Report

Problem Statement:

Determine the maximum deformation and stress developed in both two cases mentioned below in description.

Description of model:

Material Data(in both cases)

Structural steel

Tensile yield strength=250MPa Poisson ratio=0.3

Tensile ultimate strength=460MPa Young\'s Modulus=2e+5 MPa

Meshing(in both cases)

Body sizing: 5mm

Case1:

Plate dimensions:

Length=10in
Height=4in
Thickness=1.5in
Circular Hole at the center:
Diameter=2in

Boundary conditions:

left side of plate is fixed and 10000 lbs force is applied on right face of plate as shown in picture

Case2:

Plate Dimensions:

Length=10in
Height=4in
Thickness=1.0in
Circular Hole at the center:
Diameter=2in

2 Circular holes of 1in diameter of both sides 3in apart as shown

Boundary conditions:

left side of plate is fixed and 10000 lbs force is applied on right face of plate as shown in picture

Analysis Result:

Case1:

Total Deformation :

Equivalent stresses:

Case2:

Total deformation:

Equivalent stresses:

Conclusion:

Result Summery:

Here comparsion is made between both cases. It can be seen that case1 have less deformation and less maximum stress induced comparing to case2.

Case1:

	Maximum	Minimum
Total Deformation(in ft)	6.3048e-5	0
Equivalent stresses(in psf)	1.044e6	28326

Minimum FOS=5.01

Case2:

	Maximum	Minimum
Total Deformation(in ft)	1.0249e-4	0
Equivalent stresses(in psf)	1.5065e6	5195.4

Minimum FOS=3.46

There are two models are given .

From analysis standpoint, case1 is more suitable because

1. Strength of plate is primary criteria for selection between both cases . Thus acccording to strength point of view model of case1 is fulfilling the requirement of maximum strength, because it is showing maximum deformation as 6.3048e-5 ft which is less than as in case2.

2.Factor of safety in both cases are 5.01 and 3.46 respectively. Thus first model have more strength towards fluctuating loads compare to second model.

In sense of Manufacturing , case2 is more suitable because

1. Case2 has FOS= 3.46 which is suffient enough against using FOS=5.01(For case1). As ductile material have been used ,very high FOS is not required(High FOS is used in case of Brittle material) . So ,here design(case2) having FOS=3.46 should be selected.

2. As there are three holes are required in case2. Drilling is increasing total cost in case2 compare to case1, but performance of product will increase as reduction in weight which is our ultimate requirement.

3. It is seen that too much excess material had been used in case1 which is not a good practice in designing.

Removing excess material will not only reduce weight but also performance. It can be seen in disc brake there are many holes are provided in disc . These holes not only reduces weight but also increases surface area for heat interaction which ultimatelty increases performance.

In respect of material required, both will require same amount of material for manufacturing.

In respect of cost, case2 will have more cost compare to case1.

In respect of performance, design in case 2 will perform more than as in case1.

References:

Machine Elements , VB Bhandari.