07-Str-A1 · December 2018

Question 5 of 8: Influence Lines for a Two-Span Beam and for a Truss Member

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

Question 5: Influence Lines for a Two-Span Beam and for a Truss Member (18 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.

5(a) — two-span beam with an internal hinge

Given. A beam lettered A to E with a pin at A, a roller at C, an internal hinge at D and a roller at E. The dimensions are A-B = 5 m, B-C = 5 m, C-D = 2 m and D-E = 8 m, so the stations are A = 0, B = 5, C = 10, D = 12 and E = 20 m. With $r = 4$ and one condition equation the beam is determinate.

Find. Influence lines for $M_B$, for the shear immediately to the right of C, and for the reaction at C, each with the largest absolute ordinate.

Approach. Split the structure at the hinge. With the unit load on the suspended span D-E, the piece D-E behaves as a simple 8 m beam that hands a known force to D; with the load anywhere on A-C-D, that hand-over force is zero and the left piece acts alone. Two short sets of reactions then give every ordinate, and the lines are straight between the break points.

1. Reactions with the unit load on A-C-D, i.e. $0 \le x \le 12$. Segment D-E carries no load and is moment-free at D, so $R_E = 0$ and nothing crosses the hinge. The left piece is a 12 m beam with a pin at A and a roller at C plus a 2 m overhang, hence $$R_C = \frac{x}{10} \qquad R_A = 1 - \frac{x}{10}$$ Note that $R_C$ keeps rising past C and reaches $1.2$ when the load stands at the hinge.
2. Reactions with the unit load on D-E, i.e. $12 \le x \le 20$. Moments about D for the suspended span give $R_E = (x - 12)/8$, so the force delivered through the hinge to the left piece is $P_D = (20 - x)/8$ downwards. Moments about A then give $$R_C = 1.2\,P_D = 0.15(20 - x) \qquad R_A = -0.2\,P_D = -0.025(20 - x)$$ The pin reaction is now negative — it pulls down — which is why the moment influence line at B goes below the axis for loads on the far span.
3. Influence line for $M_B$. For a load to the left of B, $M_B = 5R_A - (5 - x) = 0.5x$; for a load to the right of B but still on A-C-D, $M_B = 5R_A = 5 - 0.5x$; and for a load on D-E, $M_B = 5R_A = -0.125(20 - x)$. The line is therefore a triangle peaking under B, crossing zero at C, dipping to $-1$ at the hinge and returning to zero at E: $$\text{ordinates } (0,\ 0),\ (5,\ +2.5),\ (10,\ 0),\ (12,\ -1.0),\ (20,\ 0) \qquad \boxed{|M_B|_{\max} = 2.5\ \text{m}}$$
4. Influence line for the shear just right of C. Summing the upward forces to the left of the section, $V_{C^+} = R_A + R_C - 1$ when the load is left of the section and $R_A + R_C$ when it is right of it. But $R_A + R_C = 1$ identically for any load on A-C-D, so $$V_{C^+} = 0 \ \ (0 \le x < 10) \qquad V_{C^+} = 1 \ \ (10 < x \le 12) \qquad V_{C^+} = 0.125(20 - x) \ \ (12 \le x \le 20)$$ The line is flat and zero over the whole of A-C, jumps to unity at C, stays at unity across the 2 m overhang and then falls linearly to zero at E: $$\boxed{|V_{C^+}|_{\max} = 1.00}$$ The dead flat portion is the useful observation: no load standing between A and C produces any shear just past C, because the two supports of that span between them absorb it entirely.
5. Influence line for $R_C$. Directly from steps 1 and 2, $R_C$ rises linearly from zero at A to 1.0 at C and keeps rising to $$\boxed{R_{C,\max} = 1.20 \ \text{with the load at the hinge D}}$$ before falling linearly to zero at E. The overshoot past unity is the overhang effect: with the load at D the roller at C carries more than the whole load, and the pin at A holds the beam down with 0.2.

5(b) — influence line for a truss diagonal and the governing vehicle position

Given. A Warren truss 3 m deep. The upper chord joints U₁…U₄ are at 8 m centres starting at $x = 0$, and the lower chord joints L₁, L₂, L₃ are at 8 m centres offset half a panel, at $x = 4$, 12 and 20 m. The pin is at L₁ and the roller at L₃, so the truss spans 16 m between supports and overhangs 4 m at each end. The vehicle runs on stringers at upper-chord level, so its axle loads reach the truss only at U₁…U₄. The idealised vehicle is three axles: 40 kN, then 40 kN at 2 m, then 20 kN a further 4 m behind, travelling left to right.

Find. The influence line for the force in U₂-L₁, its extreme ordinates, and the maximum compression the vehicle can produce in that member.

Approach. Because the load is delivered only at panel points, the influence line is a straight line between consecutive upper joints; so place a unit load at each of U₁…U₄ in turn, solve for the force in U₂-L₁, and join the four ordinates. Then slide the axle train along the resulting polygon to maximise the compression.

1. Reactions for a unit load at upper-chord station $x$. Moments about L₁ (at $x = 4$) give $$R_{L3} = \frac{x - 4}{16} \qquad R_{L1} = \frac{20 - x}{16}$$ so at U₁ ($x = 0$) the pin carries $1.25$ and the roller $-0.25$; at U₂ ($x = 8$) they are $0.75$ and $0.25$; at U₃ ($x = 16$) $0.25$ and $0.75$; and at U₄ ($x = 24$) $-0.25$ and $1.25$. The overhangs are what make two of these negative.
2. Use joint U₁ to dispose of the end diagonal. U₁ carries only the end diagonal U₁-L₁ and the top chord U₁-U₂. The diagonal runs $(4,\ -3)$, i.e. a 3-4-5 triangle with unit vector $(0.8,\ -0.6)$, so with a load $P_1$ at U₁ $$F_{U_1L_1} = -\frac{P_1}{0.6} = -1.6667P_1$$ and it is zero whenever the unit load stands anywhere else.
3. Resolve joint L₁ vertically for the member wanted. Three members meet the pin: the end diagonal, the bottom chord (horizontal, no vertical component) and U₂-L₁, which runs $(4,\ 3)$ from L₁ and so also has a vertical direction cosine of $0.6$. Hence $$0.6\,F_{U_1L_1} + 0.6\,F_{U_2L_1} + R_{L1} = 0 \qquad F_{U_2L_1} = -F_{U_1L_1} - \frac{R_{L1}}{0.6}$$
4. Evaluate the four ordinates. Substituting each load position in turn, $$\text{U}_1: \ 1.6667 - \frac{1.25}{0.6} = -\tfrac{5}{12} = -0.4167 \qquad \text{U}_2: \ 0 - \frac{0.75}{0.6} = -\tfrac{5}{4} = -1.2500$$ $$\text{U}_3: \ -\frac{0.25}{0.6} = -0.4167 \qquad \text{U}_4: \ +\frac{0.25}{0.6} = +0.4167$$ so the influence line is a straight-sided figure with $$\boxed{\text{minimum ordinate } -1.25 \text{ at U}_2, \quad \text{maximum ordinate } +0.4167 \text{ at U}_4}$$ Compression (negative) dominates over almost the whole span, and the member only goes into tension when the load reaches the far overhang.
5. Slide the axle train for the worst compression. The influence ordinate at a station $x$ is $-0.4167 - 0.10417x$ on U₁-U₂ and $-1.25 + 0.10417(x - 8)$ beyond U₂. Trying the train with each heavy axle in turn at the peak, and evaluating $\Sigma P_i\,y_i$, the governing arrangement puts the second 40 kN axle exactly at U₂, with the leading 40 kN 2 m behind it at $x = 6$ m and the 20 kN axle at $x = 12$ m: $$F = 40(-1.0417) + 40(-1.2500) + 20(-0.8333) = -41.67 - 50.00 - 16.67$$ $$\boxed{F_{U_2L_1} = 108.3\ \text{kN compression}}$$ Placing a 40 kN axle at U₂ with the other 40 kN ahead of it instead gives only 104.2 kN, and putting the light 20 kN axle on the peak gives 83.3 kN, so 108.3 kN governs. The same 108.3 kN is obtained if the train is taken to cross in the opposite direction, so the answer does not hinge on which axle is read as leading.