PLAIN AND CIVIL: EXAMPLE 4.3.9. OVERHANGING BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS

Determine the support reactions of the overhanging beam loaded as shown.

When a distributed load crosses over a reaction, the conversion of the load to concentrated load would need to be divided at the point of support.

$P_{1}=4\frac{kN}{m}\left ( 5m \right )=20kN$

$P_{2}=4\frac{kN}{m}\left ( 3m \right )=12kN$

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

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

$\sum M_{B}=0$

$R_{AY}\left ( 5m \right )-P_{1}\left ( 2.5m \right )+P_{2}\left ( 1.5m \right )=0$

$R_{AY}\left ( 5m \right )=20kN\left ( 2.5m \right )-12kN\left ( 1.5m \right )$

${\color{Red} \mathbf{R_{AY}=6.40kN}}$

$\sum M_{A}=0$

$P_{1}\left ( 2.5m \right )+P_{2}\left ( 5m+1.5mm \right )=R_{B}\left ( 5m \right )$

$R_{B}\left ( 5m \right )=20kN\left ( 2.5m \right )+12kN\left ( 6.5m \right )$

${\color{Red} \mathbf{R_{B}=25.60kN}}$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Wednesday 27 March 2019

EXAMPLE 4.3.9. OVERHANGING BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS

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