07-Str-A1 · December 2015

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2015, 07-Str-A1 — Elementary Structural Analysis. Three hours, closed book, approved Sharp or Casio calculator only. Six questions constitute a complete paper: answer ALL of Questions 1 to 4, then ONLY TWO of Questions 5, 6, 7 or 8. Marks are shown in the left margin (6 + 18 + 18 + 18 + 20 + 20 = 100). All eight questions are worked below, because the four alternatives exercise four different methods — moment distribution, three-hinged-frame statics, influence lines and virtual work — and the complete set is the more useful study resource.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 2 (determinacy and stability), Ch. 3–5 (trusses, internal loadings, frames), Ch. 6 (influence lines), Ch. 8–9 (deflections and virtual work), Ch. 11–12 (slope-deflection and moment distribution).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (equilibrium and determinacy), Ch. 4 (plane trusses), Ch. 5 (beams and frames), Ch. 7 (virtual work), Ch. 8–9 (influence lines), Ch. 16–17 (slope-deflection and moment distribution).
K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis, 5th ed. — a parallel treatment of the same syllabus.
C. H. Norris, J. B. Wilbur and S. Utku, Elementary Structural Analysis, 4th ed. — the classical text this exam code is named after.

Check: the roller in Question 2(c). The roller under the inclined member of Question 2(c) is drawn on a bed ruled parallel to the member, so its reaction is taken normal to the member axis. A vertical-roller reading is also arithmetically self-consistent. The distinction changes only the reaction components and the axial force: the shear and bending-moment diagrams the question asks for are identical under either reading, and both sets of reactions are reported.

Question 7: Influence lines for a truss and for 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. (a) A truss of six 4 m panels, total 24 m, with a 3 m deep raised section, carried on four supports at the bottom chord level; loads travel along the bottom chord. (b) A 20 m beam on four supports with two internal hinges, crossed by a three-axle vehicle.

Given data for Question 7
Quantity	Value
(a) Bottom chord joints	L₁ to L₇ at 0, 4, 8, 12, 16, 20, 24 m
(a) Top chord joints	U₁(8, 3), U₂(12, 3), U₃(16, 3) m
(a) Supports	Roller at L₁, pin at L₃, rollers at L₅ and L₇
(b) Beam	Supports 1, 2, 3, 4 at 0, 6, 14, 20 m; internal hinges at 8 m and 12 m
(b) Vehicle	64 kN, 64 kN, 16 kN at 2 m spacings, travelling left to right

Find. (a) the influence lines for the three named members, with the maximum tension and maximum compression coefficients; (b) the influence line for the bending moment at support 3, and the largest negative bending moment the vehicle can produce there.

7(a) — influence lines for three truss members

Figure 10 — the truss and the three influence lines. Ordinates are dimensionless force coefficients; tension is plotted positive. Between panel points every line is straight, because the travelling load reaches the truss only through the bottom-chord joints.

Approach. The structure is determinate ($m + r = 15 + 5 = 20 = 2j$), so place a unit load at each bottom-chord joint in turn, solve for the three member forces by the method of joints, and join the seven ordinates with straight lines.

The four ordinates that are zero by inspection. Four of the seven bottom-chord joints — L₁, L₃, L₅ and L₇ — are themselves supports. A unit load standing on a support passes straight into the foundation, so every member force is zero for those four positions. That fixes four of the seven ordinates of each influence line before any equilibrium is written, and it is the observation that makes the whole question short.
Unit load at L₂. The only member at L₂ with a vertical component is the diagonal L₂–U₁, whose unit vector is $(0.8,\ 0.6)$. Vertical equilibrium there gives $$ 0.6\,F_{L_2U_1} = 1 \quad\Longrightarrow\quad F_{L_2U_1} = \tfrac{5}{3}\ \text{tension}. $$ Carrying that up to U₁, where the vertical U₁–L₃ and the top chord U₁–U₂ are the only other members, $$ \sum F_y = 0 : \ F_{U_1L_3} = \boxed{-1.000}, \qquad \sum F_x = 0 : \ F_{U_1U_2} = \boxed{+1.333} $$ the first a compression of exactly the applied unit, the second a tension of $4/3$, which is simply the panel-to-depth ratio $4/3$.
Carrying the same case on to U₂. With no load at L₄ the vertical U₂–L₄ is a zero-force member, and at U₃ both members leaving towards L₅ and L₆ are unloaded, so U₂–U₃ is zero too. Joint U₂ then reduces to $$ -\tfrac{4}{3} + 1.6\,F_{U_2L_5} = 0 \quad\Longrightarrow\quad F_{U_2L_5} = \boxed{+0.833} $$ a tension of $5/6$, which is the diagonal length divided by twice the depth, $5/(2\times 3)$.
Unit load at L₄. The load is taken by the vertical U₂–L₄ as a tension of 1, and at U₂ the two diagonals down to the adjacent supports L₃ and L₅ share it symmetrically while both top chords stay at zero: $$ -1 - 1.2\,F = 0 \quad\Longrightarrow\quad F_{U_2L_3} = F_{U_2L_5} = \boxed{-0.833} $$ so U₂–L₅ is in compression, and neither U₁–U₂ nor U₁–L₃ feels the load at all.
Unit load at L₆. This is the mirror of the L₂ case about the centre line at 12 m: the diagonal U₃–L₆ takes $5/3$, the vertical U₃–L₅ takes $-1$, the top chord U₂–U₃ takes $+4/3$, and joint U₂ passes that on as $$ F_{U_2L_5} = \boxed{-0.833} $$ compression again. Members U₁–U₂ and U₁–L₃ remain at zero.
Assemble the three influence lines. Reading the seven ordinates in order from L₁ to L₇,
$$ \begin{aligned} U_1U_2 &: \quad 0,\ +1.333,\ 0,\ 0,\ 0,\ 0,\ 0 \\ U_2L_5 &: \quad 0,\ +0.833,\ 0,\ -0.833,\ 0,\ -0.833,\ 0 \\ U_1L_3 &: \quad 0,\ -1.000,\ 0,\ 0,\ 0,\ 0,\ 0 \end{aligned} $$
Each line is a set of straight segments between these values, because the travelling load is delivered to the truss only through the floor beams at the panel points.
Maximum coefficients. Member U₁–U₂ reaches $+1.333$ in tension with the load at L₂ and is never in compression — its maximum compressive coefficient is zero. Member U₂–L₅ reaches $+0.833$ in tension with the load at L₂ and $-0.833$ in compression with the load at either L₄ or L₆. Member U₁–L₃ reaches $-1.000$ in compression with the load at L₂ and is never in tension.

7(b) — influence line for the moment at support 3, and the worst vehicle position

Figure 11 — the beam, the influence line for the bending moment at support 3, and the critical position of the idealised vehicle. The influence line is a triangle with its apex −2.0 m at the right-hand hinge.

Approach. Cut at the two hinges. Moment at support 3 can only be produced by load that reaches the cantilevered end of the right-hand segment, so build the influence line by asking, for each load position, how much force arrives at the hinge at 12 m.

Load anywhere between 0 and 8 m. That length is held by the supports at 0 and 6 m, and the suspended length beyond the hinge at 8 m is unloaded and weightless, so it delivers nothing to the hinge. No force reaches the right-hand segment, and $$ \eta = 0 \qquad (0 \le a \le 8\ \text{m}). $$
Load on the suspended length, 8 m to 12 m. This 4 m length is a simple span between the two hinges, so a unit load at $a$ delivers $(a - 8)/4$ to the hinge at 12 m. That force lands 2 m to the left of support 3, so $$ \eta(a) = -\frac{a - 8}{4}\,(2) = -\frac{a - 8}{2}, \qquad \eta(12) = -2.0\ \text{m}. $$
Load on the overhang, 12 m to 14 m. Now the unit load acts directly on the right-hand segment, at a distance $14 - a$ to the left of support 3: $$ \eta(a) = -(14 - a), \qquad \eta(12) = -2.0,\quad \eta(14) = 0 . $$ The two expressions agree at the hinge, as they must.
Load between supports 3 and 4. That length is a simple span with support 3 as one of its ends, so the moment at 3 is zero: $$ \eta = 0 \qquad (14 \le a \le 20\ \text{m}). $$
The influence line. Collecting the four pieces gives a triangle: zero at 8 m, rising in magnitude linearly to $\boxed{-2.0\ \text{m}}$ at the hinge at 12 m, then falling linearly back to zero at support 3 at 14 m, and zero everywhere else. Its two slopes differ — $-0.5$ on the left of the apex and $+1.0$ on the right — which is what makes the axle placement worth checking rather than guessing.
Position the vehicle. The three axles are 64 kN, 64 kN and 16 kN at 2 m centres, spanning 4 m in all. Because the influence line is made of straight segments and the loads are concentrated, the worst case always has one axle standing exactly at the apex. Testing the three candidates: $$ \begin{aligned} \text{lead axle at 12 m} &: \ 64(2.0) + 64(0) + 16(0) = 128\ \text{kN}\cdot\text{m},\\ \text{lead axle at 10 m} &: \ 64(1.0) + 64(2.0) + 16(0) = 192\ \text{kN}\cdot\text{m},\\ \text{lead axle at 8 m} &: \ 64(0) + 64(1.0) + 16(2.0) = 96\ \text{kN}\cdot\text{m}. \end{aligned} $$
The governing value. The middle arrangement governs: the leading 64 kN at 10 m, the second 64 kN standing on the apex at 12 m and the 16 kN axle over support 3 where the ordinate is zero. Hence $$ M_3 = \boxed{-192\ \text{kN}\cdot\text{m}} $$ a hogging moment. Note that the light 16 kN axle contributes nothing in the governing case: it is the two heavy axles straddling the peak that matter, and the answer would be unchanged if the third axle were removed.

Question 7 — influence coefficients and the governing moment
Item	Maximum tension	Maximum compression
(a) U₁–U₂	+1.333 (load at L₂)	0 — never in compression
(a) U₂–L₅	+0.833 (load at L₂)	−0.833 (load at L₄ or L₆)
(a) U₁–L₃	0 — never in tension	−1.000 (load at L₂)
(b) Influence line peak at support 3	−2.0 m, at the hinge 12 m from the left end
(b) Governing negative moment at support 3	−192 kN·m, leading axle at 10 m

Topic. Influence lines for truss member forces and for the bending moment in a compound (Gerber) beam, and the placement of a train of concentrated loads.

Key relations. An influence line gives the value of one response function for a unit load at each position of travel. For a truss with a floor system, ordinates are computed at panel points and joined by straight lines. The response to a train of point loads is $R = \sum P_i\,\eta_i$, maximised by trial with one load at each vertex of the influence line in turn.

Why this works. Superposition is the whole of it: the structure is linear and determinate, so the effect of a load of any magnitude at any position is that magnitude times the ordinate there, and a set of loads adds. The straightness between panel points follows from the floor system, which distributes a load between the two adjacent panel points in inverse proportion to its distances from them — a linear rule. For the Gerber beam, the influence line is confined to the length between the two hinges and the overhang because those are the only positions from which load can reach the cantilevered arm; that is why so much of the line is exactly zero, and it is the same argument that kills four of the seven truss ordinates in part (a). Since the influence line is piecewise linear, the sum $\sum P_i \eta_i$ is a piecewise linear function of the vehicle position, so its maximum must occur with a load at a vertex — hence the three-case trial rather than a search.

Common pitfall. Assuming that the heaviest axle belongs on the apex. Here two equal 64 kN axles straddle the apex and beat that arrangement by half as much again. The second pitfall in part (a) is forgetting that a load standing on a support gives zero everywhere: without that, the question turns into seven full truss analyses instead of three.

Source. Hibbeler, Structural Analysis, 10th ed., §6.1–6.4 (beams and floor systems), §6.5 (trusses), §6.7–6.8 (series of concentrated loads); Kassimali, Structural Analysis, 6th ed., Ch. 8–9.

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