Question 8 of 8: Influence lines for a subdivided truss, and the effect of a moving distributed load (alternative 3 of 3)

Question 8: Influence lines for a subdivided truss, and the effect of a moving distributed load (alternative 3 of 3) (22 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 subdivided truss of 24 m span in four 6 m panels. The top chord U₁ to U₅ is 5.0 m above the bottom chord; the two intermediate joints M₁ and M₂ sit at mid-height, 2.5 m, over the quarter points at 6 m and 18 m; the bottom chord has joints only at L₁ (0 m), L₂ (12 m) and L₃ (24 m). The truss is pinned at L₁ and on a roller at L₃, and the moving load travels on beams spanning between the top chord panel points, so it is delivered to the truss only at U₁, U₂, U₃, U₄ and U₅.

Find. (a) influence lines for the forces in L₁–L₂, M₁–L₂ and M₁–M₂, each labelled with its largest absolute ordinate and its sense; (b) the force in M₁–M₂ under 6 kN/m covering the whole loaded chord.

Approach. Because the loading is applied only at panel points, each influence line is straight between panel points, so it is enough to place a unit load at each of U₁ to U₅ in turn, find the three member forces by joint resolution, and join the five ordinates with straight lines.

1. Note the two straight lines hidden in the web. L₁ is at $(0,0)$, M₁ at $(6,\ 2.5)$ and U₃ at $(12,\ 5)$, so those three joints are collinear; the same is true of U₃, M₂ and L₃. The truss is therefore a simple triangle L₁–U₃–L₃ whose two inclined sides have been subdivided, with M₁M₂, M₁L₂ and L₂M₂ forming the secondary system that carries load from the middle of each panel into the main triangle.
2. Unit load at U₁: every ordinate is zero. With the load directly over the support, joint U₁ has only the horizontal U₁U₂ and the vertical L₁U₁, so the end post carries the whole 1 kN down to the pin and $N_{U1U2} = 0$. At L₁ the reaction of 1 kN is then exactly cancelled by the end post, leaving nothing for the inclined L₁M₁ or for the bottom chord. All three influence coefficients are zero at U₁, and by the mirror-image argument all three are zero at U₅.
3. Unit load at U₂: work joint by joint. The reactions are $R_{L1} = 0.75$ and $R_{L3} = 0.25$. Joint U₁ is unloaded and has only two members, so both are zero. At L₁, with the end post idle, $$\begin{aligned} 0.75 + N_{L1M1}\left(\tfrac{2.5}{6.5}\right) &= 0 \;\Rightarrow\; N_{L1M1} = -1.95, \\ N_{L1L2} &= 1.95\left(\tfrac{6}{6.5}\right) = +1.80 \end{aligned}$$ At U₂ the two chord members are collinear and the 1 kN is taken entirely by the hanger, $N_{U2M1} = -1.00$, and $N_{U2U3} = 0$. Working in from the right, $R_{L3} = 0.25$ gives $N_{M2L3} = -0.65$ and $N_{L2L3} = +0.60$. Horizontal equilibrium at L₂, where the two bottom-chord forces are now known and the two inclined members must have equal and opposite vertical components, gives $$-1.80 + 0.60 - \left(\tfrac{12}{6.5}\right)N_{M1L2} = 0 \;\Rightarrow\; N_{M1L2} = \boxed{-0.65}$$ and finally horizontal equilibrium at M₁, where $N_{M1U3}$ turns out to be zero, leaves $$1.80 + \left(\tfrac{6}{6.5}\right)(-0.65) + N_{M1M2} = 0 \;\Rightarrow\; N_{M1M2} = \boxed{-1.20}$$
4. Unit load at U₃: the secondary system goes idle. Now $R_{L1} = R_{L3} = 0.5$, and the hanger U₂M₁ carries nothing because joint U₂ is unloaded and its chord members are collinear. The load is delivered straight to the apex of the main triangle, so $$\begin{aligned} N_{L1M1} &= -\frac{0.5(6.5)}{2.5} = -1.30, \\ N_{L1L2} &= +1.20 \end{aligned}$$ and at L₂ the two bottom-chord forces are equal, which forces $N_{M1L2} = 0$; at M₁ the two collinear members then carry the same $-1.30$ and leave $N_{M1M2} = 0$. All three secondary members are unstressed when the load stands over the apex.
5. Unit load at U₄: mirror the U₂ case. The truss is symmetric about its centre line, so the U₄ ordinates are the mirror images of the U₂ ones. That reverses the sign of the antisymmetric diagonal, $N_{M1L2} = +0.65$, leaves the symmetric horizontal unchanged, $N_{M1M2} = -1.20$, and gives the bottom chord $N_{L1L2} = 2.4(0.25) = +0.60$.
6. Draw the influence lines. Plotting the five ordinates and joining them with straight lines gives the three diagrams below. For L₁L₂ the maximum is $\boxed{1.80\ \text{T at } U_2}$; for M₁L₂ the two extreme ordinates are equal in magnitude, $\boxed{0.65\ \text{C at } U_2\ \text{and}\ 0.65\ \text{T at } U_4}$; and for M₁M₂ the diagram is entirely negative with equal peaks, $\boxed{1.20\ \text{C at both } U_2\ \text{and } U_4}$.
7. Part (b) — the effect of a full-length uniform load. The force produced by a distributed load is the intensity multiplied by the area under the influence line, because each elementary load $w\,dx$ contributes $w\,\eta\,dx$. The M₁M₂ diagram is made of four triangles, each 6 m long with one end at zero and the other at $-1.20$, so $$A = 4\left(\tfrac{1}{2}\right)(6)(-1.20) = -14.40\ \text{m}$$ $$N_{M1M2} = wA = 6(-14.40) = \boxed{-86.4\ \text{kN, i.e. } 86.4\ \text{kN compression}}$$ The same answer follows from statics as a check: 6 kN/m over 24 m of beams delivers 18, 36, 36, 36 and 18 kN to the five panel points, and $36(-1.20) + 36(0) + 36(-1.20) = -86.4$ kN, the two end loads contributing nothing because their influence ordinates are zero.