Rayleigh-Ritz method

The principle of minimum potential energy is used to obtain an approximate solution.

• Admissible displacements “shape functions” are assumed and their coefficients are determined.
• Admissible – satisfy the essential boundary conditions of compatibility. These are also called the geometric boundary conditions.
• This means were replace an infinite degree of freedom problem with an approximate problem that has finite degrees of freedom
• (Convergence on the solution requires completeness [a mathematical requirement] i.e., essential terms in the energy equation cannot be skipped)

• The solutions take the form: \[\begin{equation*}\begin{split} u(x,y,z) &= \; \displaystyle \sum_i a_i \, u_i(x,y,z) \\ v(x,y,z) &= \; \displaystyle \sum_i b_i \, v_i(x,y,z) \\ w(x,y,z) &= \; \displaystyle \sum_i c_i \, w_i(x,y,z) \\ \end{split}\end{equation*}\]
• Displacement functions \(u_i, v_i, w_i\) are assumed and must be admissible.
• The coefficients \(a_i, b_i, c_i\) are unknowns which must be determined.

The components of the potential energy are expressed in terms of the amplitudes \(a_i, b_i, c_i\) .

\[\begin{equation*} \Pi = U + V = f( a_i, b_i, c_i) \end{equation*}\]

• The coefficients are obtained by setting: \[\begin{equation*} {\frac{\partial \Pi}{\partial a_i}}={\frac{\partial \Pi}{\partial b_i}}={\frac{\partial \Pi}{\partial c_i}}=0 \end{equation*}\]
• Note: the assumption of a displacement field, if not admitting the actual displacement, imposes a constraint on the computed displacement.

Boundary conditions

• We’ve talked about admissible functions and essential boundary conditions, what does this mean?

• Suppose the variational method has one dependent variable.
  • Displacement for an axial bar
  • Deflection for a beam in bending
• Let \(2p\) be the highest order derivative in the dependent field of the governing differential equation.
  • Essential (geometric) boundary conditions–involve derivatives up to \(p-1\).
  • Non-essential (natural) boundary conditions–involve derivatives of \(p\) or above.
  • Admissible functions satisfy the essential boundary conditions and are \(p\) times differentiable.

Rayleigh-Ritz example one: cantilever beam with distributed end load

• Given:
  • \(p\)
  • \(M_l=p l^2\)
  • \(l\)
  • \(EI\)
• Solve for: \(v(l), \theta(l)\)

What polynomial provides a “good Rayleigh-Ritz assumption?”?

4th order ODE \(\hspace{3mm} \Longrightarrow \hspace{3mm}\) \(p=2\)

• We must assume a shape function that is at least 2 times differentiable.
• It must satisfy the essential boundary condition at x=0. \[\begin{equation*} v(x) = \displaystyle \sum_{i=2}^n b_i x^i \end{equation*}\]
• Assume we only keep the first term. \[\begin{equation*} v\left(x\right)=b_2 x^{2} \end{equation*}\] \[\begin{equation*} v{'}\left(x\right)=2 b_2 x \end{equation*}\] \[\begin{equation*} v{''}\left(x\right)=2 b_2 \end{equation*}\]

• The energy terms:
  • Strain energy: \[\begin{equation*} U=\int_{0}^{l}{\frac{ EI }{2} v{''}^{2 }\left(x\right)\;dx} \end{equation*}\] \[\begin{equation*} U=2 b_2^{2} l EI \end{equation*}\]
  • Potential of the external loads: \[\begin{equation*} V=p \int_{\frac{2 l}{3}}^{l}{v\left(x\right)\;dx}-v{'}\left(l\right) M_l \end{equation*}\] \[\begin{equation*} V=b_2 p \int_{\frac{2 l}{3}}^{l}{x^{2}\;dx}-2 b_2 l M_l \end{equation*}\] \[\begin{equation*} V=\frac{19 b_2 l^{3} p}{81}-2 b_2 l M_l \end{equation*}\]

The total potential energy: \[\begin{equation*} \Pi=V+U \end{equation*}\]

\[\begin{equation*} \Pi=2 b_2^{2} l EI -\frac{143 b_2 l^{3} p}{81} \end{equation*}\]

\[\begin{equation*} \frac{d}{d b_2} \Pi=\frac{d}{d b_2} \left(2 b_2^{2} l EI -\frac{143 b_2 l^{3} p}{81}\right) \end{equation*}\]

\[\begin{equation*} 0=4 b_2 l EI -\frac{143 l^{3} p}{81} \end{equation*}\]

\[\begin{equation*} \left[ b_2=\frac{143 l^{2} p}{324 EI } \right] \end{equation*}\]

\[\begin{equation*} v(x) = \frac{143 l^{2} p x^{2}}{324 EI } \end{equation*}\]

\[\begin{equation*} v(l) = \frac{143 l^{4} p}{324 EI } \end{equation*}\]

\[\begin{equation*} \frac{.441 l^{4} p}{ EI } \end{equation*}\]

What if we keep two terms: \[\begin{equation*} v\left(x\right)=b_3 x^{3}+b_2 x^{2} \end{equation*}\]

\[\begin{equation*} v{'}\left(x\right)=3 b_3 x^{2}+2 b_2 x \end{equation*}\]

\[\begin{equation*} v{''}\left(x\right)=6 b_3 x+2 b_2 \end{equation*}\]

The energy terms: \[\begin{equation*} U=\int_{0}^{l}{\frac{ EI }{2} v{''}^{2 }\left(x\right)\;dx} \end{equation*}\]

\[\begin{equation*} U=\frac{\int_{0}^{l}{\left(6 b_3 x+2 b_2\right)^{2}\;dx} EI }{2} \end{equation*}\]

\[\begin{equation*} V=p \int_{\frac{2 l}{3}}^{l}{v\left(x\right)\;dx}-v{'}\left(l\right) M_l \end{equation*}\]

\[\begin{equation*} V=p \int_{\frac{2 l}{3}}^{l}{b_3 x^{3}+b_2 x^{2}\;dx}-\left(3 b_3 l^{2}+2 b_2 l\right) M_l \end{equation*}\]

\[\begin{equation*} V=\left(\frac{3 b_3 l^{4}+4 b_2 l^{3}}{12}-\frac{4 b_3 l^{4}+8 b_2 l^{3}}{81}\right) p-\left(3 b_3 l^{2}+2 b_2 l\right) M_l \end{equation*}\]

The total potential energy: \[\begin{equation*} \Pi=U+V \end{equation*}\]

Apply the known values of the deflection and moment: \[\begin{equation*}\begin{split} \Pi=&\frac{\left(12 b_3^{2} l^{3}+12 b_2 b_3 l^{2}+4 b_2^{2} l\right) EI}{2} \\ & +\left(\frac{3 b_3 l^{4}+4 b_2 l^{3}}{12}-\frac{4 b_3 l^{4}+8 b_2 l^{3}}{81}\right) p-l^{2} \left(3 b_3 l^{2}+2 b_2 l\right) p \end{split}\end{equation*}\]

\[\begin{equation*}\begin{split} \frac{d}{d b_2} \Pi= & + \frac{d}{d b_2} \frac{\left(12 b_3^{2} l^{3}+12 b_2 b_3 l^{2}+4 b_2^{2} l \right) EI}{2} \\ & + \frac{d}{d b_2} \left(\frac{3 b_3 l^{4}+4 b_2 l^{3}}{12}-\frac{4 b_3 l^{4}+8 b_2 l^{3}}{81}\right) p \\ & - \frac{d}{d b_2} \left(l^{2} \left(3 b_3 l^{2}+2 b_2 l\right) p \right) \\ \end{split}\end{equation*}\]

The derivative is equal to zero: \[\begin{equation*} 0=\frac{\left(12 b_3 l^{2}+8 b_2 l\right) EI }{2}-\frac{143 l^{3} p}{81} \end{equation*}\]

Solve for the coefficient: \[\begin{equation*} \left[ b_2=-\frac{486 b_3 l EI -143 l^{2} p}{324 EI } \right] \end{equation*}\]

Now take another derivative with respect to the other coefficient: \[\begin{equation*}\begin{split} \frac{d}{d b_3} \Pi= & +\frac{d}{d b_3} \left(\frac{\left(12 b_3^{2} l^{3}+12 b_2 b_3 l^{2}+4 b_2^{2} l\right) EI}{2}\right) \\ & +\frac{d}{d b_3} \left(\left(\frac{3 b_3 l^{4}+4 b_2 l^{3}}{12}-\frac{4 b_3 l^{4}+8 b_2 l^{3}}{81}\right) p\right) \\ & -\frac{d}{d b_3} \left(l^{2} \left(3 b_3 l^{2}+2 b_2 l\right) p\right) \\ \end{split}\end{equation*}\]

The derivative is equal to zero: \[\begin{equation*} 0=\frac{\left(24 b_3 l^{3}+12 b_2 l^{2}\right) EI }{2}-\frac{907 l^{4} p}{324} \end{equation*}\]

Solve for the coefficient: \[\begin{equation*} b_3=-\frac{1944 b_2 EI -907 l^{2} p}{3888 l EI } \end{equation*}\]

We now have two equations and two unknowns. Solve for the two unknowns: \[\begin{equation*}\begin{split} b_2&=\, \frac{79 l^{2} p}{216 EI } \\ b_3&=\, \frac{49 l p}{972 EI } \\ \end{split}\end{equation*}\] or: \[\begin{equation*}\begin{split} b_2&=\, \frac{.366 l^{2} p}{ EI } \\ b_3&=\, \frac{.0504 l p}{ EI } \\ \end{split}\end{equation*}\]

Therefore: \[\begin{equation*} v(x) = \frac{49 l p x^{3}}{972 EI }+\frac{79 l^{2} p x^{2}}{216 EI } \end{equation*}\]

\[\begin{equation*}\begin{split} v(l) &=\, \frac{809 l^{4} p}{1944 EI } \\ &=\, \frac{.416 l^{4} p}{ EI } \end{split}\end{equation*}\]

Comparison

Keeping only the quadratic term: \[\begin{equation*} v(l) = \frac{.441 l^{4} p}{ EI } \end{equation*}\]

Keeping quadratic and cubic terms: \[\begin{equation*} v(l) = \frac{.416 l^{4} p}{ EI } \end{equation*}\]

Exact solution: \[\begin{equation*} v(l) = \frac{.416 l^{4} p}{ EI } \end{equation*}\]

Example two: simply supported beam with sinusoidal load

Givens:

• Section properties: \(l\), \(EI\)
• Load: \(p(x) = P \sin \left(\pi \frac{x}{l}\right)\)

Required:

• Maximum displacement: \(v_{\mathrm{max}}\)

Solution:

• First, identify the boundary conditions:
  • at \(x=0\)
    • \(v=0\)
    • \(M=0\)
  • at \(x=l\)
    • \(v=0\)
    • \(M=0\)

• Recall the strain energy in the beam due to bending loads: \[\begin{equation*} U=\int_{0}^{l}{\frac{M^2}{2\,EI}\left(x\right)\;dx} \end{equation*}\]

• Also, recall the moment-deflection relationship and thus: \[\begin{equation*} U=\int_{0}^{l}{{\frac{EI}{2}\,v^{''}}^2\left(x\right)\;dx} \end{equation*}\]

• The potential of the external loads is: \[\begin{equation*} V=-\int_{0}^{l}{p\left(x\right)\,v\left(x\right)\;dx} \end{equation*}\]

• Assume a displacement shape function of the form: \[\begin{equation*} v\left(x\right)=\displaystyle \sum_{m=1}^{\infty }{b_{m}\,\sin \left(\frac{\pi\,m\,x}{l}\right)} \end{equation*}\]

• Thus: \[\begin{equation*}\begin{split} v^{'}\left(x\right) &= \frac{\pi}{l} \,\displaystyle \sum_{m=1}^{\infty }{m\,b_{m}\,\cos \left(\frac{\pi\,m\,x}{l}\right)} \\ {v^{''}}\left(x\right) &= -\frac{\pi^2}{l^2}\,\displaystyle \sum_{m=1}^{\infty }{m^2\,b_{m}\,\sin \left(\frac{\pi\,m\,x}{l}\right)} \\ \end{split}\end{equation*}\]

• Therefore, the strain energy is: \[\begin{equation*} U=\frac{\pi^4}{2\,l^4}\,EI\,\int_{0}^{l}{\left(\displaystyle \sum_{m=1}^{\infty }{m^2\,b_{m}\,\sin \left(\frac{\pi\,m\,x}{l}\right)}\right)^2\;dx} \end{equation*}\]

• Note, it is easier to take derivatives using the chain rule prior to applying the integration: \[\begin{equation*} {\frac{d }{d b_m}} U= EI \int_{0}^{l} {{v^{''}} \left(x\right) {\frac{d }{d b_m}} v^{''} \left(x\right)\;dx} \end{equation*}\]

• Thus: \[\begin{equation*} {\frac{d }{d b_m}} U=\frac{\pi^4}{l^4}\,EI\,\int_{0}^{l}{\left(\displaystyle \sum_{m=1}^{\infty }{m^2\,\sin \left(\frac{\pi\,m\,x}{l}\right)}\right)\,\displaystyle \sum_{m=1}^{\infty }{m^2\,b_{m}\,\sin \left(\frac{\pi\,m\,x}{l}\right)}\;dx} \end{equation*}\]

• Thus load is: \[\begin{equation*} p\left(x\right)=\sin \left(\frac{\pi\,x}{l}\right)\,P \end{equation*}\]

• Thus: \[\begin{equation*} V=-\left(\int_{0}^{l}{\sin \left(\frac{\pi\,x}{l}\right)\,\displaystyle \sum_{m=1}^{\infty }{b_{m}\,\sin \left(\frac{\pi\,m\,x}{l}\right)}\;dx}\right)\,P \end{equation*}\]

  \[\begin{equation*} \frac{d}{d\,b_{m}}\,V=-\left(\int_{0}^{l}{\sin \left(\frac{\pi\,x}{l}\right)\,\sum_{m=1}^{\infty }{\sin \left(\frac{\pi\,m\,x}{l}\right)}\;dx}\right)\,P \end{equation*}\]

  \[\begin{equation*} {\frac{d }{d b_m}} \Pi = {\frac{d }{d b_m}} U + {\frac{d }{d b_m}} V \end{equation*}\]

• Thus: \[\begin{equation*}\begin{split} \tiny {\frac{d }{d b_m}} \Pi = \frac{\pi^4}{l^4} \,EI \, \left(\int_{0}^{l}{\left(\displaystyle \sum_{m=1}^{\infty }{m^2\,\sin \left(\frac{\pi\,m\,x}{l}\right)}\right)\, \tiny \displaystyle \sum_{m=1}^{\infty }{m^2\,b_{m}\,\sin \left(\frac{\pi\,m\,x}{l}\right)}\;dx}\right)& \\ \tiny - \, \left(\int_{0}^{l}{\sin \left(\frac{\pi\,x}{l}\right)\,\displaystyle \sum_{m=1}^{\infty }{\sin \left(\frac{\pi\,m\,x}{l}\right)}\;dx}\right)& \tiny = 0 \\ \end{split}\end{equation*}\]

• The orthogonality of the sin function says that the terms involving \(P\) are zero when \(m>1\). Thus: \[\begin{equation*}\begin{split} {\frac{d }{d b_m}} \Pi = \left(C_m \frac{\pi^4}{l^4} \,EI \right) \, b_m - 0 = 0 \\ b_m=0 \hspace{5mm} m>1 \\ \end{split}\end{equation*}\]

• When \(m=1\), \[\begin{equation*}\begin{split} {\frac{d }{d b_1}} \Pi = \left(\frac{\pi^4}{l^4} \,EI\right) \; b_1 - P = 0 \\ b_1 = \frac{P l^4}{\pi^4 EI} \end{split}\end{equation*}\]

• Thus, the displacement is: \[\begin{equation*} v\left(x\right)={b_{1}\,\sin \left(\frac{\pi\,x}{l}\right)} \end{equation*}\] and

  \[\begin{equation*} v_{\mathrm{max}} = \frac{P l^4}{\pi^4 EI} \end{equation*}\]

• Note, this is the exact solution to the equation since it satisfied the boundary conditions and the 4th order differential equation that governs beams.

Galerkin Method

Consider a linear differential equation of the form:

\[\begin{equation*} L u = f \end{equation*}\]

Where \(L\) is a differential operator. (The boundary conditions are homogeneous)

The right side is a forcing function. Therefore: \[\begin{equation*} (\delta W)_1 = \iiint_V f \delta u \; dV \end{equation*}\]

\(\delta u\) is a virtual displacement and the \((\delta W)_1\) is virtual work.

It must also be true (because of @ref(eq:operator)) that:

\[\begin{equation*} \iiint_V (L u) \delta u \; dV = \iiint_V f \delta u \; dV \end{equation*}\]

If we use an approximate solution to the differential equation, than the equality does not hold... however, we can use @ref(eq:intoperator) to minimize an error:

\[\begin{equation*}\begin{split} \iiint_V \left((L u) \delta u - f \delta u \right)\; dV &= \, e \\ \iiint_V \left((L u) - f \right) \delta u \; dV &= \, e \\ \iiint_V \left((L \hat{u}) - f \right) \delta \hat{u} \; dV &= \, 0 \\ \end{split}\end{equation*}\] In the above, \(\hat{u}\) is an approximate solution to the problem.

We can then assume a function of any appropriate form, for example:

\[\begin{equation*} \hat{u} = \sum_m C_m \cdot \hat{u}_m \end{equation*}\] To be appropriate, \(\hat{u}\) must satisfy the essential boundary conditions. (\(2p\) from table @ref(tab:essentialnonessential)).

Thus: \[\begin{equation*}\begin{split} G_m = \iiint_V \left((L \hat{u}) - f \right) \delta \hat{u} \; dV &= \, 0 \\ \end{split}\end{equation*}\] Since the variation (\(\delta \hat{u}\)) is arbitrary, we have as many equations (\(G_m\)) as we have coefficients; they can be determined. Determination of the coefficients provide the best possible solution of the assumed form. (Better solutions may exist, but not of the form assumed.)

Example One – Beam with distributed load on a portion

Example

• Given: \(L, a, EI, P_0\)

• Required: \(v(x)\)=?

For the B.C.s at \(x=0, L\):

• \(v=0\)
• \(M=0\) (\(EI v,_{xx}\)=0)

so assume: \[\begin{equation*}\begin{split} v_m = \sin \, \frac{m \pi x}{L} \\ v = \displaystyle \sum_{m=1}^{\infty} C_m \cdot \sin \frac{m \pi x}{L} \\ \end{split}\end{equation*}\]

The differential equation is: \[\begin{equation*} EI \frac{d^4 v}{d x^4} = p(x) \end{equation*}\] Thus: \[\begin{equation*} G_m = \int_0^L \left[ EI \frac{d^4 v}{d x^4} - p(x) \right] \cdot \sin \, \frac{m \pi x}{L} \cdot dx =0 \end{equation*}\]

Rearranging: \[\begin{equation*} EI \int_0^L \left[\frac{d^4 v}{d x^4} \right] \cdot \sin \, \frac{m \pi x}{L} \cdot dx = \int_0^L P_o \cdot \sin \, \frac{m \pi x}{L} \cdot dx \end{equation*}\]

Using the assumed function: \[\begin{equation*} \frac{d^4 v}{d x^4} = \displaystyle \sum_{j=1}^{\infty} C_j \cdot \left(\frac{j \pi}{L}\right)^4 \cdot \sin \frac{j \pi x}{L} \end{equation*}\]

Therefore: \[\begin{equation*} EI \cdot C_m \cdot \left(\frac{m \pi}{L}\right)^4 \cdot \frac{L}{2} = P_0 \cdot \int_0^a \sin \frac{m \pi x}{L} \cdot dx \end{equation*}\]

Solving for the coefficient \(C_m\): \[\begin{equation*} C_m = \frac{2 P_0}{EI L} \cdot \left(\frac{L}{m \pi}\right)^4 \cdot \int_0^a \sin \frac{m \pi x}{L} \cdot dx \end{equation*}\] We now have a set of coefficients which offer the best approximate solution of the assumed form.

If \(a=L\) (load over the whole beam): \[\begin{equation*} C_m = \frac{2 P_0}{EI L} \cdot \left(\frac{L}{m \pi}\right)^4 \cdot \int_0^L \sin \frac{m \pi x}{L} \cdot dx \end{equation*}\]

\[\begin{equation*} \int_0^L \sin \frac{m \pi x}{L} \cdot dx = \left\{ \begin{array}{cc} \frac{2 L}{m \pi}, & m \hspace{3mm} \text{odd} \\ 0, & m \hspace{3mm} \text{even} \\ \end{array} \right\} \end{equation*}\]

• Hence:
  • For m=2, 4, ... \(C_m = 0\)
  • For m=1, 3, ... \(C_m = \frac{4 P_0 L^4}{EI m^5 \pi^5}\)
• For a single term:
  • \(C_1 = \frac{4 P_0 L^4}{\pi^5 EI} = v_{\mathrm{max}}\)
• The exact solution is:
  • \(v_{\mathrm{max}} = \frac{5 P_0 L^4}{384 EI}\)
  • Ratio = \(\frac{(4/\pi^5)}{(5/384)} = 1.004\) (Error=0.4% !!)

Example Two – Choosing a Polynomial – Robust Approach

Assume a polynomial solution that is of the order of the governing differential equation: \[\begin{equation*} v\left(x\right)=C_{4}\,x^4+C_{3}\,x^3+C_{2}\,x^2+C_{1}\,x+C_{0} \end{equation*}\]

For reference, the derivatives are: \[\begin{equation*}\begin{split} {v'}\left(x\right)&=4\,C_{4}\,x^3+3\,C_{3}\,x^2+2\,C_{2}\,x+C_{1} \\ {v''}\left(x\right)&=12\,C_{4}\,x^2+6\,C_{3}\,x+2\,C_{2} \\ {v'''}\left(x\right)&=24\,C_{4}\,x+6\,C_{3} \\ {v''''}\left(x\right)&=24\,C_{4} \\ \end{split}\end{equation*}\]

The solution must satisfy the essential (geometric) boundary conditions of the problem. These are: \[\begin{equation*}\begin{split} v\left(0\right)&=0 \\ v'\left(0\right)&=0 \\ \end{split}\end{equation*}\]

This leads to: \[\begin{equation*}\begin{split} C_{0}=&\,0\cr C_{1}=&\,0\cr \end{split}\end{equation*}\]

The natural boundary conditions must also be satisfied: \[\begin{equation*}\begin{split} G_m = \iiint_V \left((L \hat{u}) - f \right) \delta \hat{u} \; dV &= \, 0 \\ \end{split}\end{equation*}\]

In this case: \[\begin{equation*}\begin{split} \iiint_V \left(L \hat{u}\right) dV &= \int \left(EI v''''(x) \right) dx \\ \delta \hat{u} &= \delta v \\ \iiint_V \left(f \hat{u} \right) dV &= P \delta v(l) \\ \end{split}\end{equation*}\] The latter of these is the virtual work associated with the point load P.

Using integration by parts: \[\begin{equation*}\begin{split} \int EI v'''' \delta v dx =&\, -\int EI v''' \delta v' dx + \left.EI v''' \delta v\right|_0^l \\ =&\, \int EI v'' \delta v'' dx + \left.EI v''' \delta v\right|_0^l - \left.EI v'' \delta v'\right|_0^l \\ =&\, \int EI v'' \delta v'' dx \\ &\, + EI v'''(l) \delta v(l) - EI v'''(0) \delta v(0) \\ &\, - EI v''(l) \delta v'(l) + EI v''(0) \delta v'(0) \\ \end{split}\end{equation*}\]

Thus: \[\begin{equation*}\begin{split} \int EI v'' \delta v'' dx \\ &\, - EI v''(l) \delta v'(l) + EI v''(0) \delta v'(0) \\ &\, + EI v'''(l) \delta v(l) - EI v'''(0) \delta v(0) = P \delta v(l)\\ \end{split}\end{equation*}\] Focusing only on the boundary terms: \[\begin{equation*}\begin{split} \int EI v'' \delta v'' dx &= 0; \\ - EI v''(l) \delta v'(l) + EI v''(0) \delta v'(0) &=0 \\ + \left(EI v'''(l) - P\right) \delta v(l) - EI v'''(0) \delta v(0)&= 0\\ \end{split}\end{equation*}\]

Since: \[\begin{equation*}\begin{split} \delta v(0) &=0 \\ \delta v(l) &\ne 0 \\ \delta v'(0) &=0 \\ \delta v'(l) &\ne 0 \\ \end{split}\end{equation*}\] We have: \[\begin{equation*}\begin{split} - EI v''(l) &=0 \\ + EI v'''(l) - P &= 0\\ \end{split}\end{equation*}\] These are the “natural” or force based boundary conditions which we have called non-essential in earlier parts of this course. The Galerkin method requires these to be satisfied as well.

Note the shear term in the above equation... it is not convenient for finding the “form” of the Galerkin approximating function. Plugging it into our assumed function yields: \[\begin{equation*}\begin{split} C_{2}=\frac{12\,C_{4}\,l^2\,EI-l\,P}{2\,EI}\cr C_{3}=-\frac{24\,C_{4}\,l\,EI-P}{6\,EI} \end{split}\end{equation*}\]

\[\begin{equation*} v(x) = \left(\frac{x^3}{6}-\frac{l\,x^2}{2}\right)\,\frac{P}{EI}+C_{4}\,\left(x^4-4\,l\,x^3+6\,l^2\,x^2\right) \end{equation*}\]

This in an inconvenient form because it involves \(P\) and doesn’t have a single leading coefficient as we required according to Equation @ref(eq:galerkinform). We will return to this equation in a second, but for now we seek a different path.

Consider that governing differential equation associates distributed load with its forth derivative. Since the distributed load is zero, we know the shear is constant, thus, we replace our last natural boundary condition and now have: \[\begin{equation*}\begin{split} - EI v''(l) &=0 \\ EI v''''(x) &= 0\\ \end{split}\end{equation*}\]

\[\begin{equation*}\begin{split} 12\,C_{4}\,l^2+6\,C_{3}\,l+2\,C_{2}&=0 \\ 24\,C_{4}&=0 \\ \end{split}\end{equation*}\]

Thus, four coefficients are: \[\begin{equation*}\begin{split} C_{0}=&\,0\cr C_{1}=&\,0\cr C_{2}=&\,-3\,C_{3}\,l\cr C_{4}=&\,0 \end{split}\end{equation*}\] and the assumed displacement field is written as: \[\begin{equation*} v\left(x\right)=C_{3}\,x^2\,\left(x-3\,l\right) \end{equation*}\]

Note, the second derivative: \[\begin{equation*} v''\left(x\right)=6\,C_{3}\,\left(x-l\right) \end{equation*}\]

There are two ways to finish the problem. We know both of these equations must be true: \[\begin{equation*}\begin{split} \int EI v'' \delta v'' dx = P \delta v(l)\\ EI v'''(l) - P = 0 \\ \end{split}\end{equation*}\]

Testing the second equation: \[\begin{equation*} 6\,C_{3}\,EI-P=0 \end{equation*}\] \[\begin{equation*} \left[ C_{3}=\frac{P}{6\,EI} \right] \end{equation*}\]

This leads to: \[\begin{equation*} v\left(x\right)=\frac{x^3\,P}{6\,EI}-\frac{l\,x^2\,P}{2\,EI} \end{equation*}\] \[\begin{equation*} v\left(x\right)=\frac{x^2\,\left(x-3\,l\right)\,P}{6\,EI} \end{equation*}\]

• This is unsatisfying since it doesn’t give confidence to the field part of the Galerkin equation.
  • It turns out it wouldn’t be this easy except that this is an analytically simple problem.
• Thus, we now examine the virtual displacement associated with our assumed field.
  • The Galerkin method assumes the virtual displacement takes the same form as the field, thus: \[\begin{equation*} {\delta v}\left(x\right)=C_{5}\,x^2\,\left(x-3\,l\right) \end{equation*}\]
• The form is the same but the constant is different.

• The derivatives: \[\begin{equation*}\begin{split} {\delta v'}\left(x\right)&=C_{5}\,x^2+2\,C_{5}\,x\,\left(x-3\,l\right) \\ {\delta v''}\left(x\right)&=2\,C_{5}\,\left(x-3\,l\right)+4\,C_{5}\,x \\ \end{split}\end{equation*}\]

• Now, plugging this into the field equation (equation @ref(eq:fieldgal): \[\begin{equation*} 12\,C_{3}\,C_{5}\,l^3\,EI=2\,C_{5}\,l^3\,P \end{equation*}\]

• Or: \[\begin{equation*} \left[ C_{3}=\frac{P}{6\,EI} \right] \end{equation*}\]

• Note \(C_{5}\) is arbitrary and thus so is the virtual displacement!!!

• This is exactly the same as we found. Both equations are satisfied and we have found the Galerkin approximation. \[\begin{equation*} v\left(x\right)=\frac{x^2\,\left(x-3\,l\right)\,P}{6\,EI} \end{equation*}\]

• Note, this happens to be the exact solution for this problem.
• Note also that this happens to be the same as equation @ref(eq:wrongright) once it is identified that \(C_{4}\) is zero.
• Ultimately, this works out only due to the simplicity of the problem.
  • You will not always find an appropriate approximate solution solely from the boundary conditions.

Example Two Revisited – Choosing a Polynomial – Shortcut Approach

We could have been a bit more sneaky... knowing that there is no distributed load (fourth derivative is zero) we could have taken a shortcut by assuming a third order function. \[\begin{equation*} v\left(x\right)=C_{3}\,x^3+C_{2}\,x^2+C_{1}\,x+C_{0} \end{equation*}\]

The derivatives: \[\begin{equation*}\begin{split} {v'}\left(x\right)&=3\,C_{3}\,x^2+2\,C_{2}\,x+C_{1} \\ {v''}\left(x\right)&=6\,C_{3}\,x+2\,C_{2} \\ {v'''}\left(x\right)&=6\,C_{3} \\ {v''''}\left(x\right)&=0 \\ \end{split}\end{equation*}\]

We really only need three boundary conditions: \[\begin{equation*}\begin{split} v\left(0\right)&=0 \\ v'\left(0\right)&=0 \\ v''\left(l\right)&=0 \\ \end{split}\end{equation*}\]

This leads to: \[\begin{equation*} v\left(x\right)=C_{3}\,x^2\,\left(x-3\,l\right) \end{equation*}\] and the solutions proceeds as above.

Example Two Revisited – Choosing a Polynomial – Longcut Approach

• The goal is to have a constant shear and meet BCs \[\begin{equation*} v\left(x\right)=C_{4}\,x^4+C_{3}\,x^3+C_{2}\,x^2+C_{1}\,x+C_{0} \end{equation*}\] \[\begin{equation*} \begin{split} {v'}\left(x\right)=4\,C_{4}\,x^3+3\,C_{3}\,x^2+2\,C_{2}\,x+C_{1} \\ {v''}\left(x\right)=12\,C_{4}\,x^2+6\,C_{3}\,x+2\,C_{2} \\ {v'''}\left(x\right)=24\,C_{4}\,x+6\,C_{3} \\ {v''''}\left(x\right)=24\,C_{4} \\ \end{split} \end{equation*}\] \[\begin{equation*} \begin{split} v\left(0\right)=0 \\ v'\left(0\right)=0 \\ v''\left(l\right)=0 \\ v''''\left(0\right)=0 \\ \end{split} \end{equation*}\]

• The last of these is a valid equation because we know the distributed load is zero everywhere.

\[\begin{equation*} \begin{split} C_{0}=0 \\ C_{1}=0 \\ 12\,C_{4}\,l^2+6\,C_{3}\,l+2\,C_{2}=0 \\ 24\,C_{4}=0 \\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} C_{0}=0 \\ C_{1}=0 \\ 12\,C_{4}\,l^2+6\,C_{3}\,l+2\,C_{2}=0 \\ 24\,C_{4}=0 \\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{pmatrix} C_{0}=0\cr C_{1}=0\cr C_{2}=-3\,C_{3}\,l\cr C_{4}=0\cr \end{pmatrix} \end{equation*}\]

Hence:

\[\begin{equation*} \begin{split} v\left(x\right)=C_{3}\,x^3-3\,C_{3}\,l\,x^2 \\ v\left(x\right)=C_{3}\,x^2\,\left(x-3\,l\right) \\ \end{split} \end{equation*}\]

Take the variation of this equation. Note the variation of the constant P related terms is zero

\[\begin{equation*} {\delta v}\left(x\right)={\delta C_3}\,x^2\,\left(x-3\,l\right) \end{equation*}\]

\[\begin{equation*} {\delta v'}\left(x\right)=3\,{\delta C_3}\,x\,\left(x-2\,l\right) \end{equation*}\]

\[\begin{equation*} {\delta v''}\left(x\right)=6\,{\delta C_3}\,\left(x-l\right) \end{equation*}\]

\[\begin{equation*} {\delta v'''}\left(x\right)=6\,{\delta C_3} \end{equation*}\]

• The first equation from the above: \[\begin{equation*} EI\,{\delta v''}\left(x\right)\,v''\left(x\right)=36\,C_{3}\,EI\,{\delta C_3}\,\left(x-l\right)^2 \end{equation*}\] \[\begin{equation*} EI\,\int_{0}^{l}{{\delta v''}\left(x\right)\,v''\left(x\right)\;dx}=-P\,{\delta v}\left(l\right) \end{equation*}\]
• The RHS of this equation is because the force is downward. \[\begin{equation*} 36\,C_{3}\,EI\,{\delta C_3}\,\int_{0}^{l}{\left(x-l\right)^2\;dx}=2\,P\,{\delta C_3}\,l^3 \end{equation*}\] \[\begin{equation*} EI\,v''\left(l\right) \end{equation*}\] \[\begin{equation*} v'''\left(0\right)\,EI-P \end{equation*}\]
• Of these, the second and third are: \[\begin{equation*} EI\,\left(12\,C_{4}\,l^2+6\,C_{3}\,l+2\,C_{2}\right) \end{equation*}\] \[\begin{equation*} 6\,C_{3}\,EI-P \end{equation*}\]

• Putting in the known coefficient values from above. \[\begin{equation*} 36\,C_{3}\,EI\,{\delta C_3}\,\int_{0}^{l}{\left(x-l\right)^2\;dx}=2\,P\,{\delta C_3}\,l^3 \end{equation*}\] \[\begin{equation*} 12\,C_{3}\,EI\,{\delta C_3}\,l^3=2\,P\,{\delta C_3}\,l^3 \end{equation*}\]

• Hence using the first equation. \[\begin{equation*} \left[ C_{3}=\frac{P}{6\,EI} \right] \end{equation*}\]

• Confirming the third equation. \[\begin{equation*} \left[ C_{3}=\frac{P}{6\,EI} \right] \end{equation*}\]

Therefore.

\[\begin{equation*} v\left(x\right)=\frac{P\,x^2\,\left(x-3\,l\right)}{6\,EI} \end{equation*}\]

\[\begin{equation*} v\left(l\right)=-\frac{P\,l^3}{3\,EI} \end{equation*}\]

Which again is exact for this problem.

Hints at choosing the Galerkin approximating function

• It is required that the form of the function you assume can satisfy all geometric and natural boundary conditions.
• For a polynomial function, this is guaranteed only if you assume a function of order \(2p\) (the highest order of the governing differential equation).
• It is possible to shortcut the problem by assuming a lower order function that can satisfy all the boundary conditions of the problem. However, you may need to manipulate it slightly to make it look like @ref(eq:galerkinform). This is not the recommended method!
• Consider also trig functions as they are easily integrated and often can be manipulated to satisfy the boundary conditions.

Reciprocal theorem

The deflection at point 1 in a given direction due to a unit load at point 2 in a second direction is equal to the deflection at the point 2 in the second direction due to a unit load at point 1 in the first direction.

Procedure 1: Apply \(F_2\) and then \(F_1\). The application is gradual such that \(F_2\) ramps and then \(F_1\) ramps. \[\begin{equation*} U_2 = {\frac{1}{2}}F_2 \delta_{22} \end{equation*}\] This is the spring energy (\(F_2 = k \delta_{22}\)).

Now apply \(F_1\) : \[\begin{equation*} \Delta U = F_2 \cdot(\delta_{2} - \delta_{22}) + {\frac{1}{2}}F_1 \cdot(\delta_{1} - \delta_{12}) \end{equation*}\] \(F_2 \cdot(\delta_{2} - \delta_{22})\) is the external work done by fixed \(F_2\) while \(F_1\) is being applied. \[\begin{equation*}\begin{split} U &= {\frac{1}{2}}F_2 \delta_{22} + F_2 \cdot(\delta_{2} - \delta_{22}) + {\frac{1}{2}}F_1 \cdot(\delta_{1} - \delta_{12}) \\ &= -{\frac{1}{2}}F_2 \delta_{22} + F_2 \delta_{2} + {\frac{1}{2}}F_1 \delta_{1} - {\frac{1}{2}}F_1 \delta_{12} \\ \end{split}\end{equation*}\]

Consider separately

• Procedure 2: Apply \(F_1\) and then \(F_2\) \[\begin{equation*} U_1 = {\frac{1}{2}}F_1 \delta_{11} \end{equation*}\]

• Now apply \(F_2\): \[\begin{equation*} \small \Delta U = F_1 \cdot(\delta_{1} - \delta_{11}) + {\frac{1}{2}}F_2 \cdot(\delta_{2} - \delta_{2}1) \end{equation*}\]

\[\begin{equation*}\begin{split} \small U &= \; {\frac{1}{2}}F_1 \delta_{11} + F_1 \cdot(\delta_{1} - \delta_{11}) + {\frac{1}{2}}F_2 \cdot(\delta_{2} - \delta_{2}1) \\ \small &= \; -{\frac{1}{2}}F_1 \delta_{11} + F_1 \delta_{1} + {\frac{1}{2}}F_2 \delta_{2} - {\frac{1}{2}}F_2 \delta_{21} \\ \end{split}\end{equation*}\] \[\begin{equation*}\begin{split} \small U &= \; -{\frac{1}{2}}F_2 \delta_{22} + F_2 \delta_{2} + {\frac{1}{2}}F_1 \delta_{1} - {\frac{1}{2}}F_1 \delta_{12} \\ \small &= \; -{\frac{1}{2}}F_1 \delta_{11} + F_1 \delta_{1} + {\frac{1}{2}}F_2 \delta_{2} - {\frac{1}{2}}F_2 \delta_{21} \small -{\frac{1}{2}}F_2 \delta_{22} + {\frac{1}{2}}F_2 \delta_{2} - {\frac{1}{2}}F_1 \delta_{12} \\ \small &= \; -{\frac{1}{2}}F_1 \delta_{11} + {\frac{1}{2}}F_1 \delta_{1} - {\frac{1}{2}}F_2 \delta_{21} \end{split}\end{equation*}\]

• These two energies must be the same because the loads are the same. \[\begin{equation*}\begin{split} F_2 (\delta_{2} - \delta_{22}) - F_1 \delta_{12} &= \; F_1 (\delta_{1} - \delta_{11})- F_2 \delta_{21} \\ \end{split}\end{equation*}\] \[\begin{equation*}\begin{split} \delta_{1} - \delta_{11} &= \; \delta_{12} \\ \delta_{2} - \delta_{22} &= \; \delta_{21} \\ \end{split}\end{equation*}\] \[\begin{equation*}\begin{split} F_2 \delta_{21} - F_1 \delta_{12} &= \; F_1 \delta_{12} - F_2 \delta_{21} \end{split}\end{equation*}\]

• Hence: \[\begin{equation*} F_2 \delta_{21} = F_1 \delta_{12} \end{equation*}\]

• This is the reciprocal theorem.

• If \[\begin{equation*} F_1=F_2 \Longrightarrow \delta_{12} = \delta_{21} \end{equation*}\]

• Application: Measure \(\delta(x)\) by measuring \(\delta\) at a single point and moving the load along \(x\)!