PLAIN AND CIVIL: EXAMPLE 4.3.5. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS AND MOMENTS

Saturday 30 March 2019

EXAMPLE 4.3.5. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS AND MOMENTS

Determine the support reactions for the beam as shown.

For beams with moments, the procedure of determining the support reactions is still the same. The summation of support reactions should be in equilibrium with the force (excluding the moment).

The way to include the moment in the equilibrium equation is to check the direction of rotation with the direction of the support reaction.

By identifying the support reactions, the FBD of the beam becomes:

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

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

Starting with the identification of the support reaction at point A, Ray, take the moment of 12kN.m as opposite the rotation for Ray, (negative moment).

$\sum M_{D}=0]$

$R_{A}\left ( 10m \right )-12kN.m-10kN\left ( 1.5m \right )=0$

${\color{Red} \mathbf{R_{A}=2.70kN\uparrow }}$

Similarly, when moment is taken about point A, notice that the moment is in the same rotation with the reaction at point D, Rd, so the moment takes the sign of the support reaction.

$\sum M_{A}=0]$ counterclockwise positive to make support reaction positive

$R_{D}\left ( 10m \right )+12kN.m-10kN\left ( 8.5m \right )=0$

${\color{Red} \mathbf{R_{D}=7.30kN\uparrow }}$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Saturday 30 March 2019

EXAMPLE 4.3.5. BEAM SUPPORT REACTIONS FOR CONCENTRATED LOADS AND MOMENTS

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