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Application of ANSYS to stress analysis

Abstract

Beams, plates, and shafts are fundamental component parts of machines and structures often encountered in daily life. These components must function properly and effectively, resisting mechanical loadings. The integrity of these parts must be ensured by adequate strength design to avoid excessive loads and deformation. The mechanics of materials can evaluate stresses and deformations induced in components of interest, but can merely cover simple problems like a slender bar subjected to bending, a plate under tension, a cylindrical bar twisted by torsional moment, etc. The finite element method, however, can make analyses of stresses and deformations in machines and structures in realistic forms subjected to complicated loadings.

The object of this chapter is to show how to carry out finite-element analyses of fundamental types of component parts such as beams, plates, shafts, etc. subjected to loadings typical for each type of component part. The validity of the results of the finite-element analyses is discussed by comparing the results obtained by experiments or by the theory of elasticity.

Keywords

Beam; Plate; Stepped bar; Bending; Tensile load; Torsion; Tensile stress; Shear stress; Circular hole; Elliptical hole; Fillet; Crack; Stress concentration; Stress intensity factor; Singularity of stress; Indentation; Hertz stress

3.1 Cantilever beam

Beams are important fundamental structural and/or machine elements; they are found in buildings and in bridges. Beams are also used as shafts in cars and trains (see Fig. 3.1), as wings in aircrafts, and as book shelves in bookstores. Arms and femurs of human beings and branches of trees are good examples of portions of living creatures that support their bodies like cantilever beams as illustrated in Fig. 3.2. Beams play important roles not only in inorganic but in organic structures.

Fig. 3.1 Modelling of an axle shaft by a simply supported beam.

Fig. 3.2 Modelling of an arm of a human being by a cantilever beam.

The mechanics of beams is one of the most important subjects in engineering.

3.1.1 Example problem: A cantilever beam

Perform an FEM analysis of a 2D cantilever beam shown in Fig. 3.3 and calculate the deflection of the beam at the loading point and the longitudinal stress distribution in the beam.

3.1.2 Problem description

Geometry: length l = 90 mm, height h = 5 mm, thickness b = 10 mm.
Material: mild steel having Young's modulus E = 210 GPa and Poisson's ratio ν = 0.3.
Boundary conditions: The beam is clamped to a rigid wall at the left end and loaded at x = 80 mm by a point load of P = 100 N.

3.1.3 Review of the solutions obtained by the elementary beam theory

Before proceeding to the FEM analysis of the beam, let us review the solutions to the example problem obtained by the elementary beam theory. The maximum deflection of the beam δ_max can be calculated by the following equation:

$δ_{\max} = \frac{{Pl}_{1}^{3}}{3 EI} (1 + \frac{3 l_{1}}{l_{2}})$

si2_e (3.1)

where l₁ (= 80 mm) is the distance of the application point of the load from the rigid wall and l₂ = l − l₁.

The maximum tensile stress σ_max(x) at x in the longitudinal direction appears at the upper surface of the beam in a cross section at x from the wall.

$σ_{\max} (x) = \{\begin{cases} \frac{P (l_{1} - x)}{I} \frac{h}{2} & (0 \leq x \leq l_{1}) \\ 0 & (0 \leq x) \end{cases}$

si3_e (3.2)

where l (= 90 mm) is the length, h (= 5 mm) is the height, b (= 10 mm) is the thickness, E (= 210 GPa) is Young's modulus, and I is the area moment of inertia of the cross section of the beam. For a beam having a rectangular cross section of a height h by a thickness b, the value of I can be calculated by the following equation:

$I = \frac{{bh}^{3}}{12}$

si4_e (3.3)

3.1.4 Analytical procedures

Fig. 3.4 shows how to make structural analyses by using FEM. In this chapter, the analytical procedures will be explained following the flowchart illustrated in Fig. 3.4.

3.1.4.1 Creation of an analytical model

(1) Creation of a beam shape to analyse

Here we shall analyse a rectangular slender beam of 5 mm (0.005 m) in height by 90 mm (0.09 m) in length by 10 mm (0.01 m) in width, as illustrated in Fig. 3.3. Fig. 3.5 shows the ANSYS Main Menu window where we can find layered command options imitating folders and files in the Microsoft Explorer folder window.

In order to prepare for creating the beam, the following operations should be made:

1. Click [A] Preprocessor to open its submenus in ANSYS Main Menu window.
2. Click [B] Modeling to open its submenus and select Create menu.
3. Click [C] Areas to open its submenus and select Rectangle menu.
4. Click to select the [D] By 2 corners menu.

After carrying out the operations above, a window called Rectangle by 2 corners appears, as shown in Fig. 3.6, for the input of the geometry of a 2D rectangular beam.

(2) Input of the beam geometry to analyse

The Rectangle by 2 corners window has four boxes for inputting the coordinates of the lower left corner point of the rectangular beam and the width and height of the beam to create. The following operations complete the creation procedure of the beam.

1. Input two 0s into [A] WP X and [B] WP Y to determine the lower left corner point of the beam on the Cartesian coordinates of the working plane.
2. Input 0.09 and 0.005 (m) into [C] Width and [D] Height, respectively, to determine the shape of the beam model.
3. Click the [E] OK button to create the rectangular area, or beam on the ANSYS Graphics window, as shown in Fig. 3.7.

Fig. 3.7 2D beam created and displayed on the ‘ANSYS Graphics’ window.

How to correct the shape of the model

When correcting the model, delete the area first, and repeat procedures (1) and (2) above. In order to delete the area, execute the following commands:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Delete → Area and Below

Then, the Delete Area and Below window opens and an upward arrow (↑) appears on the ANSYS Graphics window.

1. Move the arrow to the area to delete and click the left mouse button.
2. The colour of the area turns from light blue to pink.
3. Click the OK button and the area is deleted.

3.1.4.2 Input of the elastic properties of the beam material

Next, we specify elastic constants of the beam. In the case of isotropic material, the elastic constants are Young's modulus and Poisson's ratio. This procedure can be performed any time before the solution procedure—for instance, after setting boundary conditions. If this procedure is missed, we cannot perform the solution procedure.

[COMMAND] ANSYS Main Menu → Preprocessor → Material Props → Material Models

Then the Define Material Model Behavior window opens, as shown in Fig. 3.8.

1. Double-click the [A] Structural, Linear, Elastic and Isotropic buttons one after another.
2. Input the value of Young's modulus, 2.1e11 (Pa), and that of Poisson's ratio, 0.3, into [B] EX and [C] PRXY boxes, and click the [D] OK button of the Linear Isotropic Properties for Materials Number 1 as shown in Fig. 3.9.
3. Exit from the Define Material Model Behavior window by selecting Exit in [E] Material menu of the window (see Fig. 3.8).

Fig. 3.9 Input of elastic constants through the ‘Linear Isotropic Properties for Material Number 1’ window.

3.1.4.3 Finite-element discretisation of the beam area

Here we shall divide the beam area into finite elements. The procedures for finite-element discretisation are firstly to select the element type, secondly to input the element thickness and finally to divide the beam area into elements.

(1) Selection of the element type

[COMMAND] ANSYS Main Menu → Preprocessor → Element Type → Add/Edit/Delete

Then the Element Types window opens, as shown in Fig. 3.10.

1. Click the [A] Add … button to open the Library of Element Types window as shown in Fig. 3.11 and select the element type to use.
2. To select the 8-node isoparametric element, select [B] Structural Mass – Solid.
3. Select [C] Quad 8 node 82 and click the [D] OK button to choose the 8-node isoparametric element.
4. Click the [E] Options … button in the Element Types window as shown in Fig. 3.10 to open the PLANE82 element type options window, as depicted in Fig. 3.12. Select the [F] Plane strs w/thk item in the Element behavior box and click the [G] OK button to return to the Element Types window. Click the [H] Close button to close the window.

Fig. 3.11 ‘Library of Element Types’ window.

Fig. 3.12 ‘PLANE82 element type options’ window.

The 8-node isoparametric element is a rectangular element which has four corner nodal points and four middle points, as shown in Fig. 3.13, and can realise the finite element analysis with higher accuracy than the 4-node linear rectangular element. The beam area is divided into these 8-node rectangular #82 finite elements.

(2) Input of the element thickness

[COMMAND] ANSYS Main Menu → Preprocessor → Real Constants

Fig. 3.13 8-node isoparametric rectangular element.

Select [A] Real Constants in the ANSYS Main Menu as shown in Fig. 3.14.

1. Click the [A] Add/Edit/Delete button to open the Real Constants window as shown in Fig. 3.15 and click the [B] Add … button.
2. Then the Element Type for Real Constants window opens (see Fig. 3.16). Click the [C] OK button.
3. The Element Type for Real Constants window vanishes and the Real Constants Set Number 1. for PLANE82 window appears instead, as shown in Fig. 3.17. Input a plate thickness of 0.01 (m) in the [D] Thickness box and click the [E] OK button.
4. The Real Constants window returns with the display of the Defined Real Constants Sets box changed to Set 1, as shown in Fig. 3.18. Click the [F] Close button, which completes the operation of setting the plate thickness.
(3) Sizing of the elements

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Size Cntrls → Manual Size → Global → Size

Fig. 3.14 Setting of the element thickness from the real constant command.

Fig. 3.15 ‘Real Constants’ window before setting the element thickness.

Fig. 3.16 ‘Element Type for Real Constants’ window.

Fig. 3.17 ‘Real Constants Set Number 1. for PLANE82’ window.

Fig. 3.18 ‘Real Constants’ window after setting the element thickness.

The Global Element Sizes window opens, as shown in Fig. 3.19.

1. Input 0.002 (m) in the [A] SIZE box and click the [B] OK button.
- By the operations above, the element size of 0.002, i.e. 0.002 m or 2 mm, is specified and the beam of 5 mm by 90 mm is divided into rectangular finite elements with one side 2 mm and the other side 3 mm in length.
(4) Meshing

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Mesh → Areas → Free

Fig. 3.19 ‘Global Element Sizes’ window.

The Mesh Areas window opens, as shown in Fig. 3.20.

1. An upward arrow () appears in the ANSYS Graphics window. Move this arrow to the beam area and click this area to mesh.
2. The colour of the area turns from light blue to pink. Click the [A] OK button to see the area meshed by 8-node rectangular isoparametric finite elements, as shown in Fig. 3.21.

Fig. 3.21 Beam area subdivided into 8-node isoparametric rectangular elements.

How to modify meshing

In the case of modifying meshing, delete the elements, and repeat procedures (1)–(4) above. Repeat procedures (1), (2), or (3) to modify the element type, change the plate thickness without changing the element type, or change the element size only.

In order to delete the elements, execute the following commands:

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Clear → Areas

The Clear Areas window opens.

1. An upward arrow () appears in the ANSYS Graphics window. Move this arrow to the beam area and click this area.
2. The colour of the area turns from light blue to pink. Click the OK button to delete the elements from the beam area. After this operation, the area disappears from the display. Execute the following commands to replot the area.

[COMMAND] ANSYS Utility Menu → Plot → Areas

3.1.4.4 Input of boundary conditions

Here we shall impose constraint and loading conditions on nodes of the beam model. Display the nodes first to define the constraint and loading conditions.

(1) Display of nodes

[COMMAND] ANSYS Utility Menu → Plot → Nodes

The nodes are plotted in the ANSYS Graphics windows shown in Fig. 3.22.

(2) Zoom in the node display

An enlarged view of the models is often convenient when imposing constraint and loading conditions on the nodes. In order to zoom in the node display, execute the following commands:

[COMMAND] ANSYS Utility Menu → PlotCtrls → Pan Zoom Rotate …

The Pan-Zoom-Rotate window opens, as shown in Fig. 3.23.

1. Click [A] Box Zoom button.
2. The shape of the mouse cursor turns into a magnifying glass in the ANSYS Graphics window. Click the upper left point and then the lower right point, which enclose a portion of the beam area to enlarge, as shown in Fig. 3.24. Zoom in the left end of the beam.
3. In order to display the whole view of the beam, click the [B] Fit button in the Pan-Zoom-Rotate window.
(3) Definition of constraint conditions

Fig. 3.24 Magnification of an observation area.

Selection of nodes

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Displacement → On Nodes

The Apply U. ROT on Nodes window opens as shown in Fig. 3.25.

1. Select [A] Box button and drag the mouse in the ANSYS Graphics window so as to enclose the nodes on the left edge of the beam area with the yellow rectangular frame, as shown in Fig. 3.26. The Box button is selected to pick multiple nodes at once, whereas the [B] Single button is chosen to pick a single node.
2. After confirming that only the nodes to impose constraints on are selected, i.e. the nodes on the left edge of the beam area, click the [C] OK button.

Fig. 3.26 Picking multiple nodes by box.

How to reselect nodes: Click [D] Reset button to clear the selection of the nodes before clicking the [C] OK button in procedure (2) above, and repeat procedures (1) and (2) above. The selection of nodes can be reset either by picking selected nodes after choosing the [E] Unpick button or clicking the right mouse button to turn the upward arrow upside down.

Imposing constraint conditions on nodes

The Apply U. ROT on Nodes window (see Fig. 3.27) opens after clicking the [C] OK button in procedure (2) in Section 3.1.

1. In case of selecting [A] ALL DOF, the nodes are to be clamped, i.e. the displacements are set to zero in the directions of the x- and y-axes. Similarly, the selection of UX makes the displacement in the x-direction equal to zero and the selection of UY makes the displacement in the y-direction equal to zero.
2. Click the [B] OK button and blue triangular symbols which denote the clamping conditions appear in the ANSYS Graphics window, as shown in Fig. 3.28. The upright triangles indicate that each node to which the triangular symbol is attached is fixed in the x-direction, whereas the tilted triangles indicate the fixed condition in the y-direction.

Fig. 3.28 Imposing the clamping conditions on nodes.

How to clear constraint conditions

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Delete →Structural → Displacement → On Nodes

The Delete Node Constrai… window opens.

1. Click the Pick All button to delete the constraint conditions of all the nodes on which the constraint conditions are imposed. Select the Single button and pick a particular node by the upward arrow in the ANSYS Graphics window and click the OK button.
2. The Delete Node constraints window appears. Select ALL DOF and click the OK button to delete the constraint conditions both in the x- and y-directions. Select UX and UY to delete the constraints in the x- and y-directions, respectively.
(4) Imposing boundary conditions on nodes

Before imposing load conditions, click the Fit button in the Pan-Zoom-Rotate window (see Fig. 3.23) to get the whole view of the area and then zoom in the right end of the beam area for ease of the following operations.

Selection of the nodes

1. Pick the node at a point where x = 0.08 m and y = 0.005 m. For this purpose, click

[COMMAND] ANSYS Utility Menu → PlotCtrls → Numbering …

consecutively to open the Plot Numbering Controls window, as shown in Fig. 3.29.

2. Click the [A] NODE Off box to change it to On.
3. Click the [B] OK button to display node numbers adjacent to corresponding nodes in the ANSYS Graphics window, as shown in Fig. 3.30.
4. To delete node numbers, click the [A] NODE On box again to change it to Off.
5. Execute the following commands:

[COMMAND] ANSYS Utility Menu → List → Nodes …

Fig. 3.29 ‘Plot Numbering Controls’ window.

Fig. 3.30 Nodes and nodal numbers displayed on the results displaying window.

and the Sort NODE Listing window opens (see Fig. 3.31). Select the [A] Coordinates only button and then click the [B] OK button.

6. The NLIST Command window opens as shown in Fig. 3.32. Find the number of the node having the coordinates x = 0.08 m and y = 0.005 m, namely node #108 in Fig. 3.32.
7. Execute the following commands:

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Force/Moment → On Nodes

Fig. 3.32 ‘NLIST Command’ window showing the coordinates of the nodes; the framed portion indicates the coordinates of the node for load application.

to open the Apply F/M on Nodes window (see Fig. 3.33).

8. Pick only the #108 node having the coordinates x = 0.08 m and y = 0.005 m with the upward arrow, as shown in Fig. 3.34.
9. After confirming that only the #108 node is enclosed with the yellow frame, click the [A] OK button in the Apply F/M on Nodes window.

Fig. 3.34 Selection of a node for load application.

How to cancel the selection of the nodes of load application: Click the Reset button before clicking the OK button or click the right mouse button to change the upward arrow to the downward arrow and click the yellow frame. The yellow frame disappears and the selection of the node(s) of load application is cancelled.

Imposing load conditions on nodes

Click [A] OK in the Apply F/M on Nodes window to open another Apply F/M on Nodes window, as shown in Fig. 3.35.

1. Choose [A] FY in the Lab Direction of force/mom box and input [B] -100 in the VALUE box. A positive value for load indicates load in the upward or rightward direction, whereas a negative value load in the downward or leftward direction.
2. Click the [C] OK button to display the downward arrow attached to the #108 node indicating the downward load applied to that point, as shown in Fig. 3.36.

Fig. 3.36 Displaying the load application on a node by arrow symbol.

How to delete load conditions: Execute the following commands:

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Delete →Structural → Force/Moment → On Nodes

to open the Delete F/M on Nodes window (see Fig. 3.37). Choose [A] FY or ALL in the Lab Force/moment to be deleted and click the OK button to delete the downward load applied to the #108 node.

3.1.4.5 Solution procedures

[COMMAND] ANSYS Main Menu → Solution → Solve → Current LS

The Solve Current Load Step and /STATUS Command windows appear as shown in Figs 3.38 and 3.39, respectively.

1. Click the [A] OK button in the Solve Current Load Step window, as shown in Fig. 3.38, to begin the solution of the current load step.
2. The /STATUS Command window displays information on solution and load step options. Select the [B] File button to open the submenu and select the Close button to close the /STATUS Command window.
3. When the solution is completed, the Note window (see Fig. 3.40) appears. Click the [C] Close button to close the window.

Fig. 3.38 ‘Solve Current Load Step’ window.

3.1.4.6 Graphical representation of the results

(1) Contour plot of displacements

[COMMAND] ANSYS Main Menu → General Postproc → Plot Results → Contour Plot → Nodal Solution

The Contour Nodal Solution Data window opens, as shown in Fig. 3.41.

1. Select [A] DOF Solution and [B] Y-Component of displacement.
2. Click the [C] OK button to display the contour of the y-component of displacement, or the deflection of the beam in the ANSYS Graphics window (see Fig. 3.42). The DMX value shown in the Graphics window indicates the maximum deflection of the beam.
3. Select [D] Deformed shape with undeformed edge in the Undisplaced shape key box to compare the shapes of the beam before and after deformation.
(2) Contour plot of stresses
1. Select [A] Stress and [B] X-Component of stress, as shown in Fig. 3.43.
2. Click the [C] OK button to display the contour of the x-component of stress, or the bending stress in the beam in the ANSYS Graphics window (see Fig. 3.44). The SMX and SMN values shown in the Graphics window indicate the maximum and the minimum stresses in the beam, respectively.
3. Click the [D] Additional Options bar to open additional option items to choose. Select the [E] All applicable in the Number of facets per element edge box to calculate stresses and strains at middle points of the elements.

Fig. 3.42 Contour map representation of the distribution of displacement in the y- or vertical direction.

Fig. 3.43 ‘Contour Nodal Solution Data’ window.

Fig. 3.44 Contour map representation of the distribution of normal stress in the x- or horizontal direction.

3.1.5 Comparison of FEM results with experimental ones

Fig. 3.45 compares longitudinal stress distributions obtained by ANSYS with those by experiments and by elementary beam theory. The results obtained by three different methods agree well with one another. As the applied load increases, however, errors among the three groups of results become larger, especially at the clamped end. This tendency arises from the fact that the clamped condition can be hardly realised in the strict sense.

3.1.6 Problems to solve

Problem 3.1

Change the point of load application and the intensity of the applied load in the cantilever beam model shown in Fig. 3.3, and calculate the maximum deflection.

Problem 3.2

Calculate the maximum deflection in a beam clamped at the both ends as shown in Fig. 3.46, where the thickness of the beam in the direction perpendicular to the page surface is 10 mm.

(Answer: 0.00337 mm)

Problem 3.3

Calculate the maximum deflection in a beam simply supported at the both ends, as shown in Fig. 3.47, where the thickness of the beam in the direction perpendicular to the page surface is 10 mm.

(Answer: 0.00645 mm)

Problem 3.4

Calculate the maximum deflection in a beam shown in Fig. 3.48 where the thickness of the beam in the direction perpendicular to the page surface is 10 mm. Choose an element size of 1 mm.

(Answer: 0.00337 mm)

Note that the beam shown in Fig. 3.46 is bilaterally symmetric so that the x-component of the displacement (DOFX) is zero at the centre of the beam span. If the beam shown in Fig. 3.46 is cut at the centre of the span and the finite element calculation is made for only the left half of the beam by applying a half load of 50 N to its right end which is fixed in the x-direction but is deformed freely in the y-direction as shown in Fig. 3.48, the solution obtained is the same as that for the left half of the beam in Problem 3.2. This problem can be solved by its half model as shown in Fig. 3.48. A half model can achieve the efficiency of finite element calculations.

Problem 3.5

Calculate the maximum deflection in a beam shown in Fig. 3.49, where the thickness of the beam in the direction perpendicular to the page surface is 10 mm. This beam is the half model of the beam of Problem 3.3.

(Answer: 0.00645 mm)

Problem 3.6

Calculate the maximum value of the von Mises stress in the stepped beam, as shown in Fig. 3.50, where Young's modulus E = 210 GPa, Poisson's ratio ν = 0.3, the element size is 2 mm, and the beam thickness is 10 mm. Refer to the Appendix to create the stepped beam. The von Mises stress σ_eq is sometimes called the equivalent stress or the effective stress, and is expressed by the following formula:

$σ_{eq} = \frac{1}{\sqrt{2}} \sqrt{{(σ_{x} - σ_{y})}^{2} + {(σ_{y} - σ_{z})}^{2} + {(σ_{z} - σ_{x})}^{2} + 6 (τ_{xy}^{2} + τ_{yz}^{2} + τ_{zx}^{2})}$

si5_e (3.4)

in three-dimensional elasticity problems. It is often considered that a material yields when the value of the von Mises stress reaches the yield stress of the material σ_Y, which is determined by the uniaxial tensile tests of the material.

(Answer: 40.8 MPa)

Problem 3.7

Calculate the maximum value of the von Mises stress in the stepped beam with a rounded fillet, as shown in Fig. 3.51, where Young's modulus E = 210 GPa, Poisson's ratio ν = 0.3, the element size is 2 mm, and the beam thickness is 10 mm. Refer to the Appendix to create the stepped beam.

(Answer: 30.2 MPa)

Appendix: Procedures for creating stepped beams

A.1 Creation of a stepped beam

A stepped beam as shown in Fig. 3.50 can be created by adding two rectangular areas of different sizes:

1. Create two rectangular areas of different sizes, say 50 mm by 20 mm and 60 mm by 10 mm, following operations described in Section 3.1.4.1.
2. Select the Boolean operation of adding areas as follows to open the Add Areas window (see Fig. 3.52):

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Operate →Booleans → Add → Areas

3. Pick all the areas to add by the upward arrow.
4. The colour of the areas picked turns from light blue to pink (see Fig. 3.53). Click the [A] OK button to add the two rectangular areas to create a stepped beam area, as shown in Fig. 3.54.

Fig. 3.53 Two rectangular areas of different sizes to be added to create a stepped beam area.

Fig. 3.54 A stepped beam area created by adding two rectangle areas.

A.1.1 How to cancel the selection of areas

Click the Reset button or click the right mouse button to change the upward arrow to the downward arrow and click the area(s) picked. The colour of the unpicked area(s) turns pink to light blue and the selection of the area(s) is cancelled.

A.2 Creation of a stepped beam with a rounded fillet

A stepped beam with a rounded fillet, as shown in Fig. 3.51, can be created by subtracting a smaller rectangular area and a solid circle from a larger rectangular area:

1. Create a larger rectangular area of 100 mm by 20 mm, a smaller rectangular area of, say 50 mm by 15 mm, and a solid circular area having a diameter of 10 mm as shown in Fig. 3.55. The solid circular area can be created by executing the following operation:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Areas → Circle → Solid Circle

Fig. 3.55 A larger rectangular area, a smaller rectangular area and a solid circular area to create a stepped beam area with a rounded fillet.

to open the Solid Circular Area window. Input the coordinate of the centre ([A] WP X, [B] WP Y) and [C] Radius of the solid circle, and click the [D] OK button as shown in Fig. 3.56.

2. Select the Boolean operation of subtracting areas as follows to open the Subtract Areas window (see Fig. 3.57):

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Operate →Booleans → Subtract → Areas

3. Pick the larger rectangular area by the upward arrow as shown in Fig. 3.58 and the colour of the area picked turns from light blue to pink. Click the [A] OK button.
4. Pick the smaller rectangular area and the solid circular area by the upward arrow as shown in Fig. 3.59 and the colour of the two areas picked turns from light blue to pink. Click the [A] OK button to subtract the smaller rectangular and solid circular areas from the larger rectangular area to create a stepped beam area with a rounded fillet, as shown in Fig. 3.60.

Fig. 3.58 A larger rectangular area picked.

Fig. 3.59 A smaller rectangular area and a solid circular area picked to be subtracted from a larger rectangular area to create a stepped beam area with a rounded fillet.

Fig. 3.60 A stepped cantilever beam area with a rounded fillet.

A.2.1 How to display area numbers

Area numbers can be displayed in the ANSYS Graphics window by the following procedure.

[Commands] ANSYS Utility Menu → PlotCtrls → Numbering …

1. The Plot Numbering Controls window opens as shown in Fig. 3.29.
2. Click the AREA Off box to change it to On.
3. Click the OK button to display area numbers in corresponding areas in the ANSYS Graphics window.
4. To delete element numbers, click the AREA On box again to change it to Off.

3.2 The principle of St. Venant

3.2.1 Example problem

An elastic strip is subjected to distributed uniaxial tensile stress or negative pressure at one end and clamped at the other end.

Perform an FEM analysis of a 2D elastic strip subjected to a distributed stress in the longitudinal direction at one end and clamped at the other end, as shown in Fig. 3.61, and calculate the stress distributions along the cross sections at different distances from the loaded end in the strip.

Fig. 3.61 A 2D elastic strip subjected to a distributed force in the longitudinal direction at one end and clamped at the other end.

3.2.2 Problem description

Geometry: length l = 200 mm, height h = 20 mm, thickness b = 10 mm.
Material: mild steel having Young's modulus E = 210 GPa and Poisson's ratio ν = 0.3.
Boundary conditions: The elastic strip is subjected to a triangular distribution of stress in the longitudinal direction at the right end and clamped to a rigid wall at the left end.

3.2.3 Analytical procedures

3.2.3.1 Creation of an analytical model

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Areas → Rectangle → By 2 Corners

1. Input two 0s into the WP X and WP Y boxes in the Rectangle by 2 Corners window to determine the lower left corner point of the elastic strip on the Cartesian coordinates of the working plane.
2. Input 200 and 20 (mm) into the Width and Height boxes, respectively, to determine the shape of the elastic strip model.
3. Click the OK button to create the rectangular area, or beam on the ANSYS Graphics window.

In the procedures above, the geometry of the strip is input in millimetres. You must decide what kind of units to use in finite element analyses. When you input the geometry of a model to analyse in millimetres, for example, you must input applied loads in N (Newton) and Young's modulus in MPa, since 1 MPa is equivalent to 1 N/mm². When you use metres and N as the units of length and load, respectively, you must input Young's modulus in Pa, since 1 Pa is equivalent to 1 N/m². You can choose any system of unit you would like to, but your unit system must be consistent throughout the analyses.

3.2.3.2 Input of the elastic properties of the strip material

[COMMAND] ANSYS Main Menu → Preprocessor → Material Props → Material Models

1. The Define Material Model Behavior window opens.
2. Double-click the Structural, Linear, Elastic, and Isotropic buttons one after another.
3. Input the value of Young's modulus, 2.1e5 (MPa), and that of Poisson's ratio, 0.3, into the EX and PRXY boxes, and click the OK button of the Linear Isotropic Properties for Materials Number 1 window.
4. Exit from the Define Material Model Behavior window by selecting Exit in the Material menu of the window.

3.2.3.3 Finite-element discretisation of the strip area

(1) Selection of the element type

[COMMAND] ANSYS Main Menu → Preprocessor → Element Type → Add/Edit/Delete

1. The Element Types window opens.
2. Click the Add … button in the Element Types window to open the Library of Element Types window and select the element type to use.
3. Select Structural Mass – Solid and Quad 8 node 82.
4. Click the OK button in the Library of Element Types window to use the 8-node isoparametric element.
5. Click the Options … button in the Element Types window to open the PLANE82 element type options window. Select the Plane strs w/thk item in the Element behavior box and click the OK button to return to the Element Types window. Click the Close button in the Element Types window to close the window.

(2) Input of the element thickness

[COMMAND] ANSYS Main Menu → Preprocessor → Real Constants → Add/Edit/Delete

1. The Real Constants window opens.
2. Click [A] Add/Edit/Delete button to open the Real Constants window and click the Add … button.
3. The Element Type for Real Constants window opens. Click the OK button.
4. The Element Type for Real Constants window vanishes and the Real Constants Set Number 1. for PLANE82 window appears instead. Input a strip thickness of 10 mm in the Thickness box and click the OK button.
5. The Real Constants window returns with the display of the Defined Real Constants Sets box changed to Set 1. Click the Close button.
(3) Sizing of the elements

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Size Cntrls →Manual Size → Global → Size

1. The Global Element Sizes window opens.
2. Input 2 in the SIZE box and click the OK button.
(4) Dividing the right-end side of the strip area into two lines

Before proceeding to meshing, the right-end side of the strip area must be divided into two lines for imposing the triangular distribution of the applied stress or stress by executing the following commands.

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Operate →Booleans → Divide → Lines w/ Options

1. The Divide Multiple Lines … window opens as shown in Fig. 3.62.
2. When the mouse cursor is moved to the ANSYS Graphics window, an upward arrow () appears.
3. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the right-end side of the strip area and click the left mouse button.
4. Click the OK button in the Divide Multiple Lines … window to display the Divide Multiple Lines with Options window, as shown in Fig. 3.63.
5. Input 2 in the [A] NDIV box and 0.5 in the [B] RATIO box, and select Be modified in the [C] KEEP box.
6. Click the [D] OK button.
(5) Meshing

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Mesh →Areas → Free

1. The Mesh Areas window opens.
2. The upward arrow appears in the ANSYS Graphics window. Move this arrow to the elastic strip area and click this area.
3. The colour of the area turns from light blue to pink. Click the OK button to see the area meshed by 8-node rectangular isoparametric finite elements.

Fig. 3.62 ‘Divide Multiple Lines …’ window.

Fig. 3.63 ‘Divide Multiple Lines with Options’ window.

3.2.3.4 Input of boundary conditions

(1) Imposing constraint conditions on the left end of the strip

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Displacement → On Lines

1. The Apply U. ROT on Lines window opens and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the left-end side of the strip area and click the left mouse button.
3. Click the OK button in the Apply U. ROT on Lines window to display another Apply U. ROT on Lines window.
4. Select ALL DOF in the Lab2 box and click the OK button in the Apply U. ROT on Lines window.
(2) Imposing a triangular distribution of applied stress on the right end of the strip

Distributed load or stress can be defined by pressure on lines and the triangular distribution of applied load can be defined as the composite of two linear distributions of pressure which are antisymmetric to each other with respect to the centre line of the strip area.

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Pressure → On Lines

1. The Apply PRES on Lines window opens (see Fig. 3.64) and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the upper line of the right-end side of the strip area and click the left mouse button. Remember that the right-end side of the strip area was divided into two lines in procedure (4) in Section 3.2.3.3.
3. Another Apply PRES on Lines window opens (see Fig. 3.65). Select a Constant value in the [A] [SFL] Apply PRES on lines as a box and input [B] -10 (MPa) in the VALUE Load PRES value box and [C] 0 (MPa) in the Value box.
4. Click the [D] OK button in the window to define a linear distribution of pressure on the upper line which is zero at the upper right corner and − 10 MPa at the centre of the right-end side of the strip area (see Fig. 3.66).
5. For the lower line of the right-end side of the strip area, repeat the commands above and procedures (2)–(4).
6. Select a Constant value in the [A] [SFL] Apply PRES on lines as a box and input [B] 0 (MPa) in the VALUE Load PRES value box and [C] -10 (MPa) in the Value box as shown in Fig. 3.67. Note that the values to input in the lower two boxes in the Apply PRES on Lines window is interchanged, since the distributed pressure on the lower line of the right-end side of the strip area is antisymmetric to that on the upper line.
7. Click the [D] OK button in the window shown in Fig. 3.66 to define a linear distribution of pressure on the lower line, which is − 10 MPa at the centre and zero at the lower right corner of the right-end side of the strip area, as shown in Fig. 3.68.

Fig. 3.64 ‘Apply PRES on Lines’ window for picking the lines to which pressure is applied.

Fig. 3.65 ‘Apply PRES on Lines’ window for applying linearly distributed pressure to the upper half of the right end of the elastic strip.

Fig. 3.66 Linearly distributed negative pressure applied to the upper half of the right-end side of the elastic strip.

Fig. 3.67 ‘Apply PRES on Lines’ window for applying linearly distributed pressure to the lower half of the right end of the elastic strip.

Fig. 3.68 Triangular distribution of pressure applied to the right end of the elastic strip.

3.2.3.5 Solution procedures

[COMMAND] ANSYS Main Menu → Solution → Solve → Current LS

1. The Solve Current Load Step and /STATUS Command windows appear.
2. Click the OK button in the Solve Current Load Step window to begin the solution of the current load step.
3. Select the File button in /STATUS Command window to open the submenu and select the Close button to close the /STATUS Command window.
4. When the solution is completed, the Note window appears. Click the Close button to close the Note window.

3.2.3.6 Contour plot of stress

[COMMAND] ANSYS Main Menu → General Postproc → Plot Results → Contour Plot → Nodal Solution

1. The Contour Nodal Solution Data window opens.
2. Select Stress and X-Component of stress.
3. Click the OK button to display the contour of the x-component of stress in the elastic strip in the ANSYS Graphics window, as shown in Fig. 3.69.

Fig. 3.69 Contour of the x-component of stress in the elastic strip showing uniform stress distribution at one width or larger distance from the right end of the elastic strip to which triangular distribution of pressure is applied.

3.2.4 Discussion

Fig. 3.70 shows the variations of the longitudinal stress distribution in the cross section with the x-position of the elastic strip. At the right end of the strip, or at x = 200 mm, the distribution of the applied longitudinal stress takes the triangular shape, which is zero at the upper and lower corners and 10 MPa at the centre of the strip. The longitudinal stress distribution varies as the distance of the cross section from the right end of the strip increases, and the distribution becomes almost uniform at x = 180 mm, i.e. at one width distance from the end of the stress application. The total amount of stress in any cross section is the same, i.e. 2 kN in the strip and at any cross section at one width or larger distance from the end of the stress application, stress is uniformly distributed and the magnitude of stress becomes 5 MPa.

The above result is known as the principle of St Venant and is very useful in practice, in the design of structural components. Even if the stress distribution is very complicated at the loading points due to the complicated shape of load transfer equipment, one can assume a uniform stress distribution in the main parts of structural components or machine elements at some distance from the load transfer equipment.

3.3 Stress concentration due to elliptic holes and inclusions

3.3.1 Example problem

An elastic plate with an elliptic hole in its centre subjected to uniform longitudinal tensile stress σ₀ at one end and clamped at the other end. Perform an FEM analysis of a 2D elastic plate with an elliptic hole in its centre subjected to a uniform tensile stress σ₀ in the longitudinal direction at one end and clamped at the other end, as shown in Fig. 3.71, and calculate the maximum longitudinal stress σ_max in the plate. Calculate the stress concentration factor α = σ_max/σ₀ and observe the variation of the longitudinal stress distribution in the ligament between the foot of the hole and the edge of the plate.

Fig. 3.71 A 2D elastic plate with an elliptic hole in its centre subjected to a uniform longitudinal stress at one end and clamped at the other end.

3.3.2 Problem description

Plate geometry: length l = 400 mm, height h = 100 mm, thickness b = 10 mm.
Material: mild steel having Young's modulus E = 210 GPa and Poisson's ratio ν = 0.3.
Elliptic hole: An elliptic hole has a minor radius of 5 mm in the longitudinal direction and a major radius of 10 mm in the transversal direction.
Boundary conditions: The elastic plate is subjected to a uniform tensile stress in the longitudinal direction at the right end and clamped to a rigid wall at the left end.

3.3.3 Analytical procedures

3.3.3.1 Creation of an analytical model

Let us use a quarter model of the elastic plate with an elliptic hole as illustrated in Fig. 3.71, since the plate is symmetric about the horizontal and vertical centre lines. The quarter model can be created by a slender rectangular area from which an elliptic area is subtracted by using the Boolean operation described in Section A.1.

First, create the rectangular area by the following operation:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Areas → Rectangle → By 2 Corners

1. Input two 0s into the WP X and WP Y boxes in the Rectangle by 2 Corners window to determine the lower left corner point of the elastic plate on the Cartesian coordinates of the working plane.
2. Input 200 and 50 (mm) into the Width and Height boxes, respectively, to determine the shape of the elastic plate model.
3. Click the OK button to create the elastic plate on the ANSYS Graphics window.

Then, create a circular area with a diameter of 10 mm, and reduce its diameter in the longitudinal direction to half of the original value to obtain the elliptic area. The following commands create a circular area by designating the coordinates (UX, UY) of the centre and the radius of the circular area:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Areas → Circle → Solid Circle

1. The Solid Circular Area window opens as shown in Fig. 3.72.
2. Input two 0s into the [A] WP X and [B] WP Y boxes to determine the centre position of the circular area.
3. Input [C] 10 (mm) in the Radius box to determine the radius of the circular area.
4. Click the [D] OK button to create the circular area superimposed on the rectangular area in the ANSYS Graphics window, as shown in Fig. 3.73.

Fig. 3.73 Circular area superimposed on the rectangular area.

In order to reduce the diameter of the circular area in the longitudinal direction to half of the original value, use the following Scale → Areas operation:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Operate → Scale → Areas

1. The Scale Areas window opens, as shown in Fig. 3.74.
2. The upward arrow appears in the ANSYS Graphics window. Move the arrow to the circular area and select it by clicking the left mouse button. The colour of the circular area turns from light blue to pink. Click the [A] OK button.
3. The colour of the circular area turns to light blue and another Scale Areas window opens, as shown in Fig. 3.75.
4. Input [A] 0.5 in the RX box, select [B] Areas only in the NOELEM box and select the [C] Moved in IMOVE box.
5. Click the [D] OK button. An elliptic area appears and the circular area still remains. The circular area is an afterimage and does not exist in reality. To erase this afterimage, perform the following commands:

[COMMAND] ANSYS Utility Menu → Plot → Replot

The circular area vanishes. Subtract the elliptic area from the rectangular area in a similar manner as described in Section A.1, i.e.

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Operate →Booleans → Subtract → Areas

1. Pick the rectangular area by the upward arrow and confirm that the colour of the area picked turns from light blue to pink. Click the OK button.
2. Pick the elliptic area by the upward arrow and confirm that the colour of the elliptic area picked turns from light blue to pink. Click the OK button to subtract the elliptic area from the rectangular area to get a quarter model of a plate with an elliptic hole in its centre, as shown in Fig. 3.76.

Fig. 3.76 Quarter model of a plate with an elliptic hole in its centre created by subtracting an elliptic area from a rectangular area.

3.3.3.2 Input of the elastic properties of the plate material

[COMMAND] ANSYS Main Menu → Preprocessor → Material Props → Material Models

1. The Define Material Model Behavior window opens.
2. Double-click the Structural, Linear, Elastic, and Isotropic buttons one after another.
3. Input the value of Young's modulus, 2.1e5 (MPa), and that of Poisson's ratio, 0.3, into the EX and PRXY boxes, and click the OK button of the Linear Isotropic Properties for Materials Number 1 window.
4. Exit from the Define Material Model Behavior window by selecting Exit in the Material menu of the window.

3.3.3.3 Finite-element discretisation of the quarter plate area

(1) Selection of the element type

[COMMAND] ANSYS Main Menu → Preprocessor → Element Type → Add/Edit/Delete

1. The Element Types window opens.
2. Click the Add … button in the Element Types window to open the Library of Element Types window and select the element type to use.
3. Select Structural Mass – Solid and Quad 8 node 82.
4. Click the OK button in the Library of Element Types window to use the 8-node isoparametric element.
5. Click the Options … button in the Element Types window to open the PLANE82 element type options window. Select the Plane strs w/thk item in the Element behavior box and click the OK button to return to the Element Types window. Click the Close button in the Element Types window to close the window.
(2) Input of the element thickness

[COMMAND] ANSYS Main Menu → Preprocessor → Real Constants → Add/Edit/Delete

1. The Real Constants window opens.
2. Click the [A] Add/Edit/Delete button to open the Real Constants window and click the Add … button.
3. The Element Type for Real Constants window opens. Click the OK button.
4. The Element Type for Real Constants window vanishes and the Real Constants Set Number 1. for PLANE82 window appears instead. Input a strip thickness of 10 mm in the Thickness box and click the OK button.
5. The Real Constants window returns with the display of the Defined Real Constants Sets box changed to Set 1. Click the Close button.
(3) Sizing of the elements

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Size Cntrls → Manual Size → Global → Size

1. The Global Element Sizes window opens.
2. Input 1.5 in the SIZE box and click the OK button.
(4) Meshing

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Mesh →Areas → Free

1. The Mesh Areas window opens.
2. The upward arrow appears in the ANSYS Graphics window. Move this arrow to the quarter plate area and click this area.
3. The colour of the area turns from light blue to pink. Click the OK button to see the area meshed by 8-node isoparametric finite elements, as shown in Fig. 3.77.

Fig. 3.77 Quarter model of a plate with an elliptic hole meshed by 8-node isoparametric finite elements.

3.3.3.4 Input of boundary conditions

(1) Imposing constraint conditions on the left end and the bottom side of the quarter plate model

Due to the symmetry, the constraint conditions of the quarter plate model are UX-fixed condition on the left end and UY-fixed condition on the bottom side of the quarter plate model. Apply the constraint conditions onto the corresponding lines using the following commands:

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Displacement → On Lines

1. The Apply U. ROT on Lines window opens and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the left-end side of the quarter plate area and click the left mouse button.
3. Click the OK button in the Apply U. ROT on Lines window to display another Apply U. ROT on Lines window.
4. Select UX in the Lab2 box and click the OK button in the Apply U. ROT on Lines window.

Repeat the commands and operations (1)–(3) above for the bottom side of the model. Then, select UY in the Lab2 box and click the OK button in the Apply U. ROT on Lines window.

(2) Imposing a uniform longitudinal stress on the right end of the quarter plate model

A uniform longitudinal stress can be defined by pressure on the right-end side of the plate model as follows:

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Pressure → On Lines

1. The Apply PRES on Lines window opens and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the right-end side of the quarter plate area and click the left mouse button.
3. Another Apply PRES on Lines window opens. Select Constant value in the [SFL] Apply PRES on lines as a box and input -10 (MPa) in the VALUE Load PRES value box and leave a blank in the Value box.
4. Click the OK button in the window to define a uniform tensile stress of 10 MPa applied to the right end of the quarter plate model.

Fig. 3.78 illustrates the boundary conditions applied to the centre notched plate model by the above operations.

3.3.3.5 Solution procedures

[COMMAND] ANSYS Main Menu → Solution → Solve → Current LS

1. The Solve Current Load Step and /STATUS Command windows appear.
2. Click the OK button in the Solve Current Load Step window to begin the solution of the current load step.
3. Select the File button in the /STATUS Command window to open the submenu and select the Close button to close the /STATUS Command window.
4. When the solution is completed, the Note window appears. Click the Close button to close the Note window.

3.3.3.6 Contour plot of stress

[COMMAND] ANSYS Main Menu → General Postproc → Plot Results → Contour Plot → Nodal Solution

1. The Contour Nodal Solution Data window opens.
2. Select Stress and X-Component of stress.
3. Click the OK button to display the contour of the x-component of stress in the quarter model of the centre notched plate in the ANSYS Graphics window, as shown in Fig. 3.79.

Fig. 3.79 Contour of the x-component of stress in the quarter model of the centre notched plate.

3.3.3.7 Observation of the variation of the longitudinal stress distribution in the ligament region

In order to observe the variation of the longitudinal stress distribution in the ligament between the foot of the hole and the edge of the plate, carry out the following commands:

[COMMAND] ANSYS Utility Menu → PlotCtrls → Symbols …

The Symbols window opens, as shown in Fig. 3.80.

1. Select [A] All Reactions in the [/PBC] Boundary condition symbol buttons, [B] Pressures in the [/PSF] Surface Load Symbols box, and [C] Arrows in the [/PSF] Show pres and convect as box.
2. The distributions of the reaction force and the longitudinal stress in the ligament region (see [A] in Fig. 3.81) as well as that of the applied stress on the right end of the plate area are superimposed on the contour x-component of stress in the plate in the ANSYS Graphics window. The reaction force is indicated by the leftward red arrows, and the longitudinal stress are shown by the rightward red arrows in the ligament region.

Fig. 3.81 Applied stress on the right-end side of the plate and resultant reaction force and longitudinal stress in its ligament region.

The longitudinal stress reaches its maximum value at the foot of the hole and is decreased approaching to a constant value almost equal to the applied stress σ₀ at some distance, say, about one major diameter distance, from the foot of the hole. This tendency can be explained by the principle of St Venant, as discussed in the previous section.

3.3.4 Discussion

If an elliptical hole is placed in an infinite body and subjected to a uniform stress in a remote uniform stress of σ₀, the maximum stress σ_max occurs at the foot of the hole, i.e. Point B in Fig. 3.71, and is expressed by the following formula:

$σ_{\max} = (1 + 2 \frac{a}{b}) σ_{0} = (1 + \sqrt{\frac{a}{ρ}}) σ_{0} = α_{0} σ_{0}$

si6_e (3.5)

where a is the major radius, b is the minor radius, ρ is the local radius of curvature at the foot of the elliptic hole, and α₀ is the stress concentration factor, which is defined as

$α_{0} \equiv \frac{σ_{\max}}{σ_{0}} = (1 + 2 \frac{a}{b}) = (1 + \sqrt{\frac{a}{ρ}})$

si7_e (3.6)

The stress concentration factor α₀ varies inversely proportional to the aspect ratio of elliptic hole b/a, namely the smaller the value of the aspect ratio b/a or the radius of curvature ρ becomes, the larger the value of the stress concentration factor α₀ becomes.

In a finite plate, the maximum stress at the foot of the hole is increased due to the finite boundary of the plate. Fig. 3.82 shows the variation of the stress concentration factor α for elliptic holes having different aspect ratios with normalised major radius 2a/h, indicating that the stress concentration factor in a plate with finite width α is increased dramatically as the ligament between the foot of the hole and the plate edge becomes smaller.

From Fig. 3.82, the value of the stress intensity factor for the present elliptic hole is approximately 5.16, whereas Fig. 3.81 shows that the maximum value of the longitudinal stress obtained by the present FEM calculation is approximately 49.3 MPa, i.e. the value of the stress concentration factor is approximately 49.3/10 = 4.93. Hence, the relative error of the present calculation is approximately (4.93 − 5.16)/5.16 ≈ − 0.0446 = 4.46% which may be reasonably small.

3.3.5 Problems to solve

Problem 3.8

Calculate the value of stress concentration factor for the elliptic hole shown in Fig. 3.71 by using the whole model of the plate, and compare the result with that obtained and discussed in Section 3.3.4.

Problem 3.9

Calculate the values of stress concentration factor α for circular holes for different values of the normalised major radius 2a/h and plot the results as the α vs 2a/h diagram, as shown in Fig. 3.80.

Problem 3.10

Calculate the values of stress concentration factor α for elliptic holes having different aspect ratios b/a for different values of the normalised major radius 2a/h, and plot the results as the α vs 2a/h diagram.

Problem 3.11

For smaller values of 2a/h, the disturbance of stress in the ligament between the foot of a hole and the plate edge due to the existence of the hole is decreased and stress in the ligament approaches to a constant value equal to the remote stress σ₀ at some distance from the foot of the hole (remember the principle of St Venant in the previous section). Find at which distance from the foot of the hole the stress in the ligament region can be considered to be almost equal to the value of the remote stress σ₀.

3.4 Stress singularity problem

3.4.1 Example problem

An elastic plate with a crack of length 2a in its centre is subjected to uniform longitudinal tensile stress σ₀ at one end and clamped at the other end, as illustrated in Fig. 3.83. Perform an FEM analysis of the 2D elastic plate having a crack of length 2a in its centre subjected to a uniform tensile stress σ₀ in the longitudinal direction at one end and clamped at the other end, and calculate the value of the mode I (crack-opening mode) stress intensity factor.

Fig. 3.83 Two dimensional elastic plate with a crack of length 2a in its centre subjected to a uniform tensile stress σ₀ in the longitudinal direction at one end and clamped at the other end.

3.4.2 Problem description

Specimen geometry: length l = 400 mm, height h = 100 mm.
Material: mild steel having Young's modulus E = 210 GPa and Poisson's ratio ν = 0.3.
Crack: A crack is placed perpendicular to the loading direction in the centre of the plate and has a length of 20 mm. The centre-cracked tension plate is assumed to be in the plane strain condition in the present analysis.
Boundary conditions: The elastic plate is subjected to a uniform tensile stress in the longitudinal direction at the right end and clamped to a rigid wall at the left end.

3.4.3 Analytical procedures

3.4.3.1 Creation of an analytical model

Let us use a quarter model of the centre-cracked tension plate as illustrated in Fig. 3.83, since the plate is symmetric about the horizontal and vertical centre lines.

Here we use the singular element or the quarter point element, which can interpolate the stress distribution in the vicinity of the crack tip at which stress has the $1 / \sqrt{r}$ singularity where r is the distance from the crack tip (r/a ≪ 1). An ordinary isoparametric element which is familiar as Quad 8 node 82 has nodes at corners and also at the midpoint on each side of the element. A singular element, however, has the midpoint moved one quarter side distance from the original midpoint position to the node which is placed at the crack tip position. This is why the singular element is often called the quarter point element instead. ANSYS is equipped with a 2D triangular singular element only, not with 2D rectangular or 3D singular elements. Around the node at the crack tip, a circular area is created and is divided into a designated number of triangular singular elements. Each of these elements has its vertex placed at the crack tip position and has the quarter points on the two sides joining the vertex and the other two nodes.

In order to create the singular elements, the plate area must be created via key points set at the four corner points and at the crack tip position on the left-end side of the quarter plate area.

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Keypoints → In Active CS

1. The Create Keypoints in Active Coordinate System window opens as shown in Fig. 3.84.
2. Input the [A] key point number in the NPT box and the [B] x-, y- and z-coordinates of the key point in the three X, Y, and Z boxes, respectively. Fig. 3.84 shows the case of Key point #5 which is placed at the crack tip having the coordinates (0, 10, 0). In the present model, let us create Key point #1 to #5 at the coordinates (0, 0, 0), (200, 0, 0), (200, 50, 0), (0, 50, 0), and (0, 10, 0), respectively. Note that the z-coordinate is always 0 in 2D elasticity problems.
3. Click the [C] Apply button and create Key point2 #1 to #4 not to exit from the window and finally click the [D] OK button to create Key point #5 at the crack tip position and exit from the window (see Fig. 3.85).

Fig. 3.84 ‘Create Keypoints in Active Coordinate System’ window.

Fig. 3.85 Five key points created in the ‘ANSYS Graphics’ window.

Then create the plate area via the five key points created by the procedures above, using the following commands:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Areas → Arbitrary → Through KPs

1. The Create Area thru KPs window opens.
2. The upward arrow appears in the ANSYS Graphics window. Move this arrow to Key point #1 and click this point. Click Key points #1–#5 one by one counter-clockwise (see Fig. 3.86).
3. Click the OK button to create the plate area, as shown in Fig. 3.87.

Fig. 3.86 Clicking Key points #1 through #5 one by one counter-clockwise to create the plate area.

Fig. 3.87 Quarter model of the centre cracked tension plate.

3.4.3.2 Input of the elastic properties of the plate material

[COMMAND] ANSYS Main Menu → Preprocessor → Material Props → Material Models

1. The Define Material Model Behavior window opens.
2. Double-click the Structural, Linear, Elastic, and Isotropic buttons one after another.
3. Input the value of Young's modulus, 2.1e5 (MPa), and that of Poisson's ratio, 0.3, into the EX and PRXY boxes, and click the OK button of the Linear Isotropic Properties for Materials Number 1 window.
4. Exit from the Define Material Model Behavior window by selecting Exit in the Material menu of the window.

3.4.3.3 Finite-element discretisation of the centre-cracked tension plate area

(1) Selection of the element type

[COMMAND] ANSYS Main Menu → Preprocessor → Element Type → Add/Edit/Delete

1. The Element Types window opens.
2. Click the Add … button in the Element Types window to open the Library of Element Types window and select the element type to use.
3. Select Structural Mass – Solid and Quad 8 node 82.
4. Click the OK button in the Library of Element Types window to use the 8-node isoparametric element.
5. Click the Options … button in the Element Types window to open the PLANE82 element type options window. Select the Plane strain item in the Element behavior box and click the OK button to return to the Element Types window. Click the Close button in the Element Types window to close the window.

(2) Sizing of the elements

Before meshing, the crack tip point around which the triangular singular elements will be created must be specified using the following commands:

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Size Cntrls → Concentrate KPs → Create

1. The Concentration Keypoints window opens.
2. Display the key points in the ANSYS Graphics window by the following:

[COMMAND] ANSYS Utility Menu → Plot → Keypoints → Keypoints.

3. Pick Key point #5 by placing the upward arrow on Key point #5 and clicking the left mouse button. Then click the OK button in the Concentration Keypoints window.
4. Another Concentration Keypoints window opens, as shown in Fig. 3.88.
5. Confirming that [A] 5, i.e. the key point number of the crack tip position is input in the NPT box, input [B] 2 in the DELR box, [C] 0.5 in the RRAT box, and [D] 6 in the NTHET box, then select [E] Skewed 1/4pt in the KCTIP box in the window. Refer to the explanations of the numerical data described after the names of the respective boxes on the window. Skewed 1/4pt in the last box means that the mid-nodes of the sides of the elements which contain Key point #5 are the quarter points of the elements.
6. Click the [F] OK button in the Concentration Keypoints window.

Fig. 3.88 ‘Concentration Keypoints’ window.

The size of the meshes other than the singular elements and the elements adjacent to them can be controlled by the same procedures, as has been described in the previous sections of this chapter.

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Size Cntrls →Manual Size → Global → Size

1. The Global Element Sizes window opens.
2. Input 1.5 in the SIZE box and click the OK button.
(3) Meshing

The meshing procedures are also the same as before.

[Commands] ANSYS Main Menu → Preprocessor → Meshing → Mesh →Areas → Free

1. The Mesh Areas window opens.
2. The upward arrow appears in the ANSYS Graphics window. Move this arrow to the quarter plate area and click this area.
3. The colour of the area turns from light blue to pink. Click the OK button.
4. The Warning window appears as shown in Fig. 3.89 due to the existence of six singular elements. Click the [A] Close button and proceed to the next operation below.
5. Fig. 3.90 shows the plate area meshed by ordinary 8-node isoparametric finite elements except for the vicinity of the crack tip, where we have six singular elements.

Fig. 3.90 Plate area meshed by ordinary 8-node isoparametric finite elements and by singular elements.

Fig. 3.91 shows an enlarged view of the singular elements around the [A] crack tip, showing that six triangular elements are placed in a radial manner and that the size of the second row of elements is half the radius of the first row of elements, i.e. triangular singular elements.

3.4.3.4 Input of boundary conditions

(1) Imposing constraint conditions on the ligament region of the left end and the bottom side of the quarter plate model

Due to the symmetry, the constraint conditions of the quarter plate model are UX-fixed condition on the ligament region of the left end, i.e. the line between Key points #4 and #5, and UY-fixed condition on the bottom side of the quarter plate model. Apply these constraint conditions onto the corresponding lines by the following commands:

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply → Structural → Displacement → On Lines

1. The Apply U. ROT on Lines window opens and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the line between Key points #4 and #5, and click the left mouse button.
3. Click the OK button in the Apply U. ROT on Lines window to display another Apply U. ROT on Lines window.
4. Select UX in the Lab2 box and click the OK button in the Apply U. ROT on Lines window.

Repeat the commands and operations (1)–(3) above for the bottom side of the model. Then, select UY in the Lab2 box and click the OK button in the Apply U. ROT on Lines window.

(2) Imposing a uniform longitudinal stress on the right end of the quarter plate model

A uniform longitudinal stress can be defined by pressure on the right-end side of the plate model, as described below:

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply → Structural → Pressure → On Lines

1. The Apply PRES on Lines window opens and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the right-end side of the quarter plate area and click the left mouse button.
3. Another Apply PRES on Lines window opens. Select a Constant value in the [SFL] Apply PRES on lines as a box and input -10 (MPa) in the VALUE Load PRES value box, and leave a blank in the Value box.
4. Click the OK button in the window to define a uniform tensile stress of 10 MPa applied to the right end of the quarter plate model.

Fig. 3.92 illustrates the boundary conditions applied to the centre-cracked tension plate model by the above operations.

3.4.3.5 Solution procedures

The solution procedures are also the same as usual.

[COMMAND] ANSYS Main Menu → Solution → Solve → Current LS

1. The Solve Current Load Step and /STATUS Command windows appear.
2. Click the OK button in the Solve Current Load Step window to begin the solution of the current load step.
3. The Verify window opens, as shown in Fig. 3.93. Proceed to the next operation below by clicking the [A] Yes button in the window.
4. Select the File button in the /STATUS Command window to open the submenu and select the Close button to close the /STATUS Command window.
5. When the solution is completed, the Note window appears. Click the Close button to close this window.

3.4.3.6 Contour plot of stress

[COMMAND] ANSYS Main Menu → General Postproc → Plot Results → Contour Plot → Nodal Solution

1. The Contour Nodal Solution Data window opens.
2. Select Stress and X-Component of stress.
3. Click the OK button to display the contour of the x-component of stress, or longitudinal stress in the centre-cracked tension plate in the ANSYS Graphics window, as shown in Fig. 3.94.

Fig. 3.94 Contour of the x-component of stress in the centre-cracked tension plate.

Fig. 3.95 is an enlarged view of the longitudinal stress distribution around the crack tip showing that very high tensile stress is induced at the crack tip, whereas compressive stress occurs around the crack surface, and the crack shape is parabolic.

3.4.4 Discussion

Fig. 3.96 shows extrapolation of the values of the correction factor, or the nondimensional mode I (the crack-opening mode) stress intensity factor F_I = K_I/ $σ \sqrt{πa}$ to the point where r = 0, i.e. the crack tip position by the hybrid extrapolation method [1]. The plots in the right region of the figure are obtained by the following formula:

$K_{I} = lim_{r \to 0} \sqrt{2 πr} σ_{x} (θ = 0)$

(3.7)

$F_{I} (λ) = \frac{K_{I}}{σ \sqrt{πa}} = lim_{r \to 0} \frac{\sqrt{2 πr} σ_{x} (θ = 0)}{σ \sqrt{πa}} where λ = 2 a / λ$

si11_e (3.7′)

whereas those in the left region are obtained by the following formula:

$K_{I} = lim_{r \to 0} \sqrt{\frac{2 π}{r}} \frac{E}{(1 + ν) (κ + 1)} u_{x} (θ = π)$

si12_e (3.8)

$F_{I} (λ) = \frac{K_{I}}{σ \sqrt{πa}} = lim_{r \to 0} \sqrt{\frac{2 π}{r}} \frac{E}{(1 + ν) (κ + 1) σ \sqrt{πa}} u_{x} (θ = π)$

si13_e (3.8′)

where r is the distance from the crack tip, σ_x is the x-component of stress, σ is the applied uniform stress in the direction of the x-axis, a is the half crack length, E is Young's modulus, u_x(θ = π), κ = 3 − 4ν for plane strain condition and κ = (3 − ν)/(1 + ν) for plane stress condition, and θ is the angle around the crack tip. The correction factor F_I accounts for the effect of the finite boundary, or the edge effect of the plate. The value of F_I can be evaluated either by the stress extrapolation method (the right region) or by the displacement extrapolation method (the left region). The hybrid extrapolation method [1] is a blend of the two types of the extrapolation methods above and can bring about better solutions. As illustrated in Fig. 3.95, the extrapolation lines obtained by the two methods agree well with each other when the gradient of the extrapolation line for the displacement method is rearranged by putting r′ = r/3. The value of F_I obtained for the present CCT plate by the hybrid method is 1.0200.

Table 3.1 lists the values of F_I calculated by the following four equations [2–6] for various values of nondimensional crack length λ = 2a/h:

Table 3.1

Correction factor F_I(λ) as a function of nondimensional crack length λ = 2a/h
λ = 2a/h	Isida Eq. (3.7)	Feddersen Eq. (3.8)	Koiter Eq. (3.9)	Tada Eq. (3.10)
0.1	1.0060	1.0062	1.0048	1.0060
0.2	1.0246	1.0254	1.0208	1.0245
0.3	1.0577	1.0594	1.0510	1.0574
0.4	1.1094	1.1118	1.1000	1.1090
0.5	1.1867	1.1892	1.1757	1.1862
0.6	1.3033	1.3043	1.2921	1.3027
0.7	1.4882	1.4841	1.4779	1.4873
0.8	1.8160	1.7989	1.8075	1.8143
0.9	2.5776	–	2.5730	2.5767

Table 3.1

$F_{I} (λ) = 1 + \sum_{k = 1}^{35} A_{k} λ^{2 k}$

si14_e (3.9)

$F_{I} (λ) = \sqrt{sec (πλ / 2)} (λ \leq 0.8)$

(3.10)

$F_{I} (λ) = \frac{1 - 0.5 λ + 0.326 λ^{2}}{\sqrt{1 - λ}}$

si16_e (3.11)

$F_{I} (λ) = (1 - 0.025 λ^{2} + 0.06 λ^{4}) \sqrt{sec (πλ / 2)}$

(3.12)

Among these, Isida's solution is considered to be the best. The value of F_I(0.2) by Isida is 1.0246. The relative error of the present result is (1.0200 − 1.0246)/1.0246 = − 0.45%, which is a reasonably good result.

3.4.5 Problems to solve

Problem 3.12

Calculate the values of the correction factor for the mode I stress intensity factor for the centre-cracked tension plate having normalised crack lengths of λ = 2a/h = 0.1–0.9. Compare the results with the values of the correction factor F(λ) listed in Table 3.1.

Problem 3.13

Calculate the values of the correction factor for the mode I stress intensity factor for a double edge cracked tension plate having normalised crack lengths of λ = 2a/h = 0.1–0.9 (see Fig. 3.97). Compare the results with the values of the correction factor F(λ) calculated by Eq. (3.13) [7,8]. Note that the quarter model described in the present section can be used for the FEM calculations of this problem simply by changing the boundary conditions.

Fig. 3.97 Double edge cracked tension plate subjected to a uniform longitudinal stress at one end and clamped at the other end.

$F_{I} (λ) = \frac{1.122 - 0.561 λ - 0.205 λ^{2} + 0.471 λ^{3} - 0.190 λ^{4}}{\sqrt{1 - λ}} where λ = 2 a / h$

si18_e (3.13)

Problem 3.14

Calculate the values of the correction factor for the mode I stress intensity factor for a single edge cracked tension plate having normalised crack lengths of λ = 2a/h = 0.1–0.9 (see Fig. 3.98). Compare the results with the values of the correction factor F(λ) calculated by Eq. (3.14) [7,9]. In this case, a half plate model is clamped along the ligament and is loaded at the right end of the plate. Note that the whole plate model with an edge crack needs a somewhat complicated procedure. A quarter model described in the present section gives good solutions with an error of a few percent or less, and is effective in practice.

Fig. 3.98 Single edge cracked tension plate subjected to a uniform longitudinal stress at one end and clamped at the other end.

$F_{I} (λ) = \frac{0.752 + 2.02 λ + 0.37 {[1 - sin (πλ / 2)]}^{3}}{cos (πλ / 2)} \sqrt{\frac{2}{πλ} tan (\frac{πλ}{2})} where λ = a / h$

si19_e (3.14)

3.5 Two-dimensional contact stress

3.5.1 Example problem

An elastic cylinder with a radius of length a is pressed against a flat surface of a linearly elastic medium by a force P′. Perform an FEM analysis of an elastic cylinder of mild steel with a radius of length a pressed against an elastic flat plate of the same steel by a force P′ illustrated in Fig. 3.99 and calculate the resulting contact pressure induced in the flat plate.

Fig. 3.99 Elastic cylinder of mild steel with a radius of length a pressed against an elastic flat plate of the same steel by a force P′ and its FEM model.

3.5.2 Problem description

Geometry of the cylinder: radius a = 500 mm.
Geometry of the flat plate: width W = 500 mm, height h = 500 mm.
Material: mild steel having Young's modulus E = 210 GPa and Poisson's ratio ν = 0.3.
Boundary conditions: The elastic cylinder is pressed against an elastic flat plate of the same steel by a force of P′ = 1 kN/mm and the plate is clamped at the bottom.

3.5.3 Analytical procedures

3.5.3.1 Creation of an analytical model

Let us use a half model of the indentation problem illustrated in Fig. 3.99, since the problem is symmetric about the vertical centre line.

(1) Creation of an elastic flat plate

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Areas → Rectangle → By 2 Corners

1. Input two 0s into the WP X and WP Y boxes in the Rectangle by 2 Corners window to determine the upper left corner point of the elastic flat plate on the Cartesian coordinates of the working plane.
2. Input 500 and -500 (mm) into the Width and Height boxes, respectively, to determine the shape of the elastic flat plate model.
3. Click the OK button to create the plate on the ANSYS Graphics window.
(2) Creation of an elastic cylinder

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Areas → Circle → Partial Annulus

1. The Part Annular Circ Area window opens, as shown in Fig. 3.100.
2. Input 0 and 500 into the [A] WP X and [B] WP Y boxes, respectively, in the Part Annular Circ Area window to determine the centre of the elastic cylinder on the Cartesian coordinates of the working plane.
3. Input two 500s (mm) into the [C] Rad-1 and [E] Rad-2 boxes to designate the radius of the elastic cylinder model.
4. Input -90 and 0 (mm) into the [D] Theta-1 and [F] Theta -2 boxes, respectively, to designate the starting and ending angles of the quarter elastic cylinder model.
5. Click the [G] OK button to create the quarter cylinder model on the ANSYS Graphics window, as shown in Fig. 3.101.
(3) Combining the elastic cylinder and the flat plate models

Fig. 3.100 ‘Part Annular Circ Area’ window.

Fig. 3.101 Areas of the FEM model for an elastic cylinder pressed against an elastic flat plate.

The elastic cylinder and the flat plate models must be combined into one model so as not to allow the rigid-body motions of the two bodies.

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Operate →Booleans → Glue → Areas

1. The Glue Areas window opens, as shown in Fig. 3.102.
2. The upward arrow appears in the ANSYS Graphics window. Move this arrow to the quarter cylinder area and click this area, and then move the arrow on to the half plate area and click this area to combine these two areas.
3. Click the [A] OK button to create the contact model on the ANSYS Graphics window, as shown in Fig. 3.102.

3.5.3.2 Input of the elastic properties of the materials for the cylinder and the flat plate

[COMMAND] ANSYS Main Menu → Preprocessor → Material Props → Material Models

1. The Define Material Model Behavior window opens.
2. Double-click the Structural, Linear, Elastic, and Isotropic buttons one after another.
3. Input the value of Young's modulus, 2.1e5 (MPa), and that of Poisson's ratio, 0.3, into the EX and PRXY boxes, and click the OK button of the Linear Isotropic Properties for Materials Number 1 window.
4. Exit from the Define Material Model Behavior window by selecting Exit in the Material menu of the window.

3.5.3.3 Finite-element discretisation of the cylinder and the flat plate areas

(1) Selection of the element types

Select the 8-node isoparametric elements as usual.

[COMMAND] ANSYS Main Menu → Preprocessor → Element Type → Add/Edit/Delete

1. The Element Types window opens.
2. Click the Add … button in the Element Types window to open the Library of Element Types window and select the element type to use.
3. Select Structural Mass – Solid and Quad 8 node 82.
4. Click the OK button in the Library of Element Types window to use the 8-node isoparametric element. Note that this element is defined as a Type 1 element, as indicated in the Element type reference number box.
5. Click the Options … button in the Element Types window to open the PLANE82 element type options window. Select the Plane strain item in the Element behavior box and click the OK button to return to the Element Types window. Click the Close button in the Element Types window to close the window.

In contact problems, two mating surfaces come in contact with each other exerting great force on each other. Contact elements must be used in contact problems to prevent penetration of one object into the other.

6. Repeat-click the Add … button in the Element Types window to open the Library of Element Types window and select Contact and 2D target 169.
7. Click the OK button in the Library of Element Types window to select the target element. Note that this target element is defined as a Type 2 element, as indicated in the Element type reference number box.
8. Click the Add … button again in the Element Types window to open the Library of Element Types window and select Contact and 3 nd surf 172.
9. Click the OK button in the Library of Element Types window to select the surface effect element. Note that this target element is defined as a Type 3 element, as indicated in the Element type reference number box.
(2) Sizing of the elements

Here let us designate the element size by specifying number of divisions of lines picked. For this purpose, carry out the following commands:

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Size Cntrls →Manual Size → Lines → Picked Lines

1. The Element Size on Pick … window opens, as shown in Fig. 3.103.
2. Move the upward arrow to the upper and the right sides of the flat plate area. Click the [A] OK button and the Element Sizes on Picked Lines window opens, as shown in Fig. 3.104.
3. Input [A] 60 in the NDIV box and [B] 1/10 in the SPACE box to divide the upper and right side of the plate area into 60 subdivisions. The size of the subdivisions decreases in geometric progression as they approach the left side of the plate area. It would be preferable to have smaller elements around the contact point where high stress concentration occurs. Input 60 in the NDIV box and 10 in the SPACE box for the left and the bottom sides of the plate area, 40 and 10 for the left side and the circumference of the quarter circular area, and 40 and 1/10 for the upper side of the quarter circular area in the respective boxes.
(3) Meshing

Fig. 3.103 ‘Element Size on Pick …’ window.

Fig. 3.104 ‘Element Sizes on Picked Lines’ window.

The meshing procedures are the same as usual.

[COMMAND] ANSYS Main Menu → Preprocessor → Meshing → Mesh →Areas → Free

1. The Mesh Areas window opens.
2. The upward arrow appears in the ANSYS Graphics window. Move this arrow to the quarter cylinder and the half plate areas and click these areas.
3. The colour of the areas turns from light blue to pink. Click the OK button to see the areas meshed by 8-node isoparametric finite elements, as shown in Fig. 3.105.
4. Fig. 3.106 is an enlarged view of the finer meshes around the contact point.
(4) Creation of target and contact elements

Fig. 3.105 Cylinder and flat plate areas meshed by 8-node isoparametric finite elements.

Fig. 3.106 Enlarged view of the finer meshes around the contact point.

First select the lower surface of the quarter cylinder to which the contact elements are attached.

[Commands] ANSYS Utility Menu → Select → Entities …

1. The Select Entities window opens, as shown in Fig. 3.107.
2. Select [A] Lines, [B] By Num/Pick, and [C] From Full in the Select Entities window.
3. Click the [D] OK button in the Select Entities window and the Select lines window opens, as shown in Fig. 3.108.
4. The upward arrow appears in the ANSYS Graphics window. Move this arrow to the quarter circumference of the cylinder and click this line. The lower surface of the cylinder is selected, as shown in Fig. 3.109.

Fig. 3.109 Lower surface of the cylinder selected.

Repeat the Select → Entities … commands to select nodes on the lower surface of the cylinder.

[COMMAND] ANSYS Utility Menu → Select → Entities …

1. The Select Entities window opens, as shown in Fig. 3.110.
2. Select [A] Nodes, [B] Attached to, [C] Lines, all, and [D] Reselect in the Select Entities window.
3. Click the [E] OK button and perform the following commands:

[COMMAND] ANSYS Utility Menu → Plot → Nodes

Only nodes on the lower surface of the cylinder are plotted in the ANSYS Graphics window, as shown in Fig. 3.111.

Repeat the Select → Entities … commands again to select a smaller number of the nodes on the portion of the lower surface of the cylinder in the vicinity of the contact point.

[COMMAND] ANSYS Utility Menu → Select → Entities …

1. The Select Entities window opens, as shown in Fig. 3.112.
2. Select [A] Nodes, [B] By Num/Pick, and [C] Reselect in the Select Entities window.
3. Click the [D] OK button and the Select nodes window opens, as shown in Fig. 3.113.
4. The upward arrow appears in the ANSYS Graphics window. Select [A] Box instead of Single, then move the upward arrow to the leftmost end of the array of nodes and select nodes by enclosing them by a rectangle formed by dragging the mouse, as shown in Fig. 3.114. Click the [B] OK button to select the nodes on the bottom surface of the cylinder.

Fig. 3.114 Selection of nodes on the portion of the bottom surface of the cylinder.

Next, let us define contact elements to be attached to the lower surface of the cylinder by the commands as follows:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Elements → Elem Attributes

1. The Element Attributes window opens, as shown in Fig. 3.115.
2. Select [A] 3 CONTA172 in the TYPE box and click the [B] OK button in the Element Attributes window.

Then, perform the following commands to attach the contact elements to the lower surface of the cylinder.

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Elements → Surf / Contact → Surf to Surf

1. The Mesh Free Surfaces window opens, as shown in Fig. 3.116.
2. Select [A] Bottom surface in the Tlab box and click the [B] OK button in the window.
3. CONTA172 elements are created on the lower surface of the cylinder, as shown in Fig. 3.117.

Repeat operations similar to the above in order to create TARGET169 elements on the top surface of the flat plate. Namely, perform the Select → Entities … commands to select only nodes on the portion of the top surface of the flat plate in the vicinity of the contact point. Then, define the target elements to attach to the top surface of the flat plate using the following commands:

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Elements → Elem Attributes

1. The Element Attributes window opens, as shown in Fig. 3.118.
2. Select [A] 2 TARGET169 in the TYPE box and click the [B] OK button in the Element Attributes window.

Fig. 3.117 ‘CONTA172’ elements created on the lower surface of the cylinder.

Then, perform the following commands to attach the target elements to the top surface of the flat plate.

[COMMAND] ANSYS Main Menu → Preprocessor → Modeling → Create →Elements → Surf / Contact → Surf to Surf

3. The Mesh Free Surfaces window opens, as shown in Fig. 3.119.
4. Select [A] Top surface in the Tlab box and click the [B] OK button in the window.
5. TARGET169 elements are created on the top surface of the flat plate, as shown in Fig. 3.120.

Fig. 3.120 ‘TARGET169’ elements created on the top surface of the flat plate.

3.5.3.4 Input of boundary conditions

(1) Imposing constraint conditions on the left sides of the quarter cylinder and the half flat plate models

Due to the symmetry, the constraint conditions of the present model are in a UX-fixed condition on the left sides of the quarter cylinder and the half flat plate models, and a UY-fixed condition on the bottom side of the half flat plate model. Apply these constraint conditions to the corresponding lines by the following commands:

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Displacement → On Lines

1. The Apply U. ROT on Lines window opens and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the left sides of the quarter cylinder and the half flat plate models, and click the left mouse button on each model.
3. Click the OK button in the Apply U. ROT on Lines window to display another Apply U. ROT on Lines window.
4. Select UX in the Lab2 box and click the OK button in the Apply U. ROT on Lines window.

Repeat the commands and operations (1)–(3) above for the bottom side of the flat plate model. Then, select UY in the Lab2 box and click the OK button in the Apply U. ROT on Lines window.

(2) Imposing a uniform pressure on the upper side of the quarter cylinder model

[COMMAND] ANSYS Main Menu → Solution → Define Loads → Apply →Structural → Pressure → On Lines

1. The Apply PRES on Lines window opens and the upward arrow appears when the mouse cursor is moved to the ANSYS Graphics window.
2. Confirming that the Pick and Single buttons are selected, move the upward arrow onto the upper side of the quarter cylinder model and click the left mouse button.
3. Another Apply PRES on Lines window opens. Select a Constant value in the [SFL] Apply PRES on lines as a box and input 1 (MPa) in the VALUE Load PRES value box, and leave a blank in the Value box.
4. Click the OK button in the window to define a downward force of P′ = 1 kN/mm applied to the elastic cylinder model.

Fig. 3.121 illustrates the boundary conditions applied to the FE model of the present contact problem by the above operations.

3.5.3.5 Solution procedures

[COMMAND] ANSYS Main Menu → Solution → Solve → Current LS

1. The Solve Current Load Step and /STATUS Commands windows appear.
2. Click the OK button in the Solve Current Load Step window to begin the solution of the current load step.
3. Select the File button in /STATUS Commands window to open the submenu and select the Close button to close this window.
4. When the solution is completed, the Note window appears. Click the Close button to close this window.

3.5.3.6 Contour plot of stress

[COMMAND] ANSYS Main Menu → General Postproc → Plot Results → Contour Plot → Nodal Solution

1. The Contour Nodal Solution Data window opens.
2. Select Stress and Y-Component of stress.
3. Click the OK button to display the contour of the y-component of stress in the present model of the contact problem in the ANSYS Graphics window, as shown in Fig. 3.122.

Fig. 3.122 Contour of the y-component of stress in the present model of the contact problem.

Fig. 3.123 is an enlarged contour of the y-component of stress around the contact point, showing that very high compressive stress is induced in the vicinity of the contact point; this is known as Hertz contact stress.

3.5.4 Discussion

The open circular symbols in Fig. 3.124 show the plots of the y-component stress, or vertical stress, along the upper surface of the flat plate obtained by the present FE calculation. The solid line in the figure indicates the theoretical curve p(x) expressed by a parabola [10,11], i.e.

$p (x) = k \sqrt{b^{2} - x^{2}} = 115.4 \sqrt{{2.349}^{2} - x^{2}}$

(3.15)

where x is the distance from the centre of contact, p(x) is the contact stress at a point x, k is the coefficient given by Eq. (3.16), and b is half the width of contact given by Eq. (3.17).

$k = \frac{2 P^{'}}{π b^{2}} = 115.4 N / {mm}^{2}$

si21_e (3.16)

$b = 1.522 \sqrt{\frac{P^{'} R}{E}} = 2.348 mm$

si22_e (3.17)

The maximum stress p₀ is obtained at the centre of contact and expressed as

$p_{0} = 0.418 \sqrt{\frac{P^{'} E}{R}} = 270.9 MPa$

si23_e (3.18)

The length of contact is as small as 4.7 mm, so that the maximum value of the contact stress is as high as 271 MPa. The number of nodes that fall within the contact region is only 3, although the total number of nodes is approximately 15,000 and meshes are finer in the vicinity of the contact point. The three plots of the present results agree reasonably well with the theoretical curve, as shown in Fig. 3.124.

3.5.5 Problems to solve

Problem 3.15

Calculate the contact stress distribution between a cylinder having a radius of curvature R₁ = 500 mm pressed against another cylinder having R₂ = 1000 mm by a force P′ = 1 kN/mm, as shown in Fig. 3.125. The two cylinders are made of the same steel having Young's modulus = 210 GPa and Poisson's ratio ν = 0.3. The theoretical expression for the contact stress distribution p(x) is given by Eq. (3.19a), with the parameters given by Eqs (3.19b)–(3.19d).

Fig. 3.125 A cylinder having a radius of curvature R₁ = 500 mm pressed against another cylinder having R₂ = 1000 mm by a force P′ = 1 kN/mm.

$p (x) = k \sqrt{b^{2} - x^{2}}$

(3.19a)

$k = \frac{2 P^{'}}{π b^{2}}$

si25_e (3.19b)

$b = 1.522 \sqrt{\frac{P^{'}}{E} \cdot \frac{R_{1} R_{2}}{R_{1} + R_{2}}}$

si26_e (3.19c)

$p_{0} = 0.418 \sqrt{\frac{P^{'} E (R_{1} + R_{2})}{R_{1} R_{2}}}$

si27_e (3.19d)

Problem 3.16

Calculate the contact stress distribution between a cylinder having a radius of curvature R = 500 mm pressed against a flat plate of 1000 mm width by 500 mm height by a force of P′ = 1 kN/mm, as shown in Fig. 3.126. The cylinder is made of fine ceramics having Young's modulus E₁ = 320 GPa and Poisson's ratio ν₁ = 0.3, and the flat plate of stainless steel having E₂ = 192 GPa and ν₂ = 0.3. The theoretical curve of the contact stress p(x) is given by Eq. (3.20a), with the parameters expressed by Eqs (3.20b)–(3.20d). Compare the result obtained by the FEM calculation with the theoretical distribution given by Eqs (3.20a)–(3.20d).

Fig. 3.126 A cylinder having a radius of curvature R = 500 mm pressed against a flat plate.

$p (x) = k \sqrt{b^{2} - x^{2}}$

(3.20a)

$k = \frac{2 P^{'}}{π b^{2}}$

si25_e (3.20b)

$b = \frac{2}{\sqrt{π}} \sqrt{P^{'} R_{1} (\frac{1 - ν_{1}^{2}}{E_{1}} + \frac{1 - ν_{2}^{2}}{E_{2}})}$

si30_e (3.20c)

$p_{0} = \frac{1}{\sqrt{π}} \sqrt{\frac{P^{'}}{R_{1}} \cdot \frac{1}{(\frac{1 - ν_{1}^{2}}{E_{1}} + \frac{1 - ν_{2}^{2}}{E_{2}})}}$

si31_e (3.20d)

Problem 3.17

Calculate the contact stress distribution between a cylinder having a radius of curvature R₁ = 500 mm pressed against another cylinder having R₂ = 1000 mm by a force P′ = 1 kN/mm, as shown in Fig. 3.125. The upper cylinder is made of fine ceramics having Young's modulus E₁ = 320 GPa and Poisson's ratio ν₁ = 0.3, and the lower cylinder of stainless steel having E₂ = 192 GPa and ν₂ = 0.3. The theoretical curve of the contact stress p(x) is given by Eq. (3.21a), with the parameters expressed by Eqs (3.21b)–(3.21d). Compare the result obtained by the FEM calculation with the theoretical distribution given by Eqs (3.21a)–(3.21d).

$p (x) = k \sqrt{b^{2} - x^{2}}$

(3.21a)

$k = \frac{2 P^{'}}{π b^{2}}$

si25_e (3.21b)

$b = \frac{2}{\sqrt{π}} \sqrt{\frac{P^{'} R_{1} R_{2}}{R_{1} + R_{2}} (\frac{1 - ν_{1}^{2}}{E_{1}} + \frac{1 - ν_{2}^{2}}{E_{2}})}$

si34_e (3.21c)

$p_{0} = \frac{1}{\sqrt{π}} \sqrt{\frac{P^{'} (R_{1} + R_{2})}{R_{1} R_{2}} \cdot \frac{1}{(\frac{1 - ν_{1}^{2}}{E_{1}} + \frac{1 - ν_{2}^{2}}{E_{2}})}}$

si35_e (3.21d)

Problem 3.18

Calculate the contact stress distribution between a cylinder of fine ceramics having a radius of curvature R₁ = 500 mm pressed against a cylindrical seat of stainless steel having R₂ = 1000 mm by a force P′ = 1 kN/mm, as shown in Fig. 3.127. Young's moduli of fine ceramics and stainless steel are E₁ = 320 GPa and E₂ = 192 GPa, respectively, and Poisson's ratio of each material is the same, i.e. ν₁ = ν₂ = 0.3. The theoretical curve of the contact stress p(x) is given by Eq. (3.22a) with the parameters expressed by Eqs (3.22b)–(3.22d). Compare the result obtained by the FEM calculation with the theoretical distribution given by Eqs (3.22a)–(3.22d).

Fig. 3.127 A cylinder having a radius of curvature R₁ = 500 mm pressed against a cylindrical seat having R₂ = 1000 mm by a force P′ = 1 kN/mm.

$p (x) = k \sqrt{b^{2} - x^{2}}$

(3.22a)

$k = \frac{2 P^{'}}{π b^{2}}$

si25_e (3.22b)

$b = \frac{2}{\sqrt{π}} \sqrt{\frac{P^{'} R_{1} R_{2}}{R_{2} - R_{1}} (\frac{1 - ν_{1}^{2}}{E_{1}} + \frac{1 - ν_{2}^{2}}{E_{2}})}$

si38_e (3.22c)

$p_{0} = \frac{1}{\sqrt{π}} \sqrt{\frac{P^{'} (R_{2} - R_{1})}{R_{1} R_{2}} \cdot \frac{1}{(\frac{1 - ν_{1}^{2}}{E_{1}} + \frac{1 - ν_{2}^{2}}{E_{2}})}}$

si39_e (3.22d)

Note that Eqs (3.22a)–(3.22d) cover Eqs (3.19a)–(3.19d), (3.20a), (3.20d), and (3.21a)–(3.21d)–.

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Table of Contents for 3: Application of ANSYS to stress analysis

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3.1 Cantilever beam

3.1.1 Example problem: A cantilever beam

3.1.2 Problem description

3.1.3 Review of the solutions obtained by the elementary beam theory

3.1.4 Analytical procedures

3.1.4.1 Creation of an analytical model

3.1.4.2 Input of the elastic properties of the beam material

3.1.4.3 Finite-element discretisation of the beam area

3.1.4.4 Input of boundary conditions

3.1.4.5 Solution procedures

3.1.4.6 Graphical representation of the results

3.1.5 Comparison of FEM results with experimental ones

3.1.6 Problems to solve

Appendix: Procedures for creating stepped beams

A.1 Creation of a stepped beam

A.1.1 How to cancel the selection of areas

A.2 Creation of a stepped beam with a rounded fillet

A.2.1 How to display area numbers

3.2 The principle of St. Venant

3.2.1 Example problem

3.2.2 Problem description

3.2.3 Analytical procedures

3.2.3.1 Creation of an analytical model

3.2.3.2 Input of the elastic properties of the strip material

3.2.3.3 Finite-element discretisation of the strip area

3.2.3.4 Input of boundary conditions

3.2.3.5 Solution procedures

3.2.3.6 Contour plot of stress

3.2.4 Discussion

3.3 Stress concentration due to elliptic holes and inclusions

3.3.1 Example problem

3.3.2 Problem description

3.3.3 Analytical procedures

3.3.3.1 Creation of an analytical model

3.3.3.2 Input of the elastic properties of the plate material

3.3.3.3 Finite-element discretisation of the quarter plate area

3.3.3.4 Input of boundary conditions

3.3.3.5 Solution procedures

3.3.3.6 Contour plot of stress

3.3.3.7 Observation of the variation of the longitudinal stress distribution in the ligament region

3.3.4 Discussion

3.3.5 Problems to solve

3.4 Stress singularity problem

3.4.1 Example problem

3.4.2 Problem description

3.4.3 Analytical procedures

3.4.3.1 Creation of an analytical model

3.4.3.2 Input of the elastic properties of the plate material

3.4.3.3 Finite-element discretisation of the centre-cracked tension plate area

3.4.3.4 Input of boundary conditions

3.4.3.5 Solution procedures

3.4.3.6 Contour plot of stress

3.4.4 Discussion

3.4.5 Problems to solve

3.5 Two-dimensional contact stress

3.5.1 Example problem

3.5.2 Problem description

3.5.3 Analytical procedures

3.5.3.1 Creation of an analytical model

3.5.3.2 Input of the elastic properties of the materials for the cylinder and the flat plate

3.5.3.3 Finite-element discretisation of the cylinder and the flat plate areas

3.5.3.4 Input of boundary conditions

3.5.3.5 Solution procedures

3.5.3.6 Contour plot of stress

3.5.4 Discussion

3.5.5 Problems to solve

Table of Contents for
3: Application of ANSYS to stress analysis