Question 5 of 8: Influence Lines

Question 5: Influence Lines (9 + 11 = 20 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.

5(a) Influence lines for three truss members

Given. A 24 m truss carrying its load on the top chord. Top chord $U_1 \ldots U_5$ at 6 m centres (so the panel points are at 0, 6, 12, 18, 24 m), bottom chord $L_1$, $L_2$, $L_3$ directly below $U_2$, $U_3$, $U_4$, depth 2.5 m. Pin at $U_1$, roller at $U_5$. The web comprises the verticals $U_2L_1$, $U_3L_2$, $U_4L_3$ and the diagonals $U_1L_1$, $U_2L_2$, $L_2U_4$ and $L_3U_5$. Each diagonal spans 6 m by 2.5 m, so its length is $\sqrt{6^2+2.5^2} = 6.5$ m — a scaled 5–12–13 triangle.

Find. The influence lines for $U_3L_2$, $U_3U_4$ and $L_2U_4$, with the ordinates at the quarter, mid and three-quarter points (6 m, 12 m and 18 m).

Approach. For the vertical, inspect joint $U_3$; for the chord and the diagonal, cut the panel $U_3U_4$ and use moments about $L_2$ and vertical resolution respectively, once with the unit load left of the cut and once with it right of it.

1. The vertical $U_3L_2$ — inspection of joint $U_3$. Only three members meet at $U_3$: the collinear chord members $U_2U_3$ and $U_3U_4$, and the vertical $U_3L_2$. Neither adjacent diagonal touches $U_3$. Vertical equilibrium of the joint therefore reads $$-F_{U_3L_2} - P_{U_3} = 0 \;\Rightarrow\; F_{U_3L_2} = -P_{U_3}$$ so the member force equals minus the load standing at $U_3$ and is zero for a unit load at any other panel point. The influence line is a single triangle: $$\boxed{\text{IL}(U_3L_2):\quad 0,\ 0,\ -1.000,\ 0,\ 0 \ \text{at } U_1 \ldots U_5}$$
2. Set up the section for the other two. Cut vertically between $U_3\,(12,0)$ and $U_4\,(18,0)$. The cut severs the top chord $U_3U_4$, the diagonal $L_2U_4$ and the bottom chord $L_2L_3$ — exactly three members. The two chords are horizontal and intersect at infinity; the bottom chord and the diagonal both pass through $L_2\,(12,-2.5)$.
3. Top chord $U_3U_4$ — moments about $L_2$. With the unit load to the right of the cut, the left free body carries only $R_{U_1}$, whose lever arm about $L_2$ is 12 m, against the chord force acting at the 2.5 m depth: $$-12R_{U_1} - 2.5\,F_{U_3U_4} = 0 \;\Rightarrow\; F_{U_3U_4} = -4.8\,R_{U_1}$$ With the load to the left, the right free body gives the mirror result $F_{U_3U_4} = -4.8\,R_{U_5}$. Evaluating with $R_{U_1} = (24-x)/24$ and $R_{U_5} = x/24$: $$\boxed{\text{IL}(U_3U_4):\quad 0,\ -1.200,\ -2.400,\ -1.200,\ 0}$$ Compression throughout, as a top chord should be.
4. Diagonal $L_2U_4$ — vertical resolution on the same section. The two cut chords are horizontal, so only the diagonal carries vertical force across the cut. Its vertical direction cosine is $2.5/6.5$. Unit load left of the cut: $$R_{U_5} - \tfrac{2.5}{6.5}F_{L_2U_4} = 0 \;\Rightarrow\; F_{L_2U_4} = 2.6\,R_{U_5}$$ Unit load right of the cut: $F_{L_2U_4} = -2.6\,R_{U_1}$. Hence $$\boxed{\text{IL}(L_2U_4):\quad 0,\ +0.650,\ +1.300,\ -0.650,\ 0}$$ The sign reversal between the mid point and the three-quarter point is the diagonal changing from tension to compression as the load crosses the panel, which is exactly what an influence line for a web diagonal is for.
5. Sense check on the numbers. The three ratios are pure geometry: $4.8 = 12/2.5$ is half the span over the depth, $2.6 = 6.5/2.5$ is the diagonal length over the depth, and the vertical's peak is exactly $-1$ because a unit load standing on the joint must be carried straight down the hanger. Change the depth and only the first two move.

5(b) Shear influence line and a moving uniform load

Given. A simply supported beam of span 25 m, pin at the left support and roller at the right. Section ①–① is 5 m from the left support. The vehicle is idealised as a uniformly distributed load of $20\ \text{kN}\,\text{m}^{-1}$ over a length of 10 m, travelling from left to right.

Find. The influence line for shear at the section with its ordinates, and the maximum absolute shear the vehicle can produce there.

Approach. Write the influence line from the two support reactions, then place the 10 m load block where it captures the greatest area under the line; the shear is the load intensity times that area.

1. Ordinates of the influence line. For a unit load at distance $x$ from the left support, $R_{\text{left}} = (25-x)/25$. Taking the shear as the upward force on the left portion, $$V = -\frac{x}{25}\ \ (x < 5), \qquad V = \frac{25-x}{25}\ \ (x > 5)$$ so the line runs from 0 at the left support down to $$V_{\text{left of section}} = -\frac{5}{25} = \boxed{-0.200}$$ jumps by unity across the section to $$V_{\text{right of section}} = \frac{20}{25} = \boxed{+0.800}$$ and returns linearly to zero at the right support.
2. Governing position for positive shear. The positive lobe is the triangle from 5 m to 25 m with its peak $+0.800$ at the section, so the 10 m block should sit hard against the section, occupying 5 m to 15 m. The ordinates at its ends are $0.800$ and $(25-15)/25 = 0.400$, so the area under the line is $$A^{+} = \frac{0.800 + 0.400}{2}\,(10) = 6.00\ \text{m}$$
3. Maximum positive shear. $$V^{+}_{\max} = w A^{+} = 20(6.00) = \boxed{120.0\ \text{kN}}$$
4. Governing position for negative shear. The negative lobe is only 5 m long and its deepest ordinate is $-0.200$, so at best the vehicle covers it entirely (front axle at the section, the rear 5 m still off the span) and picks up an area of $\tfrac12(0.200)(5) = 0.50$ m: $$V^{-}_{\max} = 20(0.50) = -10.0\ \text{kN}$$ Pushing the vehicle further on only adds positive area, so this is as negative as it gets. If the vehicle is required to sit wholly on the beam, the most adverse position it can reach is 0 m to 10 m, which already gives $+60.0$ kN.
5. Answer. Comparing the two, $$|V|_{\max} = \max(120.0,\ 10.0) = \boxed{120.0\ \text{kN}}$$ with the vehicle occupying 5 m to 15 m, i.e. its rear immediately at the section.