Problem 7.6. Critical load by the Rayleigh-Ritz method

Coefficient for the critical load, $\frac{\lambda N_{\mathrm{o}}}{EI}{L}^2=$

Coefficient for the buckling length, ν=

Follow steps of Example 7.7.

Step 1.  Assume the displacement function in the form of a forth order polynomial. Set constants to satisfy the geometrical boundary conditions.

Check displacement functionHide displacement function

Step 2.  Write the potential energy function.

Check potential energyHide potential energy

The strain energy is

Eq.(7-110)

$$\begin{gathered} U=\frac{1}{2}EI{\int }_0^L{\kappa }^2\mathrm{d}x=\frac{1}{2}EI{\int }_0^L{\left(\frac{\mathrm{d}{v}^2}{\mathrm{d}x}\right)}^2\mathrm{d}x=\frac{1}{2}EI{\int }_0^L{\left(C_3\left(6x–2L\right)+C_4\left(12x^2–2L^2\right)\right)}^2\mathrm{d}x= \\ =\frac{1}{2}EI{\int }_0^L\left(C_3^2{\left(6x–2L\right)}^2+C_4^2{\left(12x^2–2L^2\right)}^2+2C_3{C}_4\left(6x–2L\right)\left(12x^2–2L^2\right)\right)\mathrm{d}x= \\ =–\frac{1}{2}{C}_3^{2}4EI{L}^3+\frac{1}{2}{C}_4^2\frac{84}{5}EI{L}^5+C_3{C}_{4}8EI{L}^4 \end{gathered}$$

The work done by the external top force on the top displacement is

See Figure 7.54 and Eq.(7-113)

$$\begin{gathered} W=–\frac{1}{2}\lambda N_{\mathrm{o}}{\int }_0^{L}v{‘}^2\left(x\right)dx=–\frac{1}{2}\lambda N_{\mathrm{o}}{\int }_0^L{\left(C_3\left(3x^2–2Lx\right)+C_4\left(4x^3–2L^{2}x\right)\right)}^{2}dx= \\ =–\frac{1}{2}\lambda N_{\mathrm{o}}\left(\frac{2}{15}{C}_3^2{L}^5+\frac{44}{105}{C}_4^2{L}^7+\frac{7}{15}{C}_3{C}_4{L}^6\right) \end{gathered}$$

After substituting the assumed deflection function the potential energy is:

$\pi =U+W=\frac{1}{2}EI\left(–C_3^{2}4L^3+\frac{1}{2}{C}_4^2\frac{84}{5}{L}^5+C_3{C}_{4}16L^4\right)–\frac{1}{2}\lambda N_{\mathrm{o}}\left(\frac{2}{15}{C}_3^2{L}^5+\frac{44}{105}{C}_4^2{L}^7+\frac{7}{15}{C}_3{C}_4{L}^6\right)$

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

Check constantsHide constants

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

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

$$\begin{gathered} \pi (C_3,C_4)=U+W=\mathrm{stat} \\ \frac{\partial \mathrm{\pi }}{\partial C_3}=C_{3}4EI{L}^3+C_{4}8EI{L}^4–\lambda N_{\mathrm{o}}\frac{2}{15}{C}_3^{}{L}^5–\lambda N_{\mathrm{o}}\frac{7}{30}{C}_4{L}^6=0 \\ \frac{\partial \mathrm{\pi }}{\partial C_4}=C_{3}8EI{L}^4+C_4\frac{84}{5}EIL–\lambda N_{\mathrm{o}}\frac{44}{105}{C}_4^{}{L}^7–2\lambda N_{\mathrm{o}}\frac{7}{30}{C}_3{L}^6=0 \end{gathered}$$

The above equations result in the following eigenvalue problem:

$\left[\begin{array}{cc}\frac{4EI}{L^2}–\lambda N_{\mathrm{o}}\frac{2}{15}& \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}\\ \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}& \frac{84}{5}\frac{EI}{L^2}–\lambda N_{\mathrm{o}}\frac{44}{105}\end{array}\right]\left\{\begin{array}{c}{C}_3\\ C_{4}L\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$

For the nontrivial solution the determinant of the coefficient matrix must be zero, and the eigenvalue, λ can be expressed:

$$\begin{gathered} \det \left[\begin{array}{cc}\frac{4EI}{L^2}–\lambda N_{\mathrm{o}}\frac{2}{15}& \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}\\ \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}& \frac{84}{5}\frac{EI}{L^2}–\lambda N_{\mathrm{o}}\frac{44}{105}\end{array}\right]=0 \\ \to \left(\frac{4EI}{L^2}–\lambda N_{\mathrm{o}}\frac{2}{15}\right)\left(\frac{84}{5}\frac{EI}{L^2}–\lambda N_{\mathrm{o}}\frac{44}{105}\right)–{\left(\frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}\right)}^2=0 \\ \to {\left(\lambda N_{\mathrm{o}}\right)}^2\left(\frac{2}{15}\frac{44}{105}–\frac{7^2}{{30}^2}\right)–\lambda N_{\mathrm{o}}\frac{EI}{L^2}\left(\frac{2}{15}\frac{84}{5}+\frac{44}{105}4–2\times 8\frac{7}{30}\right)+\frac{E{I}^2}{L^4}\left(4\frac{84}{5}–8^2\right)=0 \\ \to \lambda N_{\mathrm{o}}=\mathbf{20}\mathbf{.}\mathbf{92}\frac{EI}{L^2} \end{gathered}$$

Step 4.  Give the buckling length.

Check buckling lengthHide buckling length

Eq.(7-29)

$\lambda N_{\mathrm{o}}=20.92\frac{EI}{L^2}=\frac{{\mathrm{\pi }}^{2}EI}{{\left(\nu L\right)}^2} \to \mathbf{ }\mathit{\nu }=\frac{\mathrm{\pi }}{\sqrt{20.92}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{687}\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{7}$

Follow steps of Example 7.7.

The displacement function is assumed to be 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:

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

Now the displacement function has the following form:

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

The strain energy is

Eq.(7-110)

The work done by the external top force on the top displacement is

See Figure 7.54 and Eq.(7-113)

After substituting the assumed deflection function the potential energy is:

$\pi =U+W=\frac{1}{2}EI\left(–C_3^{2}4L^3+\frac{1}{2}{C}_4^2\frac{84}{5}{L}^5+C_3{C}_{4}16L^4\right)–\frac{1}{2}\lambda N_{\mathrm{o}}\left(\frac{2}{15}{C}_3^2{L}^5+\frac{44}{105}{C}_4^2{L}^7+\frac{7}{15}{C}_3{C}_4{L}^6\right)$

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)

The above equations result in the following eigenvalue problem:

$\left[\begin{array}{cc}\frac{4EI}{L^2}–\lambda N_{\mathrm{o}}\frac{2}{15}& \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}\\ \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}& \frac{84}{5}\frac{EI}{L^2}–\lambda N_{\mathrm{o}}\frac{44}{105}\end{array}\right]\left\{\begin{array}{c}{C}_3\\ C_{4}L\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$

For the nontrivial solution the determinant of the coefficient matrix must be zero, and the eigenvalue, λ can be expressed:

$$\begin{gathered} \det \left[\begin{array}{cc}\frac{4EI}{L^2}–\lambda N_{\mathrm{o}}\frac{2}{15}& \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}\\ \frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}& \frac{84}{5}\frac{EI}{L^2}–\lambda N_{\mathrm{o}}\frac{44}{105}\end{array}\right]=0 \\ \to \left(\frac{4EI}{L^2}–\lambda N_{\mathrm{o}}\frac{2}{15}\right)\left(\frac{84}{5}\frac{EI}{L^2}–\lambda N_{\mathrm{o}}\frac{44}{105}\right)–{\left(\frac{8EI}{L^2}–\lambda N_{\mathrm{o}}\frac{7}{30}\right)}^2=0 \\ \to {\left(\lambda N_{\mathrm{o}}\right)}^2\left(\frac{2}{15}\frac{44}{105}–\frac{7^2}{{30}^2}\right)–\lambda N_{\mathrm{o}}\frac{EI}{L^2}\left(\frac{2}{15}\frac{84}{5}+\frac{44}{105}4–2\times 8\frac{7}{30}\right)+\frac{E{I}^2}{L^4}\left(4\frac{84}{5}–8^2\right)=0 \\ \to \lambda N_{\mathrm{o}}=\mathbf{20}\mathbf{.}\mathbf{92}\frac{EI}{L^2} \end{gathered}$$

The buckling length is

Equation (7-29)