Question

The force in the member CD of the truss as shown in figure below is

• A. $P\text{(Tensile)}$
• B. $2P\text{(Tensile)}$
• C. $\text{zero}$
• D. $P\text{(Compression)}$

Answer 1

The correct option is B $2P\text{(Tensile)}$

Using method of joints,

$F_{BA}=F_{BD}=0\text{[No force acting on joint B]}$

$\text{At joint A,}$
$\sum F_H=0$
$F_{AD}=\frac{2P}{sin{45}^o}=(-2\sqrt{2}P)$
$F_{AC}=F_{AD}cos{45}^o=2\sqrt{2}P.\frac{1}{\sqrt{2}}=2P\text{(Tensile)}$

$H_E=3P(\leftarrow )$

$\sum M_F=0$

$\Rightarrow 2P\left(2a\right)+P\left(a\right)=V_E\left(a\right)$

$\Rightarrow V_E=5P(\downarrow )$

$\therefore V_F=5P(\uparrow )$

$F_{CF}cos{45}^o=F_{EF}=H_E=3P$

$\Rightarrow F_{CF}=3\sqrt{2}P\text{(Compressive)}$

$F_{CE}=V_E=5P\text{(Tension)}$

$\text{Joint C}$

$\sum F_x=0$

$\Rightarrow F_{CF}cos{45}^o+F_{CD}=P$

$\Rightarrow 3\sqrt{2}P.\frac{1}{\sqrt{2}}+F_{CD}=P$

$\Rightarrow F_{CD}=-2P\text{i.e., tensile}$

$\text{At joint D}$,
$\sum F_H=0$
$F_{CD}=2\sqrt{2}Psin{45}^0=2P\left(Tensile\right)$