Mechanical Design

3-D STRESS IN MECHANICAL DESIGN

August 2000

Copyright 2000 C. E. Knight

PURPOSE OF THE TUTORIAL This tutorial is designed to introduce and place strong emphasis on the role of 3-D stress in the process of mechanical design. Students in engineering are normally introduced to stress in its simplest one-component form defined by load divided by area of cross section.. This is a valid definition of a pure 1-D state of stress, but in many cases it seems to establish a baseline safe position for which many students don’t want to venture forth. Carrying this attitude through the mechanical design process is a recipe for failure. Everything in the mechanical design realm has solid 3-D characteristics. The same is true for the state of stress in the solid. In many simple cases the effective state of stress can be reduced to 2-D or 1-D, but only after careful consideration. In the early stages of mechanical design, the locations of most likely stress failure and the corresponding stress components acting at those locations must be identified. Once all the stress components at a given location are determined, they may then be combined to find principal stresses, maximum shear stress or other measures that are useful for predicting design success or failure. It is very important to remember that stress components for one location in a machine part should never be combined with stress components for a different location in the same part. One of the interesting developments in visualizing the combining of 2-D stress components was the creation of Mohr’s circle. This graphical representation of the 2-D stress transformation equations provides a quick, accurate and visual protrayal of the 2-D state of stress. It finds the principal stresses and a maximum shear stress (although this maximum shear stress may be quite misleading in 3-D stress). Review of Two-Dimensional Mohr’s Circle - Graphical Approach The beginning of a Mohr’s circle representation must be a stress element sketch of the 2-D state of stress as shown in the figure.

This shows all potential non-zero 2-D stress components. The graphical Mohr’s circle uses coordinate pairs of these data to make a plot. They are (σx, τxy) and (σy, τyx). These two points establish the circle diameter. By convention normal stresses, σ are positive in tension and negative in compression, however, the shear stresses, τ, in the Mohr’s circle constructions are taken as positive if they make a cw moment about the stress element. In the stress element above, τxy is ccw (-) while τyx is cw(+). This convention is useful for determining the proper orientation of principal stresses and other components relative to the x,y coordinates. As an example, assume that σx is positive and τxy is positive (cw) with σy equal zero. First sketch the normal stress axis along the horizontal and the shear stress axis along the vertical. Then plot the first coordinate pair (σx, τxy) at point A. Then plot the second pair (0, τyx ) at point B. These two points form the diameter of the circle with its center at point C. Simple geometric triangles can then determine the circle radius and all principal stress and peak shear stress values.

In a second example, assume that σx is smaller than σy , but both are posit ive, and that τxy is cw. Sketch the normal stress, σ, and shear stress, τ axes and plot the coordinate pair (σx, τxy) at point A and then (σy, τyx) at point B. Connecting these points locates the circle center at point C. Geometrical calculations finish the numerical values.

The Mohr’s circle gives a complete visual representation of the 2-D state of stress along with accurate numerical values. However, there is a highly significant factor in mechanical design that has thus far been neglected. That factor is the influence of the additional 3-D stress components on the design safety. Three-Dimensional Stress in Mechanical Design In the real world of applications all objects are 3-D. The general state of stress is pictured on the stress element below. There are six independent stress components shown in a conventient Cartesian coordinate system. It is readily seen that in the 2-D Mohr’s circle, the principal stresses are larger numerically than the cartesion components unless they are already principal stresses. The same is true in 3-D stress. A qubic equation can be solved for the three principal stress roots in the general stress case, however, in many cases of mechanical design some of the principal stresses may be determined by inspection.

3-D Mohr’s Circles Use of Mohr’s circles can again make visualization of the stress condition clearer to the designer. The definition of the three circle diagram is sketched below. Note that the principal stress values are always ordered by convention so the σ1 is the largest value in the tensile direction and σ3 is the largest value in the compressive direction. Note also that there is one dominant peak shear stress in this diagram. Be forewarned the principal stresses and this peak shear stress are going to play a strong role in determining the factor of safety in mechanical design.

What about the two 2-D examples? How do they become 3-D representations? If the stress state is only two-dimensional, then σz and all the shear stresses with z components are zero, therefore, σz = 0 is the third principal stress. Only two principal stresses were found by the Mohr’s circle transformation. Since σz = 0, it must not be important. Wrong!! Look at the modified examples below. In example 1, the second principal stress, σ2, becomes zero and the third principal stress, σ3, is negative, but the overall range is the same. In this case, there is no effect on the overall stress state. However, the second example has the same first two principal stresses σ1 and σ2, but σ3 is 0. This enlarges the outermost circle which means that the overall state of stress has increased. In this case, there is an appreciable contribution by the 3-D effect that must not be ignored in accounting for design safety.

Example 3 As a final example, use the stress conditions σx = 90 (T), τxy = 40 ccw, σy = 30 (T), and σz = - 25 (C). First sketch the normal stress and shear stress axes and then plot the coordinate pair ( σx , τxy) at A. Plot the next coordinate pair ( σy, τyx ) at B. Connect points A and B to form the diameter of the 2-D Mohr’s circle with center at C. Draw the circle and determine two of the principal stresses. The center C is located at a stress value of 60. The triangle C, σx, and τxy form a 30, 40, 50 triangle, so the circle radius is 50. The two principal stresses from the 2-D circle are 110 (T) and 10(T). Since there are no non-zero z component shear stresses, σz is the third principal stress with a value of –25 (C). The maximum shear stress is at the peak of the largest circle and is equal to half the difference between σ1 and σ3.

† Text refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations and examples with the prefix T refer to the present tutorial.

MECHANICAL ENGINEERING DESIGN TUTORIAL 6, PART A: INTRODUCTION TO STATIC FAILURE THEORIES

APPROACH Theories have been developed for the static failure of metals based upon the two classes of material failure; ductile metals yield while brittle metals fracture. Thus separate failure theories exist for ductile and brittle metals: Failure Theories for Ductile Materials

1. Maximum Shear Stress (MSS) 2. Distortion Energy (DE) 3. Ductile Coulomb-Mohr (DCM)

Failure Theories for Brittle Materials

1. Maximum Normal Stress (MNS) 2. Brittle Coulomb-Mohr (BCM)

These theories have grown out of hypotheses and experimental data in the following manner.

1. Experimental failure data is first collected through tensile tests. 2. The state of stress is correlated to the experimental data using Mohr’s circle plots. 3. A failure theory is developed from a concept of the responsible failure mechanism. 4. A design envelope is established based upon the theoretical and empirical design

equations. In light of the extensive dependence of failure theories on experimental data, we will first review the acquisition and correlation of tensile test data to failure theory. Subsequently, the criteria and application of specific failure theories will be discussed. TENSILE TEST

TEXT FIGURE 3-1: A typical tension-test specimen. Some of the standard dimensions used for d0 are 2.5, 6.25, and 12.5 mm and 0.505 in, but other sections and sizes are in use. Common gauge lengths l0 used are 10, 25, and 50 mm and 1 and 2 in.

d0

l0

Shigley, Mischke & Budynas Machine Design Tutoria

The tensile test is a standardized test (ASTM Standard E8 or E8m) and thus allows for the sharing of experimental data amongst researchers, typically in the form of stress-strain curves. Standard dimensions for the test-specimen are provided in Text Figure 3-1 while a comparison of characteristic stress-strain curves for ductile and brittle materials are shown in Text Figure 3-2. These engineering stress-strain diagrams graphically demonstrate the difference in the failure behavior of ductile and brittle metals, and the need for separate failure criteria. However, the curves do not represent true values of stress and strain; rather, they are calculated based upon the original specimen cross-sectional area, prior to loading.

20 0

0

0

4Engineering Stress (Text Eq. 3-1)

Engineering Strain (Text Eq. 3-2)

P PA d

l ll

σπ

ε

= =

−=

Referring to Text Fig. 3-2 (a), point el, the elaswhile point a represents 0.2 percent permanent (ε = 0.002). A measure of the “true” stress and strain measurements of the load and cross-sectional a

TEXT FIGURE 3-2: Stress-strain diagram obta(a) Ductile material; (b) brittle material. pl markselastic limit; y, the offset yield strength as definedmaximum or ultimate strength; and f, the fracture

u, f

Strain ε

(a)

O εu a εy εf

Su

Sf Sy

u f

yel pl

Stre

ss σ

= P

/A0

Sut

l 6: Static Failure Theories 2/5

tic limit, defines the onset of permanent set set with respect to the original gauge length

can be obtained by taking simultaneous rea during the tensile test experiment. Text

ined from the standard tensile test the proportional limit; el, the by offset strain Oa; u, the

strength.

a

Strain ε

(b)

ySy

Shigley, Mischke & Budynas Machine Design Tutorial 6: Static Failure Theories 3/5

Figure 3-4 shows a typical true stress-strain diagram for a ductile material. The curve between points u and f corresponds to a reduction in stress as the specimen necks down.

CORRELATION OF STATE OF STRESS WITH TEST DATA For design, we need to relate the expected state of stress in a part to the actual state of stress and thus, the material strength, as determined through the tensile test. We accomplish this by applying principal stresses since they characterize a state of stress independent of the original coordinate system.

σf

σu

εu

True strain

f

u

True

stre

ss

εf

TEXT FIGURE 3-4: True stress-strain diagram plotted in Cartesian coordinates.

TEXT FIGURE 3-3: Tension specimen after necking.

y

(a) State of stress for simple tension. (b) Principal stresses for simple tension.

FIGURE 6A-1: Correlation of state of stress with principal stresses for simple tension.

x0/x P Aσ =

0yx xyτ τ= =

0yσ =

0yσ =

0/x P Aσ =

0yx xyτ τ= =

1σ

2 3 0σ σ= =

1σ��


Since the tensile test generates a uniaxial state of stress, the principal stresses can be defined as,

1 axial 2 30

and 0PA

σ σ σ σ= = = =

When plotted on a Mohr’s circle diagram, these stress values translate into what looks like a single circle passing through the origin where 2σ is coincident with 3.σ Actually, there are still three circles on the Mohr’s circle diagram. Two circles, defined by principal stresses

1, 2( )σ σ and 1, 3( )σ σ , are drawn on top of each other. The third circle degenerates to a point defined by principal stresses 2, 3( )σ σ .

DEVELOPMENT OF STATIC FAILURE THEORIES Design for static loading dictates that all loading variables remain constant:

1. Magnitude of load is constant; 2. Direction of load is constant; 3. Point of application of the load is fixed.

These conditions, in conjunction with criteria specific to ductile and brittle materials, have been used in the development of the static failure theories outlined earlier. Characteristics of Ductile Materials

1. The strain at failure is, 0.05fε ≥ , or percent elongation greater than five percent. 2. Ductile materials typically have a well defined yield point. The value of the

stress at the yield point defines the yield strength, Sy. 3. For typical ductile materials, the yield strength has approximately the same value

for tensile and compressive loading ( ).yt yc yS S S≈ =

τ

σ3σ 2σ 1σ

FIGURE 6A-2: Mohr’s circle for simple tension.


4. A single tensile test is sufficient to characterize the material behavior of a ductile material, Sy and Sut.

Characteristics of Brittle Materials

1. The strain at failure is, 0.05fε ≤ or percent elongation less than five percent. 2. Brittle materials do not exhibit an identifiable yield point; rather, they fail by

brittle fracture. The value of the largest stress in tension and compression defines the ultimate strength, Sut and Suc respectively.

3. The compressive strength of a typical brittle material is significantly higher than its tensile strength, ( ).uc utS S�

4. Two material tests, a tensile test and a compressive test, are required to characterize the material behavior of a brittle material, Sut and Suc.

SUMMARY

This tutorial has attempted to provide a focused introduction to the development of static failure theory by summarizing the theories associated with specific material classifications. In addition, the experimental and analytical models, which have been employed historically to relate the experimental data to strength quantities used for static design, are presented. Subsequent tutorials, Static Failure of Ductile Materials and Static Failure of Brittle Materials, will respectively provide detailed reviews and examples, respectively, of the failure theories associated with ductile and brittle materials.

† Text refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations and examples with the prefix T refer to the present tutorial.

MECHANICAL ENGINEERING DESIGN TUTORIAL 4-14: STRESS CONCENTRATION

ORIGIN OF STRESS CONCENTRATIONS Machine members often have regions in which the state of stress is significantly greater than theoretical predictions as a result of:

1. Geometric discontinuities or stress raisers such as holes, notches, and fillets; 2. Internal microscopic irregularities (non-homogeneities) of the material created by such

manufacturing processes as casting and molding; 3. Surface irregularities such as cracks and marks created by machining operations.

These stress concentrations are highly localized effects which are functions of geometry and loading. In this tutorial, we will examine the standard method of accounting for stress concentrations caused by geometric features. Specifically, we will discuss the application of a theoretical or geometric stress-concentration factor for determination of the true state of stress in the vicinity of stress raisers. THEORETICAL (GEOMETRIC) STRESS-CONCENTRATION FACTOR, tK AND tsK

In order to predict the “actual” stress resulting from a geometric stress raiser, a theoretical stress- concentration factor is applied to the nominal stress. For a part subjected to a normal stress, the true stress in the immediate neighborhood of the geometric discontinuity is calculated as:

max 0tKσ σ= (Text Eq. 4-48)

where,

0

Theoretical stress-concentration factor Nominal normal stress

tKσ

==

Similarly, we can also estimate the highly localized amplification of shear stress in the vicinity of a geometric stress concentration,

max 0tsKτ τ=

where,

0

Theoretical stress-concentration factor for shear Nominal shear stress

tsKτ

==

The nominal stress of the above equations is typically derived from the elementary strength of materials equations, using either a net or a gross cross section.

Shigley, Mischke & Budynas Machine Design Tutorial 4-14: Stress Concentration 2/9

Characteristics of Stress-Concentration Factors

1. Function of the geometry or shape of the part, but not its size or material; 2. Function of the type of loading applied to the part (axial, bending or torsional); 3. Function of the specific geometric stress raiser in the part (e.g. fillet radius,

notch, or hole) 4. Always defined with respect to a particular nominal stress; 5. Typically assumes a linear elastic, homogeneous, isotropic material.

Determination of Kt Value

The stress-concentration factor, associated with a specific geometry and loading condition of a part, can be derived through experimentation, analysis or computational methods.

1. Experimental Methods. Optical methods, such as photoelasticity, are very

dependable and widely used for experimentally determining the stress concentration at a point on a part. However, several alternative methods have been used historically: the grid method, brittle-coating, brittle-model and strain gauge.

2. Analytical Methods. The theory of elasticity can be used to analyze certain geometrical shapes to calculate stress-concentration factors.

3. Computational Methods. Finite-element techniques provide a powerful and inexpensive computational method of assessing stress-concentration factors.

Following are comparisons of stress-concentration factors derived using experimental, analytical and computational methods for a rectangular filleted bar in tension and in pure bending. Text Figure A-15-5 provides tensile test results for the bar in simple tension while Figure T6-2-1 shows

TEXT FIGURE A-15-5: Rectangular filleted bar in tension or simple compression. 0 / , where and is the thickness.F A A dt tσ = =


FIGURE T6-2-1: Stress distribution in a rectangular filleted bar in simple tension obtained through photoelastic procedures. (S. P. Timoshenko and J. N. Goodier, "Theory of Elasticity," Third Edition, McGraw-Hill, Inc., 1969.)

the fringe pattern captured photographically from a photoelasticity experiment. Fringe patterns are indicative of the stress intensity which is directly proportional to the maximum shear stress and the principal stresses:

Intensity max 1 32 .σ τ σ σ= = −

Stress-concentration factors can be developed from these contours. Finally, Figure T6-2-2 contains the graphical results of a finite element analysis of the bar in tension. Since the bar geometry and the loads applied to the bar are symmetrical with respect to the longitudinal axis, the model only needs to incorporate the upper half of the bar; the analytical results for the lower half of the bar will be a mirror image of those in the top half. The finite element model plot contains contours of the xσ component of stress. However, since the stress-concentration factor is applied to the dominant component of the stress, xσ in this model, the finite element model can be queried for xσ to estimate the value of Kt directly

, FiniteElement

0

xtK

σσ

=

where the nominal stress must be defined for the section geometry and applied loading.


Similarly, Text Figure A-15-6, Figure T6-2-3 and Figure T6-2-4 respectfully provide results obtained by applying bending and photoelastic testing and finite element analysis to a rectangular filleted bar in pure bending.

TEXT FIGURE A-15-6: Rectangular filleted bar in pure bending. 3

0 / , where / 2, /12 and is the thickness.Mc I c d I td tσ = = =

FIGURE T6-2-2: Stress contours of xσ generated by a finite element model of one half of a rectangular filleted bar in tension.


FIGURE T6-2-3: Stress distribution in a rectangular filleted bar in pure bending obtained through photoelastic procedures. (By permission of S. P. Timoshenko and J. N. Goodier; the figure was included in, "Theory of Elasticity," Third Edition, McGraw-Hill, Inc., 1969.)

Stress-concentration factors, derived through many years of practice, have been catalogued for numerous geometric features and loading configurations in two authoritative resources:

1. Pilkey, W. D., Peterson’s Stress Concentration Factors, 2nd ed., Wiley

Interscience, 1997.

FIGURE T6-2-4: Stress contours of xσ generated by a finite element model of one half of the rectangular filleted bar in pure bending.


2. Young, W. C. and R. G. Budynas, Roark’s Formulas for Stress and Strain, 7th ed., McGraw-Hill, 2001.

Application to Ductile and Brittle Materials for Static Loading Ductile Materials. While stress concentration must be considered for fatigue and impact loading of most materials, stress-concentration factors are seldom applied to ductile materials under static loading. This design practice is justified by four points:

1. Areas of high stress caused by stress concentrations are highly localized and will not dictate the performance of the part. Rather, it is assumed that the stress state in the cross section as a whole is below the general yield condition;

2. If the magnitude of the loading is large enough to cause yielding due to the stress concentration, the localized area will plastically deform immediately upon loading;

3. Ductile materials typically work-harden (strain-strengthen) on yielding, resulting in a localized increase in material strength;

4. The static load is never cycled.

It is important to note, that even though the stress-concentration factor is not usually applied to estimate the stresses at a stress raiser in a ductile material, the higher state of stress does in fact exist.

Ductile Material Practice: max 0σ σ=

Brittle Materials. Stress-concentration factors are always required for brittle materials, regardless of the loading conditions, since brittle failure results in fracture. This type of failure is characteristic of brittle materials which do not exhibit a yielding or plastic range. As a consequence of brittle fracture, the part breaks into two or more pieces having no load carrying capability. To avoid such catastrophic failure, the design practice is to always use a stress-concentration factor for brittle materials to ensure that the state of stress is accurately represented.

Brittle Material Practice: max 0tKσ σ=


Example T6.2.1:

Problem Statement: A bar machined from an ASTM No. 20 cast iron, a brittle material, is subjected to a static axial load. Find: The critical section of the bar. Solution Methodology:

1. Assume the stress concentrations do not interact and analyze the

localized effect of each stress concentration separately. 2. Compute the actual stress in the shoulder by taking into account the

stress concentration caused by a fillet radius in a rectangular bar in tension.

3. Compute the actual stress in the region immediately adjacent to the hole by applying the stress-concentration factor associated for a bar in tension with a transverse hole.

4. Evaluate the critical section as the region having the highest actual stress.

Schematic:

Solution: 1. Material Properties: Sut = 20 kpsi

2. Actual Stress in Shoulder

a. Stress-Concentration Factor from Text Figure A-15-5:

for 2.25 in. 0.1875 in.1.5 and 0.125 in.,1.5 in. 1.5 in.

D rd d

= = = = Kt = 1.95

1000 lb D = 2 ¼ ” d = 1½”

1000 lb

¾ ” D.

¼”

3/16” R.


b. Nominal stress, as defined in the caption of Text Figure A-15-5:

00

1000 lb 2666.7 psi = 2.67 kpsi(1.5 in.)(0.25 in.)

F FA dt

σ = = = =

c. Actual stress at fillet:

max 0 1.95(2666.7 psi) 5200 psitKσ σ= = = =5.20 kpsi

TEXT FIGURE A-15-5: Rectangular filleted bar in tension or simple compression. 0 / , where and is the thickness.F A A dt tσ = =

3. Actual Stress at Hole Perimeter

a. From Text Figure A-15-1 shown on the next page:

for 0.75 in. 0.5,1.5 in.

dw

= = Kt = 2.19

b. Nominal stress, as defined in the caption of Text Figure A-15-1:

00

1000 lb( ) (1.5 in. 0.75 in.)(0.25 in.)

5333.3 psi = 5.333 kpsi

F FA w d t

σ = = =− −

=

Kt = 1.95


TEXT FIGURE A-15-1: Bar in tension or simple compression with a transverse hole. 0 / , where ( ) and where is the thickness.F A A w d t tσ = = −

c. Actual stress at hole perimeter:

max 0 2.19(5.333 kpsi)tKσ σ= = =11.68 kpsi

4. Since the actual stress at the hole is greater than the actual stress at the fillet, the

hole represents the critical section for this part.

Kt = 2.19

† Text Eq. refers to MechaR. Mischke and Richard G

MECHANICAL ENGINEERING DESIGN TUTORIAL 4 –15: PRESSURE VESSEL DESIGN

PRESSURE VESSEL DESIGN MODELS FOR CYLINDERS:

1. Thick-walled Cylinders 2. Thin-walled Cylinders

THICK-WALL THEORY

• Thick-wall theory is developed from the Theory of Elasticity which yields the state of stress as a continuous function of radius over the pressure vessel wall. The state of stress is defined relative to a convenient cylindrical coordinate system:

1. tσ — Tangential Stress 2. rσ — Radial Stress 3. lσ — Longitudinal Stress

• Stresses in a cylindrical pressure vessel depend upon the ratio of the inner radius to

the outer radius ( /o ir r ) rather than the size of the cylinder. • Principal Stresses ( 1 2 3, ,σ σ σ )

1. Determined without computation of Mohr’s Circle; 2. Equivalent to cylindrical stresses ( , ,t r lσ σ σ )

• Applicable for any wall thickness-to-radius ratio.

Cylinder under Pressure

Consider a cylinder, with capped ends, subjected to an internal pressure, pi, and an external pressure, po,

ir

ip

rσlσ

t

nica. Bu

lσ

l Engdyna

rσ

ineers; eq

tσ

ing Duatio

FI

σ

esign, 7th edition text by Joseph Edward Shigley, Charles ns and figures with the prefix T refer to the present tutorial.

GURE T4-15-1

opor

Shigley, Mischke & Budynas Machine Design Tutorial 4–15: Pressure Vessel Design 2/10

The cylinder geometry is defined by the inside radius, ,ir the outside radius, ,or and the cylinder length, l. In general, the stresses in the cylindrical pressure vessel ( , ,t r lσ σ σ ) can be computed at any radial coordinate value, r, within the wall thickness bounded by

ir and ,or and will be characterized by the ratio of radii, / .o ir rζ = These cylindrical stresses represent the principal stresses and can be computed directly using Eq. 4-50 and 4-52. Thus we do not need to use Mohr’s circle to assess the principal stresses. Tangential Stress:

22

22222 /)(

io

iooiooiit rr

rpprrrprp−

−−−=σ for i or r r≤ ≤ (Text Eq. 4-50)

Radial Stress:

22

22222 /)(

io

iooiooiir rr

rpprrrprp−

−+−=σ for i or r r≤ ≤ (Text Eq. 4-50)

Longitudinal Stress:

• Applicable to cases where the cylinder carries the longitudinal load, such as capped ends.

• Only valid far away from end caps where bending, nonlinearities and stress concentrations are not significant.

22

22

io

ooiil rr

rprp−−

=σ for i or r r≤ ≤ (Modified Text Eq. 4-52)

Two Mechanical Design Cases

1. Internal Pressure Only ( 0=op ) 2. External Pressure Only ( 0=ip ) Design Case 1: Internal Pressure Only

• Only one case to consider — the critical section which exists at irr = . • Substituting 0=op into Eqs. (4-50) and incorporating / ,o ir rζ = the

largest value of each stress component is found at the inner surface:

2 2 2

,max 2 2 21( )1

o it i t i i i ti

o i

r rr r p p p Cr r

ζσ σζ

+ += = = = =− −

(T-1)


where 2 22

2 2 2

11

o iti

o i

r rCr r

ζζ

++= =− −

is a function of cylinder geometry only.

irir prr −=== max,)( σσ Natural Boundary Condition (T-2)

• Longitudinal stress depends upon end conditions:

i lip C Capped Ends (T-3a)

lσ = 0 Uncapped Ends (T-3b)

where 21

1liCζ

=−

.

Design Case 2: External Pressure Only

• The critical section is identified by considering the state of stress at two

points on the cylinder: r = ri and r = ro. Substituting pi = 0 into Text Eqs. (4-50) for each case:

r = ri 0)( == ir rrσ Natural Boundary Condition (T-4a)

( )2 2

,max 2 2 22 2

1o

t i t o o o too i

rr r p p p Cr r

ζσ σζ

= = = − = − = −− −

(T-4b)

where, 22

2 2 2

221

oto

o i

rCr r

ζζ

= =− −

.

r = ro oror prr −=== max,)( σσ Natural Boundary Condition (T-5a)

2 2 2

2 2 21( )1

o it o o o o ti

o i

r rr r p p p Cr r

ζσζ

+ += = − = − = −− −

(T-5b)

• Longitudinal stress for a closed cylinder now depends upon external

pressure and radius while that of an open-ended cylinder remains zero:

o lop C− Capped Ends (T-6a)

lσ = 0 Uncapped Ends (T-6b)


where 2

2 1loC ζζ

=−

.

Example T4.15.1: Thick-wall Cylinder Analysis

Problem Statement: Consider a cylinder subjected to an external pressure of 150 MPa and an internal pressure of zero. The cylinder has a 25 mm ID and a 50 mm OD, respectively. Assume the cylinder is capped. Find:

1. the state of stress ( rσ , tσ , lσ ) at the inner and outer cylinder surfaces;

2. the Mohr’s Circle plot for the inside and outside cylinder surfaces; 3. the critical section based upon the estimate of maxτ .

Solution Methodology:

Since we have an external pressure case, we need to compute the state of stress ( ,rσ ,tσ lσ ) at both the inside and outside radius in order to determine the critical section.

1. As the cylinder is closed and exposed to external pressure only, Eq. (T-6a) may be applied to calculate the longitudinal stress developed. This result represents the average stress across the wall of the pressure vessel and thus may be used for both the inner and outer radii analyses.

2. Assess the radial and tangential stresses using Eqs. (T-4) and (T-5) for the inner and outer radii, respectively.

3. Assess the principal stresses for the inner and outer radii based upon the magnitudes of ( ,rσ ,tσ lσ ) at each radius.

4. Use the principal stresses to calculate the maximum shear stress at each radius.

5. Draw Mohr’s Circle for both states of stress and determine which provides the critical section.

Solution: 1. Longitudinal Stress Calculation:

OD 50 mm ID 25 mm25 mm ; 12.5 mm2 2 2 2o ir r= = = = = =

Compute the radius ratio, ζ

25 mm 2.012.5 mm

o

i

rr

ζ = = =


Then, 2 2

2 2

22

2

(2)1 (2) 1

( ) ( ) ( 150MPa)(1.3333 mm )1

lo

l i l o o o lo

C

r r r r p p C

ζζ

ζσ σζ

= = =− −

= = = = − = − = −−

21.3333 mm

MPa200−−−−====lσσσσ

2. Radial & Tangential Stress Calculations:

Inner Radius (r = ri) 2 2

2 2

2

,max 2 2

2 2(2)1 (2) 1

2( ) ( 150 MPa)(2.6667)

to

ot i t o o to

o i

C

rr r p p Cr r

ζζ

σ σ

= = =− −

= = = − = − = −−

2.6667

400 MPat iσ (r r ) Compressive= = −

0pforConditionBoundaryNatural i === 0)r(rσ ir

Outer Radius (r = ro)

2 2

2 2

2 2

,min 2 2

1 (2) 11 (2) 1

( ) ( 150 MPa)(1.6667)

ti

o it o t o o ti

o i

C

r rr r p p Cr r

ζζ

σ σ

+ += = =− −

+= = = − = − = −−

1.6667

eCompressivMPa250−−−−======== )r(rσ ot

ConditionBoundaryNaturalMPa150−−−−====−−−−======== oir p)r(rσ

3. Define Principal Stresses:

Inner Radius (r = ri ) Outer Radius (r = ro )

MPa400MPa200

MPa0

3

2

1

−==−==

==

t

l

r

σσσσσσ

MPa250MPa200MPa150

3

2

1

−==−==−==

t

l

r

σσσσσσ

4. Maximum Shear Stress Calculations:

Inner Radius (r = ri ) 1 3max

0 ( 400)( )2 2ir r σ στ − − −= = = =200 MPa

Shigley, Mischke & Budynas Machine Design Tutorial 4–15: Pressure Vessel

Outer Radius (r = ro ) 1 3max

( 150) ( 250)( )2 2or r σ στ − − − −= = = =50 MPa

5. Mohr’s Circles:

Inner Radius (r = ri ) Outer Radius (r = ro ) Critical Section

SectionCriticalrr i ⇐⇐⇐⇐======== MPa200)(maxττττ

3σ = -250 MPa

τ

2σ = -200 MPa

τ

σ

2σ = -200 MPa

3σ = -400 MPa

τ

1 0 MPaσ =

maxτ = 200 MPa

FIGURE T4-15-2

FIGURE T4-15-3

•

Design 6/10

!RadiusInsideatis

1 150 MPaσ = −

σ

max = 50 MPa


THIN-WALL THEORY

• Thin-wall theory is developed from a Strength of Materials solution which yields the state of stress as an average over the pressure vessel wall.

• Use restricted by wall thickness-to-radius ratio:

��1According to theory, Thin-wall Theory is justified for20

tr

≤

��1In practice, typically use a less conservative rule,

10tr

≤

• State of Stress Definition:

1. Hoop Stress, tσ , assumed to be uniform across wall thickness.

2. Radial Stress is insignificant compared to tangential stress, thus, 0.rσ �

3. Longitudinal Stress, lσ

��Exists for cylinders with capped ends;

��Assumed to be uniformly distributed across wall thickness;

��This approximation for the longitudinal stress is only valid far away from the end-caps.

4. These cylindrical stresses ( , , )t r lσ σ σ are principal stresses ( , , )t r lσ σ σ which

can be determined without computation of Mohr’s circle plot.

• Analysis of Cylinder Section

1

t FV

FHoop FHoop

Pressure Acting over Projected Vertical Area

di

FIGURE T4-15-4


The internal pressure exerts a vertical force, FV, on the cylinder wall which is

balanced by the tangential hoop stress, FHoop.

tpdFFF

ttAFpddppAF

tiHoopVy

ttstressedtHoop

iiprojV

σ

σσσ

� −=−==

===

===

220

)}1)({(

)}1)({(

Solving for the tangential stress,

Hoop Stress2

it

pdt

σ = (Text Eq. 4-53)

• Comparison of state of stress for cylinder under internal pressure verses external

pressure:

Internal Pressure Only

20

(Text Eq.4-55)4 2

it

r

i tl

pd Hoop Stresst

By Definitionpd Capped Case

t

σ

σσσ

=

=

= =

External Pressure Only

20

4 2

ot

r

o tl

pd Hoop Stresst

By Definitionpd Capped Case

t

σ

σσσ

=

=

= =

Example T4.15.2: Thin-wall Theory Applied to Cylinder Analysis

Problem Statement: Repeat Example T1.1 using the Thin-wall Theory (po = 150 MPa, pi = 0, ID = 25 mm, OD = 50 mm). Find: The percent difference of the maximum shear stress estimates found using the Thick-wall and Thin-wall Theories.



1. Check t/r ratio to determine if Thin-wall Theory is applicable. 2. Use the Thin-wall Theory to compute the state of stress 3. Identify the principal stresses based upon the stress magnitudes. 4. Use the principal stresses to assess the maximum shear stress. 5. Calculate the percent difference between the maximum shear stresses

derived using the Thick-wall and Thin-wall Theories.

Solution:

1. Check t/r Ratio: 101

201

21

mm25mm5.12 or

rt

�==

The application of Thin-wall Theory to estimate the stress state of this cylinder is thus not justified.

2. Compute stresses using the Thin-wall Theory to compare with Thick-wall theory estimates.

definitionbyStress Radialb.

mm)2(12.5)mm50)(MPa150(

2

wall)acrossuniformstress,(averageStressHoopa.

r 0

MPa300

=

−=−=−=

σ

σtdp oo

t

c. Longitudinal Stress (average stress, uniform across wall)

4 2o o t

lp d

tσσ −= = = −150 MPa

3. Identify Principal Stresses in terms of “Average” Stresses:

MPa300MPa150

MPa0

3

2

1

−==−==

==

t

l

r

σσσσσσ

4. Maximum Shear Stress Calculation:

MPa1502

)MPa300(02

31max +=−−=

−=

σστ

5. Percent Difference between Thin- and Thick-wall Estimates for the

Critical Section:


max,Thin max,Thick

max,Thick

% 100%

( 150) ( 200) (100%)( 200)

Differenceτ τ

τ−

= ∗

+ − += ∗ = −+

25%

�� Thin -wall estimate is 25% low! ⇐⇐⇐⇐

† Text Eq. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations with the prefix T refer to the present tutorial.

MECHANICAL ENGINEERING DESIGN TUTORIAL 4-17: PRESS AND SHRINK FITS

APPLICATION OF THICK-WALL CYLINDRICAL PRESSURE VESSEL THEORY RELATING TO STRESSES DEVELOPED FROM INTERFERENCE FITS:

1. Design application which uses the cylindrical pressure vessel Thick-Wall Theory. 2. Stresses develop between cylinders due to the contact pressure generated by an

interference fit. The interference fit is achieved by pressing a larger inside member into the smaller opening of an outside member. In the specific case of a shaft press fit into the hub of a gear, the outside diameter (OD) of the shaft is slightly larger than the inside hole diameter (ID) of the hub. The diametral difference between the shaft OD and the ID of the hub hole is referred to as the interference fit.

• The radial deformation required by the interference fit causes an interfacial

pressure, p, to develop at the nominal radius, at r = R. Consequently, radial and tangential stresses, rσ and tσ , are produced.

• Assuming uncapped ends ( 0),lσ = a biaxial state of stress exists for which two non-zero principal stresses must be considered.

• From the cylindrical pressure vessel theory, the radial and tangential stresses represent principal stresses.

• The length of the outer member is assumed to be equal to the length of the inner member.

(b) Cross-section of cylinders showing internal outside radius larger than external inside radius by a small amount of δ.

(a) End view of inner and outer members, press fit together.

ri

R

ro δ

R rori

Inner Member

Outer Member

FIGURE T4-17-1 Interference fit of two cylinders of finite length and equal lengths.

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 2/11

3. Referring to Fig. T4-17-1, the geometric features of the cylindrical parts are defined as:

the inside radius of the inner cylindernominal radius of internal outside radius and external inside radius after assemblyoutside radius of the outer cylinderradial interference

i

o

rRrδ

====

INSIDE CYLINDER

• Inner member experiences an external pressure, po = p, resulting in compressive

tangential and radial stresses. • Thick-Wall Theory may be applied with ro = R:

2 2

2 2( ) (Text Eq. 4-57)

( )

it i r R o it

i

r i r R o

R rp pCR r

p p

σ

σ

=

=

� �+= − = −� �−� �

= − = −

OUTSIDE CYLINDER

• Outer member only experiences internal pressure, pi = p, resulting in tensile tangential stress and compressive radial stress.

• Thick-Wall Theory is, as always, applicable with ri = R: 2 2

2 2( ) (Text Eq. 4-58)

( )

ot o r R i ot

o

r i r R i

r Rp pCr R

p p

σ

σ

=

=

� �+= =� �−� �

= − = −

DEFINITION OF INTERFACIAL PRESSURE

We presently have two equations and three unknowns for both the inside and outside cylinder analyses. A third equation which relates the contact pressure and the interference can be derived by examining the deformation of the members. Deflection Equation

The total radial interference may be defined as:

oi δδδ +=total where,


decrease in radius of inner cylinder increase in radius of hole

i

o

δδ

==

The deformation may also be expressed as:

total i opRK pRKδ = + (Modified Text Eq. 4-59)

where the outside member constant, Ko, is defined as,

[ ]2 2

2 21 1o

o o o oo oo

r RK CE Er R

ν ν� �� += + = +� ��

−� � � �

Using the radius ratio form defined for the cylindrical pressure vessel formulation, /o or Rζ = , we can define

2

211

oo

oC ζ

ζ+=−

Note that the Co term is a function of geometry only, while the member constant term Ko is a function of both geometry and material parameters. Similarly for the inside member constant Ki,

[ ]2 2

2 21 1i

i i i ii ii

R rK CE ER r

ν ν� �� += − = −� ��

−� � � �

with /i iR rζ = , Ci is defined as:

2

211

ii

iC ζ

ζ+=−

For the case of a solid shaft, ri = 0, /i iR rζ = = ∞ and Ci = 1.

We can now solve for the deformation for a given class of interference fits,

pRKK io ][total +=δ

Or, rearranging, the contact pressure, p, may be expressed as a function of the interference without assumptions regarding material property values:

R1 totalδ

��

��

�

+=

io KKp (Modified Text Eq. 4-60)

† Text Eq. refers to Mechanical EngineerinR. Mischke and Richard G. Budynas; equa

Example T4.17.1: Shaft & Hub Shrink Fit Problem Statement: A carbon-steel gear hub having a nominal hole diameter of 1 inch is to be shrink-fitted to a carbon-steel shaft using a class FN4 fit. The hub has a nominal thickness of ½ inch.

Find:

1. the maximum tangential and radial stresses in the hub and shaft when the loosest fit is obtained;

2. the same as (a), except using the tightest fit as a condition.


1. Using Table A-1-2 of this document, identify the diametral size ranges for the shaft and the hole based on an ANSI US Customary Standard class FN4 fit which is discussed below.

2. Calculate the interferences for the loosest and tightest fits. 3. Compute the interfacial pressure for the loosest and tightest fits. 4. For each fit, calculate the radial and tangential stresses for both the hub and

the shaft.

Schematic:

Solution:

1. Shaft and Hub Size R

This problem specifidefined by the AmerStandard Limits for C“Force fit suitable fowhere the heavy presANSI B4.1-1978 sta From Table A-1-2, fsize range is 0.95–1.inch, are:

Shaft

g Design, 7th edition text by Joseph Edward Shigley, Charles tions with the prefix T refer to the present tutorial.

anges:

es a class FN4 fit. The class FN4 fit is a specification ican National Standards Institute (ANSI), American

ylindrical Parts ANSI B4.1-1978. The FN4 fit is a r parts which can be highly stressed or for shrink fits sing forces required are impractical.” Excerpts from the

ndard are provided in Table A-1 of this tutorial.

or a nominal shaft/hole diameter of 1 in., the appropriate 19 in. The allowable tolerances, in thousandths of an

Hub


Largest Tolerance Smallest Tolerance (10-3 in.) (10-3 in.)

Hub hole +0.8 –0.0 Shaft +2.3 +1.8

The shaft and hub size ranges for a nominal 1 in. diameter are:

Largest Diameter Smallest Diameter

(in.) (in.) Hub hole 1.0008 1.0000 Shaft 1.0023 1.0018

2. Diametral and Radial Interference Calculations:

Loosest Fit Tightest Fit Hub hole 1.0008 1.0000 Shaft 1.0018 1.0023 –0.0010 –0.0023

The radial interferences are therefore δ = –0.0005 in. (loosest fit) and δ = –0.001 15 in. (tightest fit). The interference is taken as a positive number by convention. Consequently, the radial interferences for the two cases are: δ = 0.0005 in. (loosest fit) δ = 0.001 15 in. (tightest fit)

3. Interfacial Pressure

The interface pressure can be computed from Modified Eq. (4-60):

R1 totalδ

��

��

�

+=

io KKp

For this problem, the geometric features have been defined as:

1 in. 0.5 in. 1.0 in. 0.0 0.5 in.2 2 2o id dr t r R= + = + = = = =

Thus, for the hub (outside member),


[ ] ( ) [ ]

2 2

2 2

86

(1 in.) 2(0.5 in.)1 (2) 1 5 1.66671 (2) 1 3

1 1 1.667 0.292 6.5289 10 (1/psi)30 10 psi

oo

oo

o

o o oo

rR

C

K CE

ζ

ζζ

ν −

= = =

+ += = = =− −

= + = + = ××

Similarly for the shaft (inside member)

ii

Rr

ζ = = ∞

[ ] ( ) [ ]

2

2

86

1 1.01

1 1 1.0 0.292 2.360 10 (1/psi)30 10 psi

ii

i

i i ii

C

K CE

ζζ

ν −

+= =−

= − = − = ××

For the loosest fit case:

loosest fitloosest fit 8 8

1 1 0.0005R 6.5289 10 2.360 10 0.5o i

pK K

δ− −

� � � � � �= =� � � �� + × + × � �

= 11 250 psi (loosest fit)

For the tightest fit case:

tightest fittightest fit 8 8

1 1 0.00115R 6.5289 10 2.360 10 0.5o i

pK K

δ− −

� � � � � �= =� � � �� + × + × � �

= 25 875 psi (tightest fit)

4. Radial & Tangential Stress Calculations for Shaft and Hub:

Loosest Fit (p = 11 250 psi)

Hub: rσ (r R) p= = − = −11 250 psi

2 2

2 2( ) (11 250psi)(1.6667)ot o

o

r Rr R p pCr R

σ += = = = =−

18 750 psi

Shaft: rσ (r R) p= = − = −11 250 psi

2 2

2 2( ) ( 11 250 psi)(1)it i

i

R rr R p pCR r

σ += = − = − = − = −−

11 250 psi

Tightest Fit (p = 25 875 psi)


Hub: rσ (r R) p= = − = −25 875 psi

2 2

2 2( ) (25 875psi)(1.6667)ot o

o

r Rr R p pCr R

σ += = = = =−

43 125 psi

Shaft: rσ (r R) p= = − = −25 875 psi

2 2

2 2( ) ( 25 875psi)(1)it i

i

R rr R p pCR r

σ += = − = − = − = −−

25 875 psi

Thus the shaft has equal radial and tangential stress for each tightness condition, whereas, the hub’s tangential stress is consistently higher than its radial stress for both the loosest and the tightest condition.

Note: Since the shaft length is greater than the hub length, which is typical in practice, this design case violates one of the assumptions of the interfacial pressure development. In this case, there would be an increase in the interfacial pressure at each end of the hub. This condition of increased interfacial pressure would typically be accounted for by applying a stress concentration factor, Kt, for stresses calculated at points at the end of the hub such as,

actual , tangential

actual , radial

t t t

r t r

KK

σ σσ σ

=

=


Table A-1 LIMITS AND FITS FOR CYLINDRICAL PARTS† The limits shown in the accompanying tabulations are in thousandths of an inch. The size ranges include all sizes over the smallest size in the range, up to and including the largest size in the range. The letter symbols are defined as follows: RC Running and sliding fits are intended to provide a similar running performance, with suitable lubrication

allowance, throughout the range of sizes. The clearance for the first two classes, used chiefly as slide fits, increases more slowly with diameter than the other classes, so that accurate location is maintained even at the expense of free relative motion.

RC1 Close sliding fits are intended for the accurate location of parts which must assemble without perceptible play.

RC2 Sliding fits are intended for accurate location but with greater maximum clearance than class RC1. Parts made to this fit move and turn easily, but are not intended to run freely, and in the larger sizes may seize with small temperature changes.

RC3 Precision running fits are about the closest fits which can be expected to run freely and are intended for precision work at slow speeds and light journal pressures, but are not suitable where appreciable temperature differences are likely to be encountered.

RC4 Close running fits are intended chiefly for running fits on accurate machinery with moderate surface speeds and journal pressures, where accurate location and minimum play is desired.

RC5–RC6 Medium running fits are intended for higher running speeds, heavy journal pressures, or both. RC7 Free running fits are intended for use where accuracy is not essential or where large temperature variations

are likely to be encountered, or under both of these conditions. RC8–RC9 Loose running fits are intended for use where wide commercial tolerances may be necessary, together with an allowance, on the external member. L Locational fits are fits intended to determine only the location of the mating parts; they may provide rigid or

accurate location, as with interference fits, or provide some freedom of location, as with clearance fits. Accordingly, they are divided into three groups: clearance fits, transition fits, and interference fits.

LC Locational clearance fits are intended for parts which are normally stationary but which can be freely assembled or disassembled. They run from snug fits for parts requiring accuracy of location, through the medium clearance fits for parts such as ball, race, and housing, to the looser fastener fits where freedom of assembly is of prime importance.

LT Locational transition fits are a compromise between clearance and interference fits, for application where accuracy of location is important, but either a small amount of clearance or interference is permissible.

LN Locational interference fits are used where accuracy of location is of prime importance and for parts requiring rigidity and alignment with no special requirements for bore pressure. Such fits are not intended for parts designed to transmit frictional loads form one part to another by virtue of the tightness of fit, since these conditions are covered by force fits.

FN Force and shrink fits constitute a special type of interference fit, normally characterized by maintenance of constant bore pressure throughout the range of sizes. The interference therefore varies almost directly with diameter, and the difference between its minimum and maximum value is small so as to maintain the resulting pressures within reasonable limits.

FN1 Light drive fits are those requiring light assembly pressures and producing more or less permanent assemblies. They are suitable for thin sections or long fits or in cast-iron external members.

FN2 Medium drive fits are suitable for ordinary steel parts or for shrink fits on light sections. They are about the tightest fits that can be used with high-grade cast-iron external members.

FN3 Heavy drive fits are suitable for heavier steel parts or for shrink fits in medium sections. FN4–FN5 Force fits are suitable for parts which can be highly stressed or for shrink fits where the heavy pressing

forces required are impractical.

†Extracted from American Standard Limits for Cylindrical Parts ANSI B4.1-1978, with the permission of the publishers, The American Society of Mechanical Engineers, United Engineering Center, 345 East 47th Street, New York 10017. Limit dimensions are tabulated in this standard for nominal sizes up to and including 200 in. An SI version is also available.


Table A-1-1 RUNNING AND SLIDING FITS Diameter Size Range (in.) Class 0.00 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71

RC1 Hole +0.20 -0.00 +0.20 -0.00 +0.25 -0.00 +0.30 -0.00 Shaft +0.10 -0.25 -0.15 -0.30 -0.20 -0.35 -0.25 -0.45

RC2 Hole +0.25 -0.00 +0.30 -0.00 +0.40 -0.00 +0.40 -0.00 Shaft -0.10 -0.30 -0.15 -0.35 -0.20 -0.45 -0.25 -0.55

RC3 Hole +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 Shaft -0.30 -0.55 -0.40 -0.70 -0.50 -0.90 -0.60 -1.00

RC4 Hole +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 +1.00 -0.00 Shaft -0.30 -0.70 -0.40 -0.90 -0.50 -1.10 -0.60 -1.30

RC5 Hole +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 +1.00 -0.00 Shaft -0.60 -1.00 -0.80 -1.30 -1.00 -1.60 -1.20 -1.90

RC6 Hole +1.00 -0.00 +1.20 -0.00 +1.40 -0.00 +1.60 -0.00 Shaft -0.60 -1.20 -0.80 -1.50 -1.00 -1.90 -1.20 -2.20

RC7 Hole +1.00 -0.00 +1.20 -0.00 +1.40 -0.00 +1.60 -0.00 Shaft -1.00 -1.60 -1.20 -1.90 -1.60 -2.50 -2.00 -3.00

RC8 Hole +1.60 -0.00 +1.80 -0.00 +2.20 -0.00 +2.80 -0.00 Shaft -2.50 -3.50 -2.80 -4.00 -3.00 -4.40 -3.50 -5.10

RC9 Hole +2.50 -0.00 +3.00 -0.00 +3.50 -0.00 +4.00 -0.00 Shaft -4.00 -5.60 -4.50 -6.00 -5.00 -7.20 -6.00 -8.80

Diameter Size Range (in.) Class 0.71 - 1.19 1.19 - 1.97 1.97 - 3 .15 3.15 - 4.73

RC1 Hole +0.40 -0.00 +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 Shaft -0.30 -0.55 -0.40 -0.70 -0.40 -0.70 -0.50 -0.90

RC2 Hole +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 Shaft -0.30 -0.70 -0.40 -0.80 -0.40 -0.90 -0.50 -1.10

RC3 Hole +0.80 -0.00 +1.00 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft -0.80 -1.30 -1.00 -1.60 -1.20 -1.90 -1.40 -2.30

RC4 Hole +1.20 -0.00 +1.60 -0.00 +1.80 -0.00 +2.20 -0.00 Shaft -0.80 -1.60 -1.00 -2.00 -1.20 -2.40 -1.40 -2.80

RC5 Hole +1.20 -0.00 +1.60 -0.00 +1.80 -0.00 +2.20 -0.00 Shaft -1.60 -2.40 -2.00 -3.00 -2.50 -3.70 -3.00 -4.40

RC6 Hole +2.00 -0.00 +2.50 -0.00 +3.00 -0.00 +3.50 -0.00 Shaft -1.60 -2.80 -2.00 -3.60 -2.50 -4.30 -3.00 -5.20

RC7 Hole +2.00 -0.00 +2.50 -0.00 +3.00 -0.00 +3.50 -0.00 Shaft -2.50 -3.70 -3.00 -4.60 -4.00 -5.80 -5.00 -7.20

RC8 Hole +3.50 -0.00 +4.00 -0.00 +4.50 -0.00 +5.00 -0.00 Shaft -4.50 -6.50 -5.00 -7.50 -6.00 -9.00 -7.00 -10.50

RC9 Hole +5.00 -0.00 +6.00 -0.00 +7.00 -0.00 +9.00 -0.00 Shaft -7.00 -10.50 -8.00 -12.00 -9.00 -13.50 -10.00 -15.00


Table A-1-2 FORCE AND SHRINK FITS Diameter Size Range (in.) Class 0.00 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.56

FN1 Hole +0.25 -0.00 +0.30 -0.00 +0.40 -0.00 +0.40 -0.00 Shaft +0.50 +0.30 +0.60 +0.40 +0.75 +0.50 +0.80 +0.50

FN2 Hole +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 Shaft +0.85 +0.60 +1.00 +0.70 +1.40 +1.00 +1.60 +1.20

FN3 Hole Shaft

FN4 Hole +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 Shaft +0.95 +0.70 +1.20 +0.90 +1.60 +1.20 +1.80 +1.40

FN5 Hole +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 +1.00 -0.00 Shaft +1.30 +0.90 +1.70 +1.20 +2.00 +1.40 +2.30 +1.60

Diameter Size Range (in.) Class 0.56 - 0.71 0.71 - 0.95 0.95 - 1.19 1.19 - 1.58

FN1 Hole +0.40 -0.00 +0.50 -0.00 +0.50 -0.00 +0.60 -0.00

Shaft +0.90 +0.60 +1.10 +0.70 +1.20 +0.80 +1.30 +0.90FN2 Hole +0.70 -0.00 +0.80 -0.00 +0.80 -0.00 +1.00 -0.00

Shaft +1.60 +1.20 +1.90 +1.40 +1.90 +1.40 +2.40 +1.80FN3 Hole +0.80 -0.00 +1.00 -0.00

Shaft +2.10 +1.60 +2.60 +2.00FN4 Hole +0.70 -0.00 +0.80 -0.00 +0.80 -0.00 +1.00 -0.00

Shaft +1.80 +1.40 +2.10 +1.60 +2.30 +1.80 +3.10 +2.50FN5 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.60 -0.00

Shaft +2.50 +1.80 +3.00 +2.20 +3.30 +2.50 +4.00 +3.00

Diameter Size Range (in.) Class 1.58 - 1.97 1.97 - 2.56 2.56 - 3 .15 3.15 - 3.94

FN1 Hole +0.60 -0.00 +0.70 -0.00 +0.70 -0.00 +0.90 -0.00 Shaft +1.40 +1.00 +1.80 +1.30 +1.90 +1.40 +2.40 +1.80

FN2 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft +2.40 +1.80 +2.70 +2.00 +2.90 +2.20 +3.70 +2.80

FN3 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft +2.80 +2.20 +3.20 +2.50 +3.70 +3.00 +4.40 +3.50

FN4 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft +3.40 +2.80 +4.20 +3.50 +4.70 +4.00 +5.90 +5.00

FN5 Hole +1.60 -0.00 +1.80 -0.00 +1.80 -0.00 +2.20 -0.00 Shaft +5.00 +4.00 +6.20 +5.00 +7.20 +6.00 +8.40 +7.00


Table A-1-2 FORCE AND SHRINK FITS (CONTINUED)

Diameter Size Range (in.) Class 3.94 - 4.73 4.73 - 5.52 5.52 - 6.30 6.30 - 7.09

FN1 Hole +0.90 -0.00 +1.00 -0.00 +1.00 -0.00 +1.00 -0.00

Shaft +2.60 +2.00 +2.90 +2.20 +3.20 +2.50 +3.50 +2.80FN2 Hole +1.40 -0.00 +1.60 -0.00 +1.60 -0.00 +1.60 -0.00

Shaft +3.90 +3.00 +4.50 +3.50 +5.00 +4.00 +5.50 +4.50FN3 Hole +1.40 -0.00 +1.60 -0.00 +1.60 -0.00 +1.60 -0.00

Shaft +4.90 +4.00 +6.00 +5.00 +6.00 +5.00 +7.00 +6.00FN4 Hole +1.40 -0.00 +1.60 -0.00 +1.60 -0.00 +1.60 -0.00

Shaft +6.90 +6.00 +8.00 +7.00 +8.00 +7.00 +9.00 +8.00FN5 Hole +2.20 -0.00 +2.50 -0.00 +2.50 -0.00 +2.50 -0.00

Shaft +9.40 +8.00 +11.60 +10.00 +13.60 +12.00 +13.60 +12.00

† Text. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke, and Richard G. Budynas; equations and figures with the prefix T refer to the present tutorial.

MECHANICAL ENGINEERING DESIGN TUTORIAL 4-20: HERTZ CONTACT STRESSES

CHARACTERISTICS OF CONTACT STRESSES

1. Represent compressive stresses developed from surface pressures between two curved bodies pressed together;

2. Possess an area of contact. The initial point contact (spheres) or line contact (cylinders) become area contacts, as a result of the force pressing the bodies against each other;

3. Constitute the principal stresses of a triaxial (three dimensional) state of stress; 4. Cause the development of a critical section below the surface of the body; 5. Failure typically results in flaking or pitting on the bodies’ surfaces.

TWO DESIGN CASES Two design cases will be considered,

1. Sphere – Sphere Contact (Point Contact � Circular Contact Area) 2. Cylinder – Cylinder Contact (Line Contact � Rectangular Contact Area)

SPHERE – SPHERE CONTACT

TEXT FIGURE 4-42 Two Spheres in Contact

z

x

y

F

F

z

x

y

F

F

2a

1d

2d

(b) Contact stress has an elliptical distribution across contact over zone of diameter 2a.

(a) Two spheres held in contact by force F.

Shigley, Mischke & Budynas Machine Design Tutorial 4-20: Hertz Contact Stresses 2/10

Consider two solid elastic spheres held in contact by a force F such that their point of contact expands into a circular area of radius a, given as:

3

1/ 32 21 1 2 2

1 2

1 2

1 2

1 2

(Modified Text Eq. 4-72)

(1 ) / (1 ) /3where 8 (1/ ) (1/ )

applied force, Poisson's ratios for spheres 1 and 2, elastic modulii for spheres 1 and 2, diamete

a

a

a K F

E EKd d

F

E Ed d

ν ν

ν ν

=

� �− + −= � �+� �

==== rs of spheres 1 and 2

This general expression for the contact radius can be applied to two additional common cases:

1. Sphere in contact with a plane 2( );d = ∞ 2. Sphere in contact with an internal spherical surface or ‘cup’ 2( ).d d= −

Returning to the sphere-sphere case, the maximum contact pressure, maxp , occurs at the center point of the contact area.

max 23 (Text Eq. 4-73)

2Fpaπ

=

State of Stress

• The state of stress is computed based on the following mechanics:

1. Two planes of symmetry in loading and geometry dictates that ;x yσ σ= 2. The dominant stress occurs on the axis of loading: max ;zσ σ= 3. The principal stresses are yx σσσσ === 21 and zσσ =3 given 1 2 3,σ σ σ≥ ; 4. Compressive loading leads to , , and x y zσ σ σ being compressive stresses.

• Calculation of Principal Stresses

( )1

max 2

1 2

1 11 tan (1 ) (Modified Text Eq. 4-74)2 1

x aa a

y

pσ ζ νζ ζ

σ σ σ

−� �� = − − + −� �� +� � � ��

= = =


max3 2 (Modified Text Eq. 4-75)

1za

pσ σζ

−= =+

where / nondimensional depth below the surface

Poisson's ratio for the sphere examined (1 or 2)a z aζν

= ==

• Mohr’s Circle

Plotting the principal stresses on a Mohr’s circle plot results in: one circle, defined by

1 2σ σ= , shrinking to a point; and two circles, defined by 1 3,σ σ and 2 3,σ σ , plotted on top of each other. The maximum shear stress, maxτ , for the plot is calculated as:

1 3max (Modified Text Eq. 4-76)

2 2 2y zx z σ σσ σ σ στ

−− −= = =

If the maximum shear stress, maxτ , and principal stresses, 1 2 3, , andσ σ σ , are plotted as a function of maximum pressure, maxp , below the surface contact point, the plot of Fig. 4-43 is generated. This plot, based on a Poisson’s ratio of 3.0=ν , reveals that a critical section exists on the load axis, approximately 0.48a below the sphere surface. Many authorities theorize that this maximum shear stress is responsible for the surface fatigue failure of such contacting elements; a crack, originating at the point of maximum shear, progresses to the surface where lubricant pressure wedges a chip loose and thus creates surface pitting.

TEXT FIGURE 4-43: Magnitude of the stress components below the

surface as a function of maximum pressure of contacting spheres.


CYLINDER–CYLINDER CONTACT Consider two solid elastic cylinders held in contact by forces F uniformly distributed along the cylinder length l.

The resulting pressure causes the line of contact to become a rectangular contact zone of half-width b given as:

1/ 22 21 1 2 2

1 2

1 2

1 2

1 2


(1 ) / (1 ) /2where (1/ ) (1/ )

applied force, Poisson's ratios for cylinders 1 and 2, elastic modulii for cylinders 1 and 2, di

b

b

b K F

E EKl d d

F

E Ed d

ν νπ

ν ν

=

� �− + −= � �+� �

====

1 2

ameters of spheres 1 and 2 length of cylinders 1 and 2 ( assumed)l l l= =

(b) Contact stress has an elliptical distribution across contact zone of width 2b.

(a) Two right circular cylinders held in contact by forces F uniformly distributed along cylinder length l.

z

x

y

F

F

z

x

y

F

F

2b

1d

2dl

TEXT FIGURE 4-44 Two Cylinders in Contact


This expression for the contact half-width, b, is general and can be used for two additional cases which are frequently encountered:

1. Cylinder in contact with a plane, e.g. a rail 2( );d = ∞ 2. Cylinder in contact with an internal cylindrical surface, for example the race of a

roller bearing 2( ).d d= −

The maximum contact pressure between the cylinders acts along a longitudinal line at the center of the rectangular contact area, and is computed as:

max2 (Text Eq. 4-78)Fp

blπ=

State of Stress

• The state of stress is computed based on the following mechanics:

1. One plane of symmetry in loading and geometry dictates that ;x yσ σ≠ 2. The dominant stress occurs along the axis of loading: max ;zσ σ= 3. The principal stresses are equal to , , and x y zσ σ σ with zσσ =3 ; 4. Compressive loading leads to , , and x y zσ σ σ being compressive stresses.

• Calculation of Principal Stresses and Maximum Shear Stress

3 z max 2

1y

2max

2

y max 2

1 (Modified Text Eq. 4-81)1

for 0 0.436

for 0.436

where,

2 1 (Modified Text Eq. 4-79)

1 2 2 (Modified Text Eq. 4-1

b

x b

b

x b b

bb

b

p

p

p

σ σζ

σ ζσ

σ ζ

σ ν ζ ζ

ζσ ζζ

= = −+

≤ ≤��= � ≤��

� �= − + −� �

� �� +� � �= − − �� +� �

80)

/b z bζ =

Shigley, Mischke & Bud

1/ 3max

1/ 3

The maximum shear stress is thus given as:

( ) / 2 for 0 0.436( ) / 2 for 0.436

z x b

z y b

τ σ σ ζτ

τ σ σ ζ= − ≤ ≤��= � = − ≤��

When these equations are plotted as a function of maximum contact pressure up to a distance 3b below the surface contact point, the plot of Fig. 4-45 is generated. Based on a Poisson’s ratio of 0.3, this plot reveals that maxτ attains a maxima for

/ 0.786b z bζ = = and 0.3pmax.

Example T4.20.1:

Probsteel Find:
TEXT FIGURE 4-45: Magnitude of stress components belowthe surface as a function of maximum pressure for contacting cylinders.
ynas Machine Design Tutorial 4-20: Hertz Contact Stresses 6/10

lem Statement: A 6-in-diameter cast-iron wheel, 2 in wide, rolls on a flat surface carrying a 800 lbf load.

1. The Hertzian stresses 1/3, , and x y zσ σ σ τ in the cast iron wheel at

the critical section; 2. The comparative state of stress and maximum shear stress, arising

during a revolution, at point A located 0.015 inch below the wheel rim surface.



1. Compute the value of the contact half-width, b. 2. Compute the maximum pressure generated by the normal force of

the wheel. 3. Use the results of steps (1) and (2) to calculate the contact stresses

in the cast iron wheel for the critical section, z/b = 0.786. 4. Evaluate the principal stresses based upon the contact stress

calculations. 5. Calculate the maximum shear stress. 6. Compare these results with those obtained by using Fig. 4-45. 7. During a single revolution of the wheel, point A will experience a

cycle of stress values varying from zero (when point A lies well outside the contact zone) to a maximum state of stress (when A lies within the contact zone and on the line of action of the 800 lbf force.) We expect point A to “feel” the effects of a semi-elliptical contact pressure distribution as point A moves into and through the contact zone. Thus, we need to calculate the contact stresses for a depth of z = 0.015 inch, which we expect to lie within the contact zone.

Schematic:

Solution: 1. Compute contact half-width, b

61 cast iron 1 cast iron

62 steel 2 steel

Material Properties: 14.5 10 psi; 0.211

30.0 10 psi; 0.292

E E

E E

ν ν

ν ν

= = × = =

= = × = =

800 lbf 6 in

2 in

6steel

steel

30 10 psi0.292

Eν

= ×=

6cast iron

cast iron

14.5 10 psi0.211

Eν

= ×=

A ·


Dimensions: 1 26.0 in; ; 2.0 ind d l= = ∞ =

1/ 22 21 1 2 2

1 2

1/ 26 62 2

4

4


(1 ) / (1 ) /2 (1/ ) (1/ )

/(14.5 10 ) /(30.0 10 )1 (0.211) 1 (0.292)2(2.0) (1/ 6.0) (1/ )

4.291 10 in/ lbf

(4.291 10

b

b

b

b K F

E EKl d d

b K F

ν νπ

π−

−

=

� �− + −= � �+� �

� �× + ×� � � � − −� � � �= �+ ∞ �

= ×

= = × 1/2 in/ lbf )(800 lbf ) = × -21.214 10 in

2. Maximum Pressure, pmax

max 22 2(800 lbf)

(1.214 10 in)(2.0 in)Fpblπ π −= = =

×20 980 psi

3. Hertz Contact Stresses in Cast Iron Wheel

At the critical section, / = 0.786 , b z bζ =

21 max

2

2

max 2

2

2

max

2 1

2(0.211)(20 980 psi) 1+(0.786) 0.786

1 2 21

1 2(0.786)( 20 980 psi) 2(0.786)1 (0.786)

1

1

x b b

by b

b

zb

p

p

p

σ ν ζ ζ

ζσ ζζ

σζ

� �= − + −� ��

� �= − −� ��

= −

� �� +� � = − − � �+� ��

�� +� �� = − −� �� +� ��

= −

= −+

4302 psi

3895 psi

2 2

20 980 psi

1+(0.786)

−=

= −16 490 psi


Note that the small contact area involved in this type of problem gives rise to

very high pressure, relative to the applied force, and thus exceptionally high

stresses.

4. Since , , and x y zσ σ σ are all principal stresses, we can conclude:

1

2

3

3895 psi

4302 psi16 490 psi

y

x

z

σ σσ σσ σ

= = −

= = −= = −

5. Maximum Shear Stress

1 3max 1/ 3

3895 psi ( 16 490 psi)2 2 2

y zσ σσ στ τ−− − − −= = = =

= 6298 psi

6. Comparison with results based on Text Figure 4-45:

For z/b ≈ 0.75,

max

max

max

max max

0.3 6294 psi0.2 4196 psi

0.8 16 780 psi0.3 6294 psi

x

y

z

pp

pp

σσστ

≈ − = −≈ − = −

≈ − = −≈ =

Comparing these results with those calculated using a value of ν = 0.211, we

find that only xσ is a function of ν; max, , and y zσ σ τ are independent of ν

since the graphical estimates of their values are within 3 % of those obtained

from the plot which assumes a Poisson’s ratio of 0.3.

7. For a depth of 0.015 in below the cylinder surface,

20.015 in = 1.236

1.214 10 inbζ − =×

Substituting,


2

max

2

2 1

2(0.211)(20 980 psi) 1+(1.236) 1.236

x b bpσ ν ζ ζ� �= − + −� ��

� �= − −� ��

= −3133 psi

2

max 2

1 2 21

by b

b

p ζσ ζζ

� �� +� �� = − −� �� + � �

2

2

max 2 2

1 2(1.236)( 20 980 psi) 2 1.2361 (1.236)

1 20 980 psi

1 1+(1.236)z

b

pσζ

� �� +� �� = − −� � �+� � ��

= −−= − =

+= −

1652 psi

13 200 psi

1 3max

1652 ( 13 200) 2 2 2

y zσ σσ στ−− − − −= = = = 5774 psi

As expected, at a depth corresponding greater than the critical section

( z/b = 1.236 > 0.786), the magnitudes of all three principal stresses are

smaller than those calculated for z/b = 0.786. The difference between the

principal stresses is also smaller and consequently, τmax also decreases.

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