FEM

Element Stiffness Coefficients

Shape Functions

The displacement function for a flexure element is:

\[v(x) = N_1(x)v_1 + N_2(x)\theta_1 + N_3(x)v_2 + N_4(x)\theta_2 \tag{4.27a}\]

The shape functions in terms of the dimensionless coordinate $\xi = \frac{x}{L}$ are:

\[\begin{aligned} N_1(\xi) &= 1 - 3\xi^2 + 2\xi^3, \\ N_2(\xi) &= L(\xi - 2\xi^2 + \xi^3), \\ N_3(\xi) &= 3\xi^2 - 2\xi^3, \\ N_4(\xi) &= L(-\xi^2 + \xi^3) \end{aligned} \tag{4.28}\]

Second Derivatives

Second derivatives with respect to $x$:

\[\frac{d^2}{dx^2} = \frac{1}{L^2} \frac{d^2}{d\xi^2} \tag{4.46, 4.47}\]

Second derivatives of $N_i(\xi)$ with respect to $\xi$:

\[\begin{aligned} \frac{d^2 N_1}{d\xi^2} &= -6 + 12\xi, \\ \frac{d^2 N_2}{d\xi^2} &= L(-4 + 6\xi), \\ \frac{d^2 N_3}{d\xi^2} &= 6 - 12\xi, \\ \frac{d^2 N_4}{d\xi^2} &= L(-2 + 6\xi) \end{aligned} \tag{4.47a}\]

Then with respect to $x$:

\[\frac{d^2 N_i}{dx^2} = \frac{1}{L^2} \frac{d^2 N_i}{d\xi^2} \tag{4.47b}\]

Stiffness Coefficients

Element stiffness coefficients:

\[k_{mn} = k_{nm} = E I_z \int_0^L \frac{d^2 N_m}{dx^2} \frac{d^2 N_n}{dx^2} dx \tag{4.47c}\]

Substitute dimensionless derivatives ($dx = L d\xi$):

\[k_{mn} = \frac{E I_z}{L^3} \int_0^1 \frac{d^2 N_m}{d\xi^2} \frac{d^2 N_n}{d\xi^2} d\xi \tag{4.47d}\]

Step-by-Step Computation of Each Coefficient

1. $k_{11}$:

\[k_{11} = \frac{E I_z}{L^3} \int_0^1 (-6 + 12\xi)^2 d\xi = 12 \frac{E I_z}{L^3} \tag{4.48a}\]

2. $k_{12} = k_{21}$:

\[k_{12} = \frac{E I_z}{L^3} \int_0^1 (-6 + 12\xi) \, L(-4 + 6\xi) d\xi = 6 \frac{E I_z}{L^2} \tag{4.48b}\]

3. $k_{13} = k_{31}$:

\[k_{13} = \frac{E I_z}{L^3} \int_0^1 (-6 + 12\xi)(6 - 12\xi) d\xi = -12 \frac{E I_z}{L^3} \tag{4.48c}\]

4. $k_{14} = k_{41}$:

\[k_{14} = \frac{E I_z}{L^3} \int_0^1 (-6 + 12\xi) \, L(-2 + 6\xi) d\xi = 6 \frac{E I_z}{L^2} \tag{4.48d}\]

5. $k_{22}$:

\[k_{22} = \frac{E I_z}{L^3} \int_0^1 (L(-4 + 6\xi))^2 d\xi = 4 \frac{E I_z}{L} \tag{4.48e}\]

6. $k_{23} = k_{32}$:

\[k_{23} = \frac{E I_z}{L^3} \int_0^1 L(-4 + 6\xi)(6 - 12\xi) d\xi = -6 \frac{E I_z}{L^2} \tag{4.48f}\]

7. $k_{24} = k_{42}$:

\[k_{24} = \frac{E I_z}{L^3} \int_0^1 L(-4 + 6\xi) \, L(-2 + 6\xi) d\xi = 2 \frac{E I_z}{L} \tag{4.48g}\]

8. $k_{33}$:

\[k_{33} = \frac{E I_z}{L^3} \int_0^1 (6 - 12\xi)^2 d\xi = 12 \frac{E I_z}{L^3} \tag{4.48h}\]

9. $k_{34} = k_{43}$:

\[k_{34} = \frac{E I_z}{L^3} \int_0^1 (6 - 12\xi) \, L(-2 + 6\xi) d\xi = -6 \frac{E I_z}{L^2} \tag{4.48i}\]

10. $k_{44}$:

\[k_{44} = \frac{E I_z}{L^3} \int_0^1 (L(-2 + 6\xi))^2 d\xi = 4 \frac{E I_z}{L} \tag{4.48j}\]

Stiffness Coefficients

\[\begin{aligned} k_{11} &= 12 \frac{E I_z}{L^3}, & k_{12} &= 6 \frac{E I_z}{L^2}, & k_{13} &= -12 \frac{E I_z}{L^3}, & k_{14} &= 6 \frac{E I_z}{L^2}, \\ k_{22} &= 4 \frac{E I_z}{L}, & k_{23} &= -6 \frac{E I_z}{L^2}, & k_{24} &= 2 \frac{E I_z}{L}, \\ k_{33} &= 12 \frac{E I_z}{L^3}, & k_{34} &= -6 \frac{E I_z}{L^2}, & k_{44} &= 4 \frac{E I_z}{L} \end{aligned}\]

The matrix is symmetric: $k_{mn} = k_{nm}$.

Final Flexure Element Stiffness Matrix

\[[k_e] = \frac{E I_z}{L^3} \begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^2 & -6L & 2L^2 \\ -12 & -6L & 12 & -6L \\ 6L & 2L^2 & -6L & 4L^2 \end{bmatrix} \tag{4.49}\]