Question 3 of 8: Vertical deflection of an overhanging beam

Question 3: Vertical deflection of an overhanging beam (15 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. Pin $A$ at $x=1\text{ m}$, roller $C$ at $x=7\text{ m}$; $60\text{ kN}$ on the $1\text{ m}$ left overhang (tip $x=0$) and $40\text{ kN}$ on the $1\text{ m}$ right overhang (tip $x=8$). $B$ is the mid-point of span $A\!-\!C$ ($x=4\text{ m}$). $EI=15{,}000\text{ kN}\cdot\text{m}^2$.

Find. Vertical deflection at $B$.

Approach. Unit-load (virtual-work) method: $\displaystyle \delta_B=\int \frac{M\,m}{EI}\,dx$, with $M$ the real moment and $m$ the moment from a unit vertical load at $B$.

1. Reactions. $\sum M_A:\;V_C(6)=40(7)-60(1)=220\Rightarrow V_C=36.67\text{ kN}$; $V_A=100-36.67=63.33\text{ kN}$.
2. Real moment. Both overhang loads hog the beam over the supports: $M(A)=-60\text{ kN}\cdot\text{m}$, $M(B)=-50\text{ kN}\cdot\text{m}$, $M(C)=-40\text{ kN}\cdot\text{m}$ — the whole span between supports is in hogging (single curvature, concave down).
3. Virtual system. Unit load down at $B$ gives $v_A=v_C=0.5$ and a triangular $m$-diagram peaking at $m_B=+1.5\text{ m}$ under the load.
4. Integrate. $\displaystyle \delta_B=\frac{1}{EI}\int_1^7 M\,m\,dx=\frac{-225\,000}{15\,000}\;\text{kN}\cdot\text{m}^3/\text{kN}\cdot\text{m}^2 \Rightarrow \delta_B=\boxed{15.0\text{ mm}\ \uparrow}$.

Because the structure is statically determinate, both the real moment $M$ and the virtual moment $m$ follow from statics alone, so no compatibility solution is needed. The integral is conveniently evaluated span-by-span between the supports (the virtual moment is zero on the two overhangs, so those regions drop out of the calculation entirely); using the diagram-multiplication (Vereshchagin) shortcut on the two triangular $m$-segments reproduces the same $-225{,}000/EI$. It is worth checking the reactions independently — the right-overhang $40\text{ kN}$ places $M_C=-40\text{ kN}\cdot\text{m}$ and the left-overhang $60\text{ kN}$ places $M_A=-60\text{ kN}\cdot\text{m}$, both recovered above — before trusting the deflection value.