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) }