PLAIN AND CIVIL: EXAMPLE 4.3.14. BEAM SUPPORT REACTIONS FOR TRIANGULAR LOADS

Determine the support reactions for the beam shown.

To convert the distributed load, start with determining the magnitude by taking the area the load.

$P=\frac{1}{2}\omega L=\frac{1}{2}\left ( 6\frac{kN}{m} \right )\left ( 5m \right )=15kN$

After the magnitude, identify where the point load is located and that is at the centroid of the area.

For triangle:

$\bar{x}=\frac{1}{3}base$

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

${\color{Red} \mathbf{R_{CX}=0}}$

$\sum M_{C}=0$

$R_{A}\left ( 10.5m \right )-P\left ( 3.33m \right )=0$

$R_{A}\left ( 10.5m \right )=15kN\left ( 3.33m \right )$

${\color{Red} \mathbf{R_{A}=4.757kN}}$

$\sum M_{A}=0$

$P\left ( 5.5+1.67m \right )-R_{CY}\left ( 10.5m \right )=0$

$R_{CY}\left ( 10.5m \right )=15kN\left ( 7.17m \right )$

${\color{Red} \mathbf{R_{CY}=10.243kN}}$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Friday, 29 March 2019

EXAMPLE 4.3.14. BEAM SUPPORT REACTIONS FOR TRIANGULAR LOADS

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