PLAIN AND CIVIL: EXAMPLE 4.3.10. BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS AND CONCENTRATED LOADS

Determine the support reactions for the beam with the applied loads.

The FBD of the beam is as shown.

$P=4.5\frac{kN}{m}\left ( 3.5+1.25m \right )=21.375kN$

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

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

$\sum M_{D}=0$

$R_{A}\left ( 7m \right )-P\left ( 2.375+2.25m \right )-10kN\left ( 1.25+2.25m \right )=0$

$R_{A}\left ( 7m \right )=21.375kN\left ( 2.375+2.25m \right )+10kN\left ( 1.25+2.25m \right )$

${\color{Red} \mathbf{R_{A}=19.123kN}}$

$\sum M_{A}=0$

$R_{DY}\left ( 7m \right )-P\left ( 2.375m \right )-10kN\left ( 3.50m \right )=0$

$R_{DY}\left ( 7m \right )=21.375\left ( 2.375m \right )+10kN\left ( 3.50m \right )$

${\color{Red} \boldsymbol{R_{DY}=12.252kN\uparrow }}$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Wednesday, 27 March 2019

EXAMPLE 4.3.10. BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS AND CONCENTRATED LOADS

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