Boundary conditions for a 4th order beam deflection equation

In summary, the four boundary conditions for a fourth order differential equation describing the deflection of a propped cantilever beam with a uniform distributed load applied are:1. x = 0 v = 0 (no deflection at the built in support end)2. x = L v = 0 (no deflection at the simple support end)3. x = 0 dv/dx = 0 (slope of the deflection at the built in support is 0)4. x = L d^2v/dx^2 = 0 (no bending moment at the simple support end)
  • #1
Xaspire88
100
0
What would the boundary conditions be for a fourth order differential equation describing the deflection (elastic curve) of a propped cantilever beam with a uniform distributed load applied? i.e. a beam with a built in support on the left and a simple support on the right. I need 4 obviously but I am having a hard time coming up with the 4th.

So far I have

1. x = 0 v = 0 (no deflection at the built in support end)
2. x = L v = 0 (no deflection at the simple support end)
3. x = 0 dv/dx = 0 (slope of the deflection at the built in support is 0)

and for the fourth I have seen

x = L d^2v/dx^2=0

but I am having some trouble wrapping my head around that last one. Is it correct?
 
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  • #2
Xaspire88 said:
What would the boundary conditions be for a fourth order differential equation describing the deflection (elastic curve) of a propped cantilever beam with a uniform distributed load applied? i.e. a beam with a built in support on the left and a simple support on the right. I need 4 obviously but I am having a hard time coming up with the 4th.

So far I have

1. x = 0 v = 0 (no deflection at the built in support end)
2. x = L v = 0 (no deflection at the simple support end)
3. x = 0 dv/dx = 0 (slope of the deflection at the built in support is 0)

and for the fourth I have seen

x = L d^2v/dx^2=0

but I am having some trouble wrapping my head around that last one. Is it correct?
yes, at the pinned simple support which is free to rotate, there can be no bending moment, which is what that boundary condition describes (at x = L, v" = M/EI = 0)
 
  • #3
thanks, I was getting confused because some sites had "common beam" equations that were different than others.. until i realized that the supports were on different sides and thus their coordinate system was changing. now it makes sense.
 

1. What is a 4th order beam deflection equation?

A 4th order beam deflection equation is a mathematical equation used to determine the deflection of a beam under a specific load. It takes into account the beam's material properties, dimensions, and boundary conditions to calculate the amount of bending or deformation that will occur.

2. Why are boundary conditions important in a 4th order beam deflection equation?

Boundary conditions are important because they define the constraints and supports on a beam, which directly affect its deflection. Without accurately specifying the boundary conditions, the results of the 4th order beam deflection equation would not be reliable or applicable to real-world scenarios.

3. What are some examples of boundary conditions in a 4th order beam deflection equation?

Examples of boundary conditions include fixed supports, simply supported ends, and free ends. Fixed supports prevent the beam from moving or rotating at that point, while simply supported ends allow the beam to move vertically but not rotate. Free ends allow both vertical movement and rotation.

4. How does a 4th order beam deflection equation differ from lower order equations?

A 4th order beam deflection equation takes into account higher-order effects, such as the curvature of the beam and the shear deformation caused by the load. This results in a more accurate prediction of the beam's deflection compared to lower order equations, which only consider simpler factors such as the bending moment and shear force.

5. Can a 4th order beam deflection equation be used for all types of beams?

No, a 4th order beam deflection equation is typically used for beams made of linear, homogeneous materials with constant cross-sections. Beams with more complex geometries or made of non-linear or non-homogeneous materials may require different equations or methods of analysis.

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