Theoretical stress analysis of a single closed link

THEORETICAL STRESS ANALYSIS OF A SINGLE CLOSED LINK AND

VERIFICATION USING ANSYS

Trisha Gopalakrishna -1RV09ME106

Zoish Tosher Hormusjee -1RV09ME118

Neelabh Mishra- 1RV09ME121

Suraj Appachu- 1RV09ME123

Under the Guidance of P.R. VENKATESH

ProfessorMechanical Engineering, R.V. College of Engineering

Department of Mechanical EngineeringR.V College of Engineering

Bangalore

Contents

• Introductiona. Problem Statementb. Methodology

• Theory & Fundamentalsa. Theoretical Analysisb. Experimental Approach- Report from fabricatorc. ANSYS

• Results

• Conclusion

IntroductionProblem Statement

• A chain link is made of 16 mm diameter steel rod. The mean radius of the semicircular ends is 50 mm and the length of the straight portion of the link is 80 mm. Determine the maximum tensile and compressive stress when the link is subjected to a pull of 5 kN. Choose rectangular and trapezoidal cross sections accordingly and carry out the analysis.

Load= 5000NMean Radius = 50 mm

Scale=2.0

Methodology

Theoretical Analysis• Case 1- Considering Circular Cross Section

Step 1-Cross Sectional Area [A]

Step 2- Location of Neutral Axis [Rn]­­

By symmetry C₁ = C₂ =C = 8 mm

R= Ri + C₁ = 50 mmRi= 42 mm

Ro= R + C2 = 50 + 8Ro = 58 mm

Rn =

Rn= 49.678 mm

Step 3- Location of centroidal axis [R] and eccentricity [e]

R= Ri + C1 = 50 mm

Distance from the centroidal axis to the neutral axis, eccentricity, e= R- Rn = 50 – 49.768 = 0.322 mm

Step 4- Consider section A-A

Direct tensile stress when θ= 90˚ ; σd = = 0

Bending moment M=

M=

M= 94.9 x 10³ N mm

Inner fibres are subjected to compressive stress and outer fibres are subjected to tensile stress.

Therefore, net stress at the inner fibres

σi= σd- σbi

σi= =

σi= -267.96 N/mm²

Net stress at the outer fibres,

σo= σ­d + σbo

σo =

σ0 = 290.44 N/mm²

Step 5- Consider section B-B

Direct tensile stress when θ= 0˚ ; σd=

Bending moment M =

M=

M= -30.095 x 10³ Nmm

Inner fibres are subjected to tensile stress and outer fibres are subjected to compressive stress

Therefore, net stress at the inner fibresσi= σd + σbi

σi =

σi= 97.4 N/mm²

Net stress at the outer fibres,σo= σd-σbo

σo=

σo= 54.26 N/mm2

• Case 2- Considering Rectangular Cross Section

Step 1-Cross Sectional Area [A]

Step 2- Location of Neutral Axis [Rn]By symmetry C₁ = C₂ C = 8 mm

R= Ri + C₁ = 50 mmRi= 42 mm

Ro= R + C2 = 50 + 8Ro = 58 mm

Rn =

Rn= 49.57 mm

Step 3- Location of centroidal axis [R] and eccentricity [e]

R= 50 mm

Distance from the centroidal axis to the neutral axis, eccentricity, e= R- Rn = 50 – 49.57 = 0.43 mm

Step 4- Consider section A-ADirect tensile stress when θ= 90˚ ; σd = Bending moment M=

M=

M= 94.9 x 10³ N mm

Inner fibres are subjected to compressive stress and outer fibres are subjected to tensile stress.

Therefore, net stress at the inner fibresσi= σd- σbi

σi=

σi= -155.383 N/mm²

Net stress at the outer fibres,σo= σ­d + σbo

σo =

σ0 = 125.30 N/mm²

Step 5- Consider section B-B

Direct tensile stress when θ= 0˚ ; σd=

Bending moment M =

M=

M= -30.095 x 10³ Nmm

Inner fibres are subjected to tensile stress and outer fibres are subjected to compressive stress

Therefore, net stress at the inner fibresσi= σd + σbi

σi =

σi= 59.04 N/mm²

Net stress at the outer fibres,σo= σd-σbo

σo =

σo= -29.97 N/mm2

• Case 3- Considering Trapezoidal Cross Section

Step 1-Cross Sectional Area [A]

Step 2- Location of Neutral Axis [Rn]R= 50 mm

b0= b₁-b= 18-14= 4 mm

C2=

C2= 8.33 mm

C1= 16 – 8.33 = 7.66 mm

Ri= R- C1= 50-7.66=42.34 mm

Ro= R + C2= 50 + 8.33= 58.33 mm

Rn= 49.664 mm

Step 3- Location of centroidal axis [R] and eccentricity [e]

R= 50 mm

Distance from the centroidal axis to the neutral axis, eccentricity, e= R- Rn = 50 – 49.664 = 0.336 mm

Step 4- Consider section A-ADirect tensile stress when θ= 90˚ ; σd =

Bending moment M=

M= M= 94.9 x 10³ N mm

Inner fibres are subjected to compressive stress and outer fibres are subjected to tensile stress.

Therefore, net stress at the inner fibresσi= σd- σbi

σi=

σi= -190.84 N/mm²

Net stress at the outer fibres,σo= σ­d + σbo

σo =

σ0 = 163.91 N/mm²

Step 5- Consider section B-B

Direct tensile stress when θ= 0˚ ; σd=

Bending moment M =

M=

M= -30.095 x 10³ Nmm

Inner fibres are subjected to tensile stress and outer fibres are subjected to compressive stress

Therefore, net stress at the inner fibresσi= σd + σbi

σi =

σi= 70.28 N/mm²

Net stress at the outer fibres,σo= σd-σbo

σo=

σo= -42.21 N/mm2

Plan-Fabricate three specimens for analysis purpose.

Report from Fabricator/ Drawbacks-1. Time- Time required for fabrication is about 3 weeks. 2. Cost- Rs 1500/- per specimen. Hence, total cost is about Rs

4500/-3. Modification of Design- To accommodate the accessories of

the polarizer

Hence, because of financial and time constraints our guide advised us to go ahead with theoretical an FEA analysis as of now and then complete the experimental approach.It is still a part of this mini project.

Report from fabricator

ANSYSThe modeling is performed on ANSYS 14.0. By defining the key points and joining them using lines the basic structure is obtained as shown

The arcs forming the cross-section are then extruded along the lines to obtain areas as shown

After obtaining the areas, they are then converted into volumes. Different sections of the volume are then combined together using the Boolean-Add option.

Models for all three cross sections

Element Type Used:

The element type used is SOLID185 (Brick 8 Node 185). SOLID185 is used for 3-D modelling of solid structures. It is defined by eight nodes having three degrees of freedom at each node: translations in the nodal x, y, and z directions.

Material Properties:

Young’s Modulus : 2.1 x 105 Mpa

Poisson’s Ratio : 0.3

The models are then meshed using the tetrahedral mesh.

Meshed Models

Results

SECTION A-AANSYS outer fibre stress = 246.2 N/mm2

ANSYS inner fibre stress = 109.7 N/mm2

SECTION B-BANSYS outer fibre stress = 0 N/mm2

ANSYS inner fibre stress = 27.8N/mm2

SECTION A-AANSYS outer fibre stress = 124.3 N/mm2

ANSYS inner fibre stress = 70.1 N/mm2

SECTION B-BANSYS outer fibre stress = 0.45 N/mm2

ANSYS inner fibre stress = 27.7 N/mm2

SECTION A-AANSYS outer fibre stress = 182.9 N/mm2

ANSYS inner fibre stress = 91.5 N/mm2

SECTION B-BANSYS outer fibre stress = 0 N/mm2

ANSYS inner fibre stress = 27.45 N/mm2

There are deviations between the theoretical results and ANSYS results. We have tried to reduce these deviations to the maximum possible extent by 1) increasing the number of meshes and 2) refining the mesh as much as possible as after a certain point the work load was too much leading to system crashing.

Conclusion• The theoretical calculations were made and verified with ANSYS. It was found that

the rectangular cross section link has lesser stresses at the chosen sections on application of the load.

• But links of rectangular cross sections are not manufactured and mostly links of circular cross sections are manufactured mainly due to the high production costs involved in manufacturing rectangular cross section. The production machinery and techniques used in manufacturing is extrusion which is cheaper and easier for circular sections. Also, no accountability for the stress concentration is required at the edges of circular cross sections.

• In this mini project we basically understood how the stress distribution occurs in real time. We mainly understood the basics of working on ANSYS which is a powerful analysis took and knowledge of which is essential in various fields of engineering today.

• We intent to complete the experimental approach using photoelastic methods of analysis including fabrication of the given geometry. . For this purpose, a study of various photo elastic materials will be done taking into account strength, fabrication techniques, cost and feasibility.

THANK YOU