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Chapter 4

Q. 4.3

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

Step-by-Step

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