Question 5 of 8: Influence Lines for a Truss and for Shear in a Beam

Question 5: Influence Lines for a Truss and for Shear in a Beam (18 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. Part (a): bottom-chord joints $L_1$ to $L_5$ at 6 m centres, top-chord joints $U_1$, $U_2$, $U_3$ directly above $L_2$, $L_3$ and $L_4$ at a height of 4.5 m. The truss is pinned at $L_2$ and on a roller at $L_4$, so it has a 6 m overhang at each end, and the unit load travels along the bottom chord. Part (b): a beam pinned at $x = 0$, on a roller at $x = 20$ m and cantilevering 4 m beyond it; the section ①-① is at $x = 6$ m. The vehicle is three axles — 40 kN, 40 kN and 20 kN at spacings of 2 m and 4 m.

Find. The three truss influence lines with their extreme tension and compression coefficients, the ordinates of the shear influence line at ①-①, and the largest shear the vehicle can produce there.

Approach. For the truss, place the unit load at each bottom-chord joint in turn and use a section or a joint that isolates the member wanted; the influence line is then the straight-line interpolation between those five ordinates. For the beam, build the shear influence line from its definition and slide the axle group over it, testing each axle in the position of the peak ordinate.

1. Part (a), reactions. With the supports at $x = 6$ m and $x = 18$ m, a unit load at $x = z$ gives $R_{L4} = (z-6)/12$ and $R_{L2} = (18-z)/12$. Note that both go negative when the load stands on the far overhang, which is what makes the overhangs interesting.
2. Influence line for $U_1U_2$. Joint $L_1$ is a free end carrying only $L_1L_2$ and the diagonal $L_1U_1$, so whenever the unit load is anywhere but at $L_1$ both are zero. Joint $U_1$ then has $U_1U_2$, the post $U_1L_2$ and that zero diagonal, and resolving horizontally gives $F_{U_1U_2} = 0.8F_{L_1U_1}$, so the top chord is zero too. With the unit load standing on $L_1$, joint $L_1$ gives $F_{L_1U_1} = 1/0.6 = 1.667$ and hence $$\eta_{U_1U_2}(L_1) = 0.8(1.667) = \boxed{+1.333\ \text{(tension)}}$$ The influence line is therefore a single triangle over the left overhang, rising from zero at $L_2$ to $+4/3$ at $L_1$, and identically zero everywhere else. Maximum tension coefficient $+1.333$; there is no compression at any load position.
3. Influence line for the post $U_2L_3$. Joint $L_3$ carries the two collinear chord members $L_2L_3$ and $L_3L_4$ plus the vertical $U_2L_3$. Vertical equilibrium of that joint says the post force equals whatever vertical load stands on $L_3$ and nothing else, so $$\eta_{U_2L_3}(L_3) = \boxed{+1.000\ \text{(tension)}}, \qquad \eta = 0 \text{ at every other panel point}$$ The influence line is one triangle of unit height centred on $L_3$, with no compression anywhere. This is the classical zero-force-member result read as an influence line.
4. Influence line for the diagonal $U_2L_4$. Cut between $L_3$ and $L_4$; the section severs $U_2U_3$, $U_2L_4$ and $L_3L_4$, of which only the diagonal has a vertical component, of magnitude $4.5/7.5 = 0.6$. For the load to the left of the cut, work with the right-hand portion, which carries only $R_{L4}$, giving $F = -R_{L4}/0.6$; for the load to the right, the left-hand portion carries only $R_{L2}$ and gives $F = +R_{L2}/0.6$. Evaluating at the five panel points, $$\eta_{U_2L_4} = \left(+\tfrac{5}{6},\; 0,\; -\tfrac{5}{6},\; 0,\; -\tfrac{5}{6}\right) \text{ at } L_1 \ldots L_5$$ so the extreme coefficients are $\boxed{+0.833\ \text{T at } L_1}$ and $\boxed{-0.833\ \text{C at } L_3 \text{ or } L_5}$. Both supports give zero, which is the general rule that a unit load standing directly on a support produces no force in any member.

1. Part (b), ordinates of the shear influence line. For a unit load at $z$ the left reaction is $R_A = (20-z)/20$, and the shear just at the section is $V = R_A$ when the load is to the right of it and $V = R_A - 1$ when it is to the left. That gives $\eta = 0$ at the left support, $\eta = 0.7 - 1 = -0.300$ immediately to the left of the section, a unit jump to $\eta = +0.700$ immediately to its right, $\eta = 0$ at the roller, and, continuing the same straight line onto the overhang, $\eta = -0.200$ at the free end. In summary $$\boxed{\eta = 0,\ -0.300,\ +0.700,\ 0,\ -0.200 \text{ at } x = 0,\ 6^-,\ 6^+,\ 20,\ 24\ \text{m}}$$ The whole line has the single slope $-1/20$ per metre, interrupted by the unit jump at the section.
2. Position the vehicle for maximum positive shear. The peak ordinate is $+0.700$ just right of the section, so bring the leading axle of the group to that point and check the alternatives. With the first 40 kN axle at $x = 6$ m the other axles sit at 8 m and 12 m, where $\eta = 0.600$ and $0.400$: $$V = 40(0.700) + 40(0.600) + 20(0.400) = 28 + 24 + 8 = \boxed{60\ \text{kN}}$$ Placing the second 40 kN axle at the section instead throws the first onto the negative reach and yields only 30 kN, and placing the 20 kN axle there yields 10 kN, so 60 kN governs.
3. Check the negative extreme as well. The negative reaches are the 0–6 m portion and the 4 m overhang, and they are separated by the whole positive region, so the 6 m long axle group cannot occupy both. With the group entirely between the left support and the section the worst value is $-10$ kN; with the two 40 kN axles standing at 22 m and 24 m on the overhang and the trailing axle already off the deck it is $40(-0.100) + 40(-0.200) = -12$ kN. The largest shear the vehicle produces at ①-① is therefore $+60$ kN, and the design range for this section is $-12$ kN to $+60$ kN.