Question 5 of 8: Influence lines for a truss, and shear influence line under a moving vehicle (18 marks)

Question 5: Influence lines for a truss, and shear influence line under a moving vehicle (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.

(a) Truss influence lines

Given. Bottom chord L1–L5 on a 24 m baseline (4 panels @ 6 m); pin at L2 (x=6 m), roller at L4 (x=18 m); L1 and L5 are 6 m free overhangs. Top chord U1, U2, U3 (over L2, L3, L4) at 4.5 m, with fan diagonals L1-U1, U1-L3, U2-L4, L5-U3.

Find. Influence-line ordinates for U1-U2, U2-L3, U2-L4 for a unit load at each bottom-chord joint.

Approach. A truss influence line is piecewise linear between adjacent bottom-chord joints, so it is enough to place a unit load at each joint L1–L5 and solve the (still-determinate) truss by method of sections/joints each time.

1. Unit load at each joint (positive = tension). Solving the truss for a unit load placed successively at L1, L2, L3, L4, L5 (L2 and L4 are the supports, so their ordinate is always zero — a support reaction takes the whole load, none passes into the truss):

Connecting these five ordinates with straight lines gives each influence line. Maximum tension/compression coefficients:

(b) Shear influence line and moving vehicle

Given. Beam: pin at x=0, roller at x=20 m, free overhang tip at x=24 m; Section 1-1 at x=6 m. Vehicle: three axles 40, 40, 20 kN, spaced 2 m then 4 m, crossing left to right.

Find. Ordinates of the IL for shear at Section 1-1, and the maximum shear there as the vehicle crosses.

Approach. For a unit load at position x on a simply-supported span of length L=20 m with a right overhang, $$V_{1\text{-}1}(x)=-x/L$$ for x left of the section and $$V_{1\text{-}1}(x)=(L-x)/L$$ for x at or right of the section (valid on the overhang too). Slide the 3-axle group across and evaluate the sum $$\Sigma P_i\,V(x_i)$$ at each position where an axle sits exactly at the section (the only candidates for an extremum, since the IL has constant slope $$-1/20$$ apart from the unit jump at x=6).

1. IL ordinates at the labelled points: $$V(0)=0,\quad V(6^-)=-0.30,\quad V(6^+)=+0.70,\quad V(20)=0,\quad V(24)=-0.20$$
2. Governing position for maximum shear — place the rear 40 kN axle exactly at the section (ordinate +0.70), putting the middle 40 kN at x=8 m (ordinate +0.60) and the lead 20 kN at x=12 m (ordinate +0.40): $$V_{max}=40(0.70)+40(0.60)+20(0.40)=28+24+8=\boxed{60.0\text{ kN}}$$ (checked against the other two "axle-at-section" positions, 10 kN and 30 kN, which are smaller.)
3. Governing position for minimum (most negative) shear — with the lead 20 kN axle already off the far end of the beam (x>24, contributing zero) and the rear/middle 40 kN axles on the overhang at x=22 m and x=24 m: $$V_{min}=40(-0.10)+40(-0.20)+20(0)=\boxed{-12.0\text{ kN}}$$