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Bearing reactions under sloping load

This exercise shows how to calculate the bearing reactions under sloping load.

Task

A bridge on a floating bearing and a fixed bearing is loaded by a braking car as shown. Determine the bearing reactions in A and B!

Beam on two supports with sloping load
Beam on two supports with sloping load

Solution

The following video is in german language. Please scroll down for the written solution.

Lagerreaktionen bestimmen - Technische Mechanik 1, Übung 14
Free body diagram of the beam
Free body diagram of the beam
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The equilibrium of forces in the x-direction is

\[\tag{1} \sum F_x = 0 = F \cdot \sin \alpha + F_{Bx} \]

\[\tag{2} F_{Bx} = -- F \cdot \sin \alpha\]

The equilibrium of forces in the y-direction is

\[\tag{3} \sum F_y = 0 = F_{Ay} -- F \cdot \cos \alpha + F_{By} \]

The equilibrium of moments around point A is

\[\tag{4} \sum M(A) = 0 = -F \cdot \cos \alpha \cdot a + F_{By} \cdot (a+b) \]

\[\tag{5} F_{By} = \frac{F \cdot \cos \alpha \cdot a}{a+b} \]

From equation 3 follows

\[\tag{6} F_{Ay} = F \cdot \cos \alpha -- F_{By} \]

\[\tag{7} F_{Ay} = F \cdot \cos \alpha -- \frac{F \cdot \cos \alpha \cdot a}{a+b} \]

\[\tag{8} F_{Ay} = F \cdot \cos \alpha \cdot \left(1 -- \frac{a}{a+b} \right) \]

This was the basic exercise “bearing reactions under sloping load”, used to show the appliance of fixed and floating bearings. Don’t miss the other interesting exercises!