Question 2 of 3: Beam ABCD propped by a two-bar truss — nodal displacements, shear and moment diagrams

Problem 2: Beam ABCD propped by a two-bar truss — nodal displacements, shear and moment diagrams (one of three equal-value problems)

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 three-span beam of constant $EI$ built in at both ends, propped at its two interior nodes by a pair of equally inclined bars meeting at a roller-supported apex, and carrying a point load $P$ at each propped node.

Find. (a) $v$ and $\theta$ at A, B, C, D and O; (b) the shear force diagram; (c) the bending moment diagram.

[Figure not reproduced: Figure 2.1 — Figure 2 redrawn with the discretisation used below: three beam elements AB, BC, CD carrying $(v,\theta)$ at each node, and two axial-only bars OB, OC. Sign convention: $v$ positive up, $\theta$ positive counter-clockwise, sagging moment positive. See the official exam paper.]

Approach. Assemble the beam with the printed $4\times 4$ matrix, replace each inclined bar by the equivalent vertical spring it presents to the beam, exploit the symmetry of the structure and of the load to cut the system to two unknowns, then recover the element end actions to draw the two diagrams.

1. Part (a) — identify the degrees of freedom the given element library actually has. The beam element printed on the paper carries only $(v,\theta)$ at each node: it has no axial term. The bars carry only $(u,v)$. A and D are built in, so $v_{A}=\theta_{A}=v_{D}=\theta_{D}=0$; the roller at O gives $v_{O}=0$. Because the beam is continuous into two immovable walls and has no axial flexibility in this model, it is axially rigid, so $u_{B}=u_{C}=0$ and the horizontal pull of each bar is carried straight into the walls. That leaves five free degrees of freedom:
2. $$\{d\}=\left\{v_{B},\ \theta_{B},\ v_{C},\ \theta_{C},\ u_{O}\right\}^{T}$$ Replace each bar by an equivalent vertical spring. Bar OB runs from $B$ to $O$, a horizontal distance $L$ and a vertical distance $L$, so its length is $\sqrt{2}L$ and its direction cosines are $(c,s)=(1/\sqrt{2},\,1/\sqrt{2})$. With $u_{B}=u_{O}=v_{O}=0$ the elongation of the bar is $e=-v_{B}/\sqrt{2}$, so its strain energy is $\tfrac{1}{2}\left(EA/\sqrt{2}L\right)\left(v_{B}^{2}/2\right)$ and the bar behaves exactly as a vertical spring of stiffness
3. $$k_{v}=\frac{EA}{2\sqrt{2}\,L}=\frac{1}{2\sqrt{2}L}\cdot\frac{EI}{6L^{2}} =\boxed{\frac{\sqrt{2}}{24}\,\frac{EI}{L^{3}}=0.058926\,\frac{EI}{L^{3}}}$$ Use symmetry to halve the problem. The geometry, the supports and the two loads are all symmetric about the mid-point of BC, so $v_{C}=v_{B}$, $\theta_{C}=-\theta_{B}$ and $u_{O}=0$. Only $v_{B}$ and $\theta_{B}$ remain.
4. Assemble the two remaining equations. Span AB, with A fixed, contributes its lower-right $2\times 2$ block. Span BC contributes nothing to the vertical equation — under equal end deflections and equal-and-opposite end rotations the shear it transmits is zero — and contributes $(4-2)EI/L_{2}=EI/L$ to the rotation equation. Adding the spring to the vertical equation:
5. $$\begin{aligned} \left(\frac{12EI}{L^{3}}+k_{v}\right)v_{B}-\frac{6EI}{L^{2}}\theta_{B}&=-P\\ -\frac{6EI}{L^{2}}v_{B}+\left(\frac{4EI}{L}+\frac{EI}{L}\right)\theta_{B}&=0 \end{aligned}$$ Eliminate the rotation. The second equation gives $\theta_{B}=6v_{B}/5L$ directly, and substituting it into the first collapses the bending contribution from $12EI/L^{3}$ to $\left(12-\tfrac{36}{5}\right)EI/L^{3} =\tfrac{24}{5}EI/L^{3}$:
6. $$\left(\frac{24}{5}+\frac{\sqrt{2}}{24}\right)\frac{EI}{L^{3}}\,v_{B}=-P \qquad\Longrightarrow\qquad 4.858926\,\frac{EI}{L^{3}}\,v_{B}=-P$$ Solve for the nodal displacements and rotations. Writing the result as an exact fraction and then as a decimal:
7. $$\boxed{v_{B}=v_{C}=-\frac{120}{576+5\sqrt{2}}\,\frac{PL^{3}}{EI} =-0.205807\,\frac{PL^{3}}{EI}}$$ $$\boxed{\theta_{B}=-0.246968\,\frac{PL^{2}}{EI},\qquad \theta_{C}=+0.246968\,\frac{PL^{2}}{EI}}$$ and $v_{A}=\theta_{A}=v_{D}=\theta_{D}=0$ at the built-in ends, with $u_{O}=v_{O}=0$ at the apex — the roller would allow O to slide, but symmetry gives it no reason to.
8. Recover the bar forces and the reaction at the apex. The elongation of OB is $e=-v_{B}/\sqrt{2}=0.145527\,PL^{3}/EI$, so with $EA=EI/6L^{2}$
9. $$N_{OB}=N_{OC}=\frac{EA}{\sqrt{2}L}\,e=\boxed{+0.017151\,P\ \text{(tension)}}$$ Each bar therefore pulls its node upward with $N\sin 45^\circ=0.012127P$ and pulls the wall horizontally with the same amount, while the roller at O must resist $2\times 0.012127P=0.024255P$ acting downward.
10. Part (b) — element end actions for span AB. Post-multiplying the printed matrix by $\{0,\,0,\,v_{B},\,\theta_{B}\}^{T}$ gives the four end actions of AB directly:
11. $$\begin{aligned} V_{A}&=\frac{EI}{L^{3}}\left(-12v_{B}+6L\theta_{B}\right)=\left(2.469682-1.481809\right)P =0.987873P\\ M_{A}&=\frac{EI}{L^{3}}\left(-6Lv_{B}+2L^{2}\theta_{B}\right)=0.740905\,PL\\ M_{B}&=\frac{EI}{L^{3}}\left(-6Lv_{B}+4L^{2}\theta_{B}\right)=0.246968\,PL \end{aligned}$$ so the vertical reactions are $\boxed{R_{A}=R_{D}=0.987873P}$ and the fixing moments $0.740905PL$. Global equilibrium checks: $2(0.987873)+0.024255=2.000P$, exactly the applied load.
12. Build the shear diagram. Span AB carries the constant shear $+0.987873P$. At B the applied $P$ acts down and the bar pulls $0.012127P$ up, a net downward step of $0.987873P$, which takes the shear to zero; span BC therefore carries no shear at all, which is what symmetry demands. The same step at C carries the shear to $-0.987873P$ in span CD.
13. $$\boxed{V_{AB}=+0.9879P,\qquad V_{BC}=0,\qquad V_{CD}=-0.9879P}$$ Part (c) — build the bending moment diagram. With zero shear in BC the moment there must be constant, so the diagram is completely described by its two end values and the plateau between the propped nodes:
14. $$\boxed{M_{A}=M_{D}=-0.740905\,PL\ \text{(hogging)},\qquad M_{B}=M_{C}=+0.246968\,PL\ \text{(sagging)}}$$ The moment varies linearly across each outer span, so it passes through zero where $-0.740905+0.987873\,x/L=0$, that is at $x=0.750L$ from each built-in end.
15. Sanity-check against the beam without the truss. Deleting the two bars leaves a fixed-ended beam of span $4L$ with loads $P$ at $L$ and at $3L$, for which the classical result is $M_{A}=\sum Pab^{2}/L_{\text{tot}}^{2}= \left(9+3\right)PL/16=0.75PL$, $M_{B}=0.25PL$ and $v_{B}=5PL^{3}/24EI =0.208333\,PL^{3}/EI$. Our answers sit just inside those values, and the relief is only $1.21\%$ — exactly what should be expected, because $EA=EI/6L^{2}$ makes $k_{v}=0.0589EI/L^{3}$ barely one per cent of the $4.8EI/L^{3}$ the beam already offers at B.