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## Q. 4.4

A motorway warning display is supported symmetrically by two identical pin-jointed frames as shown in Fig. P.4.4. The frames must be designed to withstand a wind load of $4 kN / m ^{2}$ . Calculate the forces in the members of a frame and check your solution using a graphical method. Neglect the weight of the display.

## Verified Solution

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).