Question

A pin jointed 2D truss is loaded with a vertical force of 40 kN at joint F and another 40 kN horizontal magnitude force at joint B as shown in figure.

The force (in kN) in member BC is

• A. $0$
• B. $20$
• C. $\text{None of these}$
• D. $20\sqrt{2}$

Answer 1

The correct option is A $0$

$\sum M_A=0$

$\Rightarrow R_G\times 10=40\times 5-40\times 5$

$\Rightarrow R_G=0$

From method of section $\to$

$R_A=40kN$

Taking moment about F,

$40\times 5-40\times 5+F_{BC}\times 5=0$

$F_{BC}=0$

Alternatively,

$byinspection{R}_G=0[B{M}_E=0]$

$\therefore$ Force in member EG=0

Forces in member ED and CD is already zero.

Forces in member ED and EC is zero.

Similarly, forces in members BC and CF are zero.

Also, force in member AF=0