Question

1. FRAME Determine the support reactions at A (fixed) and at C (rocker). (Hint: You will need 2 FBDs to solve this problem.) 10 KN 4 kN/m A B 2 m 2 m 2 m

Answer 1

4 kN/m 10 KN Ax B By By B ibt Rc 2 m 2 m 2 m V By

Reaction at rocker-support C:

Take moment of forces about point-B in left FBD;

ΣMg = 0

Rc(2) - (4x2) X2 = 0

$\Rightarrow R_C = 4\;\;kN$

...(Answer)

Take equilibrium of forces for left FBD;

ΣF, = 0

$\Rightarrow B_x = 0$

and

$\sum F_y = 0$

$\Rightarrow B_y +R_C - (4\times2) = 0$

$\Rightarrow B_y +4 - (4\times2) = 0$

$\Rightarrow B_y =4\;\;kN$

Reaction at fixed-support A:

Take equilibrium of forces for right FBD;

ΣF, = 0

$\Rightarrow A_x - \frac{3}{5}(10) - B_x = 0$

$\Rightarrow A_x - \frac{3}{5}(10) - 0 = 0$

$\Rightarrow A_x=6\;\;kN$

...(Answer)

and

$\sum F_y = 0$

$\Rightarrow A_y - \frac{4}{5}(10) - B_y = 0$

$\Rightarrow A_y - \frac{4}{5}(10) - 4 = 0$

$\Rightarrow A_y =12\;\;kN$

...(Answer)

Take moment about point-A;

$\sum M_A = 0$

$\Rightarrow M_A - \left [ \frac{4}{5}(10) \right ](2) - B_y(2+2) = 0$

$\Rightarrow M_A - \left [ \frac{4}{5}(10) \right ](2) - 4(2+2) = 0$

$\Rightarrow M_A=32\;\;kN.m$

...(Answer)

________________________________________________________________________

Reaction at fixed-support A:

$\Rightarrow A_x=6\;\;kN\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\rightarrow)$

$\Rightarrow A_y =12\;\;kN\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\uparrow)$

$\Rightarrow M_A=32\;\;kN.m\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(CCW)$

Reaction at rocker-support C:

$\Rightarrow R_C = 4\;\;kN\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\uparrow)$