Engineering Analysis with ANSYS Software

Vishwambhar Sholapur

Engineering Analysis with ANSYS Software

Vishwambhar Sholapur

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Figures (617)

Comparison of the results obtained by various kinds of discrete analyses.
Figure 1.1
as shown by the dotted and the solid lines in Figure 1.1. — Comparison of the results obtained by various kinds of discrete analyses. Figure 1.1 as shown by the dotted and the solid lines in Figure 1.1.

‘igure 1.3 Discretization of the domain to analyze by finite elements and their interpolation functions.

The global simultaneous equations are obtained as

Stress states in an infinitesimal element of a two-dimensional elastic body
Figure 1.6 — Stress states in an infinitesimal element of a two-dimensional elastic body Figure 1.6

the two-dimensional Hooke’s law can be expressed in a unified form:

where a is the angle between the normal vector n and the x-axis. For free surface:
where no forces are applied, t; = 0 and ty = 0.
where ¢,; and tf; are the x- and the y-components of the traction force t*, respectively,
while the bar over ¢; and t, indicates that those quantities are prescribed on that
portion of the surface. Taking n= [cosa, sina] as the outward unit normal vector
at a point of a small element of the surface portion S,, the Cauchy relations which
represent the equilibrium conditions for surface traction forces and internal stresses
are given by the following equations: — where a is the angle between the normal vector n and the x-axis. For free surface: where no forces are applied, t; = 0 and ty = 0. where ¢,; and tf; are the x- and the y-components of the traction force t*, respectively, while the bar over ¢; and t, indicates that those quantities are prescribed on that portion of the surface. Taking n= [cosa, sina] as the outward unit normal vector at a point of a small element of the surface portion S,, the Cauchy relations which represent the equilibrium conditions for surface traction forces and internal stresses are given by the following equations:

«>

The numbers with subscript “e”, 1,, 2-, and 3¢, in the above equations are called
element nodal numbers and denote the numbers of three nodal points of the eth
element. Nodal points should be numbered counterclockwise. These three numbers
are used only in the eth element. Nodal numbers of the other type called global nodal
numbers are also assigned to the three nodal points of the eth element, being num-
bered throughout the whole model of the elastic body. The symbol A“ represents
the area of the eth element and can be expressed only by the coordinates of the nodal
points of the element, i.e., — «> The numbers with subscript “e”, 1,, 2-, and 3¢, in the above equations are called element nodal numbers and denote the numbers of three nodal points of the eth element. Nodal points should be numbered counterclockwise. These three numbers are used only in the eth element. Nodal numbers of the other type called global nodal numbers are also assigned to the three nodal points of the eth element, being num- bered throughout the whole model of the elastic body. The symbol A“ represents the area of the eth element and can be expressed only by the coordinates of the nodal points of the element, i.e.,

Figure 1.9
Procedures 5 through 7 described in Section 1.4.1 will be explained by taking an
example of the finite-element calculations for a square plate subjected to uniaxial
uniform tension as illustrated in Figure 1.9. The square plate model has a side
of unit length 1 and a thickness of unit length 1, and consists of two constant-
strain triangular elements, i.e., the model has four nodes and thus eight degrees of
freedom.
strain triangular elements, i.e., the model has four nodes and thus eight degrees of — Figure 1.9 Procedures 5 through 7 described in Section 1.4.1 will be explained by taking an example of the finite-element calculations for a square plate subjected to uniaxial uniform tension as illustrated in Figure 1.9. The square plate model has a side of unit length 1 and a thickness of unit length 1, and consists of two constant- strain triangular elements, i.e., the model has four nodes and thus eight degrees of freedom. strain triangular elements, i.e., the model has four nodes and thus eight degrees of

since the equivalent nodal forces due to initial strains ¢9 and body forces F, and F,,
{Feo} and {Fr}“) are zero. The components of the nodal displacement and force
vectors are written by the element nodal numbers. By rewriting these components by
the global nodal numbers as shown in Figure 1.9, Equation (1.105) is rewritten as — since the equivalent nodal forces due to initial strains ¢9 and body forces F, and F,, {Feo} and {Fr}“) are zero. The components of the nodal displacement and force vectors are written by the element nodal numbers. By rewriting these components by the global nodal numbers as shown in Figure 1.9, Equation (1.105) is rewritten as

In a similar way, the element stiffness equation for Element 2 is obtained as

Hence, from Equation (1.85), the element stiffness matrix for Element 1 [k©] is
calculated as — Hence, from Equation (1.85), the element stiffness matrix for Element 1 [k©] is calculated as

where the left- and right-hand sides of the equation are replaced with each other.
where the left- and right-hand sides of the equation are replaced with each other.

Let us now impose boundary conditions on the nodes. Namely, the node 1 is
clamped in both the x- and the y-directions, the node 2 is clamped only in the x-
direction, and the nodes 2 and 4 are subjected to equal nodal forces X2 = X4 = (p x
1 x 1)/2=p/2, respectively, in the x-direction. A pair of the equal nodal forces
p/2 applied to the nodes 2 and 4 in the x-direction is the finite-element model of
a uniformly distributed tension force p per unit area exerted on the side 24 in the
x-direction as illustrated in Figure 1.9. The geometrical boundary and mechanical
conditions for the present case are — where the left- and right-hand sides of the equation are replaced with each other. where the left- and right-hand sides of the equation are replaced with each other. Let us now impose boundary conditions on the nodes. Namely, the node 1 is clamped in both the x- and the y-directions, the node 2 is clamped only in the x- direction, and the nodes 2 and 4 are subjected to equal nodal forces X2 = X4 = (p x 1 x 1)/2=p/2, respectively, in the x-direction. A pair of the equal nodal forces p/2 applied to the nodes 2 and 4 in the x-direction is the finite-element model of a uniformly distributed tension force p per unit area exerted on the side 24 in the x-direction as illustrated in Figure 1.9. The geometrical boundary and mechanical conditions for the present case are

and is rewritten by using the global nodal numbers as

The results obtained by the present finite-element calculations imply that a square
plate subjected to uniaxial uniform tension in the x-direction is elongated by a uni-
form strain of p/E’ in the loading direction, whereas it is contacted by a uniform strain
of —v’p/E’ in the direction perpendicular to the loading direction and that only a uni-
form normal stress of o, = p is induced in the plate. The result that the nodal reaction
forces at the nodes 1 and 3 are equal to —p/2, i.e., X; = X3 = —p/2 implies that a uni-
form reaction force of —p is produced along the side 13. It is concluded that the above
results obtained by the FEM agree well with the physical interpretations of the present
problem. — The results obtained by the present finite-element calculations imply that a square plate subjected to uniaxial uniform tension in the x-direction is elongated by a uni- form strain of p/E’ in the loading direction, whereas it is contacted by a uniform strain of —v’p/E’ in the direction perpendicular to the loading direction and that only a uni- form normal stress of o, = p is induced in the plate. The result that the nodal reaction forces at the nodes 1 and 3 are equal to —p/2, i.e., X; = X3 = —p/2 implies that a uni- form reaction force of —p is produced along the side 13. It is concluded that the above results obtained by the FEM agree well with the physical interpretations of the present problem.

List Results menu to get displacements. Preprocessing and Solution data can be listed

Figure 2.13
are two methods to create the finite-element model: solid modeling and direct gener-
ation. With solid modeling, the geometry of shape of the model is described, and then
the ANSYS program automatically meshes the geometry with nodes and elements.
The size and shape of the elements that the program creates can be controlled. With
direct generation, the location of each node and the connectivity of each element
is manually defined. Several convenience operations, such as copying patterns of
existing nodes and elements, symmetry reflection, etc., are available.

Solved example problems in this book amply illustrate, in a step-by-step manner, — Figure 2.13 are two methods to create the finite-element model: solid modeling and direct gener- ation. With solid modeling, the geometry of shape of the model is described, and then the ANSYS program automatically meshes the geometry with nodes and elements. The size and shape of the elements that the program creates can be controlled. With direct generation, the location of each node and the connectivity of each element is manually defined. Several convenience operations, such as copying patterns of existing nodes and elements, symmetry reflection, etc., are available. Solved example problems in this book amply illustrate, in a step-by-step manner,

Modeling of an axle shaft by a simply supported beam.
Figure 3.1 — Modeling of an axle shaft by a simply supported beam. Figure 3.1

igure 3.2 Modeling of an arm of a human being by a cantilever beam.

Perform an finite-element method (FEM) analysis of a 2-D cantilever beam shown
in Figure 3.3 below and calculate the deflection of the beam at the loading point and
the longitudinal stress distribution in the beam. — Perform an finite-element method (FEM) analysis of a 2-D cantilever beam shown in Figure 3.3 below and calculate the deflection of the beam at the loading point and the longitudinal stress distribution in the beam.

Figure 3.8
Then the Define Material Model Behavior window opens as shown in
Figure 3.8:
(1) Double-click [A] Structural, Linear, Elastic, and Isotropic buttons one afte
another. — Figure 3.8 Then the Define Material Model Behavior window opens as shown in Figure 3.8: (1) Double-click [A] Structural, Linear, Elastic, and Isotropic buttons one afte another.

Input of elastic constants through the “Linear Isotropic Properties for Material Number 1”
window.
Figure 3.9 — Input of elastic constants through the “Linear Isotropic Properties for Material Number 1” window. Figure 3.9

ure 3.10 “Element Types” window.
Figure 3.10 — ure 3.10 “Element Types” window. Figure 3.10

Figure 3.18 “Real Constants” window
after setting the element thickness.
opens as shown in Figure 3.19:
Figure 3.17 — Figure 3.18 “Real Constants” window after setting the element thickness. opens as shown in Figure 3.19: Figure 3.17

Nodes and nodal numbers displayed on the ANSYS Graphics window.
Figure 3.30 — Nodes and nodal numbers displayed on the ANSYS Graphics window. Figure 3.30

“Contour Nodal Solution Data” window.
Figure 3.43 — “Contour Nodal Solution Data” window. Figure 3.43

Figure 3.45 compares longitudinal stress distributions obtained by ANSYS with those
by experiments and by the 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 the results become larger, especially at the clamped
end. This tendency arises from the fact that the clamped condition can be hardly
realized in the strict sense. — Figure 3.45 compares longitudinal stress distributions obtained by ANSYS with those by experiments and by the 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 the results become larger, especially at the clamped end. This tendency arises from the fact that the clamped condition can be hardly realized in the strict sense.

igure P3.5 A half model of the beam in Problem 3.3.
Figure P3.5
3.4 A half model of the beam in Problem 3.2.
Figure P3.4 — igure P3.5 A half model of the beam in Problem 3.3. Figure P3.5 3.4 A half model of the beam in Problem 3.2. Figure P3.4

Calculate the maximum deflection in a beam shown in Figure P3.5 where the thickness

A stepped cantilever beam subjected to a concentrated force of 100 N at the free end.
Figure P3.6
PROBLEM 3.7 — A stepped cantilever beam subjected to a concentrated force of 100 N at the free end. Figure P3.6 PROBLEM 3.7

{8 A stepped beam area created by adding two rectangle areas.
Figure 3.48 — {8 A stepped beam area created by adding two rectangle areas. Figure 3.48

Figure 3.52 The larger rectangular area picked.

ANSYS Main Menu — Preprocessor > Meshing >
Mesh — Areas — Free — ANSYS Main Menu — Preprocessor > Meshing > Mesh — Areas — Free

“Apply PRES on Lines” window for applying linearly distributed pressure to the lower half of
the right end of the elastic strip.
Figure 3.61 — “Apply PRES on Lines” window for applying linearly distributed pressure to the lower half of the right end of the elastic strip. Figure 3.61

Figure 3.64
<= 200mm, 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 center of the
strip. The longitudinal stress distribution varies as the distance of the cross section
rom the right end of the strip increases, and the distribution becomes almost uniform
it x = 180 mm, i.e., at one width distance from the end of the stress application. The
‘otal amount of stress in any cross section is the same, i.e., 1 KN in the strip and, stress
s uniformly distributed and the magnitude of stress becomes 5 MPa at any cross
section at one width or larger distance from the end of the stress application.

my. .. .... ... Fy 8B J ao ee _*. _f Ty ST. ee ee, ec ee — Figure 3.64 <= 200mm, 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 center of the strip. The longitudinal stress distribution varies as the distance of the cross section rom the right end of the strip increases, and the distribution becomes almost uniform it x = 180 mm, i.e., at one width distance from the end of the stress application. The ‘otal amount of stress in any cross section is the same, i.e., 1 KN in the strip and, stress s uniformly distributed and the magnitude of stress becomes 5 MPa at any cross section at one width or larger distance from the end of the stress application. my. .. .... ... Fy 8B J ao ee _*. _f Ty ST. ee ee, ec ee

ANSYS Main Menu — Preprocessor — Modeling — Create > Areas > Circle
Solid Circle — ANSYS Main Menu — Preprocessor — Modeling — Create > Areas > Circle Solid Circle

72 Boundary conditions applied to the quarter model of the center notched plate.

Figure 3.75
Ina finite plate, the maximum stress at the foot of the hole is increased due to the
finite boundary of the plate. Figure 3.76 shows the variation of the stress concentration
factor a for elliptic holes having different aspect ratios with normalized major radius
2a/h, indicating that the stress concentration factor in a plate with finite width h is
increased dramatically as the ligament between the foot of the hole and the plate edge
becomes smaller. — Figure 3.75 Ina finite plate, the maximum stress at the foot of the hole is increased due to the finite boundary of the plate. Figure 3.76 shows the variation of the stress concentration factor a for elliptic holes having different aspect ratios with normalized major radius 2a/h, indicating that the stress concentration factor in a plate with finite width h is increased dramatically as the ligament between the foot of the hole and the plate edge becomes smaller.

An elastic plate with a crack of length 2a in its center subjected to uniform longitudinal
tensile stress og at one end and clamped at the other end as illustrated in Figure 3.77.
Perform an FEM analysis of the 2-D elastic center-cracked tension plate illustrated
in Figure 3.77 and calculate the value of the mode I (crack-opening mode) stress
intensity factor for the center-cracked plate. — An elastic plate with a crack of length 2a in its center subjected to uniform longitudinal tensile stress og at one end and clamped at the other end as illustrated in Figure 3.77. Perform an FEM analysis of the 2-D elastic center-cracked tension plate illustrated in Figure 3.77 and calculate the value of the mode I (crack-opening mode) stress intensity factor for the center-cracked plate.

Enlarged view of the singular elements around the crack tip.
Figure 3.85 — Enlarged view of the singular elements around the crack tip. Figure 3.85

Enlarged view of the longitudinal stress distribution around the crack tip.
Figure 3.89 — Enlarged view of the longitudinal stress distribution around the crack tip. Figure 3.89

Single edge cracked tension plate subjected to a uniform longitudinal stress at one end and
clamped at the other end.
Figure P3.14 — Single edge cracked tension plate subjected to a uniform longitudinal stress at one end and clamped at the other end. Figure P3.14

Elastic cylinder of mild steel with a radius of R pressed against an elastic flat plate of the same
steel by a force P’ and the FEM model of the cylinder-plate system. — Elastic cylinder of mild steel with a radius of R pressed against an elastic flat plate of the same steel by a force P’ and the FEM model of the cylinder-plate system.

Figure 3.92 “Part Annular Circ
Area” window. — Figure 3.92 “Part Annular Circ Area” window.

Figure 3.95 “Element Size on
Pick .. .” window.
Element Size on Pick... — Figure 3.95 “Element Size on Pick .. .” window. Element Size on Pick...

Figure 3.96
The meshing procedures are the same as usual:
COMMAND — Figure 3.96 The meshing procedures are the same as usual: COMMAND

ire 3.103 Nodes on the lower surface of the cylinder plotted in the “ANSYS Graphics” window.

Figure 3.105 “Reselect nodes”
window. — Figure 3.105 “Reselect nodes” window.

Enlarged contour of the y-component of stress around the contact point.
Figure 3.115 — Enlarged contour of the y-component of stress around the contact point. Figure 3.115

Figure P3.15
5 Acylinder having a radius of curvature Rj} = 500 mm pressed against another cylinder having
R2 = 1000 mm by a force P? = 1 kKN/mm. — Figure P3.15 5 Acylinder having a radius of curvature Rj} = 500 mm pressed against another cylinder having R2 = 1000 mm by a force P? = 1 kKN/mm.

igure P3.16 A cylinder having a radius of curvature R= 500 mm pressed against a flat plate.

Calculate the contact stress distribution between a cylinder of fine ceramics having
a radius of curvature Rj =500mm and a cylindrical seat of stainless steel having
Ry = 1000 mm against which the cylinder is pressed by a force P’ = 1 KN/mmas shown
in Figure P3.18. Young’s module of fine ceramics and stainless steel are E; = 320 GPa
and E, = 192 GPa, respectively and Poisson’s ratio of each material is the same, i.e.,
Vv) = v2 =0.3. The theoretical curve of the contact stress p(x) is given by Equation
(P3.18a) with the parameters expressed by Equations (P3.18b) to (P3.18d) below. — Calculate the contact stress distribution between a cylinder of fine ceramics having a radius of curvature Rj =500mm and a cylindrical seat of stainless steel having Ry = 1000 mm against which the cylinder is pressed by a force P’ = 1 KN/mmas shown in Figure P3.18. Young’s module of fine ceramics and stainless steel are E; = 320 GPa and E, = 192 GPa, respectively and Poisson’s ratio of each material is the same, i.e., Vv) = v2 =0.3. The theoretical curve of the contact stress p(x) is given by Equation (P3.18a) with the parameters expressed by Equations (P3.18b) to (P3.18d) below.

Calculate the contact stress distribution between a cylinder having a radius of cur-
vature Rj =500 mm and another cylinder having R2 = 1000 mm against which the
smaller cylinder is pressed by a force P= 1kN/mm as shown in Figure P3.15. The
upper cylinder is made of fine ceramics having Young’s modulus E, = 320 GPa and
Poisson’s ratio v; = 0.3, and the lower cylinder stainless steel having E, = 192 GPa and
v2 = 0.3. The theoretical curve of the contact stress p(x) is given by Equation (P3.17a)
with the parameters expressed by Equations (P3.17b) to (P3.17d) below. Compare
the result obtained by the FEM calculation with the theoretical distribution given by
Equations (P3.17a)—(P3.17d). — Calculate the contact stress distribution between a cylinder having a radius of cur- vature Rj =500 mm and another cylinder having R2 = 1000 mm against which the smaller cylinder is pressed by a force P= 1kN/mm as shown in Figure P3.15. The upper cylinder is made of fine ceramics having Young’s modulus E, = 320 GPa and Poisson’s ratio v; = 0.3, and the lower cylinder stainless steel having E, = 192 GPa and v2 = 0.3. The theoretical curve of the contact stress p(x) is given by Equation (P3.17a) with the parameters expressed by Equations (P3.17b) to (P3.17d) below. Compare the result obtained by the FEM calculation with the theoretical distribution given by Equations (P3.17a)—(P3.17d).

Note that Equations (P3.18a) to (P3.18d) cover Equations (P3.15a)—(P3.15d),
(P3.16a)—(P3.16d) and (P3.17a)—(P3.17d). — Note that Equations (P3.18a) to (P3.18d) cover Equations (P3.15a)—(P3.15d), (P3.16a)—(P3.16d) and (P3.17a)—(P3.17d).

Obtain the lowest three vibration modes and resonant frequencies in the y direction
of the straight steel bar shown in Figure 4.2. — Obtain the lowest three vibration modes and resonant frequencies in the y direction of the straight steel bar shown in Figure 4.2.

igure 4.3. Analytical vibration mode and the node position.

Figure 4.9 Window of Element Type for
Real Constants. — Figure 4.9 Window of Element Type for Real Constants.

The window Element Sizes on All Selected Lines, as shown in Figure 4.20, is
opened.
The window Mesh Lines, as shown in Figure 4.22, opens. — The window Element Sizes on All Selected Lines, as shown in Figure 4.20, is opened. The window Mesh Lines, as shown in Figure 4.22, opens.

Figure 4.23 Window of Apply
U,ROT on Nodes. — Figure 4.23 Window of Apply U,ROT on Nodes.

‘igure 4.46 Window of Density for Material Number 1.

le 4.1 X,Y, and Z coordinates of keypoints for suspension

Figure 4.55 Window of Subtract
Areas. — Figure 4.55 Window of Subtract Areas.

3. ~Window of Apply U,ROT on Lines.
Figure 4.63 — 3. ~Window of Apply U,ROT on Lines. Figure 4.63

Figure 4.77 A one-axis moving table using elastic hinges.

Figure 4.82 Window of Define Material Model Behavior.

ANSYS Main Menu — Preprocessor — Modeling — Create — Arbitrary >
Through KPs
The window Add Areas opens (Figure 4.89). — ANSYS Main Menu — Preprocessor — Modeling — Create — Arbitrary > Through KPs The window Add Areas opens (Figure 4.89).

ANSYS Main Menu — Solution — Define Loads — Apply — Structural -
Displacement — On Areas
The window Apply U,ROT on Areas opens (Figure 4.104). — ANSYS Main Menu — Solution — Define Loads — Apply — Structural - Displacement — On Areas The window Apply U,ROT on Areas opens (Figure 4.104).

Analyze the flow structure of an axisymmetric conical diffuser with diffuser angle
20 = 6° and expansion ratio = 4 as shown in Figure 5.2. — Analyze the flow structure of an axisymmetric conical diffuser with diffuser angle 20 = 6° and expansion ratio = 4 as shown in Figure 5.2.

< Figure 5.10 Window of Create Area
thru KPs. — < Figure 5.10 Window of Create Area thru KPs.

Window of Element Sizes on Picked Lines.
Figure 5.14 — Window of Element Sizes on Picked Lines. Figure 5.14

Figure 5.16 Window of Mesh
Areas. — Figure 5.16 Window of Mesh Areas.

The following steps are performed to set up the analysis of FLOTRAN.

Window of Vector Plot of Predefined Vectors.
Figure 5.31 — Window of Vector Plot of Predefined Vectors. Figure 5.31

Window of Plot of Path Items on Graph.
Figure 5.38 — Window of Plot of Path Items on Graph. Figure 5.38

ible5.5 X and Y coordinates of KPs for a flow channel

Figure 5.46 Window of
Subtract Areas. — Figure 5.46 Window of Subtract Areas.

Figure 5.59
.3.5.3 DETAILED VIEW OF THE CALCULATED FLOW VELOCITY — Figure 5.59 .3.5.3 DETAILED VIEW OF THE CALCULATED FLOW VELOCITY

First select the larger area (outer brick wall) and click [A] OK button in the frame
of Figure 6.11. Next, select the smaller area (inner concrete wall) and click [A] OK
button. The smaller area is subtracted from the larger area and the outer brick wall is
produced. It is shown in Figure 6.12.

Tleing an cimilar annrnarh tha innar ranrratoa urall fa panctrurtond Lream AATCVCO
which have been created. From ANSYS Main Menu select Preprocessor > Modelling — First select the larger area (outer brick wall) and click [A] OK button in the frame of Figure 6.11. Next, select the smaller area (inner concrete wall) and click [A] OK button. The smaller area is subtracted from the larger area and the outer brick wall is produced. It is shown in Figure 6.12. Tleing an cimilar annrnarh tha innar ranrratoa urall fa panctrurtond Lream AATCVCO which have been created. From ANSYS Main Menu select Preprocessor > Modelling

Figure 6.16 Global Element Sizes.
Figure 6.16 — Figure 6.16 Global Element Sizes. Figure 6.16

Heat flow across the wall plotted as vectors.
Figure 6.33 — Heat flow across the wall plotted as vectors. Figure 6.33

Figure 6.35
onto Path. The frame shown in Figure 6.37 appears. — Figure 6.35 onto Path. The frame shown in Figure 6.37 appears.

Figure 6.53
Figure 6.52
N\ Create Cylinder by Dimensions — Figure 6.53 Figure 6.52 N\ Create Cylinder by Dimensions

ure 6.58 Quarter symmetry model of the tank—pipe intersection represented by a single volume V1.
Figure 6.58 — ure 6.58 Quarter symmetry model of the tank—pipe intersection represented by a single volume V1. Figure 6.58

Figure 6.61 Meshed quarter symmetry model of
the tank—pipe intersection.
Figure 6.60 Mesh Volumes frame. — Figure 6.61 Meshed quarter symmetry model of the tank—pipe intersection. Figure 6.60 Mesh Volumes frame.

The resulting network of elements is shown in Figure 6.61.

them. The surface load relevant in this case is convection load acting on all nodes

Figure 6.66 Select Entities.
Figure 6.67 Apply Thermal Con.
vection on Nodes. — Figure 6.66 Select Entities. Figure 6.67 Apply Thermal Con. vection on Nodes.

2 eh eee ee

Activate both [A] All DOF and TEMP and input [B] TEMP value = 232°C as
shown in Figure 6.71. Finally, click [C] OK to apply temperature constraints on
nodes at the far edge of the tank. The steps outlined above should be followed to
apply constraints at nodes located at the bottom of the tank. From Utility Menu
select Select — Entities. The frame shown in Figure 6.72 appears. — 2 eh eee ee Activate both [A] All DOF and TEMP and input [B] TEMP value = 232°C as shown in Figure 6.71. Finally, click [C] OK to apply temperature constraints on nodes at the far edge of the tank. The steps outlined above should be followed to apply constraints at nodes located at the bottom of the tank. From Utility Menu select Select — Entities. The frame shown in Figure 6.72 appears.

igure 6.71 Apply Temperature to All Nodes.
Figure 6.71 — igure 6.71 Apply Temperature to All Nodes. Figure 6.71

Figure 6.74
I4\ Create Local CS at WP Origin — Figure 6.74 I4\ Create Local CS at WP Origin

ite local CS — At WP Origin. The resulting frame is shown in Figure 6.74.
located on inner surface of the pipe. In order to
include this subset of nodes, from Utility Menu — ite local CS — At WP Origin. The resulting frame is shown in Figure 6.74. located on inner surface of the pipe. In order to include this subset of nodes, from Utility Menu

Figure 6.76
select PlotCtrls — Symbols. The frame in Figure 6.77 shows the required selection:
[A] Arrows. — Figure 6.76 select PlotCtrls — Symbols. The frame in Figure 6.77 shows the required selection: [A] Arrows.

Convection surface loads displayed as arrows.
Figure 6.78 — Convection surface loads displayed as arrows. Figure 6.78

Figure 6.85
Results — Vector Plot — Predefined. The resulting frame is shown in Figure 6.85.
In Figure 6.85, select [A] Thermal flux TF and [B] Raster Mode. Pressing [C] OK
button implements selections and produces thermal flux as vectors. This is shown in
Figure 6.86.
6.4 HEAT DISSIPATION THROUGH RIBBED SURFACE

6.4.1 PROBLEM DESCRIPTION — Figure 6.85 Results — Vector Plot — Predefined. The resulting frame is shown in Figure 6.85. In Figure 6.85, select [A] Thermal flux TF and [B] Raster Mode. Pressing [C] OK button implements selections and produces thermal flux as vectors. This is shown in Figure 6.86. 6.4 HEAT DISSIPATION THROUGH RIBBED SURFACE 6.4.1 PROBLEM DESCRIPTION

Distribution of thermal flux vectors at the intersection between the tank and the pipe.
Figure 6.86 — Distribution of thermal flux vectors at the intersection between the tank and the pipe. Figure 6.86

Figure 6.96
Figure 6.96 Subtract Areas.
Figure 6.97 — Figure 6.96 Figure 6.96 Subtract Areas. Figure 6.97

Figure 6.105
Figure 6.105 — View of the fin with mesh network.
Select all areas of the fin except the bot-
tom area and click [A] OK button. The frame
created as a result of that action is shown in
Figure 6.107. — Figure 6.105 Figure 6.105 — View of the fin with mesh network. Select all areas of the fin except the bot- tom area and click [A] OK button. The frame created as a result of that action is shown in Figure 6.107.

Clicking [A] OK button starts the solution

ArOcecc — Clicking [A] OK button starts the solution ArOcecc

Apply heat transfer coefficient and surrounding temperature.
Figure 6.107 — Apply heat transfer coefficient and surrounding temperature. Figure 6.107

igure 7.20 Model after meshing process.
Figure 7.20 — igure 7.20 Model after meshing process. Figure 7.20

Figure 7.26
Figure 7.24 Select Areas for Target.
Figure 7.25 Select Areas for Contact. — Figure 7.26 Figure 7.24 Select Areas for Target. Figure 7.25 Select Areas for Contact.

“

: By pressing [A] Pick All button, a frame shown in Figure 7.35 is called up.
As shown in Figure 7.35, [A] DOF to be constrained = UZ and the [
Displacement value = 2. Pressing [C] OK button applies selected constraints.

Figure 7.35 — “ : By pressing [A] Pick All button, a frame shown in Figure 7.35 is called up. As shown in Figure 7.35, [A] DOF to be constrained = UZ and the [ Displacement value = 2. Pressing [C] OK button applies selected constraints. Figure 7.35

Figure 7.46
Figure 7.47 appears. — Figure 7.46 Figure 7.47 appears.

Plot — Nodal Solu. In appearing frame (see Figure 7.40), the following are selected

) Configuration of the contact between cylinder and two blocks.
Figure 7.49
7.2.2.2. MODEL CONSTRUCTION — ) Configuration of the contact between cylinder and two blocks. Figure 7.49 7.2.2.2. MODEL CONSTRUCTION

Figure 7.50
Figure 7.50 A half-symmetry model.
Figure 7.51 — Figure 7.50 Figure 7.50 A half-symmetry model. Figure 7.51

Figure 7.67 Checked Hex and Sweep options.
Figure 7.68 Volume Sweeping. — Figure 7.67 Checked Hex and Sweep options. Figure 7.68 Volume Sweeping.

Figure 7.73
Figure 7.73 Select body for Target. — Figure 7.73 Figure 7.73 Select body for Target.

Figure 7.83
7.2.2.7 _POSTPROCESSING
Contour plot of von Mises stress (nodal solution) is shown in Figure 7.87. — Figure 7.83 7.2.2.7 _POSTPROCESSING Contour plot of von Mises stress (nodal solution) is shown in Figure 7.87.

This is a 3D analysis and advantage could be taken of the inherent symmetry of

Pressing Close button on MeshTool frame, ends mesh generation stage.

Pair. As a result of this selection, a frame shown in Figure 7.116 appears.
In the option [A] Facets/element edge select 2 facets/edge and click [B] OK
button as shown in Figure 7.115. — Pair. As a result of this selection, a frame shown in Figure 7.116 appears. In the option [A] Facets/element edge select 2 facets/edge and click [B] OK button as shown in Figure 7.115.

a quarter-symmetry model. Furthermore, the isometric viewing direction so far used

Quarter symmetry model with elements, constraints, and loads.
Figure 7.131 — Quarter symmetry model with elements, constraints, and loads. Figure 7.131

Linear Isotropic Properties for Material Number 1.
Figure 7.148 — Linear Isotropic Properties for Material Number 1. Figure 7.148

Figure 7.156 Define Material Model Behavior.

Real Constants (Sets No. 2, 3, and 12 shown).
Figure 7.165 — Real Constants (Sets No. 2, 3, and 12 shown). Figure 7.165

Figure 7.174
From ANSYS Utility Menu select Select — Comp/Assembly > Create Compo-
As it is seen in Figure 7.177, component name, Cname = O-ring was entered. — Figure 7.174 From ANSYS Utility Menu select Select — Comp/Assembly > Create Compo- As it is seen in Figure 7.177, component name, Cname = O-ring was entered.

Figure 7.178 Select Entities (all elements
of Type 1: HYPER74). — Figure 7.178 Select Entities (all elements of Type 1: HYPER74).

Figure 7.186
creates a frame shown in Figure 7.187. — Figure 7.186 creates a frame shown in Figure 7.187.

Figure 7.201
7.2.4.8 SOLUTION (SECOND LOAD STEP)
The second load step involves applying, in addition to the load due to the wall
movement, pressure acting over the top surface of the O-ring. Because the pressure
effectively acts over the angle from 14° to 166°, therefore, to apply it properly it is

convenient to change coordinate system from Cartesian to Polar.
hee SOE ES 9 eS es Dees ce | soe NAP RAS ees te MATT Cette: Mew Lasse whew es — Figure 7.201 7.2.4.8 SOLUTION (SECOND LOAD STEP) The second load step involves applying, in addition to the load due to the wall movement, pressure acting over the top surface of the O-ring. Because the pressure effectively acts over the angle from 14° to 166°, therefore, to apply it properly it is convenient to change coordinate system from Cartesian to Polar. hee SOE ES 9 eS es Dees ce | soe NAP RAS ees te MATT Cette: Mew Lasse whew es

Figure 7.202 von Mises stress in O-ring
(deformed shape and undeformed edge).
Figure 7.203 WP Settings (Polar
system is selected).
Selections made are shown in Figure 7.206 and are implemented by clicking
A] OK. — Figure 7.202 von Mises stress in O-ring (deformed shape and undeformed edge). Figure 7.203 WP Settings (Polar system is selected). Selections made are shown in Figure 7.206 and are implemented by clicking A] OK.

Selections made are shown in Figure 7.214. This time, when constant pressure of

Figure 7.215 von Mises stress due
to squeeze and external pressure
(deformed shape only).
Now, the solution stage ought to be initiated. From ANSYS Main Menu select
Solution > Solve > Current LS. The frame shown in Figure 7.198 appears together
with another frame, which gives a summary of solution options selected. After check-
ing the correctness of information, it should be closed by selecting File > Close.
Next, [A] OK should be pressed to initiate the solution. — Figure 7.215 von Mises stress due to squeeze and external pressure (deformed shape only). Now, the solution stage ought to be initiated. From ANSYS Main Menu select Solution > Solve > Current LS. The frame shown in Figure 7.198 appears together with another frame, which gives a summary of solution options selected. After check- ing the correctness of information, it should be closed by selecting File > Close. Next, [A] OK should be pressed to initiate the solution.

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The journal publishes original and review articles covering the concept of technical science, energy and environment, industrial engineering, quality management and other related sciences. JAES is Open-Access Journal that follows new trends and progress proven practice in listed fi elds, thus creating a unique forum for interdisciplinary or multidisciplinary dialogue. JAES is part of the electronic journal editing with a transparent editorial and review policy. Provided are:

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