PLAIN AND CIVIL: EXAMPLE 4.3.7. BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS

Determine the support reactions for the beam as shown.

The FBD of the beam is as follows:

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

$P_{2}=2\frac{kN}{m}\left ( 5m \right )=10kN$

With the loads converted, start solving for the reactions.

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

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

$\sum M_{D}=0]$

$R_{AY}\left ( 12m \right )-P_{1}\left ( 10m \right )-P_{2}\left ( 2.5m \right )=0$

$R_{AY}\left ( 12m \right )=20kN \left ( 10m \right )+10kN \left ( 2.5m \right )$

${\color{Red} \mathbf{R_{AY}=18.75kN}}$

$\sum M_{A}=0$

$P_{1}\left ( 2m \right )+P_{2}\left ( 9.5m \right )-R_{D}\left ( 12m \right )=0$

$R_{D}\left ( 12m \right )=20kN\left ( 2m \right )+10kN\left ( 9.5m \right )$

${\color{Red} \mathbf{R_{D}=11.25kN}}$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Tuesday 26 March 2019

EXAMPLE 4.3.7. BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS

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