Cantilever Beam

Cantilever beam.
Straight beam on the left side fixed by a cantiveler, on which the distributed force and the horizontal concentrated force are acting. The total length of the beam is 10 meters.

Here we have an even easier beam to calculate the support reactions, because we will not have an equation with two unknowns, as it was in a  simply supported beam. First, let’s see how I accept the return of arrowheads of support forces.

Let’s go straight to calculating the values of support reactions. A detailed description of the equation is found in the previous example of a cantiveler beam.

First, let’s do the sum of the projections of forces on the X axis (horizontal).

\Sigma _{X}=0
H_{a}-10kN=0
H_{a}=10kN

We already know the value of the support reaction Ha.

Now the sum of the projections of forces on the Y axis (vertical).

\Sigma _{Y}=0
V_{a}-3,00[kN/m]*5,00[m]=0
V_{a}=15,00kN

Easy, right? We know the next reaction value, this time Va.

We have the last reaction to counting, M_{a} (bending moment).

\Sigma M_{a}=0

-Ma + 3kN / m * 5m (the section on which the force acts) * 5.5 m (the distance of the center of gravity of the force from the point to which we count the bending moment) = 0

-M_{a}+15,00[kN]*5,50[m]=0
-M_{a}+82,50[kNm]=0
-M_{a}=-82,50[kNm] /*(-1,00)
M_{a}=82,50[kNm]

And at this point we already know all support reactions in restraint.
For training, I suggest checking if the forces are balanced.

\Sigma _{X}=0
\Sigma _{Y}=0
\Sigma M_{?}=0