Upcoming SlideShare
×

# Stress Analysis of a Rectangular Plate with a hole in the center - PDF

10,380 views

Published on

2 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
10,380
On SlideShare
0
From Embeds
0
Number of Embeds
10
Actions
Shares
0
252
0
Likes
2
Embeds 0
No embeds

No notes for slide

### Stress Analysis of a Rectangular Plate with a hole in the center - PDF

1. 1. FINITE ELEMENT STRUCTURAL ANALYSIS PROJECT – 1 (INDIVIDUAL PROBLEMS) SASI BEERA 35763829
2. 2. Aim: To analyze a symmetric plate with a hole having a unit thickness for two load cases, 1. Tensile load at the two ends of the plate 2. Bending loads at the two ends. Applying varying tensile pressure at the top half and varying compressive load at the bottom half simulates bending effect. Assumptions: The following assumptions are made in the analysis, 1. The analysis is linear static 2. The material is linear isotropic 3. The problem is considered to be plane stress. I.e. the stresses in the Z direction are ignored. 4. Geometric symmetry is exploited Units: Geometrical dimensions : mm Pressure : MPa Geometry: The dimensions for the geometry are calculated from the following set of equations based on the person number. Only the quarter symmetry is analyzed for the tensile load problem L/H = α R/H=β Where α = (K+1)/2 = 1.5 And β = 1/ (l+3) = 1/12 Calculated from the person number Nu = 0.4 Assuming R = 1 mm
3. 3. From the above, L = 18mm, H = 12mm, Nu = 0.4 BCs and loads: Constrained in x direction Pressure of 6MPa Constrained in y direction Ansys Macro For Tensile Load Case: !!! Macro for a plate with a Hole!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Finish /clear
4. 4. /title,plate_with_hole !!############ !!Total 8 key points are required to create quarter symm model !!############ k,1,0,0 !! Hole centre point k,2,2.5,0 !! Hole first point k,3,25,0 k,4,25,25 k,5,0,25 k,6,0,2.5 !! Hole last point k,7,5,0 k,8,0,5 !--------------------------------------------------------------------- larc,2,6,1,2.5 creation of arc line for hole larc,7,8,1,5 !-------------------------------------------------------- lstr,2,7 lstr,7,3 !! creation of straight line lstr,3,4 ! creation of straight line lstr,4,5 ! creation of straight line lstr,5,8 ! creation of straight line
5. 5. Pressure load of 6MPa Constrained in X direction Constrained in Y direction lstr,8,6 !------------------------------------------------------ !al,all !!creation of area by line al,p !!creation of area by graphical picking !----------------------------------------------------- et,1,42 !! element type definition 4 noded !et,1,82 !! element type defenition 8 noded !keyopt,1,3,3 !! setting keyoption to "plane stress with thickness, this is to include thickness keyopt,1,3,0 !!plane stress !r,1,2 !! defining thickness as real constant, thickness values assumed as 2mm !mp,ex,1,210000 !! material prop definition !mp,nuxy,1,0.35 mp,ex,1,289580 !! Young's Modulus for Berillium Alloy mp,nuxy,1,0.1 !! Poisson's ration for Berillium Alloy type,1 !! Activate the defined element type ( required element for meshing the current selected area) real,1 !! Activate real constant mat,1 !! Activate material property
6. 6. !lesize,all,,,12 !! define line size lesize,p mshkey,0 !! activate mesh type like mapped or free etc amesh,all !!------------------------------------------------------ !! now to apply BC's !!------------------------------------------------------ nsel,s,loc,x,0.001,-0.001 !! select nodes at the symmetry section d,all,ux,0 !! X - constarints nsel,s,loc,y,0.001,-0.001 !! select nodes at the symmetry section d,all,uy,0 !! Y - constarints !nsel,s,loc,x,9.9999,10.001 !! select nodes for load application nsel,s,loc,x,24.9999,25.001 sf,all,pres,-6 !! Apply a tensile pressure load of 6 MPa !! Solution alls !! do an all selection before entering the solution :) save,platewithhole,db !! save db, this db is loaded with mesh, Mat props and BC's :) solve !!------------END----------------------
7. 7. Results for Tensile Plane 42 Iteration 1: Max stress = 7.483 MPa Iteration 2:
8. 8. Max stress = 7.2288 MPa Iteration 3: Max stress = 11.467 MPa Iteration 4:
9. 9. Max stress = 15.918 MPa Consolidated Results: Plate under Tension (Plane 42 element) No of No of Displacement Stress Strain Stress % Error elements Nodes (mm) (MPa) Energy Concentrati in Energy (N-mm) on Factor norms 11 37 0.246E-04 7.4831 0.04956 1.643 3.456 24 69 0.258E-04 7.2288 0.04975 2.437 1.945 40 119 0.243E-04 11.467 0.04977 2.681 1.421 258 940 0.28809E-04 15.918 0.04977 2.702 0.829 The solution converged for the iteration 4 in terms of stress and strain energy.
10. 10. Stress concentration Factor = Max Stress/Nominal Stress Nominal Stress = Force/ Original Cross sectional Area The results converged with the 4th iteration in terms of stress and strain energy. Results for Tensile Plane 82 Iteration 1: Max stress = 16.139 MPa Iteration 2:
11. 11. Max stress = 17.456 MPa Iteration 3:
12. 12. Max stress = 17.828 MPa Iteration 4: Max stress = 18.091 MPa
13. 13. Consolidated Results: Plate under Tension (Plane 82 element) No of No of Displacement Stress Strain Stress % Error elements Nodes (mm) (MPa) Energy Concentrati in Energy (N-mm) on Factor norms 158 706 0.270E-04 16.139 0.04975 2.467 5.1250 638 1937 0.298E-04 17.456 0.04978 2.685 0.5714 2064 7578 0.348E-04 17.828 0.04977 2.708 0.2721 18426 64405 0.348E-04 18.091 0.04977 2.717 0.021 From the above consolidated tables, it can be clearly seen that, with a higher order element where in more number of nodes are present for the same element density gives a better accuracy in the results. Hence, it is better to use higher order elements. Only the snag with higher order elements in bigger FEA problems is that the computational time required is enormously high.
14. 14. Bending condition The same macro file shown above can be used for this case too with minor modification. Geometry and Loading: Variable tensile loading of 6 MPa Results for Bending (Plane 42)
15. 15. Max stress = 6.012 MPa Consolidated results: Plate under Bending (Plane 42 element) No of No of Displacement Stress Strain % Error in elements Nodes (mm) (MPa) Energy (N- Energy mm) norms 61 75 6.246E-04 6.121 0.02478 3.916 225 241 6.258E-04 6.019 0.02489 2.1259 502 554 6.243E-04 6.012 0.02583 1.4391 848 973 6.288E-04 6.012 0.02583 1.0130 The results converged for the fourth iteration in terms of the stress and the strain energy. The bending stresses for this case are seen on the top and the bottom fibers. Hence, the presence of the hole does not contribute to the bending stresses. Therefore the stress concentration factor will turn out to be 1 for all the cases. Bending Stress = (M*y)/I Where M = Bending moment
16. 16. Y = Distance to the max stress location I = Area moment of inertia Results for Bending (Plane 82) Max stress = 6.009 MPa
17. 17. Consolidated Results: Plate under Bending (Plane 82 element) No of No of Displacement Stress Strain % Error in elements Nodes (mm) (MPa) Energy (N- Energy mm) norms 57 209 6.046E-04 6.118 0.02477 3.916 222 774 6.058E-04 6.009 0.02489 2.1259 513 1422 6.063E-04 6.009 0.02489 1.4391 The solution converged for the iteration 3 in terms of stress and strain energy. From the above table it can be clearly seen that, with a higher order element where in more number of nodes are present for the same element density gives a better accuracy in the results and with less number of iterations. Hence, it is better to use higher order elements. Only the snag with higher order elements in bigger FEA problems is that the computational time required is enormously high. Conclusions: 1. A refined mesh gives accurate results. Once the convergence is achieved, further refinement of the mesh will produce the same results. This is illustrated by the graph below. 2. Use of higher order elements yields accurate results with less number of iterations, as is seen from the discussion in the report.