Question 5 of 8: Influence lines for a determinate three-span structure, and a moving vehicle

Question 5: Influence lines for a determinate three-span structure, and 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. Part (a): a beam ABCD with a pin at A ($x = 0$), rollers at B ($x = 8$ m), C ($x = 14$ m) and D ($x = 22$ m), and internal hinges at $x = 6$ m and $x = 16$ m. Part (b): a beam ABC with a pin at A ($x = 0$), rollers at B ($x = 8$ m) and C ($x = 14$ m) and an internal hinge at $x = 6$ m, crossed left to right by three axle loads of 100, 100 and 40 kN spaced 1.5 m and 3 m apart, the 40 kN leading.

Find. Three influence lines and their governing ordinates for part (a); the influence line for shear just right of B and the greatest shear the vehicle can produce there, for part (b).

(a) Three influence lines

Approach. Break the structure at its hinges into a primary span BC with overhangs and two suspended spans, place a unit load at a general station, and evaluate the required quantity directly; the ordinates are linear between the hinges and supports, so plotting a handful of stations is enough.

1. Identify the load paths. The middle piece runs from hinge to hinge, from $x = 6$ m to $x = 16$ m, and is carried on the rollers at B and C, so it is the primary element. The piece AB' from A to the first hinge is carried by the pin at A at one end and by the tip of the primary element at the other; the piece from the second hinge to D is carried by the primary element and by the roller at D. A unit load on a suspended span therefore reaches the primary element only as the hinge reaction $z/6$ or $(22-z)/6$.
2. (i) Bending moment at B. Everything to the left of the section at $x = 8$ m consists of the unloaded suspended span and the 2 m stub of the primary element, so for any load beyond the section the moment is exactly zero. With the unit load at $z$ on the left suspended span the reaction at A is $(6-z)/6$ and $$M_{B} = \frac{6-z}{6}(8) - (8 - z) = -\frac{z}{3}$$ which falls linearly from $0$ at A to $-2$ m at the hinge, then returns linearly to zero at B. The governing coefficient is $$\boxed{M_{B,\max} = -2.0\ \text{m at } x = 6\ \text{m}}$$ The influence line is wholly negative: no position of the load can put sagging moment at B.
3. (ii) Bending moment at mid-span of BC ($x = 11$ m). For a load inside BC the ordinate is the familiar simply supported triangle peaking at $ab/L = (3)(3)/6 = 1.5$ m under the section. Beyond the supports the overhangs reverse the sign: with the load at the left hinge the reaction at C is $-1/3$ and $M_{11} = 3(-1/3) = -1$ m, and by symmetry the right hinge gives $-1$ m as well, both tapering to zero at A and at D. Hence $$\boxed{M_{\text{mid},\max} = +1.5\ \text{m at } x = 11\ \text{m}}$$ with negative ordinates of $-1.0$ m at each hinge.
4. (iii) Reaction at A. Only load on the first suspended span can reach A, and that span behaves as a simply supported beam of 6 m between A and the hinge, so the influence line is the straight line $R_{A} = (6-z)/6$ from $1.0$ at A to zero at the hinge, and identically zero everywhere to the right of it: $$\boxed{R_{A,\max} = 1.0 \ \text{(dimensionless), with the load at A}}$$ This is a useful reminder that in a hinged multi-span beam a support can be completely shielded from load elsewhere on the structure.

(b) Shear just right of B under the moving vehicle

1. Construct the influence line. The hinge at $x = 6$ m again splits the beam into a suspended span A-hinge and a primary span from the hinge to C carried on B and C. For a unit load at $z$ on the suspended span, the reaction at A is $(6-z)/6$, the hinge delivers $z/6$ to the primary element and the reaction at B is $\tfrac{4}{3}(z/6)$, so $$\eta(z) = \frac{6-z}{6} + \frac{4}{3}\cdot\frac{z}{6} - 1 = \frac{z}{18}$$ rising from zero at A to $0.333$ at the hinge. Between the hinge and B the load is still to the left of the section, so $\eta = (8-z)/6$, falling from $0.333$ back to zero. At the section itself the ordinate jumps by unity, and from B to C it falls linearly as $\eta = (14-z)/6$ from $1.000$ to zero.
2. Read off the maximum ordinate in each span. $$\eta_{\max}\big|_{AB} = 0.333 \ \text{at the hinge}, \qquad \eta_{\max}\big|_{BC} = 1.000 \ \text{just right of B}$$ The influence line is positive everywhere, so every axle on the structure adds to the shear and the vehicle should be positioned to sit as much as possible on the tall triangle in span BC.
3. Position the vehicle. Because the influence line is piecewise linear, the maximum of $\sum P_{i}\eta_{i}$ occurs with one axle at the peak. Trying each in turn, with the rear 100 kN axle at B the other two sit at $9.5$ m and $12.5$ m, where $\eta = 0.750$ and $0.250$: $$S = 100(1.000) + 100(0.750) + 40(0.250) = 100 + 75 + 10 = 185\ \text{kN}$$ With the middle 100 kN axle at B the axles sit at $6.5$, $8.0$ and $11.0$ m and $S = 25 + 100 + 20 = 145$ kN; with the leading 40 kN axle at B, $S = 87.2$ kN. The rear axle governs.
4. Maximum shear. Hence $$\boxed{V_{\max,\,B^{+}} = 185\ \text{kN}}$$ occurring when the rear 100 kN axle stands immediately to the right of B, the second 100 kN axle is 1.5 m further along and the 40 kN axle is 4.5 m along the span. Because the vehicle travels to the right and the 40 kN axle leads, this is the last of the three critical positions the vehicle passes through.