--- title: "Iso/super/sub-parametric; Hermite shape function" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:hermite-beam-shape-isoparametric" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/hermite-beam-shape-isoparametric site: BCT Engineering --- # Iso/super/sub-parametric; Hermite shape function Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2079 Jestha, 3+7 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/hermite-beam-shape-isoparametric - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Define iso-parametric, super parametric and sub parametric elements. Derive the expression for the Hermite shape function used for the interpolation in beam elements. ## Short answer Isoparametric: same shape functions for geometry and displacement. Superparametric: geometry interpolation higher-order than displacement. Subparametric: geometry lower-order. Hermite beam shape functions (length L): N1=1-3x^2/L^2+2x^3/L^3, N2=x-2x^2/L+x^3/L^2, N3=3x^2/L^2-2x^3/L^3, N4=-x^2/L+x^3/L^2. ## Answer #### Types of parametric elements - **Isoparametric:** the same shape functions (same number of nodes) are used for the geometry and for the displacement, $x=\sum N_ix_i$, $u=\sum N_iu_i$. - **Superparametric:** the geometry is described by a higher-order interpolation (more nodes) than the displacement, e.g. quadratic geometry and linear displacement. - **Subparametric:** the geometry is described by a lower-order interpolation than the displacement, e.g. straight sides with a quadratic displacement. #### Hermite shape functions for a beam element A beam element of length $L$ has two nodes with two DOFs each: deflection $v$ and slope $\theta=dv/dx$. Both $v$ and $\theta$ must be continuous between elements ($C^1$ continuity), so a cubic (4 constants) is assumed: $$ v(x)=\alpha_1+\alpha_2x+\alpha_3x^2+\alpha_4x^3 $$ $$ \theta(x)=\frac{dv}{dx}=\alpha_2+2\alpha_3x+3\alpha_4x^2 $$ Apply the four nodal conditions: $v(0)=v_1$, $\theta(0)=\theta_1$, $v(L)=v_2$, $\theta(L)=\theta_2$: $$ \begin{Bmatrix}v_1\\\theta_1\\v_2\\\theta_2\end{Bmatrix}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\1&L&L^2&L^3\\0&1&2L&3L^2\end{bmatrix}\begin{Bmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{Bmatrix} $$ Solving: $\alpha_1=v_1$, $\alpha_2=\theta_1$, $$ \alpha_3=\frac{3(v_2-v_1)}{L^2}-\frac{2\theta_1+\theta_2}{L},\qquad \alpha_4=\frac{2(v_1-v_2)}{L^3}+\frac{\theta_1+\theta_2}{L^2} $$ Substituting and collecting the coefficients of $v_1,\theta_1,v_2,\theta_2$ gives $v=[N_1\ N_2\ N_3\ N_4]\{d\}$ with $$ \begin{aligned} N_1&=1-\frac{3x^2}{L^2}+\frac{2x^3}{L^3}\\ N_2&=x-\frac{2x^2}{L}+\frac{x^3}{L^2}\\ N_3&=\frac{3x^2}{L^2}-\frac{2x^3}{L^3}\\ N_4&=-\frac{x^2}{L}+\frac{x^3}{L^2} \end{aligned} $$ In terms of $s=x/L$: $N_1=1-3s^2+2s^3$, $N_2=L(s-2s^2+s^3)$, $N_3=3s^2-2s^3$, $N_4=L(-s^2+s^3)$. **Check:** $N_1(0)=1$, $N_1(L)=0$, $N_1'(0)=N_1'(L)=0$; $N_2'(0)=1$, $N_2(0)=N_2(L)=N_2'(L)=0$, and so on for $N_3,N_4$. Also $N_1+N_3=1$ (rigid-body translation is represented). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which describes a superparametric element? A. Geometry uses lower-order interpolation than displacement B. Geometry and displacement use the same shape functions C. Geometry uses higher-order interpolation (more nodes) than the displacement D. Displacement is not interpolated Answer: C. Superparametric: geometry is of higher order than displacement (e.g. quadratic geometry, linear displacement); subparametric is the reverse. (Key point: Superparametric element) 2. In an isoparametric element, A. x = sum Ni xi and u = sum Ni ui with the same Ni B. geometry is linear but displacement is cubic C. only the displacement is interpolated D. different numbers of nodes are used for geometry and displacement Answer: A. The same shape functions describe both geometry and displacement. (Key point: Isoparametric definition) 3. Why is a cubic polynomial assumed for the beam element displacement? A. Because the load is cubic B. There are four nodal conditions (v and slope at two nodes) and v and slope must be continuous (C1) C. Because the bending moment is linear D. Because only deflection is a DOF Answer: B. Two nodes with deflection and slope give 4 DOFs, hence 4 constants, i.e. a cubic. (Key point: Why a cubic for beams) 4. Which is true of the Hermite functions? A. N1(0) = 1 and N1'(0) = 0; N2(0) = 0 and N2'(0) = 1 B. N1(L) = 1 and N2(L) = 1 C. N1'(0) = 1 and N2(0) = 1 D. N1 + N3 = 0 Answer: A. N1 carries unit deflection at node 1 with zero slope; N2 carries unit slope at node 1 with zero deflection. 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