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1.0 Design of Static Strength 1.0 Introduction 1.1Modes of Failure 1.2Static Strength 1.2.1 Four Design Categories 1.3 Static Loads and Factor of Safety 1.4 Failure Theories 1.5 Maximum-Normal-Stress Theory 1.6 Maximum-Shear-Stress Theory 1.7 Failure of Ductile Materials under Static Loading 1.7.1 The Distortion-Energy Theory 1.8 Failure of Brittle Materials Under Static Loading 1.8.1 Even and Uneven Materials 1.8.2 The Coulomb-Mohr Theory 1.8.3 The Modified-Mohr Theory 1.9 Flow Chart for Static Failure Analysis 1.10 Important Equations Used

1/44

Introduction STRENGTH is a property or characteristic of a

material or of a mechanical element.

Static Load Sometimes a load is assumed to be static when it is known that some variation is to be expected. 2/44

Failure due to Torsion with corrosion fatigue 3/44

4/44

5/44

6/44

Impact failure on lawn mower blade

7/44

Bolt & Nut tensile failure 8/44

Fatigue failure due to bending load 9/44

Failure due to crack initiation by stress concentration 10/44

11/44

Failure caused by spring surge in an oversped engine

12/44

13/44

14/44

15/44

16/44

Modes of Failure Ductile material is one which has a relatively large tensile strain before fracture takes place. For example, steels an aluminium. Brittle materials has a relatively small tensile strain before fracture. For example, cast iron. 17/44

A

B

18/44

STATIC STRENGTH Ideally, in designing any machine element, the engineer should have at his or her disposal the results of a great many strength tests of the particular material chosen.

19/44

Four Design Categories 1.Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justified during design. 2. The part is made in large enough quantities so that a moderate series of tests is feasible.

20/44

3. The part is made in such small quantities that testing is not justified at all; or the design must be completed so rapidly that there is not enough time for testing. 4. The part has already manufactured, and tested unsatisfactory. Analysis understand why the part is what to do to improve it.

been designed, and found to be is required to unsatisfactory and

21/44

Static Loads and Factor of Safety • Factor of safety, n is defined by either of the equations Fu S or n n F

n = ns nl Fu Fp nl 22/44

Example 1.1 Two plates, subjected to a tensile force of 50 kN, are fixed together by means of three rivets as shown in Figure EX1(a). The plates and rivets are made of plain carbon steel 10C4 with a tensile yield strength of 250 N/mm2. The yield strength in shear is 57.7% of the tensile yield strength, and the factor of safety is 2.5. Neglecting stress concentration, determine: (i)

the diameter of the rivets; and

(ii)

the thickness of the plates. d

t

t x d 200

p

(A)

x

p

p

200

(B)

p

Figure EX1 (a) Riveted joint; (b) Tensile stresses in plate

23/44

Failure Theories • Internal stresses do not exceed the strength of the material • Many of the more brittle materials, such as the cast irons, do not have a yield point and so we must utilize the ultimate strength as the criterion of failure in design. • In designing parts of brittle materials it is also necessary to remember that the ultimate compressive strength is much greater than the ultimate tensile strength. • The strength of ductile materials is about the same in tension and compression. 24/44

Maximum-Shear-Stress Theory The theory states that the failure of a mechanical component subjected to biaxial or tri-axial stresses occurs when the maximum shear stress in the component become equal to the maximum shear stress in the standard specimen of the simple tension test, when yielding starts.

25/44

Maximum-Shear-Stress Theory for Ductile Materials • It predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension-test specimen of the same material when that specimen begins to yield. • Is also referred as the Tresca or Guest Theory 26/44

The MSS theory for plane stress 27/44

The Distortion-Energy Theory • Shear-Energy Theory & The von MisesHencky Theory • Best theory for ductile materials, even materials which compressive and tensile strengths are approximately the same and whose shear strengths are smaller than their tensile strengths. • These materials are considered to fail from shear stress 28/44

Distortion-Energy Theory for Ductile Materials • It predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material.

29/44

30/44

The DE theory for plane stress states 31/44

Failure of Ductile Materials under Static Loading The maximum normal-stress theory, the maximum normal-strain theory, the total strain theory, the distortion-energy (von Mises-Hencky) theory, and the maximum shear-stress theory.

32/44

Chapter 6, Problem 14. This problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the figure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F = 0.55 kN, P = 8.0 kN, and T = 30 N × m.

33/44

Chapter 6, Solution 14.

Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the DE theory to stress elements A and B with Sy = 280 MPa A: 32 Fl 4 P 32(0.55)(103 )(0.1) 4(8)(103 ) x 3 2 3 d d (0.020 ) (0.0202 ) 95.49(106 ) Pa 95.49 MPa

xy

16T 16(30) 6 19.10(10 ) Pa 19.10 MPa 3 3 d (0.020 )

3 n

2 x

Sy

2 xy

1/ 2

[95.492 3(19.1) 2 ]1/ 2 101.1 MPa

280 2.77 101.1

Ans. 34/44

B: 4P 4(8)(103 ) 6 x 3 25.47(10 ) Pa 25.47 MPa 2 d (0.020 ) 16T 4 V 16(30) 4 0.55(103 ) xy 3 3 d 3 A (0.020 ) 3 ( / 4)(0.020 2 ) 21.43(106 ) Pa 21.43 MPa

[25.47 2 3(21.432 )]1/ 2 45.02 MPa 280 n 6.22 45.02

Ans. 35/44

Maximum-Normal-Stress Theory for Brittle Materials

failure occurs whenever one of the three principal stresses equals the strength 3

-Sc -Sc St

St St -Sc

1

2

The maximumnormal-stress (MNS) hypothesis in three dimensions.

36/44

Graph of MNS theory of failure for plane stress states. Stress states that plot inside the failure locus are safe

Load line plot

37/44

max

2 = 0

1

1

1

(a) Mohr’s circle for stresses

(b) Stresses in simple tension test

Ssy = 0.5 Syt 38/44

Failure of Brittle Materials under Static Loading •Brittle materials fracture rather than yield. •Brittle fracture in tension is considered to be due to the normal tensile stress alone and thus the maximum-stress theory is applicable. •Brittle fracture in compression is due to some combination of normal compressive stress and shear stress and requires a different theory of failure.

39/44

40/44

Even and Uneven Materials • Even materials • Uneven materials max

Failure line

i

Compression test

max

Tensile test Failure line

O Compression test

Failure line

Tensil e test Failure line

(a) An even material - Suc = - Sut

(b) An uneven material

41/44

The Coulomb-Mohr Theory Ø

For brittle failure, which is adaptation of the maximum normal-stress theory.

The Modified-Mohr Theory •Uneven, brittle materials such as cast iron, are weakest in tension, these theory is the best to describes their failure. •A modified-Mohr effective stress can be calculated using the principal stresses that result from the particular combination of applied stresses at the point in question. 42/44

Static Loading Failure Analysis Materials are assumed homogeneous and isotropic

Flow Chart for Static Failure Analysis

Find all applied forces, moments, torques, etc. and draw free-body diagrams to show them applied to the part's geometry.

Based on the load distributions over the part's geometry, determine what cross sections of the part are most heavily loaded.

Determine the stress distributions within the cross sections of interest and identify locations of the highest applied and combined stresses.

Draw a stress element for each of the selected points of interest within the section and identify the stresses acting on it.

Calculate the applied stresses acting on each element and then calculate the principal stresses and maximum shear stress resulting therefrom.

A

43/44

A

If the material is ductile, then calculate the von Mises effective stress at each selected stress element based on the calculated principal stresses.

Choose a trial material and compute a safety factor based on tensile yield strength of that material.

If the material is brittle, calculate the Coulomb-Mohr effective stress at each selected stress element based on its principal stresses

Choose a trial material and compute a safety factor based on the ultimate tensile strength of that material.

If a known or suspected crack is present, calculate the stress intensity factor and compare it to the fracture toughness of the material to determine if there is any danger of a crack propagation failure.

44/44

1/44

Introduction STRENGTH is a property or characteristic of a

material or of a mechanical element.

Static Load Sometimes a load is assumed to be static when it is known that some variation is to be expected. 2/44

Failure due to Torsion with corrosion fatigue 3/44

4/44

5/44

6/44

Impact failure on lawn mower blade

7/44

Bolt & Nut tensile failure 8/44

Fatigue failure due to bending load 9/44

Failure due to crack initiation by stress concentration 10/44

11/44

Failure caused by spring surge in an oversped engine

12/44

13/44

14/44

15/44

16/44

Modes of Failure Ductile material is one which has a relatively large tensile strain before fracture takes place. For example, steels an aluminium. Brittle materials has a relatively small tensile strain before fracture. For example, cast iron. 17/44

A

B

18/44

STATIC STRENGTH Ideally, in designing any machine element, the engineer should have at his or her disposal the results of a great many strength tests of the particular material chosen.

19/44

Four Design Categories 1.Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justified during design. 2. The part is made in large enough quantities so that a moderate series of tests is feasible.

20/44

3. The part is made in such small quantities that testing is not justified at all; or the design must be completed so rapidly that there is not enough time for testing. 4. The part has already manufactured, and tested unsatisfactory. Analysis understand why the part is what to do to improve it.

been designed, and found to be is required to unsatisfactory and

21/44

Static Loads and Factor of Safety • Factor of safety, n is defined by either of the equations Fu S or n n F

n = ns nl Fu Fp nl 22/44

Example 1.1 Two plates, subjected to a tensile force of 50 kN, are fixed together by means of three rivets as shown in Figure EX1(a). The plates and rivets are made of plain carbon steel 10C4 with a tensile yield strength of 250 N/mm2. The yield strength in shear is 57.7% of the tensile yield strength, and the factor of safety is 2.5. Neglecting stress concentration, determine: (i)

the diameter of the rivets; and

(ii)

the thickness of the plates. d

t

t x d 200

p

(A)

x

p

p

200

(B)

p

Figure EX1 (a) Riveted joint; (b) Tensile stresses in plate

23/44

Failure Theories • Internal stresses do not exceed the strength of the material • Many of the more brittle materials, such as the cast irons, do not have a yield point and so we must utilize the ultimate strength as the criterion of failure in design. • In designing parts of brittle materials it is also necessary to remember that the ultimate compressive strength is much greater than the ultimate tensile strength. • The strength of ductile materials is about the same in tension and compression. 24/44

Maximum-Shear-Stress Theory The theory states that the failure of a mechanical component subjected to biaxial or tri-axial stresses occurs when the maximum shear stress in the component become equal to the maximum shear stress in the standard specimen of the simple tension test, when yielding starts.

25/44

Maximum-Shear-Stress Theory for Ductile Materials • It predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension-test specimen of the same material when that specimen begins to yield. • Is also referred as the Tresca or Guest Theory 26/44

The MSS theory for plane stress 27/44

The Distortion-Energy Theory • Shear-Energy Theory & The von MisesHencky Theory • Best theory for ductile materials, even materials which compressive and tensile strengths are approximately the same and whose shear strengths are smaller than their tensile strengths. • These materials are considered to fail from shear stress 28/44

Distortion-Energy Theory for Ductile Materials • It predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material.

29/44

30/44

The DE theory for plane stress states 31/44

Failure of Ductile Materials under Static Loading The maximum normal-stress theory, the maximum normal-strain theory, the total strain theory, the distortion-energy (von Mises-Hencky) theory, and the maximum shear-stress theory.

32/44

Chapter 6, Problem 14. This problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the figure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F = 0.55 kN, P = 8.0 kN, and T = 30 N × m.

33/44

Chapter 6, Solution 14.

Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the DE theory to stress elements A and B with Sy = 280 MPa A: 32 Fl 4 P 32(0.55)(103 )(0.1) 4(8)(103 ) x 3 2 3 d d (0.020 ) (0.0202 ) 95.49(106 ) Pa 95.49 MPa

xy

16T 16(30) 6 19.10(10 ) Pa 19.10 MPa 3 3 d (0.020 )

3 n

2 x

Sy

2 xy

1/ 2

[95.492 3(19.1) 2 ]1/ 2 101.1 MPa

280 2.77 101.1

Ans. 34/44

B: 4P 4(8)(103 ) 6 x 3 25.47(10 ) Pa 25.47 MPa 2 d (0.020 ) 16T 4 V 16(30) 4 0.55(103 ) xy 3 3 d 3 A (0.020 ) 3 ( / 4)(0.020 2 ) 21.43(106 ) Pa 21.43 MPa

[25.47 2 3(21.432 )]1/ 2 45.02 MPa 280 n 6.22 45.02

Ans. 35/44

Maximum-Normal-Stress Theory for Brittle Materials

failure occurs whenever one of the three principal stresses equals the strength 3

-Sc -Sc St

St St -Sc

1

2

The maximumnormal-stress (MNS) hypothesis in three dimensions.

36/44

Graph of MNS theory of failure for plane stress states. Stress states that plot inside the failure locus are safe

Load line plot

37/44

max

2 = 0

1

1

1

(a) Mohr’s circle for stresses

(b) Stresses in simple tension test

Ssy = 0.5 Syt 38/44

Failure of Brittle Materials under Static Loading •Brittle materials fracture rather than yield. •Brittle fracture in tension is considered to be due to the normal tensile stress alone and thus the maximum-stress theory is applicable. •Brittle fracture in compression is due to some combination of normal compressive stress and shear stress and requires a different theory of failure.

39/44

40/44

Even and Uneven Materials • Even materials • Uneven materials max

Failure line

i

Compression test

max

Tensile test Failure line

O Compression test

Failure line

Tensil e test Failure line

(a) An even material - Suc = - Sut

(b) An uneven material

41/44

The Coulomb-Mohr Theory Ø

For brittle failure, which is adaptation of the maximum normal-stress theory.

The Modified-Mohr Theory •Uneven, brittle materials such as cast iron, are weakest in tension, these theory is the best to describes their failure. •A modified-Mohr effective stress can be calculated using the principal stresses that result from the particular combination of applied stresses at the point in question. 42/44

Static Loading Failure Analysis Materials are assumed homogeneous and isotropic

Flow Chart for Static Failure Analysis

Find all applied forces, moments, torques, etc. and draw free-body diagrams to show them applied to the part's geometry.

Based on the load distributions over the part's geometry, determine what cross sections of the part are most heavily loaded.

Determine the stress distributions within the cross sections of interest and identify locations of the highest applied and combined stresses.

Draw a stress element for each of the selected points of interest within the section and identify the stresses acting on it.

Calculate the applied stresses acting on each element and then calculate the principal stresses and maximum shear stress resulting therefrom.

A

43/44

A

If the material is ductile, then calculate the von Mises effective stress at each selected stress element based on the calculated principal stresses.

Choose a trial material and compute a safety factor based on tensile yield strength of that material.

If the material is brittle, calculate the Coulomb-Mohr effective stress at each selected stress element based on its principal stresses

Choose a trial material and compute a safety factor based on the ultimate tensile strength of that material.

If a known or suspected crack is present, calculate the stress intensity factor and compare it to the fracture toughness of the material to determine if there is any danger of a crack propagation failure.

44/44