PLAIN AND CIVIL: EXAMPLE 4.3.8. BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS

Determine the support reactions for the beam shown.

The FBD of the beam can be drawn in two ways.

First method: Converting loads according to the magnitude.

$P_{1}=3\frac{kN}{m}\left (5.5m \right )=16.50kN$

$P_{2}=7.5\frac{kN}{m}\left (2.5m \right )=18.75kN$

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

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

$\sum M_{C}=0$

$R_{A}\left ( 8m \right )-P_{1}\left ( 2.75+2.5 \right )-P_{2}\left ( 1.25 \right )=0$

$R_{A}\left ( 8m \right )=16.5kN\left ( 2.75+2.5 \right )+18.75kN\left ( 1.25 \right )$

${\color{Red} \mathbf{R_{A}=13.758kN\uparrow }}$

$\sum M_{A}=0$

$P_{1}\left ( 2.75m \right )+P_{2}\left ( 5.5+1.25m \right )-R_{CY}\left ( 8m \right )=0$

$R_{CY}\left ( 8m \right )=16.5kN\left ( 2.75m \right )+18.75kN\left ( 5.5+1.25m \right )$

${\color{Red} \mathbf{R_{CY}=21.492kN}}\uparrow$

Second method: Converting loads according to location.

$P_{1}=3\frac{kN}{m}\left ( 8m \right )=24kN$

$P_{2}=4.5\frac{kN}{m}\left ( 2.5m \right )=11.25kN$

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

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

$\sum M_{C}=0$

$P_{1}\left ( 4m\right )+P_{2}\left ( 1.25m\right )-R_{A}\left ( 8m \right )=0$

$R_{A}\left ( 8m \right )=24kN\left ( 4m\right )+11.25kN\left ( 1.25m\right )$

${\color{Red} \mathbf{R_{A}=13.758kN\uparrow }}$

$\sum M_{A}=0$

$P_{1}\left ( 4m \right )+P_{2}\left ( 5.5+1.25 \right )-R_{CY}\left ( 8m \right )=0$

$R_{CY}\left ( 8m \right )=24kN\left ( 4m \right )+11.25kN\left ( 5.5+1.25 \right )$

${\color{Red} \mathbf{R_{CY}=21.492kN}}\uparrow$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

Pages

Tuesday 26 March 2019

EXAMPLE 4.3.8. BEAM SUPPORT REACTIONS FOR UNIFORMLY DISTRIBUTED LOADS

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