{ kN}\cdot\text{m}^2$$.

Find. The vertical deflection at B.

Approach. Unit-load (virtual work) method: find the real moment diagram $$M(x)$$, apply a unit vertical load at B for the virtual moment $$m(x)$$, then integrate $$\Delta_B=\displaystyle\int \frac{Mm}{EI}\,dx$$ piecewise over the three EI segments.

1. Real reactions and moments. $$\Sigma M_A=0:\ R_D(9)=81(3)+72(6)\Rightarrow R_D=\boxed{75\text{ kN}},\quad R_A=81+72-75=\boxed{78\text{ kN}}$$ $$M(x)=\begin{cases}78x,&0\le x\le3\\243-3x,&3\le x\le6\\75(9-x),&6\le x\le9\end{cases}$$ giving $$M_B=234,\ M_C=225\text{ kN}\cdot\text{m}$$.
2. Virtual moments (unit load at B). Virtual reactions $$\tfrac23$$ at A, $$\tfrac13$$ at D: $$m(x)=\begin{cases}\tfrac23 x,&0\le x\le3\\\tfrac13(9-x),&3\le x\le9\end{cases}$$
3. Integrate segment by segment. A–B $$(EI)$$: $$\int_0^3 78x\cdot\tfrac23 x\,dx=\int_0^3 52x^2dx=468$$ B–C $$(3EI)$$: $$\int_3^6(243-3x)\tfrac13(9-x)\,dx=1035$$, contributing $$1035/3=345$$ (already divided through by the extra factor of 3 in $$3EI$$). C–D $$(EI)$$: $$\int_6^9 75(9-x)\tfrac13(9-x)\,dx=225$$.
4. Sum and divide by EI. $$\Delta_B=\dfrac{468+345+225}{EI}=\dfrac{1038}{3.0\times10^4}=\boxed{0.0346\text{ m}=34.6\text{ mm}}$$ downward.