1. RAYLEIGH-RITZ METHOD Assume a deflection shape
Unknown coefficients ci and known function fi(x)
Deflection curve v(x) must satisfy displacement boundary conditions
Obtain potential energy as function of coefficients
Apply the principle of minimum potential energy to determine the coefficients
Walther Ritz, Swiss physicist,
John Willeam Strutt, 3rd Baron Rayleigh, (discovered argon, Nobel prize 1904) Physicist.

2. Application to beam problems
The differential equation for the beam displacement is
For a constant p(x)=p0, the general solution of this differential equation is
When p(x)=0, the exact solution is a cubic polynomial.
Concentrated forces and couples cause discontinuities, in third and second derivatives, respectively.

3. EXAMPLE – SIMPLY SUPPORTED BEAM
E,I,L
Assumed deflection curve
Strain energy
Potential energy of applied loads (no reaction forces)
Potential energy
PMPE:

4. EXAMPLE – SIMPLY SUPPORTED BEAM cont.
Approximate bending moment and shear force
Exact solutions

5. EXAMPLE – SIMPLY SUPPORTED BEAM cont.
Deflection
Bending moment
Shear force
Error increases

6. Quiz-like questions
Calculate the Rayleigh-Ritz approximation to the maximum displacement, bending moment and shear force for a simply supported beam under uniform load, when the displacement is assumed of the form v=Cx(L-x)
Why is the solution with the sine so much better?
The coefficient of x4 is shown in Slide 2 to be 1/(24EI). What boundary conditions does one use to calculate the coefficients of the other terms (the cubic polynomial in Slide 2)?
Answers in notes page
The calculation is given as
We can see that even the maximum displacement is more than 20% in error, the moment is constant, and the shear force is zero, so these two are in gross errors.
The sine function was so much better because it happened to satisfy also the natural boundary conditions that the bending moment is zero at the support (second derivative of the sine is the same sine times a constant).
The boundary conditions are the v=0 and M=0 at x=0,L. The zero moment implies zero second derivative. It is easy to check that the exact solution of slide 4 satisfies these boundary conditions.

7. EXAMPLE – CANTILEVERED BEAMF C
–p0
E,I,L
Assumed deflection
Need to satisfy BC
Strain energy
Potential of loads

8. EXAMPLE – CANTILEVERED BEAM cont.
Derivatives of U:
PMPE:
Data:
Solve for c1 and c2:
Deflection curve:
Exact solution:
Check exact solution

9. EXAMPLE – CANTILEVERED BEAM cont.
Deflection
Bending moment
Shear force
Error increases

10. Quiz like questions . Consider the cantilever beam in the figure
Will a cubic polynomial Ritz solution be exact?
We use the Rayleigh-Ritz method with v=Cx2.
Compare the exact and approximate shear force.
Compare the exact and approximate tip deflection.
Answers in the notes page .
With concentrated forces acting only at the boundaries and no p(x) the exact solution is a cubic polynomial. So we will use only a quadratic.
With a negative force, the exact shear force is Vy=-F. For the Rayleigh-Ritz approximation we note that
The exact tip displacement is known to be
For the Rayleigh-Ritz solution
So while the shear force is 100% off, the tip displacement is only 25% in error.