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Chapter 4

Q. 4.2

Determine the forces in the members of the truss shown in Fig. P.4.2 using the method of joints and check the forces in the members JK, JD and DE by the method of sections.

Step-by-Step

Verified Solution

Initially the support reactions are calculated. Only vertical reactions are present so that, referring to Fig. S.4.2(a), and taking moments about E

R_{ B } \times 18-30 \times 24+60 \times 6=0

i.e.

R_{ B }=20 kN \text { (upwards) }

Now resolving vertically,

R_{ E }+R_{ B }-30-60=0

which gives

R_{ E }=70 kN \quad \text { (upwards) }

All members are assumed to be in tension as shown in Fig. S.4.2(a) and for simplicity the forces in the members are designated by their joint letters. Also the diagonals of the truss are each 10 m long so that cos θ=0.6 and sin θ=0.8.
By inspection

HC =0, \quad BG =-20 kN \text { and } EK =-70 kN

Joint A:
Resolving vertically,

AG \sin \theta-30=0

i.e.

AG =+37.5 kN

Resolving horizontally,

AB + AG \cos \theta=0

i.e.

AB =-37.5 \times 0.6=-22.5 kN

Joint B:
Resolving horizontally,

BC – BA =0

i.e.

BC = BA =-22.5 kN

Joint G:
Resolving vertically,

GC \sin \theta+ GB + GA \sin \theta=0

i.e.

\begin{gathered}GC \times 0.8-20+37.5 \times 0.8=0 \\GC =-12.5 kN\end{gathered}

Resolving horizontally,

GH + GC \cos \theta- GA \cos \theta=0

i.e.

\begin{gathered}GH -12.5 \times 0.6-37.5 \times 0.6=0 \\GH =30 kN\end{gathered}

Joint H:
By inspection (or resolving horizontally)

HJ = HG = 30 kN

Joint C:
Resolving vertically,

\text { CJ } \sin \theta+C G \sin \theta=0

i.e.

\begin{gathered}CJ \times 0.8-12.5 \times 0.8=0 \\CJ =12.5 kN\end{gathered}

Resolving horizontally,

CD + CJ \cos \theta- CG \cos \theta- CB =0

i.e.

\begin{gathered}CD +12.5 \times 0.6+12.5 \times 0.6+22.5=0 \\CD =-37.5 kN\end{gathered}

Joint J:
Resolving vertically,

JD + JC \sin \theta=0

i.e.

\begin{gathered}JD +12.5 \times 0.8=0 \\JD =-10 kN\end{gathered}

Resolving horizontally,

JK – JH – JC \cos \theta=0

i.e.

\begin{gathered}JK -30-12.5 \times 0.6=0 \\JK =37.5 kN\end{gathered}

Joint D:
Resolving vertically,

DK \sin \theta+ DJ =0

i.e.

\begin{gathered}DK \times 0.8-10=0 \\DK =12.5 kN\end{gathered}

Resolving horizontally,

DE + DK \cos \theta- DC =0

i.e.

\begin{gathered}DE +12.5 \times 0.6+37.5=0 \\DE =-45 kN\end{gathered}

Joint F:
Resolving vertically,

FK \sin \theta-60=0

i.e.

\begin{gathered}FK \times 0.8-60=0 \\FK =75 kN\end{gathered}

Resolving horizontally,

FE + FK \cos \theta=0

i.e.

\begin{gathered}FE +75 \times 0.6=0 \\FE =-45 kN\end{gathered}

To check the forces in the members JK and DE take a section cutting JK, KD and DE as shown in Fig. S.4.2(b).

Taking moments about K,

\begin{gathered}ED \times 8+60 \times 6=0 \\ED =-45kN\end{gathered}

Taking moments about D,

\begin{gathered}KJ \times 8-60 \times 12+70 \times 6=0 \\KJ =37.5kN\end{gathered}

To check the force in the member JD take a section cutting JK, JD and CD as shown in Fig. S.4.2(c).

Resolving vertically,

\begin{gathered}DJ +70-60=0 \\DJ =-10 kN\end{gathered}