Beam deflection integration

In summary, the slope dy/dx = 0 at x = L/2 because at this point, the deflection curve is a flat line of constant value due to the symmetry of the diagram. This means that the derivative of the deflection curve, which is the slope curve, will also have a value of 0. The negative sign after integrating 0 = EI(d^4y/dx^4) is most likely a mistake, as the correct equation should be A = EI(d^3y/dx^3). Therefore, the value of the slope of the beam does not depend on the value of the deflection at the same location, and at x = L/2, the slope will be 0 due
  • #1
influx
164
2
62e6f5.png

Can someone explain why the slope dy/dx = 0 at x = L/2? L/2 is the midpoint and there would be a deflection here so surely the slope of the deflection curve shouldn't be 0? I'm finding it hard to visualise this.

EDIT: I think I understand the above. The slope of the deflection curve at x = L/2 will be 0 as the slope at this point is a flat line of constant value (due to the symmetry of the diagram). This is correct yes?

-----------------------------------------------------------------------

Is the part circled in red a mistake? There shouldn't be a negative sign after integrating 0 = EI(d^4y/dx^4)? As in it should be just A = EI(d^3y/dx^3) rather than A = -EI(d^3y/dx^3)?

c4cf9e.png
 
Last edited:
Engineering news on Phys.org
  • #2
influx said:
62e6f5.png

Can someone explain why the slope dy/dx = 0 at x = L/2? L/2 is the midpoint and there would be a deflection here so surely the slope of the deflection curve shouldn't be 0? I'm finding it hard to visualise this.

EDIT: I think I understand the above. The slope of the deflection curve at x = L/2 will be 0 as the slope at this point is a flat line of constant value (due to the symmetry of the diagram). This is correct yes?
The value of the slope of the beam is independent of the value of the deflection at the same location.

For the simply supported beam, the deflection at x = L/2 will be a maximum, and since the slope curve is the first derivative of the deflection curve, what value will the derivative of the deflection curve have where the deflection is a maximum? This is a basic property of derivatives from intro calculus.
 

What is beam deflection integration?

Beam deflection integration is a method used to calculate the deflection of a beam under various loads. It involves integrating the bending moment equation over the length of the beam to determine the deflection at any point along the beam.

What are the applications of beam deflection integration?

Beam deflection integration is used in structural engineering to design and analyze beams in buildings, bridges, and other structures. It is also used in mechanical engineering to design and analyze machine components such as shafts and beams.

What are the assumptions made in beam deflection integration?

The main assumptions made in beam deflection integration include the beam being homogeneous, isotropic, and elastic. It is also assumed that the beam is subject to small deflections and that the material properties of the beam remain constant.

What are some common methods used for beam deflection integration?

Some common methods used for beam deflection integration include the double integration method, the Macaulay's method, and the moment-area method. Each method has its own advantages and is suitable for different types of beams and loading conditions.

What factors affect the deflection of a beam?

The deflection of a beam is affected by various factors such as the type of beam, its length, the material properties of the beam, the type and magnitude of the load applied, and the support conditions. Other factors such as temperature changes, moisture, and creep can also affect the deflection of a beam over time.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
359
Replies
4
Views
5K
  • General Engineering
Replies
2
Views
2K
  • General Engineering
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
3K
Replies
4
Views
4K
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
3K
Back
Top