PLAIN AND CIVIL: EXAMPLE 7.0.12. FINDING REACTIONS OF INCLINED SUPPORTS

Monday, 11 March 2019

Determine the reactions of the truss with the following applied load.

The hinge at point A carries parallel and perpendicular to the surface reaction components. The roller at C carries one unknown reaction, perpendicular to the surface.

Summation of horizontal forces $\sum F_{X}=0]\rightarrow$ does not apply as the first procedure because both supports do have horizontal forces.

$\sum M_{C}=0]$

$Ry_{A}\left ( 4m \right )+3kN\left ( 1.5m \right )-4kN\left ( 2m \right )=0$

${\color{Red} \mathbf{Ry_{A}=0.875kN}}$ downwards

By taking the moment about point A, the reaction Rc will be determined. However, since this reaction is inclined, it would be difficult to measure the moment arm, thus we use the components of Rc.

$Ry_{C}=R_{C}cos30^{o}$

$\sum M_{A}=0]$

$Ry_{C}\left ( 4m \right )=3kN\left ( 1.5m \right )+4kN\left ( 2m \right )$

$R_{C}cos30^{o}=\frac{12.5kN\cdot m}{4m}$

${\color{Red} \mathbf{R_{C}=3.61kN}}$ inclined upwards

Then we can solve for the horizontal reaction at point A.

$\sum F_{X}=0]\rightarrow$

$Ry_{A}=3kN-Rx_{C}$

$Ry_{A}=3kN-R_{C}sin30^{o}$

$Ry_{A}=3kN-3.61kNsin30^{o}$

${\color{Red} \mathbf{Ry_{A}=1.195kN}}$ to the left

RELATED TOPIC:

7.0. EQUILIBRIUM OF FORCE SYSTEMS

PLAIN AND CIVIL

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Monday, 11 March 2019

EXAMPLE 7.0.12. FINDING REACTIONS OF INCLINED SUPPORTS

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