Question

The figure below shows a pin jointed frame.


The algebric sum of forces in members, BE, CE and CD is ______kN.
[Take tension force as positive and compression force as negative]

• A. -23.82

Answer 1

The correct option is A -23.82
Calculating support reactions

$\Sigma M_A=0$ $10\times 2+20\times 4=V_D\times 4$
$V_D=25kN(\uparrow )$
$\Sigma M_D=0$ $20\times 4-10\times 2=V_A\times 4$
$V_A=15kN(\downarrow )$

$tan\alpha =\frac{4}{4}$
$\Rightarrow \alpha ={45}^o$
$tan\beta =\frac{4}{2}$
$\Rightarrow \beta ={63.43}^o$
$AtjointA,F_{AB}=\frac{15}{sin\alpha }=\frac{15}{sin{45}^o}=21.21kN\left(T\right)$
$20=F_{AB}cos\alpha +F_{AE}$
$\Rightarrow F_{AE}=5kN\left(T\right)$
$AtjointD{F}_{AE}=0$
$F_{DC}=-25kN\left(C\right)$

$AtjointE,F_{CE}=\frac{5}{cos\beta }=11.18kN\left(T\right)$

$F_{BE}=-11.18sin\beta =-10kN\left(C\right)$


$AtjointB{F}_{AB}=F_{BC}=21.21kN$

Hence the forces on members are shown below.

$\therefore$ Algebraic sum of forces in BE, CE and CD
= (-10) + 11.18 + (-25)
= -23.82 kN