Question 6 of 8: Influence lines for a truss and for shear in a beam

Question 6: Influence lines for a truss and for shear in a beam (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.

Given. (a) A 24 m Warren truss whose panel points on the loaded (top) chord are $U_{1}(0,4.5)$, $U_{2}(12,4.5)$ and $U_{3}(24,4.5)$, with bottom joints $L_{1}(6,0)$ and $L_{2}(18,0)$; pin at $U_{1}$, roller at $U_{3}$; loads travel along the top chord. (b) A 19 m beam with a pin 2 m from the left end and a roller 2 m from the right end, section A–A 6 m to the right of the pin; the vehicle is 100 kN, 100 kN and 50 kN at spacings of 1.5 m and 3 m.

Find. The two truss influence lines with their governing coefficients and senses, the shear influence line at A–A with its maximum and minimum ordinates, and the greatest shear the vehicle can produce there.

Approach. Cut the truss between $L_{1}$ and $U_{2}$ and use moments about $L_{1}$ for the top chord and vertical equilibrium for the diagonal, expressing the panel-point distribution of a unit load at any station; for the beam, write the shear influence ordinate directly from the reaction and place the wheel group where the ordinates are largest.

6(a) — truss influence lines

1. Set up the section. A vertical cut anywhere between $L_{1}$ and $U_{2}$ severs exactly three members: the top chord $U_{1}-U_{2}$, the diagonal $L_{1}-U_{2}$ and the bottom chord $L_{1}-L_{2}$. Two of those three meet at $L_{1}$, so moments about $L_{1}$ isolate the top chord, while the two chords are horizontal, so vertical equilibrium isolates the diagonal.
2. Unit load at station $a$ from $U_{1}$. The reactions are $R_{U_{1}} = (24 - a)/24$ and $R_{U_{3}} = a/24$. A load anywhere in the panel $U_{1}U_{2}$ is shared by the stringer as $(12 - a)/12$ at $U_{1}$ and $a/12$ at $U_{2}$.
3. Top chord $U_{1}-U_{2}$. Taking moments about $L_{1}$ for the left-hand portion, with the chord acting on a 4.5 m lever arm and $U_{1}$ 6 m to the left of $L_{1}$, for $0 \le a \le 12$ m $$-6\left[\frac{24 - a}{24} - \frac{12 - a}{12}\right] - 4.5\,F = 0 \;\Rightarrow\; F = -\frac{a}{18}$$ and for $12 \le a \le 24$ m the whole load is on the right portion, giving $F = -(24 - a)/18$. The influence line is therefore a triangle with its apex under $U_{2}$: $$F_{\max} = -\frac{12}{18} = \boxed{-0.667\ \text{(0.667 C)}}$$
4. Diagonal $U_{2}-L_{2}$. Cut between $U_{2}$ and $L_{2}$; the two chords crossed are horizontal, so the vertical component of the diagonal alone balances the net vertical force on the left portion. The diagonal runs from $(12,4.5)$ to $(18,0)$, a length of 7.5 m, so its vertical component is $4.5/7.5 = 0.6$. For $0 \le a \le 12$ m the entire unit load sits on the left portion: $$\frac{24 - a}{24} - 1 - 0.6F = 0 \;\Rightarrow\; F = -\frac{a}{14.4}$$ and for $12 \le a \le 24$ m the panel $U_{2}U_{3}$ splits the load, giving $F = -(24 - a)/14.4$. Both branches meet at $a = 12$ m, where $$F_{\max} = -\frac{12}{14.4} = \boxed{-0.833\ \text{(0.833 C)}}$$
5. Interpret the shapes. Both influence lines are wholly negative, so no position of a downward travelling load can put either member into tension: the top chord and this diagonal are permanently in compression, and the design load case is simply the heaviest wheel group standing at $U_{2}$.

6(b) — shear at section A–A

1. Influence ordinates. Measure $p$ from the left-hand end. With the pin at 2 m and the roller at 17 m, the span is 15 m and $R_{\text{pin}} = (17 - p)/15$. For a unit load left of the section, $V_{A} = R_{\text{pin}} - 1 = (2 - p)/15$; for a unit load right of it, $V_{A} = (17 - p)/15$. Hence $$V_{A}(0) = +0.133, \quad V_{A}(2) = 0, \quad V_{A}(8^{-}) = \boxed{-0.400}, \quad V_{A}(8^{+}) = \boxed{+0.600}$$ $$V_{A}(17) = 0, \quad V_{A}(19) = -0.133$$ The unit step across the section is the signature of a shear influence line.
2. Place the vehicle for maximum positive shear. The positive branch falls from $0.600$ at the section to zero at the roller at a rate of $1/15$ per metre, so the group should sit as far left as possible with all three wheels on that branch. Putting the leading 100 kN exactly at the section places the wheels at 8.0, 9.5 and 12.5 m, where the ordinates are $0.600$, $0.500$ and $0.300$: $$V_{A} = 100(0.600) + 100(0.500) + 50(0.300) = 60 + 50 + 15 = \boxed{125\ \text{kN}}$$
3. Test the alternatives. Advancing the group so that the second 100 kN sits at the section puts the first 100 kN on the negative branch and gives only 48.3 kN. Crossing in the opposite direction, so that the 50 kN leads, gives at best $50(0.600) + 100(0.400) + 100(0.300) = 100$ kN. Neither beats 125 kN.
4. Maximum negative shear. Now the wheels must sit on the negative branch, which runs from zero at the pin to $-0.400$ just left of the section. The group crossing the other way, with the 50 kN leading, places the wheels at 3.5, 6.5 and 8.0 m, where the ordinates are $-0.100$, $-0.300$ and $-0.400$: $$V_{A} = 50(-0.100) + 100(-0.300) + 100(-0.400) = -75.0\ \text{kN}$$
5. Design value. The section must therefore be designed for $\boxed{V_{A} = +125\ \text{kN}}$ and $-75.0$ kN, the positive case governing. A single sweep of the group across the beam at 0.5 mm steps confirms that no other position exceeds either figure.