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Chapter 4

Q. 4.5

If the railway track of P.3.21 is supported at the lower chord joints of the truss, calculate the forces in the members BE, BF, EF, and EG with the head of the train at mid-span.

Step-by-Step

Verified Solution

Referring to Fig. P.3.21, the angle of the diagonals to the horizontal is \tan ^{-1}(1.5 / 2)=36.9^{\circ}.
The total weight of the engine = 4 × 200 = 800 kN
Then the load at B, F, and C due to engine weight = 800/6 = 133.3 kN.
The total weight of the freight wagons on the bridge = 50 × 2 = 100 kN.
The load due to the freight wagons at C and D is therefore = 100/4 = 25 kN.
The total vertical loads on the truss due to both engine and freight wagons are then

\begin{aligned}&\text { At B : } 133.3 kN \\&\text { At F : } 133.3 kN \\&\text { At C: } 133.3+25=158.5 kN \\&\text { At D : } 25 kN\end{aligned}

Now taking moments about A

R_{D} \times 12-1333 \times 6-133.3 \times 8 \times 158.3 \times 10-25 \times 12=0

from which

R_{ D }=312.5 kN

By inspection

F_{ BE }=133.3 kN \text { (tension) }

Take a section through BF, EF, and EG as shown in Fig. S.4.5.

Resolving forces vertically

F_{ EF } \sin 36.9^{\circ}-133.3-158.3-25+312.5=0

which gives

F_{ EF }=6.8 kN \text { (tension) }

Taking moments about F

F_{ EG } \times 1.5-158.3 \times 2-25 \times 4+312.5 \times 4=0

from which

F_{ EG }=-555.6 kN (\text { compression })

Resolving forces horizontally

F_{ BF }+F_{ EG }+F_{ EF } \cos 36.9^{\circ}=0

which gives

F_{ BF }=550.2 kN \text { (tension) }