Question 5 of 8: Influence lines

Question 5: Influence lines (20 marks)

Part (a) — determinate Gerber beam

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. Supports at points 1 ($x=2$), 2 ($x=8$) and 3 ($x=14$); an internal hinge at $x=10$; free overhangs at both ends. Find. Three influence lines and their peak coefficients.

Approach. Place a unit load at position $p$, solve the determinate beam (using $M=0$ at the hinge as the fourth equation), and read the required response — the ordinate at each $p$ traces the influence line. Every influence line is piecewise-linear with break points at the supports, the hinge and the section.

1. (i) Moment at point 2. Zero while the load is anywhere between supports 1 and 2, it grows linearly to $-2.0$ when the load sits over the hinge ($x=10$), returns to zero at support 3, and reaches $+1.0$ at the free right end. Peak $|\eta|=\boxed{2.0\ \text{m}}$.
2. (ii) Moment at mid-span of 1–2 ($x=5$). A triangle peaking at $+1.5$ m under the section, negative ($-1.0$) over the overhangs and the hinge. Peak $|\eta|=\boxed{1.5\ \text{m}}$.
3. (iii) Shear just left of support 2. Rises to $-1.0$ as the load approaches the section from the span side and falls off over the rest of the beam. Peak $|\eta|=\boxed{1.0}$.

Part (b) — moving vehicle on a truss

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. Six bottom joints at $4$ m spacing (span $20$ m); the diagonal $U_1L_3$; a three-axle group $18$–$18$–$8$ kN with $4$ m then $2$ m spacing. Find. IL ordinates and the maximum $U_1L_3$ force.

1. Ordinates. Placing a unit load successively at each bottom joint and solving for $U_1L_3$: $\eta_{L_1}=0,\ \eta_{L_2}=-\tfrac{7}{12}=-0.583,\ \eta_{L_3}=+0.500,\ \eta_{L_4}=+0.333,\ \eta_{L_5}=+0.167,\ \eta_{L_6}=0$; the line is linear between joints.
2. Position for maximum tension. Slide the group so the two 18 kN axles straddle the peak: $18$ kN at $L_3$ ($\eta=0.500$), $18$ kN at $L_4$ ($\eta=0.333$) and $8$ kN at $x=14$ ($\eta=0.250$).
3. Maximum force. $F_{\max}=18(0.500)+18(0.333)+8(0.250)=9+6+2=\boxed{17.0\ \text{kN (T)}}$. The largest compression the group can produce (over the negative lobe at $L_2$) is only $-10.5$ kN, so tension governs.