PLAIN AND CIVIL: EXAMPLE 4.3.1. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS

Tuesday, 26 March 2019

EXAMPLE 4.3.1. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS

Determine the support reactions for the beam with concentrated loads as shown.

To start solving for the reactions, one should be able to figure out the number of reactions required in each support.

The hinge requires one perpendicular-to-the-surface reaction, $R_{AY}$ . And also, one parallel-to-the-surface reaction, $R_{AX}$ .
The roller only requires one perpendicular-to-the-surface reaction, $R_{DY}$ .

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

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

Using summation of vertical forces will not determine any of the vertical reactions because there will be two unknown reactions in the equation. Use summation of moments instead. Have all forces rotating clockwise from the pivot point as positive and counterclockwise as negative.

$\sum M_{D}=0$

$R_{AY}\left ( 50m \right )-60kN\left ( 35m \right )-40kN\left ( 10m \right )=0$

${\color{Red} \mathbf{R_{AY}=50kN}}$

$\sum M_{A}=0$

$60kN\left ( 15m \right )+40kN\left ( 40m \right )-R_{D}\left ( 50m \right )=0$

${\color{Red} \mathbf{R_{D}=50kN}}$

RELATED TOPIC:

SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Tuesday, 26 March 2019

EXAMPLE 4.3.1. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS

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