Problem 6.10. Rayleigh-Ritz method

 Derive deflection function, v(x).

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Step 1.  Assume the solution (deflection function of the beam) in the form of a forth order polynomial. Set constants to satisfy the geometrical boundary conditions.

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Step 2.  Write the potential energy function.

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The potential energy is given as

$\pi =U+W$

Eq.(6-25)

where the strain energy is
$$\begin{gathered} U=\frac{1}{2}EI{\int }_0^L{\kappa }^2\mathrm{d}x=\frac{1}{2}EI{\int }_0^L{\left(\frac{\mathrm{d}{v}^2}{\mathrm{d}x}\right)}^2\mathrm{d}x=\frac{1}{2}EI{\int }_0^L{\left(C_3\left(6x–2L\right)+C_4\left(12x^2–2L^2\right)\right)}^2\mathrm{d}x= \\ =\frac{1}{2}EI{\int }_0^L\left(C_3^2{\left(6x–2L\right)}^2+C_4^2{\left(12x^2–2L^2\right)}^2+2C_3{C}_4\left(6x–2L\right)\left(12x^2–2L^2\right)\right)\mathrm{d}x= \\ =–\frac{1}{2}{C}_3^{2}4EI{L}^3+\frac{1}{2}{C}_4^2\frac{252}{15}EI{L}^5+C_3{C}_{4}8EI{L}^4 \end{gathered}$$

and the work of the external forces is

$$\begin{gathered} W=–{\int }_0^{L}pv\left(x\right)dx=–p{\int }_0^L\left(C_3\left(x^3–L{x}^2\right)+C_4\left(x^4–L^2{x}^2\right)\right)dx= \\ =C_3\frac{1}{12}p{L}^4+C_4\frac{2}{15}p{L}^5 \end{gathered}$$

After substituting the assumed deflection function the potential energy results in:

$\pi =–\frac{1}{2}{C}_3^{2}4EI{L}^3+\frac{1}{2}{C}_4^2\frac{252}{15}EI{L}^5+C_3{C}_{4}8EI{L}^4+C_3\frac{1}{12}p{L}^4+C_4\frac{2}{15}p{L}^5$

Step 3.  Determine the remaining constants from the stationary condition of the potential energy.

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The derivatives of the potential energy with respect to the constants of the deflection function must be zero:

Eqs.(6-97) and (6-98)

$$\begin{gathered} \pi (C_3,C_4)=\mathrm{stat} \\ \frac{\partial \mathrm{\pi }}{\partial C_3}=C_{3}4EI{L}^3+C_{4}8EI{L}^4+\frac{1}{12}p{L}^4=0 \\ \frac{\partial \mathrm{\pi }}{\partial C_4}=C_{3}8EI{L}^4+C_4\frac{252}{15}EI{L}^5+\frac{2}{15}p{L}^5=0 \end{gathered}$$

Solving the above linear equation system, the constants, C₃ and C₄ can be expressed:

$EI\left[\begin{array}{cc}4& 8\\ 8& \frac{252}{15}\end{array}\right]\left\{\begin{array}{c}{C}_3/L\\ C_4\end{array}\right\}=p\left\{\begin{array}{c}–\frac{1}{12}\\ –\frac{2}{15}\end{array}\right\} \to \left\{\begin{array}{c}{C}_3\\ C_4\end{array}\right\}= \left\{\begin{array}{c}–\frac{5}{48}\frac{pL}{EI}\\ \frac{1}{24}\frac{p}{EI}\end{array}\right\}$

Step 4.  Give the approximate deflection function. Compare the result with the exact solution.

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$\mathit{v}\mathbf{\left(}\mathit{x}\mathbf{\right)}=–\frac{5}{48}\frac{pL}{EI}\left(x^3–L{x}^2\right)+\frac{1}{24}\frac{p}{EI}\left(x^4–L{x}^2\right)\mathbf{=}\frac{\mathit{p}}{\mathit{E}\mathit{I}}\left(\frac{\mathbf{3}}{\mathbf{48}}{\mathit{L}}^{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{–}\frac{\mathbf{5}}{\mathbf{48}}\mathit{L}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{24}}{\mathit{x}}^{\mathbf{4}}\right)$

The Rayleigh-Ritz method assuming a fourth order polynomial results in the exact solution.

See Problem 3.5.

The solution (deflection function of the beam) is assumed in the form of a forth order polynomial. 

$v\left(x\right)=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{x}^4$

From the geometrical boundary conditions, the following constants can be determined:

$$\begin{gathered} v\left(0\right)=0 \to C_0=0 \\ \frac{\mathrm{d}v}{\mathrm{d}x}=0 \to C_1=0 \\ v\left(L\right)=0 \to C_2=–C_{3}L–C_4{L}^2 \end{gathered}$$

The deflection function is:

$v\left(x\right)=C_3\left(x^3–L{x}^2\right)+C_4\left(x^4–L^2{x}^2\right)$

The potential energy is given as

$\pi =U+W$

Eq.(6-25)

where the strain energy is
$$\begin{gathered} U=\frac{1}{2}EI{\int }_0^L{\kappa }^2\mathrm{d}x=\frac{1}{2}EI{\int }_0^L{\left(\frac{\mathrm{d}{v}^2}{\mathrm{d}x}\right)}^2\mathrm{d}x=\frac{1}{2}EI{\int }_0^L{\left(C_3\left(6x–2L\right)+C_4\left(12x^2–2L^2\right)\right)}^2\mathrm{d}x= \\ =\frac{1}{2}EI{\int }_0^L\left(C_3^2{\left(6x–2L\right)}^2+C_4^2{\left(12x^2–2L^2\right)}^2+2C_3{C}_4\left(6x–2L\right)\left(12x^2–2L^2\right)\right)\mathrm{d}x= \\ =–\frac{1}{2}{C}_3^{2}4EI{L}^3+\frac{1}{2}{C}_4^2\frac{252}{15}EI{L}^5+C_3{C}_{4}8EI{L}^4 \end{gathered}$$

and the work of the external forces is

$$\begin{gathered} W=–{\int }_0^{L}pv\left(x\right)dx=–p{\int }_0^L\left(C_3\left(x^3–L{x}^2\right)+C_4\left(x^4–L^2{x}^2\right)\right)dx= \\ =C_3\frac{1}{12}p{L}^4+C_4\frac{2}{15}p{L}^5 \end{gathered}$$

After substituting the assumed deflection function the potential enegry results in:

$\pi =–\frac{1}{2}{C}_3^{2}4EI{L}^3+\frac{1}{2}{C}_4^2\frac{252}{15}EI{L}^5+C_3{C}_{4}8EI{L}^4+C_3\frac{1}{12}p{L}^4+C_4\frac{2}{15}p{L}^5$

The remaining constants are determined from the stationary condition of the potential energy. The derivatives of the potential energy with respect to the constants of the deflection function must be zero:

Eqs.6-97) and (6-98)

Solving the above linear equation system, the constants, C₃ and C₄ can be expressed:

$EI\left[\begin{array}{cc}4& 8\\ 8& \frac{252}{15}\end{array}\right]\left\{\begin{array}{c}{C}_3/L\\ C_4\end{array}\right\}=p\left\{\begin{array}{c}–\frac{1}{12}\\ –\frac{2}{15}\end{array}\right\} \to \left\{\begin{array}{c}{C}_3\\ C_4\end{array}\right\}= \left\{\begin{array}{c}–\frac{5}{48}\frac{pL}{EI}\\ \frac{1}{24}\frac{p}{EI}\end{array}\right\}$

The approximate deflection function becomes

$\mathit{v}\mathbf{\left(}\mathit{x}\mathbf{\right)}=–\frac{5}{48}\frac{pL}{EI}\left(x^3–L{x}^2\right)+\frac{1}{24}\frac{p}{EI}\left(x^4–L{x}^2\right)\mathbf{=}\frac{\mathit{p}}{\mathit{E}\mathit{I}}\left(\frac{\mathbf{3}}{\mathbf{48}}{\mathit{L}}^{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{–}\frac{\mathbf{5}}{\mathbf{48}}\mathit{L}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{24}}{\mathit{x}}^{\mathbf{4}}\right)$

Comparing the result of the Rayleigh-Ritz method to the exact solution we see, that assuming a fourth order polynomial the Rayleig-Ritz approximation results in the exact solution.

See the exact solution in Problem 3.5.