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Chapter 6
Finite Elements for Two‐Dimensional Solid Mechanics

6.1 INTRODUCTION

All real‐life structures are three–dimensional. It is engineers who make the approximation as a one‐dimensional (e.g., beam) or a two‐dimensional structure (e.g., plate or plane solid). In chapter 5 we explained in detail the conditions under which such approximation could be made. When the stresses on a plane normal to one of the axes are approximately zero, then we say that the solid is in the state of plane stress. Similarly, when the corresponding strains are zero, the solid is in the state of plane strain. A two‐dimensional solid is also called a plane solid. Some examples of plane solids are: (1) a thin plate subjected to in‐plane forces; or (2) a very thick solid with a constant cross‐section in the thickness direction. In this chapter, we will discuss when an engineering problem can be assumed to be two‐dimensional and how to solve such a problem using two‐dimensional finite elements. We will introduce three different types of two‐dimensional problems and corresponding two‐dimensional elements. Every element has its own characteristics. In order to use the finite element method appropriately, a thorough understanding of the capabilities and limitations of each element is required.

In general, two‐dimensional elasticity problem can be expressed by a system of coupled second‐order partial differential equations. Based on the constraints imposed in the thickness direction, a two‐dimensional problem can be a plane stress or plane strain problem. The main difference is in the stress‐strain relations. Yet another type of problem that can be classified as two‐dimensional is the axisymmetric problem where the structure has symmetry about an axis and it deforms only in the radial and axial directions.

In an elasticity problem, the main variables are the displacements in the coordinate directions. After solving for the displacements, stresses and strains can be calculated from the derivatives of displacements. The displacements are calculated using the fact that the structure is in equilibrium when the total potential energy has its minimum value. This will yield a matrix equation similar to the beam problem in chapter 3.

6.2 TYPES OF TWO‐DIMENSIONAL PROBLEMS

6.2.1 Governing Differential Equations

In two‐dimensional problems, the stresses and strains are independent of the coordinate in the thickness direction, (usually the z‐axis). By setting all the derivatives with respect to the z‐coordinate in eq. (5.69) in chapter 5 to zero, we obtain the governing differential equations for plane problems as

(6.1)

where b_x and b_y are the body forces.

Let u and v be the displacement in the x‐ and y‐direction, respectively. From eqs. (5.39)–(5.45), the strain components in a plane solid are defined as

(6.2)

In addition, the stresses and strains are related by the following constitutive relation:

(6.3)

Substituting the stress‐strain relations in eq. (6.3) and strain‐displacement relations in eq. (6.2) in the equilibrium equations in eq. (6.1), we obtain a pair of second‐order partial differential equations in two variables u(x,y) and v(x,y). The explicit form of the equations is available in textbooks on elasticity, for instance, Timoshenko and Goodier¹.

The differential equation must be accompanied by boundary conditions. Two types of boundary conditions can be defined. The first one is the boundary in which the values of displacements are prescribed (essential boundary condition). The other is the boundary in which the tractions are prescribed (natural boundary condition). The boundary conditions can be formally stated as

(6.4)

where S_g and S_T, respectively, are the boundaries where the displacement and traction boundary conditions are prescribed. The objective is to determine the displacement field u(x,y) and v(x,y) that satisfies the differential equation (6.1) and the boundary conditions in eq. (6.4). Now, we will discuss the stress‐strain relations in eq. (6.3) for the two different plane problems—plane stress and plane strain. The third type of 2D problem, namely axisymmetric problems, will be discussed later in the chapter.

6.2.2 Plane Stress Problems

Plane stress conditions exist when the thickness dimension (usually the z‐direction) is much smaller than the length and width dimensions of a solid. Since stresses at the two surfaces normal to the z‐axis are zero, it is assumed that stresses in the normal direction are zero throughout the body, that is, σ_zz = τ_xz = τ_yz = 0. In such a case, the structure can be modeled in two dimensions. An example of the plane stress problem is a thin plate or disk with applied in‐plane forces (see figure 6.1).

– Nonzero stress components: σ_xx, σ_yy, τ_xy.
– Nonzero strain components: ε_xx, ε_yy, γ_xy, ε_zz.

3D thin plate with in-plane applied forces indicated by arrows labeled fy and fx. — **Figure 6.1** Thin plate with in‐plane applied forces

Under plane stress conditions, for linear isotropic materials, the stress‐strain relation can be written as (see section 5.3)

(6.5)

where [C_σ] is the stress‐strain matrix or elasticity matrix for the plane stress problem. It should be noted that the normal strain ε_zz in the thickness direction is not zero; it can be calculated from the following relation:

(6.6)

6.2.3 Plane Strain Problems

A state of plane strain will exist in a solid when the thickness dimension is much larger than other two dimensions. When the deformation in the thickness direction is constrained, the solid is assumed to be in a state of plane strain even if the thickness dimension is small. A proper assumption is that strain components with z subscript are zero, that is, ε_zz = γ_xz = γ_yz = 0. In such a case, it is sufficient to model a slice of the solid with unit thickness. Some examples of plane strain problems are the retaining wall of a dam and long cylinder such as a gun barrel (see figure 6.2).

– Nonzero stress components: σ_xx, σ_yy, τ_xy, σ_zz.
– Nonzero strain components: ε_xx, ε_yy, γ_xy.

Dam structure with plane strain assumption, illustrated by 3D and 2D trapezoids labeled Plane strain model. — **Figure 6.2** Dam structure with plane strain assumption

For linear isotropic materials, the stress‐strain relations under plane strain conditions can be written as (see section 5.3)

(6.7)

where [C_ε] is the stress‐strain matrix for the plane strain problem. It should be noted that the transverse stress σ_zz is not equal to zero in the plane strain case, and it can be calculated from the following relation:

(6.8)

6.2.4 Equivalence between Plane Stress and Plane Strain Problems

Although plane stress and plane strain problems are different by definition, they are quite similar from the computational viewpoint. Thus, it is possible to use the plane strain formulation and solve the plane stress problem. In such case, two material properties, E and ν, need to be modified. Similarly, it is also possible to convert the plane stress formulation into the plane strain formulation. Table 6.1 summarizes the conversion relations.

6.3 CONSTANT STRAIN TRIANGULAR (CST) ELEMENT

In finite element analysis, a plane solid can be divided into a number of contiguous elements. In chapter 4, we introduced the 3‐node triangular element for heat transfer problems. In this section, we will use the same element for plane solid problems. Figure 6.3 shows a typical mesh consisting of triangular elements. The three nodes of the element, as shown in figure 6.3, have two displacement components, u and v, respectively in the x‐ and y‐direction for two‐dimensional solid mechanics problems. The first node of an element can arbitrarily be chosen. However, the sequence of the nodes 1, 2, and 3 should be in the counterclockwise direction.

A typical mesh consisting of triangular elements, one of which is shaded. An arrow from the mesh points to a shaded triangle with two displacement components, u and v, respectively in the x- and y-direction. — **Figure 6.3** Constant strain triangular (CST) element

The nodal values of the displacement components are interpolated within the element using shape functions. In the polynomial approximation, the displacement has to be a linear function in x and y because displacement information is available only at three points (nodes), and a linear polynomial has three unknown coefficients. Since displacement is a linear function, strain and stress are constant within an element, and that is why a triangular element is called a constant strain triangular element.

6.3.1 Displacement and Strain Interpolation

The first step in deriving the finite element matrix equation is to interpolate the displacement function within the element using the nodal values of displacements. Let the x‐ and y‐directional displacements be u(x,y) and v(x,y), respectively. Since the two coordinates are perpendicular to each other (orthogonal), u(x,y) and v(x,y) are independent of each other. Hence, u(x,y) needs to be interpolated in terms of u₁, u₂, and u₃, and v(x,y) in terms of v₁, v₂, and v₃. It is obvious that the interpolation function must be a three‐term polynomial in x and y. Since we must have rigid body displacements (constant displacements) and constant strain terms in the interpolation function, the displacement interpolation must be of the following form:

(6.9)

where α’s and β’s are constants. In finite element analysis, we would like to replace the constants by the nodal displacements. Let us consider x‐directional displacements, which are u₁, u₂, and u₃. At node 1, for example, x and y take the values of x₁ and y₁, respectively, and the nodal displacement is u₁. If we repeat this for other two nodes, we obtain the following three simultaneous equations:

(6.10)

In matrix notation, the above equations can be written as

(6.11)

If the three points (x₁, y₁), (x₂, y₂), and (x₃, y₃) are not on a straight line, then the inverse of the above coefficient matrix exists. Thus, we can calculate the unknown coefficients as

(6.12)

where A is the area of the triangle and

(6.13)

The area A of the triangle can be calculated from

(6.14)

Note that the determinant in eq. (6.14) is zero when three nodes are collinear. In such a case, the area of the triangular element is zero, and we cannot uniquely determine the three coefficients.

A similar procedure can be applied to the y‐directional displacement v(x, y), and the unknown coefficients β_i, (i = 1, 2, 3) are determined using the following equation:

(6.15)

After calculating α_i and β_i, the displacement interpolation can be written as

(6.16)

where the shape functions are defined by

(6.17)

Note that N₁, N₂, and N₃ are linear functions of x‐ and y‐coordinates. Thus, interpolated displacement varies linearly in each coordinate direction.

In order to make the derivations simple, we will rewrite the interpolation relation in eq. (6.16) in matrix form. Let {u} = {u, v}^T be the displacement vector at any point (x, y). The interpolation can be written in the matrix notation by

(6.18)

Equation (6.18) is the critical relationship in finite element approximation. When a point (x, y) within a triangular element is given, the shape function [N] is calculated at this point. Then, the displacement at this point can be calculated by multiplying this shape function matrix with the nodal displacement vector {q}. Thus, if we solve for the nodal displacements, we can calculate the displacement everywhere in the element. Note that the nodal displacements will be evaluated using the principle of minimum total potential energy in the following section.

After calculating displacement within an element, it is possible to calculate the strain by differentiating the displacement with respect to x and y. From the expression in eq. (6.18), it can be noted that the nodal displacements are constant, but the shape functions are functions of x and y. Thus, the strain can be calculated by differentiating the shape function with respect to the coordinates. For example, ε_xx can be written as

(6.19)

Note that u₁, u₂, and u₃ are nodal displacements, and they are independent of coordinate x. Thus, only the shape function is differentiated with respect to x. A similar calculation can be carried out for ε_yy and γ_xy. Using the matrix notation, we have

(6.20)

It may be noted that the [B] matrix is constant and depends only on the coordinates of the three nodes of the triangular element. Thus, one can anticipate that if this element is used, then the computed strains will be constant over a given element and will depend only on nodal displacements. Hence, this element is called the Constant Strain Triangular Element or CST element. The interpolation of displacement in eq. (6.18) and the expression for strain in eq. (6.20) are used for approximating the strain energy and potential energy of applied loads.

6.3.2 Properties of the CST Element

Before we derive the strain energy, it may be useful to study some interesting aspects of the CST element. Since the displacement field is assumed to be a linear function in x and y, one can show that the triangular element deforms into another triangle when forces are applied. Furthermore, an imaginary straight line drawn within an element before deformation becomes another straight line after deformation.

Let us consider the displacements of points along one of the edges of the triangle. Consider the points along the edge 1–2 in figure 6.4. These points can be conveniently represented by a coordinate ξ. The coordinate ξ = 0 at node 1 and ξ = a at node 2. Along this edge, x and y are related to ξ. By substituting this relation in the displacement functions, one can express the displacements of points on the edge 1–2 as a function of ξ. It can be easily shown that the displacement functions, for both u and v, must be linear in ξ, that is,

where γ’s are constants to be determined. Since the variation of displacement is linear, it might be argued that the displacements should depend only on u₁ and u₂, and not on u₃. Then, the displacement field along the edge 1‐2 takes the form:

(6.21)

where H₁ and H₂ are shape functions defined along the edge 1‐2, and a is the length of edge 1‐2. One can also note that a condition called inter‐element displacement compatibility is satisfied by triangular elements. This condition can be described as follows: any point should have a unique displacement including along shared edges between elements. After the loads are applied and the solid is deformed, the displacements at any point in an element can be computed from the nodal displacements of that particular element and the interpolation functions in eq. (6.18). Consider a point on a common edge of two adjacent elements. This point can be considered as belonging to either of the elements. Then the nodes of either triangle can be used in interpolating the displacements of this point. However, one must obtain a unique set of displacements independent of the choice of the element. This can be true only if the displacements of the points depend only on the nodes common to both elements. In fact, this will be satisfied because of eq. (6.21). Thus, the CST element satisfies the inter‐element displacement compatibility.

2 Triangles labeled Element 1 (top) and Element 2 (bottom) with vertices indicated by circles 1, 2, and 3. An arrow between the 2 elements is labeled ξ. — **Figure 6.4** Inter‐element displacement compatibility of constant strain triangular element

EXAMPLE 6.1 Interpolation in a triangular element

Consider the two triangular elements shown in figure 6.5. The nodal displacements are given as {u₁, v₁, u₂, v₂, u₃, v₃, u₄, v₄} = {−0.1, 0, 0.1, 0, −0.1, 0, 0.1, 0}. Calculate displacements and strains in both elements.

SOLUTION

Element 1 has nodes 1, 2, and 4. Then, using the nodal coordinates, we can derive the shape functions as shown below:

In addition, the area of the element is 0.5. Thus, from eq. (6.17) the shape functions can be derived as

Then, the displacements in element 1 can be interpolated as

Strains can be calculated from eq. (6.20), or directly differentiating the above expressions for displacements, as

Element 2 connects nodes 2 − 3 − 4. Thus, using the nodal coordinates, we can build the shape functions, as

In addition, the area of the element is 0.5. Thus, from eq. (6.17), the shape functions can be obtained as

Then, the displacements of element 2 can be interpolated as

Strains can be calculated from eq. (6.20), or directly differentiating the above expressions for displacements, as

Figure 6.5 Interpolation of displacements in triangular elements

Note that the displacements are linear and the strains are constant in each element. From the given nodal displacements, it is clear that the top edge has strain ε_xx = −0.2, while the bottom edge has ε_xx = 0.2. The strain varies linearly along the y‐coordinate. However, the triangular element cannot represent this change, and provides constant values of ε_xx = 0.2 for element 1 and ε_xx = −0.2 for element 2. In general, if a plane solid is under the constant strain states, the CST element will provide accurate solutions. However, if the strain varies in the solid, then the CST element cannot represent it accurately. In such a case, a dense mesh with many elements should be used to approximate it as a series of step functions. Note that the strains along the interface between two elements are discontinuous.

6.3.3 Strain Energy

Let us calculate the strain energy in a typical triangular element, say element e. In eq. (5.72), the strain energy of the plane solid was derived in terms of strains and the elasticity matrix [C]. Substituting for strains from eq. (6.20), we obtain

(6.22)

where [k^(e)] is the element stiffness matrix of the triangular element. The column vector {q^(e)} contains the displacements of the three nodes that belong to the element. The dimension of [k^(e)] is 6 × 6. In the case of the triangular element, all entries in matrices [B] and [C] are constant and can be integrated easily. After integration, the element stiffness matrix takes the form

(6.23)

where A^(e) is the area of the plane element. Using the expression for [B] in eq. (6.20) and stress–strain relation in eq. (6.5), the element stiffness matrix can be calculated.

One may note that in the case of truss and frame elements, we used a transformation matrix [T] in deriving the element stiffness matrix. However, in the present case, we have used the global coordinate system in the derivation of [k^(e)], and that is the reason for not using a transformation matrix. In some cases, however, it is required to define the element in a local coordinate system. For example, if the material is not isotropic, it will have a specific directional property. In such a case, [k^(e)] is first derived in a local coordinate system and transformed to the global coordinates by multiplying by appropriate transformation matrices.

The strain energy of the entire solid is simply the sum of the element strain energies. That is,

(6.24)

where NEL is the number of elements in the model. The superscript (e) in q^(e) implies that it is the vector of displacements or DOFs of element e. The summation in the above equation leads to the assembling of the element stiffness matrices into the global stiffness matrix.

(6.25)

where {Q_s} is the column vector of all displacements in the model and [K_s] is the structural stiffness matrix obtained by assembling the element stiffness matrices.

6.3.4 Potential Energy of Applied Loads

Concentrated forces at nodes:

The next step is to calculate the potential energy of external forces. We will consider three different types of applied forces. The first type is concentrated forces at nodes. It may be noted that the model expects two forces, one in the x‐direction and the other in the y‐direction, at each node. In general, the potential energy of concentrated nodal forces can be written as

(6.26)

where is the vector of applied nodal forces, and ND is the number of nodes in the solid. The contribution of a particular node to the potential energy will be zero if no force is applied at the node, or the displacement of the node becomes zero. The above potential energy of concentrated forces does not include a supporting reaction because the displacement at those nodes will be zero.

Distributed forces along element edges:

The second type of applied force is the distributed force (traction) on the side surface of the plane solid. In the plane solid, the traction is assumed to be a constant through the thickness. Let the surface traction force {T} = {T_x, T_y}^T is applied on the element edge 1‐2 as shown in figure 6.6. The unit of the surface traction is Pa (N/m²) or psi. Since the force is distributed along the edge, the potential energy of the surface traction force must be defined in the form of integral as

(6.27)

where is the vector of displacements along the edge 1‐2, is the vector of applied tractions along the edge 1‐2, is the vector of displacements of nodes 1 and 2, and

is the matrix of shape functions defined in eq. (6.21). The integration can be performed in a closed form if the specified surface tractions (T_x and T_y) are simple functions of s. We will modify eq. (6.27) to include all the six DOFs of the element and rewrite as

(6.28)

We have used the complete shape function matrix in eq. (6.28):

(6.29)

A mesh consisting of triangular elements, one of which is shaded (bottom part). An arrow from the mesh points to a shaded triangle with arrows labeled Ty and Tx in the bottom edge. — **Figure 6.6** Applied surface traction along edge 1‐2

If the last expression in eq. (6.28) is examined carefully, it is possible to note that the force vector {f_T} is nodal force vector that is equivalent to the distributed force applied on the edge of the element. This is also called the work‐equivalent nodal force vector. For a constant surface traction T_x and T_y, we can calculate the equivalent nodal force, as

(6.30)

For the uniform surface traction force, the equivalent nodal forces are obtained by simply dividing the total force equally between the two nodes on the edge.

The potential energy of distributed forces of all elements whose edge belongs to the traction boundary S_T must be assembled to build the global force vector of distributed forces:

(6.31)

where NS is the number of elements whose edge belongs to S_T.

Body forces:

The body forces are distributed over the entire element (e.g., centrifugal forces, gravitational forces, inertia forces, or magnetic forces). For simplification of the derivation, let us assume that a constant body force b = {b_x, b_y}^T is applied to the whole element. The potential energy of body force becomes

(6.32)

where

(6.33)

The resultant of body forces is hAb_x in the x‐direction and hAb_y in the y‐direction. Equation (6.33) equally distributes these forces to the three nodes. Similar to the distributed force, {f_B} is the equivalent nodal force corresponding to the constant body force.

The potential energy of body forces of all elements must be assembled to build the global force vector of body forces:

(6.34)

where NEL is the number of elements.

6.3.5 Global Finite Element Equations

Since the strain energy and potential energy of applied forces are now available, let us go back to the potential energy of the triangular element. The discrete version of the potential energy becomes

(6.35)

The principle of minimum potential energy in chapter 2 states that the structure is in equilibrium when the potential energy is minimum. Since the potential energy in eq. (6.35) is a quadratic form, the displacement vector {Q_s}, we can differentiate Π to obtain

(6.36)

The stationary condition of the potential energy yields the global finite element matrix equations.

The assembled structural stiffness matrix [K_s] is singular due to the rigid body motion. After constructing the global matrix equation, the boundary conditions are applied by removing those DOFs that are fixed or prescribed. After imposing the boundary condition, the global stiffness matrix becomes nonsingular and it can be inverted to solve for the nodal displacements.

6.3.6 Calculation of Strains and Stresses

Once the nodal displacements are calculated, strains and stresses in individual elements can be calculated. First, the nodal displacement vector {q^(e)} for the element of interest needs to be extracted from the global displacement vector. Then, the strains and stresses in the element can be obtained from

(6.37)

and

(6.38)

where [C] = [C_σ] for the plane stress problems and [C] = [C_ε] for the plane strain problems.

As discussed before, stress and strain are constant within an element because matrices [B] and [C] are constant. This property can cause difficulties in interpreting the results of the finite element analysis. When two adjacent elements have different stress values, it is difficult to determine the stress value at the interface. Such discontinuity is not caused by the physics of the problem but by the inability of the triangular element in describing the continuous change of stresses across element boundary. In fact, most finite elements cannot maintain continuity of stresses across the element boundary. Most programs average the stress at the element boundaries in order to make the stress look continuous. However, as we refine the model using smaller‐size elements, this discontinuity can be reduced. The following example illustrates discontinuity of stress and strain between two adjacent CST elements.

EXAMPLE 6.2 Cantilevered plate

Consider a cantilevered plate as shown in figure 6.7. The plate has the following properties: h = 0.1 in., E = 30 × 10⁶ psi and ν = 0.3. Model the plate using two CST elements to determine the displacements and stresses.

SOLUTION

This problem can be modeled as plane stress because the thickness of the plate is small compared to the other dimensions.

Element 1: Nodes 1–2–3
Using nodal coordinates, we can calculate the constants defined in eq. (6.13) as

In addition, from the geometry of the element, the area of the triangle A₁ = 0.5 × 10 × 10 = 50. Matrix [B] in eq. (6.20) and [C_σ] can be written as

and

Using the above two matrices, the element stiffness matrix can be obtained as
Element 2: Nodes 1–3–4
By following similar procedures as for element 1, the constants in eq. (6.13) for element 2 can be written as

The area of element 2 is twice of element 1: A₂ = 0.5 × 20 × 10 = 100. The strain–displacement matrix [B] can be obtained as

By following the same procedure, the stiffness matrix for element 2 can be computed as
Global finite element matrix equations
The two element stiffness matrices are assembled to form the global stiffness matrix. Since there are four nodes, the model has eight DOFs: each node has two DOFs. Thus, the global matrix has a dimension of 8 × 8. After assembly, the global matrix equation can be written as

where R_x1, R_y1, R_x4, and R_y4 are unknown reaction forces at nodes 1 and 4.
Applying boundary conditions
The displacement boundary conditions are: u₁ = v₁ = u₄ = v₄ = 0. Thus, we remove the first, second, seventh, and eighth rows and columns. After removing those rows and columns, we obtain the following reduced matrix equation:

Note that the stiffness matrix of the above equation is nonsingular and, therefore, the unique solution can be obtained.
Solution
The above matrix equation can be solved for unknown nodal displacements:
Strain and stress in element 1:
After calculating the nodal displacements, strain and stress can be calculated at the element level. First, the displacements for those nodes that belong to the element need to be extracted from the global nodal displacement vector. Since nodes 1, 2, and 3 belong to element 1, the nodal displacements will be {q} = {u₁,v₁,u₂,v₂,u₃,v₃}^T = {0,0,–2.147 × 10^–3,–4.455 × 10^–2,1.891 × 10^–2,–2.727 × 10^–2}^T. Then, strain in eq. (6.20) can be calculated using

The stresses in the element are obtained from eq. (6.5)
Strains and stresses in element 2:
Element 2 has nodes 1, 3, and 4. Thus, the nodal displacements will be {q} = {u₁, v₁, u₃, v₃, u₄, v₄}^T = {0, 0, 1.891 × 10^–2,–2.727 × 10^–2, 0, 0}^T. Using the element displacements, the strains and stresses in the element can be obtained as

and

If the stresses in the two elements are examined, one can note that the stress value changes suddenly across the element boundary. For example, σ_xx in element 1 is –24,709 psi, whereas in element 2, it is 62,354 psi. Such a drastic change in stresses is an indicator that the finite element analysis results from the current model are not accurate and further mesh refinement is necessary.

Figure 6.7 Cantilevered plate

From example 6.2, we can conclude the following:

– Stresses are constant over the individual element.
– The solution is not accurate because there are large discontinuities in stresses across element boundaries.
– With only two elements, the mesh is very coarse, and we obviously cannot expect very good results.

6.4 FOUR–NODE RECTANGULAR ELEMENT

6.4.1 Lagrange Interpolation for a Rectangular Element

A rectangular element is composed of four nodes and eight DOFs (see figure 6.8). It is a part of a plane solid that is composed of many rectangular elements. Each element shares its edge and two corner nodes with an adjacent element, except for those on the boundary. The four vertices of a rectangle are the nodes of that element as shown in figure 6.8. The first node of an element can arbitrarily be chosen. However, the sequence of the nodes 1, 2, 3, and 4 should be in the counterclockwise direction. Each node has two displacements, u and v, respectively in the x‐ and y‐direction.

Image described by caption and surrounding text. — **Figure 6.8** Four–node rectangular element

Since all edges are parallel to the coordinate directions, this element is not practical, but it is useful, as it is the basis for the quadrilateral element discussed in the following chapter. In addition, the behavior of the rectangular element is similar to that of the quadrilateral element. Shape functions can be calculated using procedures similar to that of CST element, but it is more instructive to use the Lagrange interpolation functions in the x‐ and y‐direction.

Consider the rectangular element in figure 6.8. From the geometry, it is clear that x₃ = x₂, y₄ = y₃, x₄ = x₁, and y₂ = y₁. We will use a polynomial in x and y as the interpolation function. Since there are four nodes, we can apply four boundary conditions, and hence the polynomial should have four terms as follows:

(6.39)

Let us calculate unknown coefficients α_i using the x‐directional displacement u:

It is obvious that we need to invert the 4 × 4 matrix in order to calculate the interpolation coefficients.

Instead of matrix inversion method, we use the Lagrange interpolation method to interpolate u and v. The goal is to obtain the following expression:

(6.40)

where N₁, …, N₄ are the interpolation functions. In order to do that, let us first consider displacement along edge 1‐2 in figure 6.8. Along edge 1‐2, y = y₁ (constant); therefore, shape functions must be functions of x only as shown below:

(6.41)

Using the one‐dimensional Lagrange interpolation formula, the shape functions can be obtained as

(6.42)

This is the same procedure that was used in chapter 2. Next, since y = y₃ = y₄ along edge 4‐3 in figure 6.8, the displacement can be interpolated as

(6.43)

Again from the one‐dimensional Lagrange interpolation formula, we have

(6.44)

Equations (6.41) and (6.43) represent interpolation of displacements at the top and bottom of the element, respectively. So far, we have interpolated displacements in the x‐direction only. Now, we can extend the interpolation in the y‐direction between u_I(x,y₁) and u_II(x,y₃) using the same Lagrange interpolation method. By considering u_I(x,y₁) and u_II(x,y₃) as nodal displacements, we have the following interpolation formula:

(6.45)

where

(6.46)

are the Lagrange interpolations in the y‐direction. By substituting eqs. (6.41) and (6.43) into eq. (6.45), we have the following formula:

(6.47)

Thus,

(6.48)

Comparing the above expression with eq. (6.40) we can define shape functions. N₁, …, N_4. In the rectangular element, it is enough to use the coordinates of two nodes, because x₁ = x₄, y₁ = y₂, and so forth. We will use the coordinates of nodes 1 and 3. Using the property that the area of the element is A = (x₃ – x₁)(y₃ – y₁), we obtain

(6.49)

Note that the shape functions for rectangular elements are the product of Lagrange interpolations in the two coordinate directions. Let us discuss the properties of the shape functions. It can be easily verified that N₁(x, y) is:

– 1 at node 1 and 0 at other nodes
– a linear function of x along edge 1‐2 and linear function of y along edge 1‐4 (bilinear interpolation)
– zero along edges 2‐3 and 3‐4

Other shape functions have similar behavior. Because of these characteristics, the i‐th shape function is considered associated with node i of the element.

In order to make the derivations simple, we rewrite the interpolation relation in eq. (6.40) in matrix form. Let {u} = {u, v}^T be the displacement vector at any point (x, y). The interpolation can be written using the matrix notation by

(6.50)

Note that the dimension of the shape function matrix is 2 × 8.

Since the shape functions are given as a function of x‐ and y‐coordinates, we can use an approach similar to that of CST element to obtain the strain–displacement relations. Thus, the strain can be calculated by differentiating the shape function with respect to the coordinates. For example, ε_xx can be written as

(6.51)

Note that u₁, u₂, u₃, and u₄ are nodal displacements, and they are independent of coordinate x. Thus, only the shape function is differentiated with respect to x. Similar calculation can be carried out for ε_yy and γ_xy. Then, we have

(6.52)

Note that matrix [B] is a linear function of x and y. Thus, the strain will change linearly within the element. For example, ε_xx will vary linearly in the y‐direction, while it is constant with respect to x. Thus, the element will have approximation error, if the actual strains vary in the x‐direction.

6.4.2 Element Stiffness Matrix

The element stiffness matrix can be calculated from the strain energy of the element. By substituting for strains from eq. (6.52) into the expression for strain energy in eq. (6.52) we have

(6.53)

where [k^(e)] is the element stiffness matrix. Calculation of the element stiffness matrix requires two‐dimensional integration. We will discuss numerical integration in the next chapter. When the element is square and the problem is plane stress, analytical integration of the strain energy yields the following form of element stiffness matrix:

(6.54)

It is interesting to note that the element stiffness matrix does not depend on the actual element dimensions, but it is a function only of material properties (E and ν) and thickness h.

The strain energy of the entire solid can be obtained using eq. (6.24), which involves the assembly process.

6.4.3 Potential Energy of Applied Loads

In the CST element, we discussed three different types of applied loads: concentrated forces at nodes, distributed forces along element edges, and body force. The first two types are independent of the element used. Thus, the same forms in eqs. (6.26) and (6.31) can be used for the potential energies of concentrated force and distributed force, respectively.

In the case of the body force, the element shape functions are used to calculate equivalent nodal forces. When a constant body force b = {b_x, b_y}^T acts on a rectangular element, the potential energy of body force becomes

(6.55)

where

(6.56)

Equation (6.56) equally divides the total magnitude of the body force to the four nodes. {f_b} is the equivalent nodal force corresponding to the constant body force. The potential energy of body forces of all elements must be assembled to build the global force vector of body forces as in eq. (6.34).

Using the principle of minimum total potential energy in eq. (6.36), a similar global matrix equation for the rectangular elements can be obtained. Applying boundary conditions and solving the matrix equations are identical to those of the CST element. After solving for nodal displacements, strains and stresses in each element can be calculated using eqs. (6.37) and (6.38), respectively.

EXAMPLE 6.4 Simple shear deformation of a square element

A square element shown in figure 6.11 is under a simple shear deformation. Material properties are given as E = 10 GPa, ν = 0.25, and thickness is h = 0.1 m. When a distributed force f = 100 kN/m² is applied at the top edge, calculate stress and strain components. Compare the results with the exact solution.

SOLUTION

Since the problem consists of one element, we do not need assembly process. The element has eight DOFs: {Q_s} = {u₁, v₁, u₂, v₂, u₃, v₃, u₄, v₄}^T. From the boundary condition given in figure 6.11, only two DOFs are nonzero: u₃ and u₄. Thus, from the element stiffness matrix given in eq. (6.54), all fixed DOFs are deleted to obtain

The total distributed load of 10,000 N at the top edge will be equally divided into two nodes: 4 and 3. Thus, the global matrix equation becomes

The above equation can be solved for unknown nodal displacements, as u₃ = u₄ = 0.025 mm. Then, from eq. (6.52), the strain components can be obtained, as

Note that the shear strain is the only nonzero strain. Thus, the rectangular element can accurately represent the simple shear condition. Figure 6.12 shows the deformed shape of the solid. Note that the deformation is magnified for illustration purposes.

Using the stress–strain relation in eq. (6.5) for plane stress, the stress components can be obtained, as

Since the distributed force f = 10 kN/m² is applied at the top edge, the above shear stress is exact.

Figure 6.11 A square element under a simple shear condition

Figure 6.12 Simple shear deformation of a square element

EXAMPLE 6.5 Pure bending deformation of a square element

A pure bending condition can be achieved by applying a couple in the case of a beam (see chapter 3). For the plane solid, the effect of a couple can be achieved by applying equal forces in opposite directions. A square element shown in figure 6.13 is under a pure bending condition. Material properties are given as E = 10 GPa, ν = 0.25, and thickness is h = 0.1 m. When an equal and opposite force f = 100 kN is applied at nodes 2 and 3, calculate stress and strain components. Compare the results with the exact solutions from the beam theory.

SOLUTION

Analytical solution: If we consider the above plane solid as a cantilevered beam, the moment of inertia I = 8.333 × 10⁻³ m⁴ and the applied couple M = 100 kN⋅m. Thus, the maximum stress will occur at the bottom edge with the magnitude of
The minimum stress will occur at the top edge with the same magnitude but in compression. Since the stress varies linearly along the y‐coordinate, we have

All other stress components are zero.
Numerical solution: Since we use only one element, we do not need the assembly process. The element has eight DOFs: {Q_s} = {u₁, v₁, u₂, v₂, u₃, v₃, u₄, v₄}^T. From the boundary condition given in figure 6.13, only four DOFs are nonzero: u₂, v₂, u₃, and v₃. Thus, from the element stiffness matrix given in eq. (6.54), all fixed DOFs are deleted to obtain
Using the applied nodal forces, the global matrix equation becomes

The above equation can be solved for unknown nodal displacements as

Then, from eq. (6.52), the strain components can be obtained as

Using the stress–strain relation in eq. (6.5) for plane stress, the stress components can be obtained as

The deformed shape of the element is shown in figure 6.14. In a plane solid, the applied couple produces a curvature, but the rectangular element is unable to produce deformation corresponding to the curvature because the displacement can only change linearly within the element. The rectangular shape deforms to the trapezoidal shape, and as a result, nonzero shear stress is produced. Note that the maximum stress (σ_xx)_max is only 73% (4.364/6.0) of the exact solution. In addition, σ_yy and τ_xy have nonzero values. The applied couple is supported by other stress components, σ_yy and τ_xy, and as a result, the element shows smaller (σ_xx)_max. In a sense, the element shows a stiff behavior.

Figure 6.13 A square element under pure bending condition

Figure 6.14 Pure bending deformation of a square element

6.5 AXISYMMETRIC ELEMENT

Axisymmetric problems are also classified as two‐dimensional solid mechanics problems. For such problems, the structure has a geometry that is symmetric about an axis. For instance, cylinders and cones are axisymmetric geometries. Any structure where the geometry can be described as the volume swept by a two‐dimensional profile or section revolved about an axis is a structure with symmetry about that axis. If the applied loads and constraints on such geometry are also symmetric about the same axis, then the structure would deform in an axisymmetric manner. Even though all real structures are three‐dimensional, we can state the problem as two‐dimensional only if all the quantities of interest are constant in the direction normal to a plane and the displacement field has only two components that are parallel to this plane. In this case, the domain of the analysis is an area on this plane. For plane stress and plane strain problems, the thickness in the z‐direction is constant and we are able to easily convert volume integrals into area integrals because the field variables of interest are constant in the z‐direction.

Axisymmetric problems can also be stated as two‐dimensional problems when viewed with respect to a cylindrical coordinate system as shown in figure 6.15 where the geometry in (a) is obtained by revolving the section in (b). The section plane is on the r‐z plane and is the plane on which the displacement vector must lie. If the z‐axis is the axis of symmetry, then points on the r‐z plane must remain on this plane as the structure deforms in order for it to be considered a two‐dimensional deformation. Any twisting about the z‐axis will result in a three‐dimensional deformation.

To summarize, the conditions under which we have axisymmetric deformation are:

The geometry of the structure must be symmetric about an axis.
The applied load and boundary conditions must also be symmetric about the same axis.
The structure should not be subjected to loads that produce a torque about the axis of symmetry. In other words, the structure should not twist about the axis.

For an axisymmetric problem, the displacement field has two components (u_r, u_z) which are the radial and axial components in a cylindrical coordinate system. The component of displacement in the tangential or circumferential direction must be zero (). Any displacement in the circumferential direction would cause twisting about the axis, which is the reason for assuming that there is no torque applied about the axis. Even though there is no displacement in the θ ‐direction, there is a stress in this direction, often referred to as the hoop stress. To understand the reason for this stress, imagine a circle in the structure on a plane normal to the axis with its center at the axis of symmetry of the structure as shown in figure 6.16. During the deformation, if there is any displacement u_r in the radial direction, this will cause the radius of the circle to increase, which implies that the circumference will increase resulting in a strain in the circumferential/tangential direction, which is the hoop strain, ε_θθ. This strain is an extensional strain and can be defined as the change in circumference over the original circumference of this circle as shown in the following equation.

(6.57)

Circumferential strain due to radial displacement illustrated by 2 concentric circles with radii labeled r (inner circle) and r + ur (dashed outer circle). — **Figure 6.16** Circumferential strain due to radial displacement

From the above discussion, it is clear that there are four strain components for axisymmetric problems. Similarly, there are four stress components: σ_rr the radial stress, σ_zz the axial stress, σ_θθ or the hoop stress, and finally τ_rz the shear stress. The strain components, placed in a column matrix, can be defined as:

(6.58)

For finite element analysis, within each element, we can compute the strains using the interpolated displacement components. For convenience we will use the notation: and . The nodal values of the displacement components can then be denoted as . We will assume that when modeling the two‐dimensional cross section, we will create the model on the xy‐plane such that the r‐ and z‐axis of the cylindrical coordinate system align with the x‐ and y‐axis of the Cartesian coordinates. Therefore, in the subsequent discussion, we will assume that and . The strain components can be numerically computed within each element as:

The preceding equation is valid for any two‐dimensional element with n nodes. The first two columns of the [B] matrix are multiplied by or the degrees of freedom of the first node, and the same pattern repeats for every node. The third row corresponds to the hoop strain, where the denominator is the distance of the point at which the strain is being computed from the axis of revolution. It appears as though the hoop strain could be undefined at the axis because the distance from the axis is zero along the axis. Indeed it would be infinite unless the numerator is also zero or in other words, points along the axis cannot have any displacement in the radial direction regardless of the applied loads. This makes sense when you consider that displacement of points at the axis in the radial direction is equivalent to a hole being created at the axis. Therefore, for axisymmetric problems, it is not necessary to add displacement boundary conditions in the radial direction on any edge of the domain that is along the axis of symmetry.

The stress‐strain relation for axisymmetric problems can be obtained by eliminating from the three‐dimensional stress‐strain relations the two rows and columns that correspond to the two shear strains that are zero. For linear isotropic materials, the stress‐strain relations under axisymmetric conditions can be written as:

(6.59)

The stiffness matrix computation for axisymmetric elements is very similar to other two‐dimensional elements with the main difference being that the integration is carried out in a cylindrical coordinate system. The real geometry of a triangular axisymmetric element is the triangle revolved around the axis of symmetry. For any element, the stiffness matrix is computed as:

The integration in the preceding equation is over the volume of an element V^(e). Performing this volume integration in cylindrical coordinates, we can restate the stiffness matrix computation as:

(6.60)

The volume integral is converted to an area integral by first integrating in the θ direction, noting that the integrand does not vary in the θ direction and therefore can be treated as a constant while integrating with respect to θ. Here we have integrated from 0 to 2π, though we could have integrated from 0 to 1 radian instead. Either way, the resultant equations will be the same as long as all the volume integrals in the energy equation are computed in the same way because the constant 2π will then occur in all the terms and will cancel out. For axisymmetric problems, we do not need to specify a thickness as we did earlier for plane stress problems.

EXAMPLE 6.6 Triangular axisymmetric element

Consider an annular ring with a rectangular cross section expanded by radial pressure. The radial pressure p is applied uniformly along its entire inner radius. For a triangular element in the mesh that is along the inner boundary as shown in figure 6.17: (i) Compute the equivalent nodal forces to be applied at the nodes of the element. (ii) Assuming that the computed nodal displacement vector for this element is: , compute the components of strain in this element.

SOLUTION

The geometry and the applied pressure is symmetric about an axis, and therefore we can use an axisymmetric model for this problem. The model of the annular ring is shown in figure 6.17(a), where the rectangular region is the domain of the analysis and it represents the cross section of the ring. The axis of symmetry is the y‐axis, so the distance of the left edge of the rectangle from the axis must be equal to the inner radius (R_i) of the ring, and the right edge corresponds to the outer surface of the ring. The real geometry of the ring can be obtained by revolving this rectangle around the axis. The height of the rectangle corresponds to the thickness of the ring, h. The figure shows one element labeled (e) in the mesh with the local node numbering shown in figure 6.17(b).

The equivalent nodal forces acting on the edge can be computed as:
In the preceding equation, S_e is the surface of the element on which the load is applied. The integral over this surface can be decomposed into integrals in the circumferential direction and the axial direction. The integrand is constant in the circumferential direction and therefore can be carried out analytically resulting in the term 2πr, and for the axial direction, we use a local coordinate s attached to node 1 and along the edge of the element as shown in the figure.

As the triangular element is linear, we can use linear shape functions derived earlier for one‐dimensional elements

Evaluating the integral on the length of the edge of the triangle l_e we get,
The strain components in the element can be computed by first computing the displacement field within the element and then computing the required derivatives. In the finite element method, the strain components are typically computed using the [B] matrix as follows:

where, f_i, b_i, and c_i are the coefficients of the shape function defined in eq. (6.13) and A is the area of the triangle. Note that for this element, the hoop strain is not constant within the element while all the other components are constant.

Figure 6.17 Triangular axisymmetric element; (a) axisymmetric model of ring, (b) element e

6.6 FINITE ELEMENT MODELING PRACTICE FOR SOLIDS

The most important step in finite element analysis is selecting the right model to use for solving a given problem. There may be more than one way to correctly model a problem. For example, a beam bending problem can be solved using beam elements, but it is also a plane stress problem. The geometry of the structure plays a big role in determining the appropriate model, but as we have seen, the applied load and boundary conditions also can change the nature of the problem and therefore the type of analysis. Very often the right model also depends on the purpose of the analysis, that is, the question you are trying to answer through the analysis. For example, to find the maximum deflection of a frame‐like structure, a model using beam/frame elements is often the most appropriate, but if one needs to calculate stress concentration at a joint of the frame, a different type of model is needed. To make the right decision, it is important to understand the underlying theory and the assumptions used for various types of analysis and elements. In this sections, we will study the nature of the solution obtained by the finite element method using some examples solved using commercial software. We have selected examples that can be modeled correctly in more than one way.

EXAMPLE 6.7 Cantilevered beam

Model a cantilever beam with a concentrated tip load whose length‐to‐height ratio is equal to 10 using plane stress elements, and compare the solution to analytical and beam element solutions. Assume that the length is L = 100 mm, the cross section is 10 × 10 mm², the applied concentrated load is P = 100 N, the material properties are Young’s modulus Pa and Poisson’s ratio .

SOLUTION

A cantilever beam is best modeled using beam elements if it is slender, that is, if the length‐to‐height ratio is greater than 10. Otherwise, if this ratio is less than 10, then the shear strain energy is no longer negligible, and therefore a plane stress model is more appropriate. Using Euler‐Bernoulli beam theory, the analytical solution for a cantilever beam with a concentrated load at the tip is:

When the beam is subjected to concentrated forces, the analytical solution for the beam deflection is a cubic function. Beam elements use cubic shape functions and therefore, a model using a single beam element is sufficient to yield the exact solution for the beam deflection in this example. The normal stresses in beam elements are computed using beam theory (see chapter 3). The normal stress on any section of the beam is a linear function of the distance from the neutral axis. When using software for analysis, one has to provide the height of the cross section so that the maximum stress can be computed.

In contrast, the two‐dimensional plane stress model requires a planar rectangular geometry that is meshed using triangular or quadrilateral elements. In this chapter, we described the 3‐node constant strain triangular (CST) elements. The strain is approximated to be constant within this element, and therefore it is not a very accurate element. A relatively high‐density mesh is needed to get acceptable results since the normal stress and strain vary linearly through the height of the cross section of the beam. To reasonably approximate this linear variation using elements with constant stress and strain, we will need a lot of elements through the height of the beam. Better elements will be discussed in a subsequent chapter that use higher‐order polynomials for the interpolation.

Figure 6.18 shows the plane stress model of the beam constructed using triangular elements. To clamp the beam at the left end, a sliding boundary condition has been applied along the left edge so that the displacement is fixed only in the x‐direction while the y‐component of the displacement is not fixed. If the y‐component were also fixed, then shrinkage or expansion in that direction that occurs due to Poisson’s effect will be prevented causing some artificial stresses. It is important to apply sufficient boundary conditions to prevent any rigid body motion. To prevent rigid body motion in the y‐direction, a point along the left edge is fixed as shown in the figure. Ideally, if a node is available at the exact midpoint of the left edge, then this node should be fixed. However, with automatic mesh generation, this is not guaranteed. In some software packages, a node can be placed there during model definition, but these details vary widely between programs. The load, which is typically modeled as a concentrated force in theoretical and beam models, is better defined as a distributed load in a plane stress model. So the load is applied at the right end as a traction load that is distributed evenly over the right face of the beam. This option is better than applying a concentrated force at one of the vertices, which could create stress concentration at the point of application that is unrealistic. Figure 6.19 shows the deformed shape of the beam computed using the plane stress model. As seen in the figure, the maximum deflection computed is very close to the analytical solution indicating that the mesh density was sufficient for this example if computing the deflection of the beam was the main intent. Compared to beam elements, however, this is clearly a less efficient model that requires a lot more elements and computation.

Figure 6.18 Beam model using plane stress CST elements

Figure 6.19 Beam deflection computed using CST elements

The normal strain computed using CST elements is displayed in figure 6.20. The linear variation through the beam height is approximated by constant values within each triangle. Notice that the strain is not perfectly symmetrical about the neutral axis as one would expect. The scale indicates that the maximum and minimum do not have exactly the same magnitude. This is due to the fact that the mesh is not symmetric about the neutral axis, and unlike displacements, the strains computed using this model is not as accurate. In fact, the analytical solution for the maximum strain is .

Figure 6.20 Computed normal strain component without smoothing

The plot for stresses and strains can be improved by smoothing the computed result. A very simple method for smoothing is to average the computed values from all the elements attached to a node and then using these averaged nodal values to interpolate the stresses or strains. Figure 6.21 shows the normal stress plot after smoothing the results. Now the stress variation through the thickness appears more realistic because the stress is interpolated using average nodal values and is therefore not constant within elements. The maximum normal stress should be Pa. The results can be improved by further refining the mesh using much smaller elements.

Figure 6.21 Normal stress component after smoothing

Even though the results do not exactly match the analytical solution, the CST element models do give a reasonable approximation of the solution and are able to tell us where the maximum and minimum stresses would occur in a structure. As the beam height decreases, the stiffness of the beam is overestimated for this type of 2D elements leading to a phenomenon called shear locking for very slender beams if modeled using plane stress elements. This happens because the typical finite element solution has errors due to the fact that the shear does not reduce to zero as the beam height decreases, and this leads to overestimation of stiffness. For such slender beams, the beam element is more appropriate and is, in fact, able to predict the theoretical solution if the applied loads are all concentrated forces. On the other hand, for short beams where the shear strain is not negligible, plane stress models are preferred. In general, the accuracy of the solution improves with increasing mesh density. With increasing mesh density, the solutions should converge, meaning that the change in the solution should decrease with increasing mesh density. The solution is said to have converged when the solution does not change significantly with further increase in mesh density.

EXAMPLE 6.8 Thick‐walled cylinder

A thick‐walled tube subjected to internal pressure is studied here as illustrated in figure 6.22(a) where its cross section is shown. The inner and outer radii are 0.3 and 0.5 in., respectively. The inner surface is subjected to a pressure of 500 psi. The material is assumed to be made of steel with a modulus of elasticity, E = 30 x10⁶ psi and a Poisson’s ratio of 0.3. Using two‐dimensional models, compute the radial displacement and the stress distribution in the tube.

SOLUTION

This problem can be modeled in a finite element software in two different ways. As the tube has a cylindrical geometry and the load is symmetric about its axis, it is a problem that can be modeled as axisymmetric. On the other hand, if the tube is very long compared to the diameter of its cross section, then we can assume that the internal pressure will not cause a change in length (or axial strain). Therefore, it can also be modeled as a plane strain problem. Obviously, the geometry of the analysis domain will be very different depending on the model.

To model the cylinder as a plane strain problem, we model only a quarter of the cross section of the cylinder as shown in figure 6.22(b). The displacement boundary conditions are applied on the straight edges of the quarter such that only radial displacement is allowed while tangential displacement is restricted. For the axisymmetric model, as shown in figure 6.22(c), the domain of analysis is a rectangle placed such that its inner (left‐side) boundary is at a distance equal to the inner radius from the axis. An arbitrary length of the cylinder can be modeled if in fact the tube is very long and we expect no axial displacements/strains. In this case, since the model length is significantly less than the real length, it is important to also apply displacement boundary conditions such that there is no axial displacement. The internal pressure is applied along the inner radius as shown in the figure.

Figure 6.22 Thick‐walled cylinder; (a) cross‐section, (b) plane strain model, (c) axisymmetric model

Results obtained from the two models are compared in figure 6.23. When the thick‐walled cylinder is modeled using plain strain elements, the maximum displacement was computed to be . whereas the same cylinder (same dimensions and loads) when modeled using axisymmetric elements yields displacement equal to ., which is approximately the same with a difference of less than 0.3%. The two models will converge toward the analytical solution as the element size is decreased. Similarly, the von Mises stress computed by the two models are also approximately equal. Triangular elements linearly interpolate nodal values of displacement, and therefore most of the strain and stress components are computed to be constant within each element as noted in the previous example. The stress plot in figure 6.23 does not show a discontinuity at the edges of the elements because stresses are smoothed before plotting. Smoothing the results in this manner improves the accuracy of the solutions since the exact solution for stresses is indeed continuous throughout the domain.

Figure 6.23 Comparison of results using plane strain and axisymmetric models for the thick‐walled cylinder; (a) displacement magnitude with plane strain model, (b) displacement magnitude with axisymmetric model, (c) von Mises stress with plane strain model, (d) von Mises stress with axisymmetric model

6.7 PROJECT

Project 6.1 Accuracy and convergence analysis of a cantilever beam

In this project, we want to compare the finite element results of plane solid elements with that of uniaxial bar and beam elements. Consider a cantilever beam shown in figure 6.24 under horizontal and transverse forces at the tip. The beam has a square cross section of 0.1 m × 0.1 m, length of L = 1 m, Young’s modulus E = 207 GPa, and Poisson’s ratio ν = 0.3.

Figure 6.24 Cantilever beam model

PART I

Consider the case of F₁ = 100 N and F₂ = 0. Solve the problem using a uniaxial bar element to find the elongation u(x). Calculate ε_xx and σ_xx. Assume that σ_yy = σ_zz = τ_xy = τ_yz = τ_xz = 0. Compare the results with analytical solution.
Consider the case of F₁ = 0 and F₂ = 500 N. Solve the problem using a beam element to find the deflection w(x). Calculate ε_xx and σ_xx. Assume that σ_yy = σ_zz = τ_xy = τ_yz = τ_xz = 0. Compare the results with analytical solution. Plot σ_xx as a function of y at x = L/2.

PART II

Consider the case of F₁ = 100 N and F₂ = 0. Solve the problem using: (i) 20 CST elements and (ii) 10 rectangular elements to find the elongation u(x). Calculate ε_xx and σ_xx. Compare the results with those from part I. Explain the results using an interpolation scheme.
Consider the case of F₁ = 0 and F₂ = 500 N. Solve the problem using: (i) 20 CST elements and (ii) 10 rectangular elements to find the deflection w(x). Calculate ε_xx and σ_xx. Compare the results with those of part I. Explain the results using an interpolation scheme.
Consider the case of F₁ = 0 and F₂ = 500 N. Perform convergence study by gradually decreasing the element size, and show the deflection and stress converge to the exact solution.

6.8 EXERCISES

Answer the following descriptive questions.
1. What are nonzero stress and strain components for plane stress problems?
2. What are nonzero stress and strain components for plane strain problems?
3. What are nonzero stress and strain components for axisymmetric problems?
4. When would a 3‐node triangular element be invalid?
5. If only nonzero displacements of a 3‐node triangular element are u₃ and v₃, what would be the displacement along the edge of nodes 1 and 2?
6. How do the strains, ε_xx, ε_yy, and γ_yx, vary within a 3‐node triangular element?
7. How do the strains, ε_xx, ε_yy, and γ_xy, vary within a 4‐node rectangular element?
8. If a gravitational force is applied in the y‐coordinate direction for a 3‐node triangular element, what would the equivalent nodal forces be? Assume that the total gravitational force is ρhAg, where ρ is density, h is thickness, A is area, and g is gravitational acceleration.
9. For a four‐node rectangular plane element, we use u(x,y) = a₀ + a₁x + a₂y + a₃xy as the form of solution. What would be the problem if we use u(x,y) = a₀x² + a₁xy + a₂y² + a₃x²y² instead?
10. For a rectangular element, plot the shape function N₁(x,y).
11. Define inter‐element displacement compatibility.
Repeat example 6.2 with the following element connectivity:
- Element 1: 1–2–4
- Element 2: 2–3–4
Does the different element connectivity change the results?
Solve example 6.2 using a commercial finite element analysis program.
Using two CST elements, solve the simple shear problem depicted in the figure and determine whether the CST elements can represent the simple shear condition accurately or not. Material properties are given as E = 10 GPa, ν = 0.25, and thickness is h = 0.1 m. The distributed force f = 100 kN/m² is applied at the top edge.
Solve problem 4 using a commercial finite element analysis program.
A structure shown in the figure is modeled using one triangular element. Plane strain assumption is used.
1. Calculate the strain–displacement matrix [B].
2. When nodal displacements are given by {u₁, v₁, u₂, v₂, u₃, v₃} = {0, 0, 2, 0, 0, 1}, calculate element strains.
Calculate the shape function matrix [N] and strain–displacement matrix [B] of the triangular element shown in the figure
The nodal coordinates and corresponding displacements in a plane triangular element are given in the table below. Calculate the u‐displacement at a point given by (1,1).

Node number (x,y) (mm) u (mm) v (mm)

1 (0,0) 0 0

2 (3,0) 1 1

3 (0,3) 2 0
The nodal displacements of the triangular element are given in the table below.

Node (x, y) u v

1 (0, 0) 0 6

2 (3, 0) 3 5

3 (0, 4) 8 4
1. It is intended to use the polynomial to interpolate the u displacements in the triangular element. Calculate the constants a₁, a₂, and a₃.
2. What is the strain ε_xx at the centroid of the triangle?
3. Derive an expression for u displacements of points on the edge 2–3 as a function of ξ. That is, derive an expression for u(ξ).
Hint: Along the edge 2–3, .
The coordinate of the nodes and corresponding displacements in a triangular element are given in the table. Calculate the displacement u and v and strains ε_xx, ε_yy, and γ_xy at the centroid of the element given by the coordinates (1/3, 1/3)

Node x (m) y (m) u (m) v (m)

1 0 0 0 0

2 1 0 0.1 0.2

3 0 1 0 0.1
A 2 × 2 × 1 mm³ square plate with E = 70 GPa and ν = 0.3 is subjected to a uniformly distributed load as shown in the left figure. Due to symmetry, it is sufficient to model one‐quarter of the plate with artificial boundary conditions as shown in the right figure. Use two triangular elements to find the displacements, strains, and stresses in the plate. Check the answers using simple calculations from mechanics of materials.
A beam problem under the pure bending moment is solved using CST finite elements, as shown in the figure. Assume E = 200 GPa and ν = 0.3. The thickness of the beam is 0.01 m. In order to simulate the pure bending moment, two opposite forces F = ±100,000 N are applied at the end of the beam. Using any available finite element program, calculate the stresses in the beam along the neutral axis and top and bottom surfaces. Compare the numerical results with the elementary beam theory. Provide an element stress contour plot for σ_xx.
For a rectangular element shown in the figure, displacements at four nodes are given by {u₁, v₁, u₂, v₂, u₃, v₃, u₄, v₄} = {0.0, 0.0, 1.0, 0.0, 2.0, 1.0, 0.0, 2.0}. Calculate displacement (u, v) and strain ε_xx at point (x, y) = (2, 1).
The figure shows the assemblage of two square elements, elements 23 and 24. The coordinates and displacements of all six nodes are given in the table below.
1. Consider point P given by the global coordinates (2, 1). Calculate the displacements u and v at this point. As this point belongs to both elements, one can calculate the displacements u and v at this point using the nodal displacements of either element 23 or element 24. Will they be same? Explain.
2. Repeat the above question (a) including discussion for calculating the strain ε_xx at point P.
Node 1 2 3 4 5 6

x 0 2 4 0 2 4

y 0 0 0 2 2 2

u 0 0.1 0.3 0 0.2 0.4

v 0 ‐0.05 ‐0.2 0 ‐0.5 ‐0.2
The FE model of a plane solid consists of two square elements of dimensions 1 × 1 as shown in the figure below. The nodal coordinates and corresponding displacements are given in the table.
1. Calculate the displacement u at (x,y) = (1, 0.5);
2. Estimate the strain ε_xx at (x,y) = (1, 0.5) using element 1 and element 2.
Node (x, y) u v

1 (0,0) 0 0

2 (0,1) 0.1 −0.1

3 (1,0) 0.1 0.1

4 (1,1) −0.2 0.2

5 (2,0) 0.15 −0.15

6 (2,1) 0.2 0
For the 4‐node plane stress finite element shown in the figure, finite element computation yields the displacements at the nodes as follows.

u₁ v₁ u₂ v₂ u₃ v₃ u₄ v₄

0.005 0.003 0.006 0.005 0.0 0.0 0.0 0.0
1. What is the displacement field u(x,y) and v(x,y) within the element?
2. Compute the strain displacement matrix [B] for this element to express strain as ?
3. Compute the displacements (u,v) and the strain components at the origin.
4. Show that you can use this element to represent a constant state of strain.
A rectangular element with thickness 1.0 m is under gravity. Express strain ε_yy in terms of vertical nodal displacement –v at the top. (v₁ = v₂ = 0, v₃ = v₄ = –v). Explain if the calculated strain ε_yy is exact or not. (If you need, use g: gravitational acceleration, ρ: density, E: elastic modulus).
For the 4‐node element shown in the figure, a linearly varying pressure p is applied along the edge. The finite element method converts the distributed force into an equivalent set of nodal forces {F_e} such that
where {T} is the traction (force per unit area) on the surface S and [N] is a 2 × 8 matrix of shape functions. The applied pressure in the above figure is normal to the surface (in the x‐direction), therefore the traction can be expressed as where p can be expressed as . L_e is the length of the edge. Integrate the left‐hand side of the equation above to compute the work‐equivalent nodal forces {F_e} (8 × 1 vector).
Six rectangular elements are used to model the cantilevered beam shown in the figure. Sketch the graph of σ_xx along the top surface that a finite element analysis would yield. There is no need to actually solve the problem, but use your knowledge of shape functions for rectangular elements.
A rectangular element as shown in the figure is used to represent a pure bending problem. Due to the bending moment M, the element is deformed as shown in the figure with displacement {q} = {u₁, v₁, u₂, v₂, u₃, v₃, u₄, v₄}^T = {−1, 0, 1, 0, −1, 0, 1, 0}^T.
1. Write the mathematical expressions of strain component ε_xx, ε_yy, and γ_xy, as functions of x and y.
2. Does the element satisfy pure bending condition? Explain your answer.
3. If two CST elements are used by connecting nodes 1–2–4 and 4–2–3, what will be ε_xx along line A–B?
Five rectangular elements are used to model a plane beam under pure bending. The element in the middle has nodal displacements as shown in the figure. Using the bilinear interpolation scheme, calculate the shear strain along the edge AB and compare it with the exact shear strain.
A uniform beam is modeled by two rectangular elements with thickness b. Qualitatively, and without performing calculations, plot σ_xx and τ_xy along the top edge from A to C, as predicted by FEA. Also, plot the exact stresses according to beam theory.
A beam problem under the pure bending moment is solved using five rectangular finite elements, as shown in the figure. Assume E = 200 GPa and ν = 0.3 are used. The thickness of the beam is 0.01 m. In order to simulate pure bending moment, two opposite forces F = ± 100,000 N are applied at the end of the beam. Using a commercial FE program, calculate strains in the beam along the bottom surface. Draw graphs of ε_xx and γ_xy with the x–axis being the beam length. Compare the numerical results with the elementary beam theory. Provide an explanation for the differences, if any. Is the rectangular element stiff or soft compared to the CST element?
Normally, a commercial finite element program provides stress and strain at the nodes of the element by averaging the stresses computed in the adjacent elements. Thus, you may use nodal displacement data from FE code to calculate strains along the bottom surface of the element. Calculate the strains at about ten points in each element for plotting purposes. Make sure that the commercial program uses the standard Lagrange shape function.

Repeat the above procedure when an upward vertical force of 200,000 N is applied at the tip of the beam. Use boundary conditions similar to the clamped boundary conditions of a cantilevered beam.