Question 2 of 8: Overhanging beam — tip force for a 1 mm deflection limit

Question 2: Overhanging beam — tip force for a 1 mm deflection limit

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. A beam pinned at C, on a roller at B (span CB = L), with an overhang BA = L; the free tip A carries an unknown vertical force P and a clockwise couple M.

Find. the magnitude and direction of P so the tip A deflects exactly 1 mm downward (the limiting case).

Approach. The beam is statically determinate; obtain the bending-moment diagram from the reactions, then use the unit-load (virtual work) method for the tip deflection and solve for P.

1. Reactions. With C at $x=0$, B at $x=L$, A at $x=2L$, moment about C and vertical equilibrium give $$B_y=2P+\frac{M}{L},\qquad C_y=-\left(P+\frac{M}{L}\right).$$
2. Real bending moment. Measuring $x$ from C: $$M_r(x)=\begin{cases}-\left(P+\tfrac{M}{L}\right)x, & 0\le x\le L\\[4pt] Px-2PL-M, & L\le x\le 2L\end{cases}$$
3. Unit-load moment. A unit downward load at A gives $m(x)=-x$ on CB and $m(x)=x-2L$ on BA. Applying $\delta_A=\tfrac1{EI}\int M_r\,m\,dx$ over both segments, $$\delta_A=\frac{1}{EI}\left[\frac{2}{3}PL^3+\frac{5}{6}ML^2\right]\quad(\text{downward positive}).$$
4. Deflection from the couple alone. Setting $P=0$ with $EI=200\,000\times850\times10^6=1.70\times10^{14}\ \text{N}\cdot\text{mm}^2$, $$\delta_{A,M}=\frac{5ML^2}{6EI}=\frac{5(1.8\times10^7)(10^4)^2}{6(1.70\times10^{14})}=\boxed{8.82\ \text{mm (down)}}$$ The couple alone already drives A far past the 1 mm limit, so P must act upward.
5. Solve for P. Impose $\delta_A=+1$ mm: $$1=\frac{1}{EI}\left[\frac{2}{3}PL^3+\frac{5}{6}ML^2\right]\;\Rightarrow\; P=\frac{EI-\tfrac{5}{6}ML^2}{\tfrac{2}{3}L^3}=-1.99\times10^{3}\ \text{N}.$$ The negative sign (against the assumed downward direction) means $$\boxed{P\approx 1.99\ \text{kN, directed upward}}$$