Week 1 Stress Concentration on a Plate with hole

Aim :

To perform static structural analysis on plate with hole using Ansys workbench to analyze stress concentration.

Objectives :

• Determine the maximum deformation and stress developed in the model.
• Compare the deformation and maximum stress developed on a structural steel model for the 2 cases.
• From an analysis standpoint, which of the design will you select.
• From a manufacturing standpoint, which of the design will you select.

Case 1 :

Design a plate with a hole model similar to what we worked in the class, but with increased thickness.
Determine the maximum deformation and stress developed in the model.

Plate dimensions:

1. Length=300mm
2. Height=120mm
3. Thickness=30mm
4. Circular Hole at the center: Diameter=60mm
   
   

Case 2 :

Smaller holes 90mm away from the center: Diameter=30mm

Boundary Conditions:

1. Plate fixed at left face.
2. Load of 500N on the right face.

Procedure :

A) Ansys-workbench setup and material selection :

• Open Ansys-workbench and select static structural and then click on Engineering data to select material.
• From engineering data select material as structural steel.

B) Geometry :

• Open spaceclaim and create geometry for both cases respectively.
• Assign thickness to the geometry using Pull tool.

Case 1 :

Case 2 :

C) Meshing :

• Go to mesh and insert method as tetrahedrones after selecting geometry.

• Give element size as 2 mm and generate mesh.

D) Boundary conditions :

• Go to static structural.
• Insert fixed support and select the face.

• Again go to static structural.
• Insert Force and select the face.
• Give load value as 500 N.

Give similar meshing and boundary conditions for Case 2 setup

E) Solution :

• Go to solution and insert total deformation.
• Similarly insert Equivalent stress and Equivalent strain.
• Go to Static structural and solve to get results.
• All results are to be at true scale.

Results :

Case 1 : Plate with a hole.

Total Deformation :

Minimum : 0 mm

Maximum : 2.7492 e-4 mm

Average : 1.3736 e-4 mm

Equivalent Stress :

Minimum : 8.2318 e-3 Mpa

Maximum : 0.60606 Mpa

Average : 0.17438 Mpa

Equivalent Strain :

Minimum : 4.305 e-8

Maximum : 3.0303 e-6

Average : 8.7428 e-7

Case 2 : Plate with three hole.

Total Deformation :

Minimum : 0 mm

Maximum : 2.9728 e-4 mm

Average : 1.4486 e-4 mm

Equivalent Stress :

Minimum : 1.1789 e-3 Mpa

Maximum : 0.5827 Mpa

Average : 0.17255 Mpa

Equivalent Strain :

Minimum : 1.2465 e-8

Maximum : 2.9135 e-6

Average : 8.6661 e-7

Result Table :

Sr. No.	Maximum Values of solution	Case 1	Case 2
1	Total Deformation (mm)	2.7492 e-4	2.9728 e-4
2	Equivalent Stress (Mpa)	0.60606	0.5827
3	Equivalent Strain	3.0303 e-6	2.9135 e-7

Conclusion :

• The plate with three holes (Case 2) has more total deformation than plate with one hole (Case 1).
• Case 1 has more maximum equivalent stress and strain values than Case 2.

From an analysis standpoint :

• One should go with Case-2 cause which helps in reducing stress concentration.
• And this helps in improving the components life without compromising its functionality.
• In further iteration larger dia holes holes can be replaced by smaller dia holes which give more relief from stress concentration.

From a manufacturing standpoint :

• One should go with Case-1 because it consumes less time and less cost to manufacture.
• Since in Case-2 there are additional holes which cause increase in overall labour cost, material and time to manufature this component.