Question 2 of 3: Beam propped by two inclined truss bars

Question 2: Beam propped by two inclined truss bars (25 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. A beam clamped at both ends, node 1 at the left wall, node 3 at the right wall and node 2 an interior node a distance $L$ from node 1 and $L/2$ from node 3. Bending rigidity $EI$; axial rigidity infinite. Two truss bars, each of length $L$ and axial rigidity $EA = 24EI/L^{2}$, run from node 2 up to pinned supports at nodes 4 and 5, each making 30° with the beam axis. A downward point load $P_0$ acts at node 2. The element stiffness matrix quoted above is used for both beam elements, with $L$ replaced by that element's own length.

Find. (2.1) the active degrees of freedom; (2.2) the assembled system stiffness matrix on those degrees of freedom; (2.3) the displacements; (2.4) the axial force in each truss bar.

[Figure not reproduced: Figure 2 redrawn: beam clamped at nodes 1 and 3, interior node 2 at L from the left clamp and L/2 from the right. Two hinged truss bars of length L rise from node 2 at 30° to pinned supports 4 and 5. The load P₀ acts downward at node 2. See the official exam paper.]

Approach. Count the surviving degrees of freedom first — the inextensible beam between two clamps removes every horizontal movement, which is what makes the problem a 2×2 one. Then add the two beam element blocks and the vertical part of the two truss bars at node 2, solve, and recover the bar forces from the vertical movement of node 2.

1. Part 2.1 — identify the active degrees of freedom. Start from the full list. Nodes 4 and 5 are pinned to the foundation, so $u_4=v_4=u_5=v_5=0$. Nodes 1 and 3 are clamped, so $v_1=\theta_1=v_3=\theta_3=0$ and both are held horizontally as well. That leaves node 2, in principle with $u_2$, $v_2$ and $\theta_2$. But the beam is stated to be inextensible: the distance from node 1 to node 2 cannot change, and node 1 is fixed horizontally, so $u_2 = 0$ identically. Hence $$\boxed{\ \text{active DOF} = \{v_2,\ \theta_2\}\ \text{(two degrees of freedom)}\ }$$ It is worth noticing what this throws away. The two truss bars do have horizontal stiffness, $2\frac{EA}{L}\cos^{2}30^\circ = 1.5\,EA/L$, but with $u_2$ suppressed that stiffness never does any work; the bars serve the structure only as vertical props. The same cancellation kills the coupling term $\sum \frac{EA}{L}CS$, because the two bars lean symmetrically and their $CS$ products are equal and opposite.
2. Part 2.2 — write the two beam element contributions. Element A spans nodes 1–2 with length $\ell_A = L$; node 2 is its second node, so the relevant block is the lower-right 2×2 of the printed matrix, $$[k_A]_{22} = \frac{EI}{L^{3}}\begin{bmatrix} 12 & -6L \\ -6L & 4L^{2}\end{bmatrix}.$$ Element B spans nodes 2–3 with length $\ell_B = L/2$; node 2 is its first node, so the relevant block is the upper-left 2×2 with $L$ replaced by $L/2$: $$[k_B]_{11} = \frac{EI}{(L/2)^{3}} \begin{bmatrix} 12 & 6(L/2) \\ 6(L/2) & 4(L/2)^{2}\end{bmatrix} = \frac{EI}{L^{3}}\begin{bmatrix} 96 & 24L \\ 24L & 8L^{2}\end{bmatrix}.$$ Halving the span multiplies the translational stiffness by eight — the short right span is far stiffer than the long left one, and that is what governs the answer.
3. Add the truss bars and assemble. A bar of length $L$ inclined at $\beta = 30^\circ$ contributes $\frac{EA}{L}\sin^{2}\beta$ to the vertical stiffness of node 2, and nothing to its rotation because the ends are hinged. For the pair, $$K^{\text{truss}}_{vv} = 2\,\frac{EA}{L}\sin^{2}30^\circ = \frac{EA}{L}\cdot\frac12 = \frac{1}{2}\cdot\frac{24EI}{L^{2}}\cdot\frac{1}{L} = \frac{12EI}{L^{3}} .$$ The substitution $EA = 24EI/L^{2}$ has been chosen by the examiner so that the two props are worth exactly as much vertical stiffness as one full-length beam element. Summing all three contributions on $\{v_2,\theta_2\}$, $$[K] = \frac{EI}{L^{3}}\begin{bmatrix} 12+96+12 & -6L+24L \\ -6L+24L & 4L^{2}+8L^{2} \end{bmatrix} = \frac{EI}{L^{3}}\begin{bmatrix} 120 & 18L \\ 18L & 12L^{2}\end{bmatrix}.$$
4. Part 2.3 — solve for the displacements. The only applied action is the downward force at node 2, so $\{R\} = \{-P_0,\ 0\}^{T}$ (no applied couple). Inverting the 2×2 with $\det[K] = (120)(12L^{2}) - (18L)^{2} = 1116\,L^{2}\,(EI/L^{3})^{2}$, $$\begin{Bmatrix} v_2 \\ \theta_2 \end{Bmatrix} = \frac{L^{3}}{1116\,EI\,L^{2}}\begin{bmatrix} 12L^{2} & -18L \\ -18L & 120 \end{bmatrix}\begin{Bmatrix} -P_0 \\ 0\end{Bmatrix},$$ which gives $$\boxed{\begin{aligned} v_2 &= -\frac{12P_0L^{3}}{1116\,EI} = -\frac{P_0L^{3}}{93\,EI} = -0.010753\,\frac{P_0L^{3}}{EI} \\ \theta_2 &= +\frac{18P_0L^{2}}{1116\,EI} = +\frac{P_0L^{2}}{62\,EI} = +0.016129\,\frac{P_0L^{2}}{EI} \end{aligned}}$$ The negative $v_2$ is a downward deflection, as expected. The positive $\theta_2$ says the section at node 2 rotates counter-clockwise, i.e. the beam is still descending to the left of node 2 when it passes through it — the lowest point of the deflected shape lies inside the long left span, not at the load. That is a direct consequence of the eight-fold stiffness of the short right span.
5. Part 2.4 — recover the axial force in each truss bar. Node 2 has no horizontal movement, so the elongation of a bar running from node 2 to its fixed support is $e = -\sin\beta\; v_2$, positive because $v_2$ is negative and the props are being stretched as the beam sags away from them. Hence $$N = \frac{EA}{L}\,\sin\beta\,(-v_2) = \frac{24EI}{L^{3}}\cdot\frac12\cdot\frac{P_0L^{3}}{93EI},$$ $$\boxed{\ N_{2\text{-}4} = N_{2\text{-}5} = \frac{4P_0}{31} = 0.1290\,P_0 \ \text{(tension in both)}\ }$$
6. Check global equilibrium. Back-substituting $\{v_2,\theta_2\}$ into each element gives the clamp reactions $V_1 = \tfrac{7}{31}P_0 = 0.2258P_0$ and $M_1 = \tfrac{3}{31}P_0L = 0.0806P_0L$ at the left wall, $V_3 = \tfrac{20}{31}P_0 = 0.6452P_0$ and $M_3 = -\tfrac{6}{31}P_0L$ at the right wall, while the two props between them carry $2N\sin 30^\circ = \tfrac{4}{31}P_0$ vertically. The three shares sum to $\tfrac{7}{31}+\tfrac{20}{31}+\tfrac{4}{31} = 1$, so vertical equilibrium closes exactly, and the two beam elements deliver equal and opposite moments $\pm\tfrac{4}{31}P_0L$ to node 2, satisfying moment equilibrium there. The short right span attracts almost two-thirds of the load; the hangers take about 13 per cent.