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# Bearing reactions under sloping load

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

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!

## Solution

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

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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!