Question 4 of 9: Influence lines for a wall-braced pin-jointed truss (16 marks)

Question 4: Influence lines for a wall-braced pin-jointed truss (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 loaded only at its bottom-chord panel points.

Find. The influence line for the force in U4U5, U2L3, U1U2 and U4L5, and the maximum ordinate on each.

Approach. Confirm determinacy, notice that the reaction at U1 must act along U1U2 and that the line U1–U2–L3 is straight, then read every ordinate off one moment equation and two section cuts.

1. Check that the truss is determinate. There are $m=15$ members, $n=10$ joints and $r=5$ reaction components (two pins and one roller), so $$m+r-2n=15+5-20=0.$$ The count only works because there is no chord between U2 and U4 — the sawtooth panel L2–L4 is unbraced along the top, which on its own would leave the truss one member short of stability. The wall joint U1, its single member U1U2 and the two reaction components there restore exactly that one degree, which is why the question asks for the force in U1U2.
2. Use joint U1 to fix the direction of the wall reaction. Only one member meets U1, so the two equilibrium equations there force the reaction to be collinear with U1U2. That member runs from $(0,8)$ to $(4,4)$, i.e. along $(1,-1)/\sqrt{2}$ — and extending that line one more panel reaches $(8,0)$, which is the joint L3. The reaction at U1 therefore passes through L3, and so do the members U2L3 and L3U4.
3. Take moments about L3 on the left free body. Cut the truss anywhere in the sawtooth panel: only two members are severed (a bottom chord and a diagonal), and both of their lines pass through L3, as does the U1 reaction. Moments about L3 for everything to the left therefore contain only $V_{L1}$ and the loads: $$8\,V_{L1}=\sum_{\text{loads left of L3}} P\,(8-x_{P}).$$ A unit load at L1 gives $V_{L1}=1$, at L2 gives $V_{L1}=0.5$, and anywhere at or right of L3 gives $V_{L1}=0$.
4. And about L3 for the whole truss. Repeating the same moment equation for the complete structure and subtracting the previous one leaves only the roller: $$12\,V_{L6}=\sum_{\text{loads right of L3}} P\,(x_{P}-8),$$ so a unit load gives $V_{L6}=0$ at L1, L2, L3, then $\tfrac13$ at L4, $\tfrac23$ at L5 and $1$ at L6.
5. (c) Influence line for U1U2. Vertical equilibrium of the whole truss gives the vertical component of the wall reaction, $R_{U1,y}=1-V_{L1}-V_{L6}$, and because that reaction is at $45^{\circ}$ the member force is $$\boxed{\eta_{U1U2}=\sqrt{2}\,\bigl(1-V_{L1}-V_{L6}\bigr)}$$ Ordinates: $0$, $+0.707$, $+1.414$, $+0.943$, $+0.471$, $0$ at L1…L6, so the member is always in tension and the maximum ordinate is $+\sqrt{2}=+1.414$ with the load at L3.
6. (b) Influence line for U2L3. Cut between L2 and L3 and take vertical equilibrium of the left free body. The bottom chord is horizontal and contributes nothing, so the diagonal alone balances the vertical forces: $$\boxed{\eta_{U2L3}=\sqrt{2}\,\bigl(R_{U1,y}+V_{L1}-P_{\text{left}}\bigr)}$$ giving $0$, $0$, $+1.414$, $+0.943$, $+0.471$, $0$. The ordinate is zero for a load at L2 because the load, the pin reaction and the wall reaction then balance without the diagonal; the peak is again $+1.414$ (tension) with the load at L3.
7. (a) Influence line for U4U5. Cut between L4 and L5, which severs the top chord, the diagonal U4L5 and the bottom chord. Take moments about L5 on the right free body: both the diagonal and the bottom chord pass through L5, leaving the top chord with its 4 m lever arm: $$4\,F_{U4U5}+4\,V_{L6}+\sum_{\text{right}}P\,(x_{P}-16)=0\;\Rightarrow\;\boxed{\eta_{U4U5}=-V_{L6}-\tfrac14\!\!\sum_{\text{right}}\!(x_{P}-16)}$$ Ordinates: $0,\,0,\,0,\,-0.333,\,-0.667,\,0$. The member is in compression, maximum ordinate $-0.667$ with the load at L5.
8. (d) Influence line for U4L5. Same cut, vertical equilibrium of the right free body: $$\boxed{\eta_{U4L5}=-\sqrt{2}\,\bigl(V_{L6}-P_{\text{right}}\bigr)}$$ giving $0,\,0,\,0,\,-0.471,\,+0.471,\,0$. This diagonal reverses: $0.471$ compression with the load at L4 and $0.471$ tension with it at L5, so both ordinates must be shown.
9. Draw the lines. Because the load travels on the bottom chord and is transferred only at the panel points, every influence line is a straight segment between consecutive ordinates. Note that all four lines are zero at L1 and at L6: a load sitting directly over a support passes straight into it. Note also that U1U2 and U2L3 share the same ordinates from L3 rightwards — the wall reaction and the diagonal are then carrying the same panel force.