PLAIN AND CIVIL: EXAMPLE 4.3.17. BEAM SUPPORT REACTIONS FOR TRAPEZOIDAL LOADS

Saturday, 30 March 2019

Determine the support reactions for the beam with the trapezoidal loading as shown.

Because the distributed load is just above a support, the load needs to be broken down into parts in order to clarify how their effect is to the left and right side of the support. With a rectangular load, it is divided into two parts - one for the left side and the other for the right. Doing the same principle, the trapezoidal load will be divided on the left and the right of the support. However, the moment arm for trapezoidal load is difficult to determine directly, the two loads needed to be broken down further to two basic shapes - rectangular and triangular loads.

The FBD for the beams becomes:

In order to start with the conversion of distributed loads into concentrated loads, identify the magnitude of load just on the support itself, y, in order to determine magnitudes of individual load parts.Since w2 is constant from beginning to end of the load, the y solved is only a part of the triangle above.

By using ratio and proportion,

$\frac{y}{4.50m}=\frac{\left ( 12-8 \right )}{6m}$ ; $y=3\frac{kN}{m}$

So the converted concentrated loads become:

$P_{1}=\frac{1}{2}\left ( 12\frac{kN}{m} -8\frac{kN}{m}-y\right )\left ( 1.50m \right )$

$P_{1}=\frac{1}{2}\left ( 12\frac{kN}{m} -8\frac{kN}{m}-3\frac{kN}{m}\right )\left ( 1.50m \right )=0.75kN$

$P_{2}=\left ( 8\frac{kN}{m}+y \right )\left ( 1.50m \right )$

$P_{2}=\left ( 8\frac{kN}{m}+3\frac{kN}{m} \right )\left ( 1.50m \right )=16.50kN$

$P_{3}=\frac{1}{2}y\left ( 4.5m \right )$

$P_{3}=\frac{1}{2}\left ( 3\frac{kN}{m} \right )\left ( 4.5m \right )=6.75kN$

$P_{4}=\omega _{2}L=8\frac{kN}{m}\left ( 4.50m \right )=36kN$

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

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

$\sum M_{D}=0]$

$R_{AY}\left ( 9.0m \right )+P_{3}\left ( 1.50m \right )+P_{4}\left ( 2.25m \right )-P_{1}\left ( 1.0m \right )-P_{2}\left ( 0.75m \right )=0$

$R_{AY}\left ( 9.0m \right )=0.75kN\left ( 1.0m \right )+16.50kN\left ( 0.75m \right )-6.75kN\left ( 1.50m \right )-36kN\left ( 2.25m \right )$

${\color{Red} \mathbf{R_{AY}=-8.667kN=8.667kN\downarrow }}$

$\sum M_{A}=0]$

$P_{1}\left ( 9-1.0m \right )+P_{2}\left ( 9-0.75m \right )+P_{3}\left ( 9+1.5m \right )+P_{4}\left ( 9+2.25m \right )-R_{C}\left ( 9.0m \right )=0$

$0.75kN\left ( 9-1.0m \right )+16.50kN\left ( 9-0.75m \right )+6.75kN\left ( 9+1.5m \right )+36kN\left ( 9+2.25m \right )-R_{C}\left ( 9.0m \right )=0$

${\color{Red} \mathbf{R_{C}=68.667kN}}\uparrow$

RELATED TOPIC:

4.0. SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

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Saturday, 30 March 2019

EXAMPLE 4.3.17. BEAM SUPPORT REACTIONS FOR TRAPEZOIDAL LOADS

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