The integration method allows us to obtain the slope and deflection at a particular point on the beam. These information are crucial to the design of beams and shafts to ensure they meet the safe design criteria.

To get our slope and deflection, we start with this relation:

- EI is called the flexural rigidity. It is the Young’s modulus E multiplied by the moment of inertia I.
- ν is the deflection (units: m or mm)
- M(x) is the internal bending moment, expressed as a function of x, which is the distance along the beam.

The integration method allows us to obtain the slope and deflection at a particular point on the beam. These information are crucial to the design of beams and shafts to ensure they meet the safe design criteria.

To get our slope and deflection, we start with this relation:

- EI is called the flexural rigidity. It is the Young’s modulus E multiplied by the moment of inertia I.
- ν is the deflection (units: m or mm)
- M(x) is the internal bending moment, expressed as a function of x, which is the distance along the beam.

Then, using the relation above, you integrate once to get dν/dx (slope) and another time to get ν (deflection):

You can then shift the EI term over the right-hand-side to get your slope (dν/dx) or deflection (ν).

Notice that integrating twice produced two constants C_{1} and C_{2}. How do we resolve these? That’s where our *boundary conditions (BCs)* come in =)

Boundary conditions are usually taken at the supports. If the beam is symmetrical, BCs can be taken at the point of symmetry as well:

There are 2 constants C_{1} and C_{2}, and therefore we need to apply 2 BCs to solve for both C_{1} and C_{2}.

The integration method might seem long and complex, but don’t worry, it’s easier than it looks. The best way to learn is by example.

Then, using the relation above, you integrate once to get dν/dx (slope) and another time to get ν (deflection):

You can then shift the EI term over the right-hand-side to get your slope (dν/dx) or deflection (ν).

Notice that integrating twice produced two constants C_{1} and C_{2}. How do we resolve these? That’s where our *boundary conditions (BCs)* come in =)

Boundary conditions are usually taken at the supports. If the beam is symmetrical, BCs can be taken at the point of symmetry as well:

There are 2 constants C_{1} and C_{2}, and therefore we need to apply 2 BCs to solve for both C_{1} and C_{2}.

The integration method might seem long and complex, but don’t worry, it’s easier than it looks. The best way to learn is by example.