Question 5 of 8: Influence lines

Question 5: Influence lines (20 marks)

5(a) — Influence lines for a propped, hinged frame (9 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. Top beam $A(0)\!-\!B(1)\!-\!C(3)\!-\!D(9\text{ m})$ with a free left overhang $A\!-\!B$, an internal hinge at $C$ and a pin (roller-type vertical support) at $D$; a $2\text{ m}$ column $B\!-\!E$ rigidly built into the beam at $B$ and pinned at $E$. Find. the three influence lines.

Approach. With the hinge at $C$ and pins at $D$ and $E$ the frame is determinate, so each influence line is piecewise-linear; place a unit load at the control points ($A,B,C,D$) and solve by statics.

1. Moment just right of $B$. Zero on the overhang and at $B$ and $D$; it peaks when the load sits at the hinge $C$ (lever $C\!-\!B=2\text{ m}$): $$\boxed{(\text{IL }M)_{\max}=+2.0\text{ m at }C}$$
2. Shear just right of $B$. Zero for load on the overhang, a unit step at $B$, then $+1$ across $B\!-\!C$ falling linearly to zero at $D$: $$(\text{IL }V)_{\max}=+1.0.$$
3. Horizontal reaction at $E$. The overhang bends the column ($H_E=-0.5$ with the load at $A$); the maximum occurs with the load at $C$, where the moment $2\text{ kN}\cdot\text{m}$ over the $2\text{ m}$ column gives $$\boxed{(\text{IL }H_E)_{\max}=1.0\;(\text{load at }C)}.$$

5(b) — Moving vehicle on a truss (11 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. Deck truss, five bottom panels of $6\text{ m}$ (pin $L_1$, roller $L_5$), height $5\text{ m}$; the long diagonal $L_1\!-\!U_2$ crosses the vertical $U_1\!-\!L_2$ without connecting. Find. IL ordinates and the extreme member forces.

1. IL ordinates. A vertical section between $L_2$ and $L_3$ cuts the horizontal chords and $L_1\!-\!U_2$ (vertical component $5/13$); $\Sigma F_y$ of the left free body gives, panel by panel: $$\boxed{\eta_{L_1},\eta_{L_2},\eta_{L_3},\eta_{L_4},\eta_{L_5}=0,\;+0.65,\;-1.30,\;-0.65,\;0}$$ (tension +). The line crosses zero at $x=8\text{ m}$.
2. Maximum compression. Cluster the two $100\text{ kN}$ loads over the negative lobe near $L_3$ (at $x=12$ and $14\text{ m}$) with the $50\text{ kN}$ trailing at $20\text{ m}$: $$F=100(-1.30)+100(-1.083)+50(-0.433)=\boxed{-260\text{ kN}\;(260\text{ kN compression})}.$$
3. Maximum tension. Place the two $100\text{ kN}$ loads over the positive lobe near $L_2$ ($x=4$ and $6\text{ m}$), the $50\text{ kN}$ off the effective span: $$F=100(0.433)+100(0.65)\approx\boxed{+108\text{ kN}\;(108\text{ kN tension})}.$$