Question 4 of 9: Influence lines for a pin-jointed truss loaded on the top chord

Question 4: Influence lines for a pin-jointed truss loaded on the top chord (16 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 pin-jointed truss of four 5 m panels (20 m overall) and 5 m depth, pinned at $L_1$ and on a roller at $L_5$. Both chords are continuous; verticals connect every pair $U_kL_k$; the web diagonals are $U_1L_2$, $U_1L_3$, $U_5L_4$ and $U_5L_3$, and the long diagonals cross the verticals $U_2L_2$ and $U_4L_4$ without being connected to them. With 17 members, 10 joints and 3 reaction components, $m+r=17+3=20=2n$, so the truss is statically determinate; a rank check on the joint equilibrium equations confirms that it is also stable. A unit load travels along the top chord.

Find. The influence lines for the axial forces in $U_1L_2$, $U_1L_3$ and $L_2L_3$, with the maximum and minimum ordinates marked (tension positive).

Approach. Place the unit load at each top-chord panel point in turn, find the reaction, and take a section or isolate a joint that exposes the wanted member; because the load is delivered through stringers at the panel points, the influence lines are straight between those points.

1. Reactions. With the load a distance $x$ from $L_1$ on a 20 m span, $$R_{L1}=\frac{20-x}{20},\qquad R_{L5}=\frac{x}{20},$$ so the reactions at the five load positions $U_1\ldots U_5$ are $1,\ 0.75,\ 0.50,\ 0.25,\ 0$ and $0,\ 0.25,\ 0.50,\ 0.75,\ 1$ respectively.
2. (c) $L_2L_3$ first — the section that unlocks the panel. Cut a vertical section between $U_2$ and $U_3$. It severs exactly three members: the top chord $U_2U_3$, the long diagonal $U_1L_3$ and the bottom chord $L_2L_3$. The top chord and the diagonal both pass through the joint $U_1$ at $(0,\,5)$, so taking moments about $U_1$ for the left-hand free body isolates $L_2L_3$: $$5\,F_{L_2L_3} + \sum \left(\text{moments of the left-hand loads about } U_1\right)=0 .$$ The reaction $R_{L1}$ acts vertically through the abscissa of $U_1$, so it contributes no moment about $U_1$ for any load position.
3. (c) Evaluate the ordinates. When the load stands at $U_2$ it lies to the left of the section, 5 m to the right of $U_1$, and $$5\,F_{L_2L_3}-5(1)=0\;\Longrightarrow\;F_{L_2L_3}=+1.000 .$$ For the load at $U_1$ the lever arm is zero, and for the load anywhere to the right of the section the left-hand free body carries only $R_{L1}$, whose moment about $U_1$ also vanishes. Hence $$\boxed{\eta_{L_2L_3}=\bigl(0,\;+1.000,\;0,\;0,\;0\bigr)\ \text{at } U_1\ldots U_5 ,}$$ a single triangle peaking at $U_2$ — maximum $+1.000$ (tension), minimum $0$. The member is completely unstressed for every load position beyond $U_3$.
4. (b) $U_1L_3$ by vertical equilibrium of the same free body. Of the three cut members only the diagonal has a vertical component. Its direction cosines follow from its projection, 10 m horizontally and 5 m vertically, so $L=\sqrt{125}=11.180\text{ m}$ and $c_y=5/11.180=0.4472$. Vertical equilibrium of the left-hand free body gives $$R_{L1}-\left(\text{loads left of the section}\right)-0.4472\,F_{U_1L_3}=0 .$$
5. (b) Evaluate the ordinates. With the load at $U_2$ (left of the section) $0.75-1-0.4472F=0$, so $F=-0.559$; with the load at $U_3$, $U_4$ or $U_5$ (right of the section) $F=R_{L1}/0.4472$, giving $+1.118$, $+0.559$ and $0$; and with the load at $U_1$ the reaction and the load cancel, so $F=0$: $$\boxed{\eta_{U_1L_3}=\bigl(0,\;-0.559,\;+1.118,\;+0.559,\;0\bigr) .}$$ The ordinates are $\mp\sqrt5/4$ and $+\sqrt5/2$ exactly. Maximum $+1.118$ (tension, load at $U_3$); minimum $-0.559$ (compression, load at $U_2$).
6. (a) $U_1L_2$ by isolating two joints. A vertical section through the first panel cuts four members, so use joints instead. At $U_2$ the only member with a vertical component is the vertical $U_2L_2$ (the two chords are collinear and horizontal, and the long diagonal misses the joint), so a unit load at $U_2$ puts $F_{U_2L_2}=-1.000$ (compression) and every other load position leaves it at zero. At $L_2$ the vertical component of $U_1L_2$ must then balance it; that diagonal rises 5 m over 5 m, so $c_y=1/\sqrt2$ and $$-1.000+\frac{1}{\sqrt2}F_{U_1L_2}=0\;\Longrightarrow\;F_{U_1L_2}=+\sqrt2=+1.414 .$$
7. (a) Complete the influence line. For every other load position $F_{U_2L_2}=0$, and joint $L_2$ then requires $F_{U_1L_2}=0$ as well: $$\boxed{\eta_{U_1L_2}=\bigl(0,\;+1.414,\;0,\;0,\;0\bigr) ,}$$ maximum $+1.414$ (tension), minimum $0$. Like $L_2L_3$, this diagonal responds only to load in the first two panels.
8. Sanity check the shapes. All three influence lines vanish at both supports, as they must for a load standing directly over a support of a determinate truss. The two members that are activated only by load at $U_2$ are precisely those that hang the second panel point off the end joint $U_1$; the long diagonal $U_1L_3$ is the only web member crossing the second and third panels, which is why it alone carries a full-length influence line with a sign reversal.