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# Beam with joint and offset load

This exercise is about the following questions:

• How to calculate the forces in a joint?
• How to calculate the bearing loads if a force acts with an offset?

A force F acts with an offset on a beam. The beam is fixed on one side and connected to a floating bearing with a joint. Find the reaction forces in A, B and C.

## Solution

The following video is in german language.

The beam is divided into two sections to determine the bearing reactions and the joint forces.

### Section 1 Free-cut section 1 of the beam $\require{cancel}$

The equilibrium of forces in the x-direction is

$\tag{1} \sum F_x = 0 = F + F_{Bx}$

$\tag{2} F_{Bx} = -F$

The equilibrium of forces in the y-direction is

$\tag{3} \sum F_y = 0 = F_{Ay} + F_{By}$

The equilibrium of moments around point A gives

$\tag{4} \sum M(A) = 0 = -F \cdot a + F_{By} \cdot 2 \cdot a$

$\tag{5} F_{By} = \frac{F \cdot \bcancel{a}}{2 \cdot \bcancel{a}}$

So the result from equation 3 is

$\tag{6} F_{Ay} = - \frac{F}{2}$

### Section 2 Free-cut section 2 of the beam $\require{cancel}$

The equilibrium of forces in the x-direction is

$\tag{7} \sum F_x = 0 = - F_{Bx} + F_{Cx}$

$\tag{8} F_{Cx} = -F$

The equilibrium of forces in the y-direction is

$\tag{9} \sum F_y = 0 = - F_{By} + F_{Cy}$

$\tag{10} F_{Cy} = \frac{F}{2}$

The equilibrium of moments around point C gives

$\tag{11} \sum M(C) = 0 = F_{By} \cdot 2a + M_C$

$\tag{12} M_C = - F \cdot a$