PLAIN AND CIVIL: EXAMPLE 4.3.15. BEAM SUPPORT REACTIONS FOR TRIANGULAR LOADS

Determine the support reactions for the beam as shown.

For distributed load over a support, the rotation of moments is different on the load on the left side of the support from the right side. Because of this rotation change, converting the distributed load directly to point load will result to wrong calculations.

The load needs to be broken down to parts - left and right of the support.

The distributed load would have to be divided at point C. To get the magnitude of the loads P1, P2, P3, use ratio and proportion to have the magnitude of load at point C.

By ratio and proportion:

$\frac{height_{B}}{5.50m}=\frac{height_{C}}{3.00m}$

$\frac{3kN/m}{5.50m}=\frac{height_{C}}{3.00m}$

$height_{C}=\frac{3kN/m}{5.50m}3.00m=1.64\frac{kN}{m}$

Having known the load at point C, continue determining P1, P2, and P3.

$P_{1}=\frac{1}{2}\left ( 3-1.64\frac{kN}{m} \right )\left (2.5m \right )=1.70kN$

$P_{2}=1.64\frac{kN}{m}\left ( 2.50m \right )=4.10kN$

$P_{3}=\frac{1}{2}\left ( 1.64\frac{kN}{m} \right )\left (3.0m \right )=2.46kN$

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

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

$\sum M_{C}=0$

$R_{AY}\left ( 6.5m \right )+P_{3}\left ( 1.0m \right )-P_{1}\left ( 2.5-0.83m \right )-P_{2}\left ( 1.25m \right )=0$

$R_{AY}\left ( 6.5m \right )=1.70kN\left ( 2.5-0.83m \right )+4.10kN\left ( 1.25m \right )-2.46kN\left ( 1.0m \right )$

${\color{Red} \boldsymbol{R_{AY}=0.847kN}}$

$\sum M_{A}=0$

$P_{1}\left ( 6.5-2.5+0.83m \right )+P_{2}\left ( 6.5-1.25m \right )+P_{3}\left ( 6.5+1m \right )-R_{C}\left ( 6.5m \right )=0$

$R_{C}\left ( 6.5m \right )=1.70kN\left ( 6.5-2.5+0.83m \right )+4.10kN\left ( 6.5-1.25m \right )+2.46kN\left ( 6.5+1m \right )$

${\color{Red} \mathbf{R_{C}=7.413kN}}$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Friday 29 March 2019

EXAMPLE 4.3.15. BEAM SUPPORT REACTIONS FOR TRIANGULAR LOADS

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