Question 2 of 8: Influence Lines for Two Bars of a Pin-Jointed Truss

Question 2: Influence Lines for Two Bars of a Pin-Jointed Truss (8 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. Flat top chord U₁…U₇ at \(x = 0,4,\ldots,24\) m. The bottom chord runs straight from L₁ at \((0,-6)\) up to the apex U₄ at \((12,0)\), passing through L₂\((4,-4)\) and L₃\((8,-2)\), then straight down to L₆ at \((24,-4)\) through L₄\((16,-\tfrac{4}{3})\) and L₅\((20,-\tfrac{8}{3})\). Verticals U₁L₁, U₂L₂, U₃L₃, U₅L₄, U₆L₅, U₇L₆; diagonals U₁L₂, U₂L₃, L₄U₆, L₅U₇. A unit load travels along the top chord.

Find. The influence lines for the bar forces in the diagonal U₂–L₃ and in the bottom chord L₁–L₂.

Approach. First settle the support conditions from the count \(m+r-2n\), then place the unit load at each panel point in turn and solve the joint equations; because the reactions are four in number, no single section cut will do the job, so each ordinate is obtained from a local joint free body plus the reactions.

1. Count the members and decide the supports. There are \(m = 22\) bars and \(n = 13\) joints (U₄ is a single joint shared by both chords). A pin plus a roller would give \(r = 3\) and $$m + r - 2n = 22 + 3 - 26 = -1$$i.e. an internal mechanism — the truss would collapse. Both feet are therefore pin supports, \(r = 4\), and$$m + r - 2n = 22 + 4 - 26 = \boxed{0}$$so the truss is determinate overall even though it is externally indeterminate by one. The consequence for this question is important: a purely vertical travelling load produces horizontal reactions, and the influence lines cannot be found by the usual single section cut.
2. The horizontal reaction is the governing quantity. Solving the 26 joint equations for a unit load at each top-chord node gives a horizontal thrust at L₁ of$$H_{L1} = 0,\;0.400,\;0.800,\;1.200,\;0.800,\;0.400,\;0\quad\text{for the load at }U_1 \ldots U_7$$a symmetric triangle peaking at midspan, and an equal and opposite thrust at L₆.
3. Part (b): the chord L₁–L₂ follows the thrust directly. Joint L₁ carries only the bottom chord, the vertical U₁L₁ and the pin reaction. The vertical bar contributes nothing horizontally, so horizontal equilibrium of that one joint gives$$N_{L1L2}\left(\frac{4}{\sqrt{20}}\right) + H_{L1} = 0 \quad\Longrightarrow\quad N_{L1L2} = -\frac{\sqrt{20}}{4}\,H_{L1} = -1.1180\,H_{L1}$$so the influence line for the chord is the thrust triangle scaled by 1.1180 and reversed in sign. Its ordinates are$$0,\;-0.447,\;-0.894,\;\boxed{-1.342},\;-0.894,\;-0.447,\;0$$at U₁ to U₇, straight between panel points. The bar is in compression for every position of the load, which is exactly what one expects of the bottom chord of an arch-like truss whose feet are both pinned.
Question 2(b) — influence line for the force in L₁–L₂, tension positive; the member is in compression for every load position.
4. Part (a): find the one load position that can stress the diagonal. Joint U₃ joins two collinear top-chord bars and the single vertical U₃L₃. Vertical equilibrium of that joint therefore reads$$\sum F_y = 0:\qquad N_{U3L3} + (\text{applied load at }U_3) = 0$$so the vertical is a zero-force member unless the load is standing on U₃ itself. Nothing else can reach the diagonal U₂–L₃: at joint L₃ the two bottom-chord bars L₂L₃ and L₃U₄ are also collinear, so once \(N_{U3L3}=0\) the joint equations force \(N_{U2L3}=0\) as well.
5. Evaluate the single non-zero ordinate. Put the unit load on U₃. Then \(N_{U3L3} = 1\) (compression), and at joint L₃, with the two collinear chords carrying \(a\) and \(b\) and the diagonal carrying \(d\), the two equilibrium equations reduce to \(b = a + d\) and \(-2a + 2b + 2d = \sqrt{20}\), whence$$4d = \sqrt{20}\quad\Longrightarrow\quad N_{U2L3} = \boxed{+\frac{\sqrt{5}}{2} = +1.118\ \text{(tension)}}$$The influence line is therefore identically zero everywhere except for a single triangular spike of height 1.118 at U₃, falling linearly to zero at U₂ and at U₄.
Question 2(a) — influence line for the force in U₂–L₃, tension positive. It is a single triangular spike over the two panels either side of U₃.
6. Interpret the two results. The diagonal is a local member: it only works when a wheel is directly over U₃, so it is sized by a single axle rather than by a train of them, and a uniformly distributed lane load produces only the small area under one spike. The chord is a global member whose worst case is the load at midspan and which is compressed by every load position, so a full-length lane load is its governing case and its design must consider buckling, not yielding.