In Fig. S.4.4(a), the length of BC is obtained by similar triangles.

i.e.

\frac{ BC }{2.5}=\frac{3}{5.25}

which gives

BC =1.43 m

Further

\alpha=\tan ^{-1} \frac{3}{5.25}=29.7^{\circ}

and

\beta=\tan ^{-1} \frac{1.43}{2.75}=27.5^{\circ}

The total wind load on the display is 4 × 5 × 4 = 80 kN. Therefore, the horizontal forces at A and B are 20 kN.
Resolving forces horizontally at B

F_{ BC }+20=0

so that

F_{ BC }=-20 kN \text { (compression) }

Resolving forces horizontally at A

F_{ AC } \sin 29.7^{\circ}+20=0

from which

F_{ AC }=-40.4 kN \text { (compression) }

Resolving forces vertically at A

F_{ AB }+F_{ AC } \cos 29.7^{\circ}=0

which gives

F_{ AB }=35.1 kN \text { (tension) }

Resolving forces perpendicular to AC at C

F_{ CD } \cos 32.8^{\circ}+F_{ AC } \cos 29.7^{\circ}=0

so that

F_{ CD }=20.7 kN \text { (tension) }

Resolving forces parallel to CE at C

F_{ CE }-F_{ CA }-F_{ CB } \cos 60.3^{\circ}+F_{ CD } \cos 57.2^{\circ}=0

which gives

F_{ CE }=-61.5 kN \text { (compression) }

Finally, resolving forces horizontally at E

F_{ ED }+F_{ CE } \cos 60.3^{\circ}=0

giving

F_{ ED }=30.5 kN \text { (tension) }

The graphical solution is obtained by first numbering the spaces between the forces as shown in Fig. S.4.4(a) and then completed as shown in Fig. S.4.4(b).