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Chapter 4

Q. 4.7

The roof truss shown in Fig. P.4.7 is comprised entirely of equilateral triangles; the wind loads of 6 kN at J and B act perpendicularly to the member JB. Calculate the forces in the members DF, EF, EG and EK.

Step-by-Step

Verified Solution

There will be horizontal and vertical reactions at A and a vertical reaction at B. Then, resolving horizontally,

R_{ A , H }-2 \times 6 \cos 30^{\circ}=0

i.e.

R_{ A , H }=10.4 kN

Taking moments about B,

R_{ A , V } \times 12-5 \times 36 \times 6-6 \times 3=0

i.e.

R_{ A , V }=91.5 kN

From symmetry at K, the forces in the members KE and KG are equal. Then, assuming they are tensile and resolving vertically,

2 \times KE \cos 30^{\circ}+36=0

i.e.

KE =-20.8 kN (= KG )

Knowing KE (and KG) we can now take a section cutting through KG, EG, EF and DF as shown in Fig. S.4.7.
Resolving vertically,

EF \cos 30^{\circ}+ KG \cos 30^{\circ}+3 \times 36-91.5=0

i.e.

\begin{gathered}EF \times 0.866-20.8 \times 0.866+108-91.5=0 \\EF =1.7 kN\end{gathered}

Taking moments about E,

\begin{gathered}DF \times 1.5 \tan 60^{\circ}- KG \times 1.5 \tan 60^{\circ}-36 \times 1.5+36 \times 3-91.5 \times 4.5 \\+10.4 \times 1.5 \tan 60^{\circ}=0\end{gathered}

i.e.

\begin{gathered}DF \times 2.6+20.8 \times 2.6-54+108-411.8+10.4 \times 2.6=0 \\DF =106.4 kN\end{gathered}

Resolving horizontally,

EG + EF \cos 60^{\circ}+ KG \cos 60^{\circ}+ DF +10.4=0

i.e.

\begin{gathered}EG +1.7 \times 0.5-20.8 \times 0.5+106.4+10.4=0 \\EG =-107.3 kN\end{gathered}