Question 7 of 8: Influence lines for a truss and for a hinged frame under a moving vehicle

Question 7: Influence lines for a truss and for a hinged frame under a moving vehicle (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.

Find. (a) Three influence lines with the largest influence coefficient of each. (b) The influence line for bending moment immediately left of C, and the largest negative bending moment the vehicle can produce there.

Approach. For the truss, place the unit load at each bottom-chord joint in turn and analyse; because the load is transferred at panel points the influence line is straight between them. For the frame, work out the ordinate as a function of the load position from the two free bodies the hinges create, then slide the 4 m block to the position that maximises the area of the negative lobe.

1. Part (a) — how the truss carries a unit load. The pin at $L_2$ and the roller at $L_4$ are 12 m apart with a 6 m overhang beyond each. For a unit load at bottom joint $L_i$ the reactions follow from moments about the far support, so a load on an overhang lifts the far reaction. For example a unit load at $L_1$ gives $V_{L_2} = 1.5$ upward and $V_{L_4} = 0.5$ downward, while a load at $L_3$ splits equally, 0.5 at each support.
2. Part (a) — influence line for the top chord $U_1U_2$. Cut just to the right of $L_2$, severing $U_1U_2$, $L_2U_2$ and $L_2L_3$, and take moments about $L_2$, which lies on the line of the other two cut members. With the top chord 4 m above the bottom chord, its lever arm about $L_2$ is 4 m. With the load on the span the left part carries nothing but the reaction at $L_2$, which has no lever arm, so the ordinate is zero at $L_2$ and everywhere to its right. With the unit load at $L_1$, $$4\,F_{U_1U_2} = 1\,(6) \;\Rightarrow\; \eta = +1.500$$ The influence line is therefore a straight fall from $+1.500$ at $L_1$ to zero at $L_2$ and zero thereafter, so $$\eta_{\max} = \boxed{+1.500\ \text{at}\ L_1\ \text{(tension)}}$$
3. Part (a) — influence line for the bottom chord $L_2L_3$. Cut between $L_2$ and $L_3$ and take moments about $U_2$ at $(9,\,4)$, the point where the top chord and the diagonal $L_2U_2$ meet. The lever arm of the bottom chord about $U_2$ is 4 m. Placing the unit load in turn at each joint gives ordinates $$\eta_{L_1} = -1.125, \quad \eta_{L_2} = 0, \quad \eta_{L_3} = +0.375, \quad \eta_{L_4} = 0, \quad \eta_{L_5} = -0.375$$ so the largest coefficient is $$\eta_{\max} = \boxed{-1.125\ \text{at}\ L_1\ \text{(compression)}}$$ The chord goes into compression whenever the load is on either overhang and into tension when it is inside the span, which is the signature of a bottom chord in a beam with two cantilevers.
4. Part (a) — influence line for the diagonal $U_2L_3$. Move the cut to the right of $U_2$, between $x = 9$ m and $x = 12$ m, where it severs $U_2U_3$, $U_2L_3$ and $L_2L_3$. Both chords are horizontal, so vertical equilibrium of the left part isolates the diagonal alone. It runs from $(9,\,4)$ to $(12,\,0)$, so its vertical direction cosine is $4/5$ and the panel shear divided by $0.8$ gives the force. Evaluating at each joint, $$\eta_{L_1} = +0.625, \quad \eta_{L_2} = 0, \quad \eta_{L_3} = +0.625, \quad \eta_{L_4} = 0, \quad \eta_{L_5} = -0.625$$ so the largest absolute coefficient is $$|\eta|_{\max} = \boxed{0.625}$$ reached in tension with the load at $L_1$ or at $L_3$ and in compression with the load at $L_5$.
5. Part (b) — what the hinges do. The beam is in three pieces: A to B on the roller at A and hanging from the hinge at B, the rigid tee B to C to D carried by the built-in column, and D to E on the hinge at D and the roller at E. A load between D and E, or between C and D, is carried entirely by parts that lie to the right of the section just left of C, so it produces no moment there at all. Only a load between A and C can register.
6. Part (b) — build the ordinate. With the unit load at distance $a$ from A, the piece A to B is a simple 6 m span between the roller and the hinge, so $$\begin{aligned} R_{A} &= \frac{6 - a}{6} \quad (0 \le a \le 6) \\ R_{A} &= 0 \quad (a > 6) \end{aligned}$$ Taking the free body to the left of the section just left of C, which is 8 m from A, $$\begin{aligned} \eta &= 8R_{A} - (8 - a) \quad (a < 8) \\ \eta &= 8R_{A} \quad (a \ge 8) \end{aligned}$$ Substituting the two ranges, $$\begin{aligned} \eta &= -\frac{a}{3} \quad (0 \le a \le 6) \\ \eta &= -(8 - a) \quad (6 \le a \le 8) \\ \eta &= 0 \quad (a \ge 8) \end{aligned}$$
7. Part (b) — the influence line. The result is a triangle, entirely negative, running from zero at A down to $$\eta_{B} = \boxed{-2.00\ \text{m}}$$ at the hinge B and back to zero at C, and flat at zero from C to E. The peak sits at the hinge, not at the section, because the hinge is where the left-hand piece hands its load to the cantilevering tee.
8. Part (b) — position the vehicle. A uniform load of intensity $q$ covering a length of the influence line produces $q$ times the area beneath that length, so the worst 4 m position is the one that maximises the negative area. Differentiating the area with respect to the position of the leading wheel shows the maximum occurs where the ordinates at the two ends of the block are equal, that is where $$-\frac{s}{3} = -(8 - s - 4) \;\Rightarrow\; s = 3\ \text{m}$$ so the vehicle occupies 3 m to 7 m from A, with $\eta = -1.00$ m at each end.
9. Part (b) — the largest negative moment. The area under that 4 m of influence line is one trapezium from 3 m to 6 m and a second from 6 m to 7 m: $$A = -\tfrac{1}{2}(1.00 + 2.00)(3) - \tfrac{1}{2}(2.00 + 1.00)(1) = -4.50 - 1.50 = -6.00\ \text{m}^{2}$$ so the largest negative bending moment immediately left of joint C is $$M = qA = 10\,(-6.00) = \boxed{-60.0\ \text{kN}\cdot\text{m}}$$ For comparison, a vehicle long enough to cover the whole triangle would give $10(-8.00) = -80$ kN·m, so the 4 m length captures three quarters of the worst possible effect.