Question

Two steel truss members, AC and BC, each having cross sectional area of $100m{m}^2,$ are subjected to a horizontal force F as shown in figure. All the joints are hinged.

If F = 1 kN, the magnitude of the vertical reaction force developed at the point B in kN is

• A. 0.63
• B. 0.32
• C. 1.26
• D. 1.46

Answer 1

The correct option is A 0.63

Method I:
Using Lami's theorem
$\frac{T_1}{\sin {120}^{\circ }}=\frac{T_2}{\sin {135}^{\circ }}=\frac{F}{\sin {105}^{\circ }}$

$T_1=0.8965F$
$T_2=0.732F$

Vertical reaction at B,
$R_B=T_2\cos {30}^{\circ }=0.732\cos {30}^{\circ }$
$R_B=0.634kN$

Method II:
$\sum F_x=0$
$(F_{AC}{)}_x+(F_{BC}{)}_x=F.....\left(i\right)$

$\Rightarrow F_{AC}\cos {45}^{\circ }+F_{BC}\cos {60}^{\circ }=F$

$\sum F_y=0,$

$F_{AC}\sin {45}^{\circ }=F_{BC}sin{60}^{\circ }$

$F_{AC}=\frac{F_{BC}\sin {60}^{\circ }}{\sin {45}^{\circ }}=1.224F_{BC}$

$\Rightarrow 1=0.865F_{BC}+0.5F_{BC}$

$\therefore F_{BC}=\frac{1}{1.365}=0.732kN$

Vertical force:
$atB,(R_B{)}_v=F_{BC}\sin {60}^{\circ }$
$=0.732\sin {60}^{\circ }=0.634kN$