Question 5 of 8: Influence lines for a compound beam and for a truss diagonal

Question 5: Influence lines for a compound beam and for a truss diagonal (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.

(a) Influence lines for the compound beam

Given. A 16 m beam with free ends. Support 1 is a pin at $x = 2$ m, support 2 a roller at $x = 8$ m, an internal hinge lies at $x = 10$ m, and support 3 is a roller at $x = 14$ m. The dimension string is 2 + 6 + 2 + 4 + 2 m.

Find. The influence lines for the bending moment at support 2, the bending moment at mid-span of span 1–2 ($x = 5$ m), and the shear immediately to the left of support 2, together with the largest absolute ordinate of each.

Approach. The hinge splits the beam into two rigid lengths: body I from the left end to the hinge (carrying supports 1 and 2) and body II from the hinge to the right end (carrying support 3). Place the unit load in each region in turn, find the reactions, and evaluate the required quantity from the free body on one side of the section.

1. Reactions with the unit load on body I ($0 \le x \le 10$ m). Body II is then unloaded, so it carries nothing and the hinge transmits no force. Body I is simply supported by 1 and 2, giving $$ R_2 = \frac{x-2}{6}, \qquad R_1 = 1 - R_2, \qquad R_3 = 0 . $$
2. Reactions with the unit load on body II ($10 \le x \le 16$ m). Body II is carried by the hinge at 10 m and the roller at 14 m, so moments about support 3 give the hinge force $$ V_h = -\frac{x-14}{4} , \qquad R_3 = 1 - V_h , $$ $V_h$ being the upward force that body I applies to body II. Body I then carries a downward force $V_h$ at its right-hand end, and $R_2 = \tfrac{8}{6}V_h$, $R_1 = V_h - R_2$.
3. (i) Bending moment at support 2. Take the free body to the right of $x = 8$ m, which contains only the roller reaction $R_3$ and the unit load if it lies beyond the section, so $M_2 = 6R_3 - \langle x-8\rangle$. For any load position between 0 and 8 m the whole right-hand length is unloaded and $M_2 = 0$; the line is a flat zero over that entire stretch. Between 8 m and the hinge it falls linearly to $$ \boxed{M_2 = -2.00\ \text{m at the hinge}} $$ recovers to zero at support 3, and rises to $+1.00$ m at the right free end. The largest absolute ordinate is $2.00$ m.
4. (ii) Bending moment at mid-span of 1–2 ($x = 5$ m). Taking the free body to the left, $M = 3R_1 - \langle 5-x \rangle$. The line starts at $-1.00$ m at the left free end, crosses zero at support 1, peaks under the section itself at $$ \boxed{M_{5} = +1.50\ \text{m}} $$ (the familiar $ab/L = 3\times3/6$ for a simply supported span), returns to zero at support 2, falls to $-1.00$ m at the hinge, returns to zero at support 3 and rises to $+0.50$ m at the right end.
5. (iii) Shear immediately left of support 2. Here $V = R_1 - \langle 1 \rangle$ for a load to the left of the section, so the line begins at $+\tfrac{1}{3}$ at the left free end, passes through zero at support 1 and falls to $$ \boxed{V_{8^-} = -1.00} $$ immediately left of the section. It steps to zero just to the right of the support, falls to $-\tfrac{1}{3}$ at the hinge, returns to zero at support 3 and rises to $+\tfrac{1}{6}$ at the right end. The governing ordinate is $1.00$.

(b) Influence line for the truss diagonal and the vehicle

Given. A 20 m truss in five 4 m panels, pinned at L₁ and on a roller at L₆. The top chord is at 3 m above the bottom chord at U₁ and U₄ and at 4 m at U₂ and U₃. The load travels on floor beams at bottom-chord level. The vehicle is three point loads — 18 kN, 18 kN and 8 kN — spaced 4 m and 2 m apart, the 8 kN axle leading.

Find. The ordinates of the influence line for the force in U₁–L₃ and the maximum force the vehicle can produce in that member.

Approach. Cut the panel L₂–L₃, which severs exactly three members, and take moments about the point where the other two cut members meet. Because the top chord is not parallel to the bottom chord, that point is not at infinity; locating it is the whole of the work.

1. Locate the moment centre. A vertical cut between L₂ and L₃ severs the top chord U₁–U₂, the diagonal U₁–L₃ and the bottom chord L₂–L₃. The top chord runs from $(4, 3)$ to $(8, 4)$, so its axis produced meets the bottom chord ($y=0$) where $0 = 3 + (x-4)/4$, that is at $$ \boxed{O = (-8,\ 0)} $$ 8 m outside the left support. Both chords pass through $O$, so a moment equation about $O$ contains only the diagonal.
2. Unit load to the right of the cut. Keep the left-hand portion, on which the only external force is $R_{L_1} = (20-x)/20$ acting 8 m from $O$. The diagonal acts at U₁, whose position vector from $O$ is $(12, 3)$, with unit vector $(0.8, -0.6)$ towards L₃; its moment arm about $O$ works out to $9.6$ m. Hence $$ 8R_{L_1} = 9.6\,T \;\Rightarrow\; T = \tfrac{5}{6}R_{L_1} = \frac{20-x}{24} $$ which gives $+0.500$ at L₃, $+0.333$ at L₄, $+0.167$ at L₅ and zero at L₆.
3. Unit load to the left of the cut. Now keep the right-hand portion, on which the only external force is $R_{L_6} = x/20$, acting 28 m from $O$, while the diagonal acts at L₃, 16 m from $O$. Hence $$ 28R_{L_6} + 9.6\,T = 0 \;\Rightarrow\; T = -\tfrac{35}{12}\,\frac{x}{20} $$ so the ordinate at L₂ is $$ \boxed{\eta_{L_2} = -\tfrac{7}{12} = -0.583} $$ and zero at L₁. Between L₂ and L₃ the load is carried by the floor beam of that panel and the line is straight, which is the break in the drawn shape.
4. Assemble the influence line. The ordinates at the six bottom-chord joints are $$ 0,\quad -0.583,\quad +0.500,\quad +0.333,\quad +0.167,\quad 0 $$ at L₁ to L₆ respectively, joined by straight lines. This is exactly the shape printed on the examination paper: a dip below the axis under L₂, a peak above it at L₃ and a long straight fall to zero at the far support.
5. Position the vehicle for maximum tension. The peak ordinate is at L₃, so place the two heavy axles at L₃ and L₄, which the 4 m spacing allows exactly; the 8 kN axle is then 2 m beyond L₄, where the ordinate is $\tfrac{1}{3} - \tfrac{1}{2}\left(\tfrac{1}{3}-\tfrac{1}{6}\right) = 0.250$. Hence $$ T = 18(0.500) + 18(0.333) + 8(0.250) = 9 + 6 + 2 = \boxed{17.0\ \text{kN tension}} $$
6. Position it for maximum compression. The negative peak is at L₂. Placing the rear 18 kN axle at L₁, where the ordinate is zero, puts the second 18 kN axle on the peak and the 8 kN axle 2 m beyond it, at an ordinate of $-\tfrac{1}{24}$, giving $$ T = 18(0) + 18(-0.583) + 8(-0.0417) = \boxed{-10.8\ \text{kN (compression)}} $$ A search over all positions confirms these two as the extremes, so the greatest force the vehicle can cause in U₁–L₃ is 17.0 kN tension.