scripts of youtube video 'How to use the integral formulas of Mechanics of Materials to calculate beam deflections'

 Slide 1

This is a simply-supported beam. X is the beam axis before deformation. This curve is its axis after deformation under force P. O, C, C1 are points on the beam axis, and C1 is the position of point C after force P is exerted. Point O is the origin of the coordinate system. The horizontal coordinate of Point C is x.

W(x) is the deflection at point C, which equals to the length of line CC1. If the beam at point C moves downward under force P, the deflection at C w(x) is positive. Otherwise, negative.

θ(x) is the rotation angle of beam cross section at point C under force P. If the beam cross section at point C rotates clockwise under force P, rotation angle of beam cross section at point C θ(x) is positive, otherwise negative.

Before force P exerts, the cross section at point C is vertical. Under force P, it’s slanted. The section turns in this direction. That’s clockwise. So the rotation angle of beam cross section at point C θ(x) is positive.

 

Slide 2

These are integral formulas of beams with constant cross sections to calculate beam deflections. The formulas belong to college curriculum of mechanics of materials. This integral formula about θ(x) is formula 1. This is formula 2. We will use them a little later in the examples.

E is material elastic module, I, moment of inertia of cross section; M(x), bending moment at point C; C and D constants; and x, horizontal coordinate of point C. 

If bending moment M(x) makes beam bottom tensile, it’s positive. Otherwise, negative. M(x) is inner force in beam cross section caused by load P.

You might notice there is a negative sign here, and in some other books, there is no such sign. This is because of the difference of sign rules between these books. For example, some books might consider upward deflection is positive and they omit the negative sign here.

To stipulate two constants C and D, we need two boundary conditions.

 

Slide 3

This is the left part of a simply supported beam, i.e., a detachment. This is M(x) in this cross section. This moment makes the bottom tensile and the top compressive. So the moment is positive.

 

Slide 4

These are boundary conditions of the simply-supported beam.

B is the point where force P exerts. L is beam span.

At points O and A, the beam cannot move vertically. So w(x) is zero at these points. Since the beam is continuous at point B, we can calculate θ(x) and w(x) at point B from parts OB and BA of the beam respectively, and the calculations should be equal. We can use these conditions to solve the constants of integral equations.

  

Slide 5

These are boundary conditions of the cantilever beam.

L is the distance between left end D and where force P exerts.

At point D, the beam can neither move vertically or rotate. So both w(x) and θ(x) are zeros at point D.

 

Slide 6

This is a simply supported beam.

Force P exerts at point B which is the central point of the beam span. The span is 3 meters. The question is to solve the beam deflection at point B.

 

Slide 7

Here is the solution to the question. For part AB of the beam, we can calculate its M(x). Using formula one, we can get the equation about its θ(x). Then we can get EIθ(x) at point B.

For part BC of the beam, we do the same.

 

Slide 8

Since the beam is continuous at point B, EIθ(x) calculated in parts AB and BC is the same. So we get equation 3.

Back to part AB of the beam, using formula 2, we get EIw(x) equation.

Using the boundary condition at point A, both its x and w(x) are zeros, we get D1 equals zero.

We also get EIw(x) equation at point B.

 

Slide 9

Here we do the same things for part BC.

 

Slide 10

Since the beam is continuous at point B, EIw(x) calculated in both parts AB and BC is the same. So we get this equation 10.

Combine equations 3, 5, 8, 10, we get the equation system.

 

Slide 11

From the equation system, we solve C1.

Put the values of C1 and D1 into equation 4, we have equation 11 and we calculate the beam deflection at point B with this equation.

Here we use equation 11, which we get from equation 4, to calculate the deflection at point B.

You might ask if we can use E*I*w(x) equation of part BC of the beam, i.e., equation 7, to calculate the deflection at point B as well. Yes you can because point B is at the boundary of parts AB and BC. It belongs to the both parts.

But how about the point whose x equals 1000mm? In this case, we can only use equation 11 to calculate its deflection since it belongs to part AB of the beam only.

For the same reason, we can only use equation 7 to calculate the beam deflection at the point whose x equals 2000mm since it’s only in part BC.

 

Slide 12

This example is a cantilever beam.

Its material, dimensions and force are all the same as in example 1.

Point A is its left end. Point B is where force P exerts. The distance between points A and B is three thousand millimeters.

The deflection of point B is asked.

 

Slide 13

The solution process is the same as in example 1 except for this example 2 is easier.

Here we get its M(x) and EIθ(x) expressions.

Using the boundary condition, we get constant C.

With formula 2, we have this equation 2.

Using the boundary condition, we get constant D value.

Put C, D values into equation 2, we have equation 3.

Then we calculate the deflection at point B using equation 3.