PLAIN AND CIVIL: EXAMPLE 4.3.16. BEAM SUPPORT REACTIONS FOR TRAPEZOIDAL LOADS

Determine the support reaction of the beam shown.

Identify how many reactions are required for the combination of support; convert the distributed loads into concentrated loads; and locate these point loads to have clear moment arms.

The FBD of the beam becomes:

$P_{1}=\frac{1}{2}\left ( \omega _{1}-\omega _{2} \right )L$

$P_{1}=\frac{1}{2}\left ( 15-9 \right )\left ( 6m \right )=18kN$

$P_{2}=\omega L=9\frac{kN}{m}\left ( 6m \right )=54kN$

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

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

$\sum M_{D}=0]$

$R_{AY}\left ( 12m \right )-P_{1}\left ( 7m \right )-P_{2}\left ( 6m \right )=0$

$R_{AY}\left ( 12m \right )=18kN\left ( 7m \right )+54kN\left ( 6m \right )$

${\color{Red} \mathbf{R_{AY}=37.50kN\uparrow }}$

$\sum M_{A}=0]$

$P_{1}\left ( 5m \right )+P_{2}\left ( 6m \right )-R_{D}\left ( 12m \right )=0$

$R_{D}\left ( 12m \right )=18\left ( 5m \right )+54\left ( 6m \right )$

${\color{Red} \mathbf{R_{D}=34.50kN}}\uparrow$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Saturday 30 March 2019

EXAMPLE 4.3.16. BEAM SUPPORT REACTIONS FOR TRAPEZOIDAL LOADS

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