Question 7 of 8: Influence Lines for a Truss and for a Gerber Beam

Question 7: Influence Lines for a Truss and for a Gerber 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.

(a) Influence lines for three truss members

Given. A truss whose top chord runs U₁ to U₇ in six panels of 4 m, a total of 24 m, with a bottom chord 3 m below carrying only three joints.

Find. The influence lines for U₂-U₃, L₁-U₃ and L₂-L₃, with the maximum tension and compression coefficients stated for each.

Approach. Place a unit load at each of the seven top-chord panel points in turn and solve the truss; join the resulting ordinates by straight lines, which is valid because the loads reach the truss through the floor beams at those panel points only. Two shortcuts remove most of the work.

1. Confirm the truss is determinate despite four supports. With $m = 15$, $j = 10$ and $r = 5$, $$m + r = 20 = 2j$$ so it is exactly determinate. Four supports under a determinate truss looks wrong at first sight but is quite regular here, because the bottom chord carries only three joints and the web is correspondingly sparse.
2. Use the standing-on-a-support rule. A unit load placed directly on a support joint is carried straight into the ground by that support and produces no force anywhere in the truss. Loads at U₁ and U₇ therefore give zero in every member, and four of the seven ordinates of each influence line are known before any equilibrium is written.
3. Use the two-member-joint rule at U₁ and U₇. Joint U₁ carries only the horizontal member U₁-U₂ and a vertical roller reaction. Vertical equilibrium gives $V_{U1} = 0$ for every load position, and horizontal equilibrium then gives $F_{U1U2} = 0$. The same argument at U₇ gives $F_{U6U7} = 0$ throughout.
4. Deduce the whole influence line for U₂-U₃. At joint U₂ the members are U₁-U₂ (always zero), the horizontal U₂-U₃ and the diagonal U₂-L₁ with direction cosines $(0.8,\ -0.6)$. With no load at U₂, vertical equilibrium forces $F_{U2L1} = 0$ and hence $F_{U2U3} = 0$. Only a load standing at U₂ itself can produce anything, and then $$-0.6\,F_{U2L1} - 1 = 0 \;\Rightarrow\; F_{U2L1} = -\tfrac{5}{3}, \qquad F_{U2U3} = 0.8 \times \tfrac{5}{3} = \boxed{+\tfrac{4}{3}}$$ The influence line is a single triangle peaking at U₂ with an ordinate of $+1.333$ and zero everywhere else.
5. Solve the remaining positions numerically. For loads at U₃, U₄, U₅ and U₆ the joint pattern is no longer reducible by inspection, so the $20 \times 20$ joint-equilibrium system is assembled and solved once per load position. A load standing at U₃ passes straight down the vertical L₁-U₃ into the pin below it, giving an ordinate of $-1.000$ there, and by the same argument a load at U₅ passes down U₅-L₃ into that support.
6. Collect the influence line for L₁-U₃. The ordinates at U₁ through U₇ are $$0,\; -0.500,\; \boxed{-1.000},\; -0.500,\; 0,\; \boxed{+0.500},\; 0$$ so the greatest compression coefficient is $1.000$, with the load at U₃, and the greatest tension coefficient is $0.500$, with the load at U₆.
7. Collect the influence line for L₂-L₃. By the mirror image of the U₂-U₃ argument — the truss is symmetric about U₄ for vertical loading — the ordinates are zero everywhere except at U₆, where $$F_{L2L3} = \boxed{-\tfrac{4}{3}} = -1.333$$ The member can therefore never be put into tension by a load travelling along the top chord; its maximum compression coefficient is $1.333$.

(b) Influence line for shear in a Gerber beam, and the critical vehicle position

Given. A beam pinned at A (x = 0), on a roller at B (x = 8 m), with an internal hinge at x = 10 m and a roller at C (x = 16 m); the dimension string reads 8 m + 2 m + 6 m. The idealized vehicle is three point loads travelling to the right: 64 kN, a second 64 kN 2 m ahead of it, and 20 kN a further 3.2 m ahead, so the 20 kN axle leads.

Find. The influence line for shear immediately to the left of B, and the largest such shear force the vehicle can produce.

Approach. Build the influence line piece by piece, treating the beam as a main span A-B with a 2 m cantilever to the hinge, carrying a suspended span from the hinge to C; then position the axle group so that a heavy axle stands on the peak ordinate and test each axle in that position.

1. Identify the determinate arrangement. The four reaction components (pin at A, rollers at B and C) are matched by three equations of equilibrium plus the one condition the internal hinge supplies, so the beam is determinate. The part from the hinge to C is a suspended span carried at one end by the cantilever tip of the main part and at the other by C.
2. Load on the main part, to the left of B. With the unit load at $a \le 8$ m the suspended span is unloaded and transmits nothing, so $R_A = (8 - a)/8$ and $$\eta(a) = R_A - 1 = -\frac{a}{8}$$ which falls linearly from $0$ at A to $\boxed{-1.000}$ just left of B.
3. Load between B and the hinge. Now the unit load is to the right of the section, so $\eta = R_A = (8 - a)/8$, which is negative because $R_A$ itself has reversed: $$\eta(8^{+}) = 0, \qquad \eta(10) = \frac{8 - 10}{8} = \boxed{-0.250}$$ The influence line therefore jumps by $+1$ across B, as an influence line for shear always does.
4. Load on the suspended span. A unit load at $a$ between 10 m and 16 m is shared between C and the hinge, the hinge taking $(16 - a)/6$. That force acts on the cantilever tip at $x = 10$ m, so $$\eta(a) = -0.25 \times \frac{16 - a}{6}$$ which runs linearly from $-0.250$ at the hinge to zero at C. Every ordinate on the whole beam is negative or zero, so the critical shear is negative.
5. Try the trailing 64 kN axle on the peak. Placing it at $8^{-}$ puts the other 64 kN at $x = 10$ m and the 20 kN at $x = 13.2$ m, with ordinates $-1.000$, $-0.250$ and $-0.117$: $$V = 64(1.000) + 64(0.250) + 20(0.117) = 82.3\ \text{kN}$$
6. Try the middle 64 kN axle on the peak. Now the trailing 64 kN sits at $x = 6$ m and the 20 kN at $x = 11.2$ m, with ordinates $-0.750$, $-1.000$ and $-0.200$: $$V = 64(0.750) + 64(1.000) + 20(0.200) = 48 + 64 + 4 = \boxed{116\ \text{kN}}$$
7. Try the leading 20 kN axle on the peak, and conclude. That puts the two 64 kN axles at $x = 4.8$ m and $x = 2.8$ m, ordinates $-0.600$ and $-0.350$, giving $20 + 38.4 + 22.4 = 80.8$ kN. Because the influence line is piecewise linear with its extreme ordinate at $8^{-}$, the maximum must occur with some axle standing there, and all three have now been tested. The largest shear immediately left of B is therefore $$V_{\max} = \boxed{116\ \text{kN}} \quad \text{(negative shear)}$$ with the middle 64 kN axle at B, the trailing 64 kN axle 2 m back at $x = 6$ m and the leading 20 kN axle at $x = 11.2$ m.