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1. Objective: To Perform a static analysis on two models of a plate with holes. (Axial Loading). CASE-1 CASE-2 Length=300mm Length=300mm Height=120mm Height=120mm Thickness=30mm Thickness=30mm Circular Hole at the center Diameter=60mm Circular Hole at the center Diameter=60mm -----------------NA-------------------…
Neeraj Dixit
updated on 12 Aug 2022
To Perform a static analysis on two models of a plate with holes. (Axial Loading).
CASE-1 |
CASE-2 |
Length=300mm |
Length=300mm |
Height=120mm |
Height=120mm |
Thickness=30mm |
Thickness=30mm |
Circular Hole at the center |
Circular Hole at the center |
-----------------NA------------------- |
Two holes of 30 mm diameter are placed 90mm away from the central hole on both sides. |
1.1 Boundary Condition:
Plate Fixed at Left Side
Load 500N applied on the right Face.
Various methodologies are developed to get the best advantages from the material to produce an optimized product. Stress calculation becomes a vital part of this process to design a product that won't break or deform. Geometrical discontinuities such as notches, holes should be considered as they make the component weaker. These geometrical discontinuities in the component lead to stress concentration in the localized area.
2.1 Geometric Irregularities:
2.2 Concept of Stress Concentration:
When an axial force is applied to a member, it creates a complex stress distribution within the localized region of the point of load application. Not only do complex stress distributions arise just under a concentrated loading, but they can also arise at sections where the member’s cross-sectional area changes. For example, consider the bar in Fig.1 a, which is subjected to an axial force P. Here the once horizontal and vertical grid lines deflect into an irregular pattern around the hole centered in the bar. The maximum normal stress in the bar occurs in section a – a, which is taken through the bar’s smallest cross-sectional area. Provided the material behaves in a linear-elastic manner, the stress distribution acting on this section can be determined either from mathematical analysis, using the theory of elasticity, or experimentally by measuring the strain normal to section a – a and then calculating the stress using Hooke’s law, [σ=E⋅ε]. Regardless of the method used, the general shape of the stress distribution will be as that shown in Fig.1 b. Similarly, if the bar has a reduction in its cross-section, achieved using shoulder fillets as in Fig.2 a, then again the maximum normal stress in the bar will occur at the smallest cross-sectional area, section a – a, and the stress distribution will look like that shown in Fig.2 b. In engineering practice, the actual stress distributions in Fig.1 b and Fig.2 b do not have to be determined. Instead, only the maximum stress at these sections must be known, and the member is then designed to resist this stress when the axial load P is applied.
Specific values of this maximum normal stress can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity. The results of these investigations are usually reported in graphical form using a stress-concentration factor K. We define K as a ratio of the maximum stress to the average normal stress acting at the cross-section; i.e.,
Specific values of K are generally reported in handbooks related to stress analysis. Note that K is independent of the bar’s material properties; rather, it depends only on the bar’s geometry and the type of discontinuity. As the size r of the discontinuity is decreased, the stress concentration is increased. the maximum normal stress will be three times greater than the average normal stress on the smallest cross-section.
2.3 The theoretical stress concentration factor:
The theoretical stress concentration factor can be calculated with the handbook data.
the data is based on the diameter of the hole and the width of the plate.
this data can be used for a plate with a hole subjected to uniaxial loading.
The stress-concentration factors given in were determined on the basis of static loading, with the assumption that the stress in the material does not exceed the proportional limit. If the material is very brittle, the proportional limit may be at the fracture stress, and so for this material, failure will begin at the point of stress concentration. Essentially a crack begins to form at this point, and a higher stress concentration will develop at the tip of this crack. This, in turn, causes the crack to propagate over the cross-section, resulting in a sudden fracture. For this reason, it is very important to use stress concentration factors in design when using brittle materials. On the other hand, if the material is ductile and subjected to a static load, it is often not necessary to use stress-concentration factors since any stress that exceeds the proportional limit will not result in a crack. Instead, the material will have reserve strength due to yielding and strain hardening. In the next section, we will discuss the effects caused by this phenomenon.
Stress concentrations are also responsible for many failures of structural members or mechanical elements subjected to fatigue loadings. For these cases, a stress concentration will cause the material to crack if the stress exceeds the material’s endurance limit, whether or not the material is ductile or brittle. Here, the material localized at the tip of the crack remains in a brittle state, and so the crack continues to grow, leading to a progressive fracture. As a result, one must seek ways to limit the amount of damage that can be caused by fatigue.
2.4 Methods to reduce stress concentration:
2.5 Examples:
1. Failure of this steel pipe in tension occurred at its smallest cross-sectional area, which is through the hole. Notice how the material yielded around the fractured surface.
2. This saw blade has grooves cut into it in order to relieve both the dynamic stress that develops within it as it rotates and the thermal stress that develops as it heats up. Note the small circles at the end of each groove. These serve to reduce the stress concentrations that develop at the end of each groove.
3.1 Formula:
Stress concentration factor can be calculated based on the following formula.
[Kt=σmaxσnom]
Where:-
Kt = Theoretical Stress concentration factor
σmax = Maximum stress
σnom = Nominal stress is Average stress in a component having uniform cross-sectional maximum stress- stress occurred due to discontinuity.
The stress is calculated based on the elastic theory is called nominal stress. where in the elastic theory we assume that the material obeys the hooks law.
3.2 Calculating the maximum stress through the theoretical stress concentration factor:
we know that [Kt=σmaxσnom]
hence, σmax=Kt⋅σnom
To calculate the max stress we need to calculate first the nominal stress based on the following formula.
σnom=FA
Where:-
A=(w−d)⋅t)
F = load
W = width
d=dia for a hole
t = thickness of the plate
3.2.1 The calculation for Case-1:
a. Calculating Nominal Stress:
Because σnom=F(w−d)⋅t
σnom=500(120−60)⋅30
σnom=0.277 Mpa
b. Calculating D/W ratio:
we need to find out Value to stress concentration factor from Graph.
Therefore, we need a D/W ratio.
Hence,
So, from the graph, the value of stress concentration factor for 0.5 D/W ratio is 2.1.
Hence Kt=2.1
c. Calculating Theoretical Maximum Stress:
because σmax=Kt⋅σnom becomes
hence, σmax=(2.1)⋅(0.277)=0.5833
3.2.2 The calculation for Case-2:
a. Calculating Nominal Stress:
Because σnom=F(w−d)⋅t
therefore for three holes,
σnom=500((120−60)+(120−30)+(120−30))⋅30
σmax=0.0694 Mpa
b. Calculating D/W ratio:
we need to find out Value to stress concentration factor from Graph as we discussed in case-1.
Therefore, we need a D/W ratio.
Hence, DW=30120=14=0.25
So, from the graph, the value of stress concentration factor for 0.25 D/W ratio is 2.4.
Hence Kt=2.4
But in this case, we have three holes in the plate out of which the dimension of a larger hole is the same as the case-1 so the theoretical stress concentration factor for this hole is the same as before. i.e Kt=2.1 . and for remaining, 2 holes have Kt=2.4
Hence overall theoretical stress concentration factor is as follows.
Kt=2.1+2.4+2.4=6.9
c. Calculating Theoretical Maximum Stress:
because σmax=Kt⋅σnom becomes
hence, σmax=(6.9)⋅(0.0694)=0.4761 Mpa
3.3 Analytical Method:
Other than the graphical method we can calculate the theoretical stress concentration value using the Analytical Method as discussed below.
Kt=3−3.14⋅(DW)+3.667⋅(DW)2−1.527⋅(DW)3
1. for case -1:
Kt=3−3.14(0.5)+3.667(0.5)2−1.527(0.5)3
Kt=2.115
hence,
σmax=(2.115)⋅(0.277)=0.5858 Mpa
1. for case -2:
As we know we have three holes in the plate out of which the major hole has the same diameter as case-1. so that stress concentration value for this hole is Kt=2.115.
we need to calculate the stress concentration value for the other two holes.
hence, stress concentration value for the 30mm diameter holes having D/W ratio is 0.25.
Kt=3−3.14⋅(0.25)+3.667⋅(0.25)2−1.527(0.25)3
Kt=2.419890
therefor, Overall value is Kt=2.155+2.419890+2.419890=6.9947
Hence, σmax=(6.9947)⋅(0.0694)=0.485763 Mpa
4.1 Consideration for both Cases:
i. Material
The Material is considered stainless steel whose properties are shown below.
ii. Mesh Size: 5 mm
iii. Method: Tetrahedrons with patch confirming algorithm.
iv. Boundary Condition: Plate Fixed on Left Side and Load 500N applied on the right Face.
v. Analysis: 3D Static Structural.
4.2 Case-1:
Step-1: Sketching
First, we are selecting the static structural analysis system to do the analysis. after defining the materials as stainless steel from the Engineering data sources.
Now we are using space claim to generate the 3D Geometry to do analysis. hence, first Right click on the Geometry tab and select the space claim and draw the sketch as shown below.
Step-2: converting Sketch into 3D
Step-3 Defining Material
Step-4 Defining Mesh
Step-5 Defining Boundary condition
Step-6 Defining Solution
Step-7 Final Result
4.3 Case-2:
Step-1: Sketching
Step-2 converting Sketch into 3D
Step-3 Defining Material
Step-4 Defining Mesh
Step-5 Defining Boundary condition
Step-6 Defining Solution
Step-7 Final Result
Comparing the deformation and maximum stress developed on a structural steel model for the 2 cases.
|
Stress N/mm^2 |
Displacement (mm) |
Case-1 |
0.59932 |
0.00028 |
Case-2 |
0.57599 |
0.00030 |
Analytical |
Stress N/mm^2 |
|
Case-1 |
0.5833 | |
Case-2 |
0.4761 |
In case-1 the stress concentration is uniform away from the hole and it is non-uniform in the middle of the cross-section. i.e at the hole. the same result is applied to case-2 also, but the number of holes increased which causes more non-uniform stress.
The Displacement result is approximately the same in both cases. also, the Equivalent stress in case 2 is less than that in case 1
We know that Stress concentrations occur at sections where the cross-sectional area suddenly changes. The more severe the change, the larger the stress concentration.
For design or analysis, it is only necessary to determine the maximum stress acting on the smallest cross-sectional area. This is done using a stress concentration factor, K, that has been determined through experiment and is only a function of the geometry of the specimen.
Normally the stress concentration in a ductile specimen that is subjected to a static loading will not have to be considered in design; however, if the material is brittle, or subjected to fatigue loadings, then stress concentrations become important.
The stress concentration factor is independent of the material it only depends upon the Geometry in many materials but for some materials, the stress concentration factor depends on the type of material also.
In a ductile material, the stress concentration is neglected because when material yields (i.e permanently deforms) it will move stress away from the stress concentration zone around the hole and forces the material around the cross-section of the hole to carry more stress. i.e the material which was holding the fewer stresses is now compensated to carry the stresses and equalized all over the material.
Note: In case 2 we need to refine the mesh because the Hand Calculation and Ansys Results are having more errors.
Based on the above discussion we can say that the machining work required in case 1 is less than in case 2. while the material requirement is less in case 2.
Hence, from the manufacturing point of view, if the time and machining are important then we can consider Case 1 while if the material weight is important then we can consider case 2.
Reference:-
mechanics of materials rc hibbeler
https://mechanicalc.com/calculators/stress-concentration/
The Flow of Report:-
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