Question 2.7.2: Fig. 2.7-5 shows a bow-string girder in which the panel poin...

Since the upper panel points lie on a parabola

DE = 8 – (l/4)(l/4)8 = 7.5 m

The intersection of CD and FE produced is at I such that AI = 48 m. The forces may be determined using the method of sections. If the truss is cut by an imaginary plane, passing through CD, DF and FE, both portions must be in equilibrium under the action of the externally applied loads and the forces in the cut members. If moments were taken about I the forces in CD and EF would be eliminated from consideration, and that in DF determined explicitly. Similarly, moments taken about F would eliminate the forces in DF and EF and permit the determination of the force in CD.

IJ=IF\sin I\overset{\frown}{F} J=64\sin (\tan ^{-1}(7.5/4))=56.47  m

FG=2\left(\frac{Area\triangle DCF}{DC} \right) =\frac{32}{√4^{2}+(0.5)^{2}}=7.94  m

Now, defining the position of the load by its distance x, from A.
If 0 < x < 12—considering the equilibrium of that portion of the frame to the right of the imaginary cutting plane, and taking moments about I

IJ  F_{DF}+80R_{B}=0

F_{DF}=-(80/56.47)R_{B}=-1.417R_{B}

and taking moments about F

F_{DC} GF+16R_{B}=0
F_{x}=-(16/7.94)R_{B}=-2.02R_{B}

If 16 < x < 32, considering the equilibrium of that portion of the frame to the left of the imaginary cutting plane, and taking moments about I

IJ  F_{DF}=48R_{A}
F_{DF}=(48/56.47)R_{A}=0.85R_{A}

and taking moments about F

16R_{A}+FG  F_{CD}=0
F_{CD}=-2.02R_{A}

The influence lines shown in Fig. 2.7-5(c) and (d) have been completed by straight lines joining the ordinates for x = 12 m and 16 m. It will be seen that the two portions for F_{CD} in the range 0-16 m are co-linear. The influence line for F_{DF} shows that stress reversal will occur as a load rolls across the span.