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## 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.

## 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}

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

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

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

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

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

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