Plate, Shell, and Solid Structures

Abstract

In this chapter the finite element method is considered as a general method for the analysis of plate and solid structures. While the method for truss and beam structures – as it has been treated so far – is exact, for plate and solid structures it can no longer be formulated as an exact method, i.e. here only an approximate solution is obtained by a finite element analysis. The topic of the first part of the chapter is the basic finite element theory of plate, shell and solid structures. After a short historical outline the chapter begins by explaining the approximate character of the finite element method for the example of a tapered truss element. Many characteristic properties of a finite element analysis even for surface and solid structures can be understood studying this basic element. It is followed by the derivation of a rectangular plane stress element used in the analysis of plates in plane stress. Further formulations of finite elements for plates in plane stress are discussed. The next sections deal with plates in bending. To explain the basic concept, a shear flexible rectangular plate element is derived. Advanced formulations for shear flexible and shear stiff plate elements and their properties are discussed. Elements for shells and three-dimensional solids including axisymmetric elements are treated in the following. A particular section addresses the problem of the connection of different types of elements as e.g. beam with plate elements. The second also very comprehensive part of the chapter is dedicated to the modelling of structural elements and buildings with finite elements. General aspects such as mesh generation, the required mesh size, the influence of singularities, the edge effect in bending plates are discussed in detail and illustrated by examples. In the same way the modelling of columns in deep beams and slabs, of RC beams partly integrated in slabs and of mounting parts are treated. An own section deals with the modelling of soil for foundation slabs. Continuum models of layered soils as well as subgrade reaction and two-parameter models are surveyed and compared concerning their practical application. The finite element analysis of three-dimensional building models is the topic of another section. The influence of taking into account construction stages is investigated in detail. The final section of the chapter deals with quality assurance of finite element analyses. It includes some aspects of estimations of the discretization error. The chapter concludes with some remarks on the documentation of finite elements computations.

Exercises

Problems

4.1

A stiffness matrix based on a displacement shape function has been determined for a truss element with 4 nodal points. The element has a linearly variable cross section. With which accuracy are the displacements, the normal stresses and the strains in the element represented, if a polynomial distribution is assumed?

4.2

The cross-sectional area A(x) of a truss element has an arbitrary distribution with x. Derive the stiffness matrix of a truss member with 2 nodes using the principle of virtual displacements

4.3

A truss member has a cross-sectional area with a quadratic distribution, which is described as $A(x) = A_{1} + \dfrac{x}{l}\left( { - 3A_{1} + 4A_{m} - A_{2} } \right) + \dfrac{{x^{2} }}{{l^{2} }}\left( {2A_{1} - 4A_{m} + 2A_{2} } \right)$ where $A_{1} ,\;A_{2} ,\;A_{m}$ are the cross-sectional areas at points 1, 2 and at midpoint m of the element ($x = l/2$) , respectively. Determine the stiffness matrix of the element analytically by means of the result of problem 4.2. Check your result using the stiffness matrix determined in Example 4.2 for a member with the following parameters:

$$ l = 500\;{\text{cm,}}\; E = 1000\;{{\text{kN}} / {\text{cm}}}^{ 2} ,\; A_{1} = 500\;{\text{cm}}^{ 2} ,\; A_{m} = 300\;{\text{cm}}^{ 2} ,\; A_{2} = 100\;{\text{cm}}^{ 2} $$

4.4

A truss member with quadratically variable cross-sectional area (cf. Problems 4.2, 4.3) has the parameters

$l = 500\;{\text{cm,}}\quad E = 1000\;{{\text{kN}} / {\text{cm}}}^{ 2} ,$

$A_{1} = 900\;{\text{cm}}^{ 2} ,\quad A_{m} = 400\;{\text{cm}}^{ 2} ,\quad A_{2} = 100\;{\text{cm}}^{ 2}$

Determine the stiffness matrix of the truss element

• (a) analytically with the stiffness matrix according to Problem 4.3

• (b) numerically with a Gaussian-2-point integration

4.5

A bar subjected to tension is modeled with isoparametric plane stress elements. Draw in row 0 of the table the exact distribution of the normal stresses for the loadings A, B, and C and give their mathematical equation. The rows 1 to 4 show different finite element models. Indicate with “exact” or “approximate” if the FE models give the exact solution or not when loaded according to columns A, B, and C

		A	B	C
0	Exact solution of stress distribution
1
2
3
4

4.6

The stiffness matrix of a FE model for plane stress is considered. Which are the degrees of freedom of the given FE model in plane stress? Show the structure of the stiffness matrix, label the degrees of freedom and indicate for each term of the stiffness matrix if it is equal to zero (“0”) or unequal to zero (“x”)

4.7

What is the order n × n of the stiffness matrix of the two solid elements shown?

4.8

Solve the integral with numerical Gaussian 3-point integration:

$${ \int_{0.5}^{1} {\dfrac{\sqrt[\scriptstyle 3] {\sin (3{\cdot}{x})}} {{x}^2 +30} \, {\rm d}x }}$$

4.9

A force $F_{x}$ is applied on a conforming, rectangular 4-nodal plane stress element. Determine the equivalent nodal forces

4.10

A finite element program has calculated the nodal displacements of a plane stress element given below. The element used is a conformal rectangular element with a bilinear shape function. How large are the displacements, stresses, and strains at point p?

Nodal displacements:

$E = 3 {\cdot} 10^{7} \;{\text{kN/m}}^{ 2}, \quad \mu = 0.2$

Node	u [mm]	v [mm]
1	0.1690	−0.2811
2	0.1596	−0.2808
3	0.1530	−0.2871
4	0.1693	−0.2875

4.11

Verify that the bar element subjected to torsion fulfills the rigid body motion condition required for finite elements

4.12

Verify that the truss member inclined at the angle α fulfills the rigid body displacement conditions required for finite elements

4.13

The finite element system shown consists of a plane stress element and a spring. Determine all displacements, the stresses in the middle of the element, and the force in the spring.

Parameters:

Plane stress element: $t = 0.20\,{\text{m}}, \quad E = 30000\,{\text{MN}}/{\text{m}}^{2}$

Spring: $k = 1000\,{\text{MN}}/{\text{m}}$

Load: $F = 0.500\,{\text{MN}}$

4.14

A plate is modeled with two plane stress elements. Determine the nodal displacements and the stresses (all components) of point 3 by the FEM. Assess the vertical displacement and the vertical stress in point 3 by a simple hand calculation. How do you assess the accuracy of the FEM calculation? Give the displacements in the midpoint of element 1. The thickness of the plate is t = 0.10 m.

4.15

A system consisting of four plane stress elements is investigated. Determine the nodal displacements of point 5 and the stresses (all components) in the midpoint of element 1.

$E = \;3 {\cdot} 10^{7} \;{\text{kN/m}}^{2}, \quad \mu = 0.2, \quad t = 0.30\;{\text{m}}, \quad F = 10000\;{\text{kN}}$

4.16

A square 4-node element for plates in bending with bilinear displacement shape functions is loaded with a distributed load

$p(x,y) = N_{1} (x,y) \cdot p_{1}$ with $N_{1} = \dfrac{1}{4} - \dfrac{1}{2a}x - \dfrac{1}{2b}y + \dfrac{1}{ab}xy$

Determine the equivalent nodal forces using a numerical 2-point Gaussian integration.

$a = 1.00\;{\text{m}},\quad b = 1.00\;{\text{m}}, {\quad} p_{1} = 1.0\;{\text{kN/m}}^{ 2}, \quad p_{2} = p_{3} = p_{4} = 0$

4.17

A plate clamped on three sides with a free edge and a line load is discretized into two nonconforming, shear rigid plate elements. At the clamped edges, the translational, as well as all rotational degrees of freedom, are fixed. Determine the deflection and rotational angles in the middle of the free edge.

$E = 3.10^{7}\, \text{kN}/\text{m}^{2}, \quad \mu = 0.2, \quad t = 0.3\, {\rm m}, \quad p_z = 10 \, \text{kN}/\text{m}$

Solutions

4.1

Displacements are exactly represented up to 3rd order polynomials, stresses, and strains up to 2nd order polynomials.

4.2

$$ \begin{aligned} & u(x) = u_{1} + \frac{x}{l} \cdot \left( {u_{2} - u_{1} } \right) = \left[ {\begin{array}{*{20}c} {1 - \dfrac{x}{l}} & {\dfrac{x}{l}} \\ \end{array} } \right] \cdot \left[ {\begin{array}{*{20}c} {u_{1} } \\ {u_{2} } \\ \end{array} } \right]\qquad \varepsilon = \frac{{{\text{d}}u}}{{{\text{d}}x}} = \underline{B} \cdot \underline{u} \\ & \underline{B} = \left[ {\begin{array}{*{20}c} { - \dfrac{1}{l}} & {\dfrac{1}{l}} \\ \end{array} } \right]\qquad \underline{u} = \left[ {\begin{array}{*{20}c} {u_{1} } \\ {u_{2} } \\ \end{array} } \right]\qquad \underline{D} = \left[ E \right] \\ & \underline{K} = \int {\underline{B}^{\text{T}} } \cdot \underline{D} \cdot \underline{B} {\text {d}}V = \int\limits_{0}^{l} {\underline{B}^{\text{T}} \cdot \underline{D} \cdot \underline{B} } \cdot A(x) \;\; {\text{d}}x = \frac{E}{{l^{2} }} \cdot \left[ {\begin{array}{*{20}r} 1 & { - 1} \\ { - 1} & 1 \\ \end{array} } \right] \cdot \int\limits_{0}^{l} {A(x) \;\; {\text{d}}x} \\ \end{aligned} $$

4.3

General solution: $\underline{K} = \dfrac{E}{l} \cdot \left[ {\begin{array}{*{20}r} 1 & { - 1} \\ { - 1} & 1 \\ \end{array} } \right] \cdot \left( {\dfrac{1}{6}A_{1} + \dfrac{2}{3}A_{m} + \dfrac{1}{6}A_{2} } \right)$

Solution with the numerical values given: $\underline{K} = 600 \cdot \left[ {\begin{array}{*{20}r} 1 & { - 1} \\ { - 1} & 1 \end{array} } \right] \; \dfrac{\textrm{kN}}{\text{cm}}$

4.4

(a) and (b): $\underline{K} = 866.7 \cdot \left[ {\begin{array}{*{20}r} 1 & { - 1} \\ { - 1} & 1 \end{array} } \right] \; \dfrac{\textrm{kN}}{\text{cm}}$

4.5

		A	B	C
0	Exact solution for the bar	constant	linear	quadratic
1		Exact	Approximate	Approximate
2		Exact	Approximate	Approximate
3		Exact	Exact	Approximate
4		Exact	Exact	Exact

4.6

4.7

(a) 12 × 12,

(b) 60 × 60.

4.8

$\int {} = 0.014$

4.9

$F_{\text{L}x1} = 0.05\;F_{x}, \quad F_{\text{L}x2} = 0.20\;F_{x}, \quad F_{\text{L}x3} = 0.60\;F_{x}, \quad F_{\text{L}x4} = 0.15\;F_{x}$

4.10

$$ {\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!} \begin{aligned} u_{p} & = 0.158\;{\text{mm}},\quad v_{p} = - 0.286\;{\text{mm}},\\ \varepsilon_{x} & = - 1.215 \cdot 10^{ - 4}, \quad \varepsilon_{y} = - 0.791 \cdot 10^{ - 4}, \quad \gamma_{xy} = - 0.578 \cdot 10^{ - 4} \\ \sigma_{x} & = - 4290\;{\text{kN/m}}^{ 2}, \quad \sigma_{y} = - 3230\;{\text{kN/m}}^{ 2}, \quad \tau_{xy} = - 723\;{\text{kN/m}}^{ 2}\end{aligned} $$

4.11

Nodal moments when both nodes are rotated with $\varphi_{\text{rigid}}$:

$\dfrac{{G \cdot I_{\rm T} }}{\ell } \cdot \left[ {\begin{array}{*{20}r} 1 & { - 1} \\ { - 1} & 1 \\ \end{array} } \right] \cdot \left[ \begin{aligned} \varphi_{\text{rigid}} \hfill \\ \varphi_{\text{rigid}} \hfill \\ \end{aligned} \right] = \left[ \begin{aligned} 0 \hfill \\ 0 \hfill \\ \end{aligned} \right]$

i.e. a rigid body rotation does not generate nodal moments.

4.12

Nodal forces when both nodes are subjected to the nodal displacements $u_{\text{rigid}}$, $v_{\text{rigid}}$ ($s = \sin \alpha ,\;c = \cos \alpha$):

$\dfrac{E \cdot A}{l} \cdot \left[ {\begin{array}{*{20}r} \hfill {c^{2} } & \hfill {s \cdot c} & \hfill { - c^{2} } & \hfill { - s \cdot c} \\ \hfill {s \cdot c} & \hfill {s^{2} } & \hfill { - s \cdot c} & \hfill { - s^{2} } \\ \hfill { - c^{2} } & \hfill { - s \cdot c} & \hfill {c^{2} } & \hfill {s \cdot c} \\ \hfill { - s \cdot c} & \hfill { - s^{2} } & \hfill {s \cdot c} & \hfill {s^{2} } \\ \end{array} } \right] \cdot \left[ {\begin{array}{*{20}c} {u_{\text{rigid}} } \\ {v_{\text{rigid}} } \\ {u_{\text{rigid}} } \\ {v_{\text{rigid}} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 \\ 0 \\ 0 \\ 0 \\ \end{array} } \right]$

i.e. a rigid body displacement does not generate nodal forces.

4.13

$u_{1} = 0.044\;{\text{mm}},\;\;\;v_{1} = - 0.186\;{\text{mm,}}\quad F_{\text{spring}} = - 44.5\;{\text{kN}}\quad \left( {\text{compression}} \right),$

$\sigma_{x} = 57\;{\text{kN}}/{\text{m}}^{ 2}, \quad \sigma_{y} = - 1381\;{\text{kN}}/{\text{m}}^{ 2}, \quad \tau_{xy} = - 441\;{\text{kN}}/{\text{m}}^{ 2}$

4.14

Nodal point 3:

$u_{3} = - 0.636\;{\text{mm}},\quad v_{3} = - 3.959\;{\text{mm}}$

Element 1:

$\sigma_{x}^{(1)} = - 992\;{\text{kN}}/{\text{m}}^{ 2}, \quad \sigma_{y}^{(1)} = - 2838\;{\text{kN}}/{\text{m}}^{ 2}, \quad \tau_{xy}^{(1)} = - 1277\;{\text{kN}}/{\text{m}}^{ 2}$

Element 2:

$\sigma_{x}^{(2)} = - 992\;{\text{kN}}/{\text{m}}^{ 2}, \quad \sigma_{y}^{(2)} = - 2838\;{\text{kN}}/{\text{m}}^{ 2}, \quad \tau_{xy}^{(2)} = - 1277\;{\text{kN}}/{\text{m}}^{ 2}$

Nodal point 3:

$\sigma_{x} = - 658\;{\text{kN}}/{\text{m}}^{ 2}, \quad \sigma_{y} = - 5411\;{{\text{kN}} \mathord{\left/ {\vphantom {{\text{kN}} {{\text{m}}^{ 2} }}} \right. \kern-0pt} {{\text{m}}^{ 2} }},\quad \tau_{xy} = 746\;{\text{kN}}/{\text{m}}^{ 2}$

Approximate solution:

$\begin{aligned} v_{3, \text{approx}} = \dfrac{F \cdot l}{E \cdot A} & = \dfrac{500 \cdot 3 \cdot 3}{{0.5 \cdot (2 \cdot 10^{6} + 6 \cdot 10^{6} ) \cdot 0.1 \cdot 3}} \\ & = 3.75 \cdot 10^{ - 3} {\rm m} = 3.75\;{\rm mm} \;\; {\text{(downwards)}}\end{aligned}$

$\begin{aligned}\sigma_{y,\text{approx}} & = - \frac{F}{A} = - \frac{500}{0.1} = - 5000\;{\text{kN}}/{\text{m}}^{ 2}, \\ \sigma_{x,\text{approx}} & = \mu \cdot \sigma_{y,\text{approx}} = - 1000\;{\text{kN}}/{\text{m}}^{ 2}\end{aligned}$

Displacements in the midpoint of element 1:

$u_{m}^{(1)} = - 0.159\;{\text{mm}}, \quad v_{m}^{(1)} = - 0.990\;{\text{mm}}$

4.15

$\begin{aligned} & u_{5} = 0.404\;{\text{mm}},\quad v_{5} = 0.404\;{\text{mm,}} \\ &\sigma_{x,m}^{(1)} = 7576\;{\text{kN}}/{\text{m}}^{ 2}, \quad \sigma_{y,m}^{(1)} = 7576\;{{\text{kN}} \mathord{\left/ {\vphantom {{\text{kN}} {{\text{m}}^{ 2} }}} \right. \kern-0pt} {{\text{m}}^{ 2} }},\quad \tau_{xy,m}^{(1)} = 5051\;{\text{kN}}/{\text{m}}^{ 2} \end{aligned}$

4.16

$F_{z,1} = 0.111\;{\text{kN}},\quad F_{z,2} = 0.056\;{\text{kN}},\quad F_{z,3} = 0.028\;{\text{kN}},\quad F_{z,4} = 0.056\;{\text{kN}},{\text{e}}.{\text{g}}.$

$\begin{aligned} F_{z,2} & = \,\frac{1}{4} \cdot ( N_{2} ( - 0.289, - 0.289) \cdot N_{1} ( - 0.289, - 0.289) \\ & \qquad \qquad + N_{2} (0.289, - 0.289 \cdot N_{1} (0.289, - 0.289) + \hfill \\ & \qquad \qquad + N_{2} (0.289, - 0.289 \cdot N_{1} (0.289, - 0.289) \\ & \qquad \qquad + N_{2} ( - 0.289, - 0.289 \cdot N_{1} ( - 0.289, - 0.289) \hfill) \\ & = 0.056\;{\text{kN}} \\ \end{aligned}$

4.17

$u_{5} = 0.647\;{\text{mm}},\quad \varphi_{x,5} = 0.247 \cdot 10^{ - 3}, \quad \varphi_{y,5} = 0\quad \left( {\text{due to symmetry}} \right)$