PLAIN AND CIVIL: EXAMPLE 7.0.9. FINDING REACTIONS

Saturday, 9 March 2019

Determine the reactions at pin A and rocker B on the beam.

There are two reaction components in the hinge at A and there is one perpendicular to the surface reaction at B.

Taking summation of forces at this point will lead nowhere because the reaction of the roller is inclined (meaning there are x- and y- components of this reaction). Take summation of moments instead, to cancel out Rb.

$\sum M_{B}=0$

$Ry_{A}\left ( 8m \right )-4kN\left ( 2m \right )=0$

${\color{Red} \mathbf{Ry_{A}=1kN}}$

Summation of moments at point A should also give value to Rb; however, the it's the reaction's y- component and not Rb.

$\sum M_{A}=0$

$4kN\left ( 6m \right )-Ry_{B}\left ( 8m \right )=0$

$Ry_{B}=3kN$

$R_{B}=\frac{Ry_{B}}{cos30^{o}}$

$R_{B}=\frac{3kN}{cos30^{o}}$

${\color{Red} \mathbf{R_{B}=3.46kN}}$

Then lastly, take $\sum F_{X}=0\rightarrow$ to compute for the value of Rax.

$Rx_{A}-Rx_{B}=0$

$Rx_{A}=R_{B}sin30^{o}$

$Rx_{A}=3.46sin30^{o}$

${\color{Red} \mathbf{Rx_{A}=1.73kN}}$

RELATED TOPIC:

7.0. EQUILIBRIUM OF FORCE SYSTEMS

PLAIN AND CIVIL

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Saturday, 9 March 2019

EXAMPLE 7.0.9. FINDING REACTIONS

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