PLAIN AND CIVIL: EXAMPLE 4.3.3. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS AND MOMENTS

Tuesday, 26 March 2019

Determine the support reactions of the beam with concentrated loads and moments.

The FBD of beam would look like as follows, assuming vertical reactions direct upwards.

The moment at D will be considered from the point D when producing the equilibrium equation. From point D the moment at D will have the same rotation as the assumed force at A, this means that the moment will same as the moment at A.

$\sum M_{E}=0$

$R_{A}\left ( 18m \right )+10kN-m-60kn\left ( 14m \right )-30kN\left ( 9m \right )=0$

${\color{Red} \mathbf{R_{A}=61.111kN}}$

$\sum M_{A}=0$

$60kN\left ( 4m \right )+30kN\left ( 9m \right )-10kN-m-R_{EY}\left ( 18m \right )=0$

${\color{Red} \mathbf{R_{EY}=27.778kN}}$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Tuesday, 26 March 2019

EXAMPLE 4.3.3. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS AND MOMENTS

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