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## Q. 4.3

Calculate the forces in the members of the truss shown in Fig. P.4.3. ## Verified Solution

Referring to Fig. S.4.3 the reactions at A and B are each 15 kN from symmetry.

Also sin $\theta=4 / 8=0.5 \text { so that } \theta=30^{\circ}$, the remaining angles follow. By inspection (or by resolving vertically) BJ=-15 kN. Further, by inspection, FB=0. All members are assumed to be in tension.
Joint A:
Resolving vertically,

$AC \sin \theta+15=0$

i.e.

$\begin{gathered}AC \times 0.5+15=0 \\AC =-30 kN\end{gathered}$

Resolving horizontally,

$AP+AC\cos \theta =0$

i.e.

$\begin{gathered}AP -30 \times 0.866=0 \\AP =26.0 kN\end{gathered}$

Joint C:
Resolving parallel to CP,

$CP +10 \cos 30^{\circ}=0$

i.e.

$CP =-8.7 kN$

Resolving parallel to CE,

$CE – CA -10 \sin 30^{\circ}=0$

i.e.

$\begin{gathered}CE +30-10 \times 0.5=0 \\CE =-25 kN\end{gathered}$

Joint P:
Resolving vertically,

$PE \cos 30^{\circ}+ PC \cos 30^{\circ}=0$

i.e.

$PE =- PC =8.7 kN$

Resolving horizontally,

$PF + PE \cos 60^{\circ}- PC \cos 60^{\circ}- PA =0$

i.e.

$\begin{gathered}PF +8.7 \times 0.5+8.7 \times 0.5-26.0=0 \\PF =17.3 kN\end{gathered}$

Joint F:
Resolving vertically,

$FE \cos 30^{\circ}+ FH \cos 30^{\circ}=0$

i.e.

FE = -FH

Resolving horizontally,

$\begin{gathered}FH \cos 60^{\circ}- FE \cos 60^{\circ}- FP =0( FB =0) \\FH \times 0.5+ FH \times 0.5-17.3=0 \\FH =- FE =17.3 kN\end{gathered}$

Joint G:
Resolving parallel to GH,

$GH +10 \cos 30^{\circ}=0$

i.e.

$GH =-8.7 kN$

Resolving parallel to GE,

$GJ – GE -10 \cos 60^{\circ}=0$

i.e.

$GJ – GE -10 \times 0.5=0$                              (i)

Joint H:
Resolving perpendicularly to HJ,

$HE \cos 30^{\circ}+ HG \cos 30^{\circ}=0$

i.e.

$HE =- HG =8.7 kN$

Resolving parallel to HJ,

$HJ – HF + HG \cos 60^{\circ}- HE \cos 60^{\circ}=0$

i.e.

$\begin{gathered}HJ -17.3-8.7 \times 0.5-8.7 \times 0.5=0 \\HJ =26.0 kN\end{gathered}$

Joint J:
Resolving vertically,

$JG \cos 60^{\circ}+ JH \cos 30^{\circ}+ JB =0$

i.e.

$\begin{gathered}JG \times 0.5+26.0 \times 0.866-15=0 \\JG =-15.0 kN\end{gathered}$

Substituting for JG in Eq. (i) gives

GE = -20.0 kN 