Question 2 of 3: Two-span beam with a seating gap at the middle support (Problem 2)

Question 2: Two-span beam with a seating gap at the middle support (Problem 2)

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 two-span beam, built in at node 1 and at node 3, with a rigid support seat placed a small distance below the beam at the intermediate node 2, so that the support can only push upward and only after the beam has deflected onto it. The uniform load acts on span 2-3 alone; span 1-2 carries no applied load at all.

Find. The shear force and bending moment diagrams of span 1-2 under $q = 10\ \text{kN/m}$, and then the procedure — not the full solution — that would be followed if the load were raised to $q = 12\ \text{kN/m}$.

[Figure not reproduced: Figure 2 redrawn: fixed at nodes 1 and 3, with a seating gap of 5 mm below the beam at node 2. The uniform load acts on span 2-3 only. See the official exam paper.]

Approach. The gap makes this a contact problem rather than an ordinary linear analysis, so it is solved in two stages: first assume the support is not touched and solve the beam as if node 2 were free, then compare the computed deflection with the gap and, only if the gap is exceeded, re-solve with the vertical displacement at node 2 prescribed at $-\Delta$ and recover the contact force.

1. Part 1 — section properties. For the rectangular section, with $b$ the width and $h$ the height, $$I = \frac{b\,h^{3}}{12} = \frac{0.120 \times 0.300^{3}}{12} = 2.70 \times 10^{-4}\ \text{m}^{4} ,$$ so the flexural rigidity is $$EI = 200 \times 10^{6}\ \text{kPa} \times 2.70 \times 10^{-4}\ \text{m}^{4} = 5.40 \times 10^{4}\ \text{kN}\cdot\text{m}^{2} .$$
2. Discretise and identify the free degrees of freedom. Use two beam elements, 1-2 and 2-3, each of length $L$, with the element matrix printed on the paper. Nodes 1 and 3 are built in, so $v_1 = \theta_1 = v_3 = \theta_3 = 0$ and only the two degrees of freedom at node 2 survive. Adding the lower-right block of element 1-2 to the upper-left block of element 2-3, $$\mathbf{K}_{ff} = \frac{EI}{L^{3}} \begin{bmatrix} 12 & -6L \\ -6L & 4L^{2} \end{bmatrix} + \frac{EI}{L^{3}} \begin{bmatrix} 12 & 6L \\ 6L & 4L^{2} \end{bmatrix} = \frac{EI}{L^{3}} \begin{bmatrix} 24 & 0 \\ 0 & 8L^{2} \end{bmatrix} .$$ The coupling terms cancel because the two spans are equal and lie on opposite sides of the node, so the vertical and rotational equations decouple: $K_{vv} = 24EI/L^{3} = 6000\ \text{kN/m}$ and $K_{\theta\theta} = 8EI/L = 72\,000\ \text{kN}\cdot\text{m/rad}$.
3. Form the work-equivalent nodal loads. Only element 2-3 is loaded. Its consistent load vector for a downward uniform load is $\left[-qL/2,\; -qL^{2}/12,\; -qL/2,\; +qL^{2}/12\right]$, of which the first two entries act at node 2: $$F_{v2} = -\frac{qL}{2} = -30\ \text{kN}, \qquad F_{\theta 2} = -\frac{qL^{2}}{12} = -30\ \text{kN}\cdot\text{m} .$$ These are the fixed-end reactions of the loaded span, reversed and applied at the node.
4. Stage one — solve with the gap assumed open. With no contact force at node 2 the two decoupled equations give $$v_2 = \frac{F_{v2}}{24EI/L^{3}} = -\frac{qL^{4}}{48EI} = -\frac{10 \times 6^{4}}{48 \times 54\,000} , \qquad \theta_2 = \frac{F_{\theta 2}}{8EI/L} = -\frac{qL^{3}}{96EI} ,$$ so that $$\boxed{\,v_2 = -5.000\ \text{mm}, \qquad \theta_2 = -0.4167 \times 10^{-3}\ \text{rad}\,}$$ the negative signs meaning downward and clockwise.
5. Apply the contact check — this is the whole point of the problem. The beam would have to descend 5.000 mm to reach the seat, and the gap is exactly $\Delta = 5\ \text{mm}$. The load is therefore precisely the threshold value: rearranging the deflection expression, the uniform load that just closes the gap is $$q_{\text{cr}} = \frac{48\,EI\,\Delta}{L^{4}} = \frac{48 \times 54\,000 \times 0.005}{6^{4}} = 10.00\ \text{kN/m} .$$ Since $q = q_{\text{cr}}$, the beam just touches the seat without pressing on it, and $$\boxed{\,R_2 = 0\ \text{kN}\,}$$ The structure to be analysed for the diagrams is therefore the single fixed-fixed beam of total length $2L = 12\ \text{m}$ carrying the uniform load over its right-hand half; the middle support may be ignored.
6. Recover the end forces of span 1-2. Span 1-2 carries no applied load, so its member end forces follow from the element matrix alone, acting on the displacement vector $\{0,\ 0,\ v_2,\ \theta_2\}$: $$\mathbf{f}^{(1\text{-}2)} = \frac{EI}{L^{3}} \begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^{2} & -6L & 2L^{2} \\ -12 & -6L & 12 & -6L \\ 6L & 2L^{2} & -6L & 4L^{2} \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ -0.005 \\ -0.4167\times 10^{-3} \end{Bmatrix} = \begin{Bmatrix} 11.25 \\ 37.5 \\ -11.25 \\ 30.0 \end{Bmatrix} ,$$ in kN and kN·m, with $EI/L^{3} = 250\ \text{kN/m}$. Reading the entries: the element is pushed up by 11.25 kN at node 1 and down by 11.25 kN at node 2, and carries end moments of 37.5 and 30.0 kN·m.
7. Convert to diagram ordinates. Converting the nodal moments to the sagging-positive convention used for plotting, the moment at node 1 is hogging and the moment at node 2 is sagging: $$\boxed{\;V_{1\text{-}2} = +11.25\ \text{kN (constant)}, \qquad M_1 = -37.5\ \text{kN}\cdot\text{m}, \qquad M_2 = +30.0\ \text{kN}\cdot\text{m}\;}$$ Because the span carries no distributed load, $dV/dx = 0$ and the shear is constant, while $dM/dx = V$ makes the moment vary linearly. The point of contraflexure sits at $$x_0 = \frac{37.5}{11.25} = 3.333\ \text{m}$$ from node 1, that is at $0.556L$.
8. Check the result independently. Treating the beam as fixed-fixed over $2L = 12\ \text{m}$ and integrating the standard point-load fixed-end formulae $M_A = \int P a b^{2} / L_t^{2}$ and $R_A = \int P b^{2}(3a+b)/L_t^{3}$ over the loaded half returns $M_1 = 37.5\ \text{kN}\cdot\text{m}$, $M_3 = 82.5\ \text{kN}\cdot\text{m}$ and $R_1 = 11.25\ \text{kN}$, matching the element result exactly. Span 1-2 also closes on its own: $M_1 + M_2 - V L = 37.5 + 30.0 - 11.25 \times 6 = 0$. The agreement is exact rather than approximate because the cubic Hermite functions are the exact solution of $EI\,v'''' = 0$, which is the governing equation of the unloaded span.

Procedure if the load increases to q = 12 kN/m

Raising the load past the threshold changes the character of the problem: the support becomes active and the beam becomes continuous over three supports. The gap makes the response piecewise linear, so the procedure below is a two-stage linear analysis rather than an iteration.

1. Repeat the gap-open analysis at the new load. Keep the same $\mathbf{K}_{ff}$ and rescale the consistent load vector. The free deflection at node 2 becomes $v_2 = -qL^{4}/(48EI) = -6.000\ \text{mm}$.
2. Compare with the gap and decide the contact state. Because $6.000\ \text{mm} \gt \Delta = 5\ \text{mm}$, the beam lands on the seat. The gap-open solution is inadmissible and must be discarded — it would have the beam pass through a rigid support.
3. Re-solve with a prescribed displacement. Impose $v_2 = -\Delta = -5\ \text{mm}$ as a support settlement and leave $\theta_2$ free. The remaining scalar equation is $$K_{\theta\theta}\,\theta_2 = F_{\theta 2} - K_{\theta v}\,v_2 ,$$ which for equal spans has $K_{\theta v} = 0$ and therefore $\theta_2 = -qL^{3}/(96EI) = -0.500 \times 10^{-3}\ \text{rad}$.
4. Recover the contact force and confirm the state is admissible. From the vertical equation, $$R_2 = K_{vv}\,v_2 + K_{v\theta}\,\theta_2 - F_{v2} = 6000 \times (-0.005) + 0 + 36 = +6.00\ \text{kN} .$$ The sign test is the whole check: a unilateral support may only push, so a positive (upward) $R_2$ confirms contact. Had $R_2$ come out negative, the correct answer would have been the gap-open solution of step 1.
5. Redraw the diagrams from the new end forces. With $v_2$ and $\theta_2$ known, the span 1-2 end forces follow from the element matrix exactly as in step 6 above, giving $$\boxed{\;V_{1\text{-}2} = +10.5\ \text{kN}, \qquad M_1 = -36.0\ \text{kN}\cdot\text{m}, \qquad M_2 = +27.0\ \text{kN}\cdot\text{m}\;}$$ with the point of contraflexure moving out to $36.0/10.5 = 3.429\ \text{m}$. The shape of the diagrams is unchanged — constant shear and a straight moment line — because span 1-2 is still unloaded; only the ordinates change.

The step that carries the marks is the second one. Increasing the load by twenty per cent does not increase the span 1-2 moments by twenty per cent; it reduces them slightly, from 37.5 to 36.0 kN·m at the fixed end, because the newly active support takes 6.00 kN of the load and relieves span 1-2. A gap support therefore makes the structure non-linear in the loading even though every material remains elastic, and superposition may not be used across the contact threshold. In a general problem, where the several gap supports do not all close at the same load, the same two stages are repeated support by support in the order in which the gaps close.