Chapter 3 Load and Stress Analysis -...

Chapter 3

Load and Stress Analysis

September 28, 2013 Dr. Mohammad Suliman Abuhaiba, PE 1

Shear Force and Bending

Moments in Beams

Dr. Mohammad Suliman Abuhaiba, PE

September 28, 2013 2

Change in shear force from A to B = area of load diagram between xA &

xB.

Change in moment from A to B = area of shear-force diagram between

xA & xB.

Distributed Load on Beam

Distributed load q(x):

load intensity



Shear-

Moment

Diagrams


Fig. 3−5


Stress element



Shear stress is often resolved into perpendicular components

First subscript indicates direction of surface normal

Second subscript indicates direction of shear stress

In most cases, “cross shears” are equal

Plane stress occurs when stresses on one surface are zero

Plane-Stress Transformation Equations

Cutting plane stress element at an arbitrary angle and balancing

stresses gives plane-stress transformation equations


Fig. 3−9


Principal Stresses for Plane Stress

Differentiating Eq. (3-8) with respect to f and setting equal to zero maximizes s and gives

The two values of 2fp are the principal directions.

Stresses in the principal directions are the principal stresses.

The principal direction surfaces have zero shear stresses.

There is a third principal stress, equal to zero for plane stress.



Extreme-value Shear Stresses for Plane Stress

Performing similar procedure with shear stress in Eq. (3-

9), max shear stresses are found to be on surfaces that

are ±45º from principal directions.

The two extreme-value shear stresses are

Three extreme-value shear stresses.

If principal stresses are ordered so that s1 > s2 > s3,

then tmax = t1/3



Mohr’s Circle

Diagram


Fig. 3−10



Example 3-4

Fig. 3−11

x-y

orientation

Principal stress

orientation

Max shear

orientation

September 28, 2013


11

Example 3-4

Example 3-4


General Three-Dimensional Stress


Fig. 3−12


Principal stresses are ordered such that

s1 > s2 > s3, in which case tmax = t1/3

Hooke’s law

For axial stress in x direction,

For a stress element undergoing sx, sy, and sz,

simultaneously,

Elastic Strain



Elastic Strain

Hooke’s law for shear:

Shear strain g is the change in a right angle of a stress

element when subjected to pure shear stress.

G is the shear modulus of elasticity or modulus of rigidity.

For a linear, isotropic, homogeneous material,



Uniformly Distributed Stresses

For tension and compression,

For direct shear (no bending present),



Normal Stresses for Beams in Bending Straight beam in positive bending

x axis is neutral axis

xz plane is neutral plane

Neutral axis is coincident with the

centroidal axis of cross section


Fig. 3−13



Example 3-5

Dimensions in

mm

Fig. 3−15

Transverse Shear Stress Transverse shear stress is always accompanied with bending stress


Fig. 3−18


Maximum Values of Transverse Shear Stress




Example 3-7

Fig. 3−20


Example 3-7

Torsion


Fig. 3−21



Example 3-9

Fig. 3−24


Example 3-9

Stress Concentration Localized increase of stress near discontinuities

Kt is Theoretical (Geometric) Stress Concentration Factor



Theoretical Stress Concentration Factor



Stress Concentration for Static and

Ductile Conditions

With static loads and ductile materials

Highest stressed fibers yield (cold work)

Load is shared with next fibers

Cold working is localized

Overall part does not see damage unless ultimate strength

is exceeded

Stress concentration effect is commonly ignored for static

loads on ductile materials




Example 3-13

Fig. 3−30

The 2-mm-thick bar shown in Fig. 3–30 is loaded axially with a

constant force of 10 kN. The bar material has been heat treated and

quenched to raise its strength, but as a consequence it has lost most

of its ductility. It is desired to drill a hole through the center of the 40-

mm face of the plate to allow a cable to pass through it. A 4-mm hole

is sufficient for the cable to fit, but an 8-mm drill is readily available.

Will a crack be more likely to initiate at the larger hole, the smaller

hole, or at the fillet?

Since material is brittle, the effect of stress concentrations near discontinuities

must be considered. Dealing with the hole first, for a 4-mm hole, nominal

stress is

The theoretical stress concentration factor, from Fig. A–15–1, with d/w = 4/40

0.1, is Kt = 2.7. The maximum stress is


Example 3-13

Fig. A−15 −1

Similarly, for an 8-mm hole,

With d/w = 8/40 = 0.2, then Kt 2.5, and the maximum stress is

Though stress concentration is higher with the 4-mm hole, increased nominal

stress with the 8-mm hole has more effect on maximum stress. For the fillet,


Example 3-13

From Table A–15–5, D/d =

40/34 = 1.18, and r/d = 1/34 =

0.026. Then Kt = 2.5.

crack will most likely occur with

8-mm hole, next the 4-mm

hole, and least likely at fillet

Chapter 4

Deflection and Stiffness

September 28, 2013


39

Spring Rate

Relation between force and deflection, F = F(y)

Spring rate

For linear springs, k is constant, called spring constant



Axially-Loaded Stiffness Total extension or contraction of a uniform bar in tension

or compression

Spring constant, with k = F/d



Torsionaly-Loaded Stiffness

Angular deflection (in radians) of a uniform solid or hollow

round bar subjected to a twisting moment T

Torsional spring constant for round bar



Deflection Due to Bending



Example 4-1


Fig. 4–2

Example 4-1


The boundary conditions for the simply supported beam are y = 0 at x = 0 and

l. Applying the first condition, y = 0 at x = 0, to Eq. (2) results in C2 = 0.

Applying the second condition to Eq. (2) with C2 = 0,

Strain Energy

External work done on elastic member in deforming it is

transformed into strain energy, or potential energy.

Strain energy equals product of average force and

deflection.

For axial loading, applying k = AE/l from Eq. (4-4),



Some Common Strain Energy Formulas

For torsional loading, applying k = GJ/l from Eq. (4-7),



For transverse shear loading,

C is a modifier dependent on the cross sectional shape.



Some Common Strain Energy Formulas


Example 4-8

Castigliano’s Theorem

When forces act on elastic systems subject to small

displacements, displacement corresponding to any force,

in the direction of force, is equal to the partial derivative

of total strain energy with respect to that force.

For rotational displacement, in radians,



Example 4-9


Utilizing a Fictitious Force

Castigliano’s method can be used to find a deflection at a

point even if there is no force applied at that point.

Apply a fictitious force Q at the point, and in the direction,

of the desired deflection.

Set up the equation for total strain energy including

energy due to Q.

Take derivative of total strain energy with respect to Q.

Once derivative is taken, Q is no longer needed and can

be set to zero.




Common Deflection Equations


Example 4-10

Fig. 4–10


Example 4-10

Compression Members Column – A member loaded in compression such that

either its length or eccentric loading causes it to

experience more than pure compression

Four categories of columns

Long columns with central loading

Intermediate-length columns with central loading

Columns with eccentric loading

Struts or short columns with eccentric loading



Long Columns with Central Loading

When P reaches

critical load, column

becomes unstable

and bending

develops rapidly

Critical load

depends on end

conditions


Fig. 4–18


Euler Column Formula For pin-ended column, critical load is given by Euler

column formula,

Applies to other end conditions with addition of constant C

for each end condition


(4-42)

(4-43)


Recommended Values for End Condition Constant


Table 4–2


Long Columns with Central Loading

Using I = Ak2, where A is the area and k is the radius of

gyration, Euler column formula can be expressed as

l/k is slenderness ratio, used to classify columns

according to length categories.

Pcr/A is critical unit load, load per unit area necessary

to place column in a condition of unstable equilibrium.



Euler Curve Plotting Pcr/A vs l/k, with C = 1 gives curve PQR

Dr. Mohammad Suliman Abuhaiba, PE Fig. 4–19


vulnerability to failure near point Q

Since buckling is sudden and catastrophic, a

conservative approach near Q is desired

Point T is usually defined such that Pcr/A =

Sy/2, giving

For (l/k) > (l/k)1, use Euler equation

For (l/k) ≤ (l/k)1, use a parabolic curve

between Sy and T

Intermediate-Length Columns with Central Loading

For intermediate-length columns, where (l/k) ≤ (l/k)1,

use a parabolic curve between Sy and T

General form of parabola

If parabola starts at Sy, then a = Sy

If parabola fits tangent to Euler curve at T, then

Results in parabolic formula, also known as J.B. Johnson

formula



Columns with Eccentric Loading For eccentrically loaded column

with eccentricity e,

M = -P(e+y)

Substituting into d2y/dx2=M/EI,

Solving with boundary

conditions

y = 0 at x = 0 and at x = l


Fig. 4–20


At midspan where x = l/2

Max compressive stress includes axial and

bending

Substituting Mmax from Eq. (4-48)



Columns with Eccentric Loading

Using Syc as max value of sc, and solving for P/A, we obtain secant column

formula

Secant Column Formula

Secant Column

Formula

ec/k2 is eccentricity

ratio

Design charts of

secant column

formula for various

eccentricity ratio can

be prepared for a

given material

strength


Fig. 4–21


Example 4-16


Chapter 5

Failures Resulting from

Static Loading

September 28, 2013


70

Maximum Shear Stress Theory (MSS) Yielding begins when max shear stress in a stress element exceeds

max shear stress in a tension test specimen of the same material

when that specimen begins to yield.

For a tension test specimen, max shear stress is s1 /2.

At yielding, when s1 = Sy, max shear stress is Sy /2 .

For any stress element, use Mohr’s circle to find max shear stress.

Compare max shear stress to Sy/2.

Ordering principal stresses such that s1 ≥ s2 ≥ s3,



consider a plane stress state

Let sA & sB represent two non-zero principal stresses,

then order them with the zero principal stress such that

s1 ≥ s2 ≥ s3

Assuming sA ≥ sB there are three cases to consider

Case 1: sA ≥ sB ≥ 0 , s1 = sA and s3 = 0

Eq. (5–1) reduces to sA ≥ Sy

Case 2: sA ≥ 0 ≥ sB , s1 = sA and s3 = sB

Eq. (5–1) reduces to sA − sB ≥ Sy

Case 3: 0 ≥ sA ≥ sB , s1 = 0 and s3 = sB

Eq. (5–1) reduces to sB ≤ −Sy



Maximum Shear Stress Theory (MSS)

Plot three cases on

principal stress axes

Case 1: sA ≥ sB ≥ 0

sA ≥ Sy

Case 2: sA ≥ 0 ≥ sB

sA − sB ≥ Sy

Case 3: 0 ≥ sA ≥ sB

sB ≤ −Sy

Other lines are

symmetric cases

Inside envelope is

predicted safe zone


Fig. 5–7



Comparison to experimental

data

Conservative in all quadrants




Distortion Energy (DE) Failure Theory Also known as: Octahedral Shear Stress, Shear Energy, Von Mises

Ductile materials stressed hydrostatically (equal principal stresses) exhibited

yield strengths greatly in excess of expected values.

If strain energy is divided into hydrostatic volume changing energy and

angular distortion energy, yielding is primarily affected by distortion energy.

Yielding occurs when distortion strain energy per unit volume reaches

distortion strain energy per unit volume for yield in simple tension or

compression of the same material.



Hydrostatic stress is average of principal stresses

Strain energy per unit volume,



Distortion Energy (DE) Failure Theory

Strain energy for producing only volume change is obtained by

substituting sav for s1, s2, and s3

Obtain distortion energy by subtracting volume changing energy, Eq.

(5–7), from total strain energy, Eq. (b)

Tension test specimen at yield has s1 = Sy and s2 = s3 =0

DE theory predicts failure when distortion energy exceeds distortion

energy of tension test specimen



Distortion Energy (DE) Failure Theory

Von Mises Stress

For plane stress, simplifies to

In terms of xyz components, in three dimensions

In terms of xyz components, for plane stress



Distortion Energy Theory With Von Mises Stress

Von Mises Stress: a single, equivalent, or effective stress

for the entire general state of stress in a stress element.

DET simply compares von Mises stress to yield strength.

Introducing a design factor,

Expressing as factor of safety,


ySn

s


Octahedral Stresses Octahedral stresses are identical on 8 surfaces symmetric to the

principal stress directions.

Octahedral stresses allow representation of any stress situation with

a set of normal and shear stresses.


Principal stress element with single

octahedral plane showing

All 8 octahedral planes showing


Octahedral normal stresses are normal to the octahedral surfaces,

and are equal to the average of the principal stresses.

Octahedral shear stresses lie on the octahedral surfaces.


Fig. 5–10


Octahedral Stresses

Yielding begins when octahedral

shear stress in a stress element

exceeds octahedral shear stress

in a tension test specimen at

yield.

For a tension test specimen at

yielding, s1 = Sy , s2 = s3 = 0,

Octahedral Shear Stress Failure Theory



Shear Strength Predictions For pure shear loading, Mohr’s circle shows

that sA = −sB = t

Plotting this equation on principal stress

axes gives load line for pure shear case

Intersection of pure shear load line with

failure curve indicates shear strength has

been reached

Each failure theory predicts shear strength

to be some fraction of normal strength

For MSS theory, intersecting pure shear

load line with failure line [Eq. (5–5)] results

in

For DE theory, intersection pure shear load

line with failure curve [Eq. (5–11)] gives


Fig. 5–9


DE theory predicts shear

strength as

Example 5-1


Example 5-1


Fig. 5−11

Mohr Theory Some materials have compressive strengths different from tensile

strengths

Mohr theory is based on three simple tests: tension, compression,

and shear

Plotting Mohr’s circle for each, bounding curve defines failure

envelope


Fig. 5−12


Coulomb-Mohr Theory Curved failure curve is difficult

to determine analytically

Coulomb-Mohr theory

simplifies to linear failure

envelope using only tension

and compression tests

(dashed circles)


Fig. 5−13


For ductile material, use tensile

and compressive yield strengths

For brittle material, use tensile

&compressive ultimate strengths

To plot on principal stress axes, consider three cases

Case 1: sA ≥ sB ≥ 0 For this case, s1 = sA and s3 = 0

Eq. (5−22) reduces to

Case 2: sA ≥ 0 ≥ sB For this case, s1 = sA and s3 = sB

Eq. (5-22) reduces to

Case 3: 0 ≥ sA ≥ sB For this case, s1 = 0 and s3 = sB

Eq. (5−22) reduces to



Coulomb-Mohr Theory

Plot three cases on principal stress axes

Similar to MSS theory, except with different strengths for

compression and tension


Fig. 5−14


Coulomb-Mohr Theory

Intersect the pure shear

load line with the failure

line to determine the

shear strength

Since failure line is a

function of tensile and

compressive strengths,

shear strength is also a

function of these terms.

Example 5-2


Failure Theories for Brittle Materials Experimental data

indicates some

differences in failure for

brittle materials.

Failure criteria is

generally ultimate

fracture rather than

yielding

Compressive strengths

are usually larger than

tensile strengths


Fig. 5−19


Brittle Coulomb-Mohr


Quadrant condition Failure criteria

Fig. 5−14


Brittle Failure Experimental Data

Coulomb-Mohr is

conservative in 4th

quadrant

Modified Mohr

criteria adjusts to

better fit the data in

4th quadrant


Fig. 5−19


Modified-Mohr

Shigley’s Mechanical

Engineering Design

Quadrant condition Failure criteria

Example 5-5


Fig. 5−16

Example 5-5


Selection of Failure Criteria in Flowchart Form


Fig. 5−21


Chapter 6

Fatigue Failure Resulting

from Variable Loading

The McGraw-Hill Companies © 2012

September 28, 2013


100

Stages of Fatigue Failure

Stage I – Initiation of micro-

crack due to cyclic plastic

deformation

Stage II – Progresses to

macro-crack that repeatedly

opens and closes, creating

bands called beach marks

Stage III – Crack has

propagated far enough that

remaining material is

insufficient to carry the load,

and fails by simple ultimate

failure


Fig. 6–1


Fatigue-Life Methods

Three major fatigue life models

Methods predict life in number of cycles to failure, N, for a specific

level of loading

1. Stress-life method

Least accurate, particularly for low cycle applications

Most traditional, easiest to implement

2. Strain-life method

Detailed analysis of plastic deformation at localized regions

Several idealizations are compounded, leading to uncertainties in

results

3. Linear-elastic fracture mechanics method

Assumes crack exists

Predicts crack growth with respect to stress intensity



Stress-Life Method Test specimens are subjected to repeated stress while counting

cycles to failure

specimen Subjected to pure bending with no transverse shear

As specimen rotates, stress fluctuates between equal magnitudes of

tension and compression, completely reversed stress cycling


Fig. 6–9


Strength corresponding to the

knee is called endurance limit

Se

S-N Diagram for Nonferrous Metals Nonferrous metals often do not have an endurance limit.

Fatigue strength Sf is reported at a specific number of cycles

Figure 6–11 shows typical S-N diagram for aluminums


Fig. 6–11


• Stress levels

below Se predict

infinite life

• Between 103 &

106 cycles, finite

life is predicted

• Below 103 cycles

is known as low

cycle, and is

often considered

quasi-static.

Yielding usually

occurs before

fatigue in this

zone.

Strain-Life Method Fatigue failure begins at a

local discontinuity

When stress at

discontinuity exceeds

elastic limit, plastic strain

occurs

Cyclic plastic strain can

change elastic limit,

leading to fatigue

Fig. 6–12 shows true

stress-true strain

hysteresis loops of first five

stress reversals


Fig. 6–12


Relation of Fatigue Life to Strain Figure 6–13: relationship of fatigue life to true-strain amplitude

Fatigue ductility coefficient e'F : true strain corresponding to fracture in one

reversal (point A in Fig. 6–12)

Fatigue strength coefficient s'F : true stress corresponding to fracture in one

reversal (point A in Fig. 6–12)


Fig. 6–13


Fatigue ductility exponent c: slope of plastic-strain line, and is power to which life 2N must be raised to be proportional to true plastic-strain amplitude. 2N stress reversals corresponds to N cycles.

Fatigue strength exponent b: slope of elastic-strain line, and is power to which life 2N must be raised to be proportional to true-stress amplitude.

Total strain amplitude is half the total strain range

Equation of the plastic-strain line in Fig. 6–13

Equation of the elastic strain line in Fig. 6–13

Applying Eq. (a), the total-strain amplitude is

Known as Manson-Coffin relationship between fatigue life and total strain

Some values of coefficients and exponents given in Table A–23



Relation of Fatigue Life to Strain

The Endurance Limit Simplified estimate of endurance limit for steels for the rotating-beam

specimen, S'e


Fig. 6–17


Fatigue Strength

For design, an approximation of the idealized S-N

diagram is desirable.

To estimate the fatigue strength at 103 cycles, start with

Eq. (6-2)

Define the specimen fatigue strength at a specific

number of cycles as

Combine with Eq. (6–2),



At 103 cycles,

f is the fraction of Sut represented by

Solving for f,

The SAE approximation for steels with HB ≤ 500 may be used.

To find b, substitute the endurance strength and corresponding cycles

into Eq. (6–9) and solve for b

Eqs. (6–11) and (6–12) can be substituted into Eqs. (6–9) and (6–

10) to obtain expressions for S'f and f


310( )fS


Fatigue Strength

Fatigue Strength Fraction f Plot Eq. (6–10) for fatigue strength fraction f of Sut at 103 cycles

Use f from plot for S'f = f Sut at 103 cycles on S-N diagram

Assumes Se = S'e= 0.5Sut at 106 cycles


Fig. 6–18


Equations for S-N Diagram


Write equation for S-N line

from 103 to 106 cycles

Two known points

At N =103 cycles, Sf = f Sut

At N =106 cycles, Sf = Se

Equations for line:

Fig. 6–10

Equations for S-N Diagram


If a completely reversed stress srev is given, setting Sf = srev

in Eq. (6–13) and solving for N gives,

Note that the typical S-N diagram is only applicable for

completely reversed stresses

For other stress situations, a completely reversed stress with

the same life expectancy must be used on the S-N diagram

Low - cycle Fatigue

Low-cycle fatigue is defined for fatigue failures in the

range 1 ≤ N ≤ 103

On the idealized S-N diagram on a log-log scale, failure

is predicted by a straight line between two points (103, f

Sut) and (1, Sut)




Example 6-2

Endurance Limit Modifying Factors

Endurance limit S'e is for carefully prepared and tested specimen

If warranted, Se is obtained from testing of actual parts

A set of Marin factors are used to adjust the endurance limit



Surface Factor ka

Stresses tend to be high at the surface

Surface finish has an impact on initiation of cracks at localized

stress concentrations

Higher strengths are more sensitive to rough surfaces.



Size Factor kb Larger parts have greater surface area at high stress levels

Likelihood of crack initiation is higher

For bending & torsion loads, trend of the size factor data is given by

Applies only for round, rotating diameter

For axial load, there is no size effect, so kb = 1



Size Factor kb For parts that are not round and rotating, an equivalent round rotating

diameter is obtained.

Equate the volume of material stressed at and above 95% of max

stress to the same volume in the rotating-beam specimen.

Lengths cancel, so equate areas.

For a rotating round section, the 95% stress area is the area of a ring,

Equate 95% stress area for other conditions to Eq. (6–22) and solve

for d as the equivalent round rotating diameter

For non-rotating round,

Equating to Eq. (6-22) and solving for equivalent diameter,



Size Factor kb


Table 6–3

A95s for common non-

rotating structural shapes



Example 6-4

Loading Factor kc

Accounts for changes in endurance limit for different types of fatigue loading.

Only to be used for single load types. Use Combination Loading method (Sec. 6–14) when more than one load type is present.



Temperature Factor kd

Endurance limit appears to maintain same relation to ultimate

strength for elevated temperatures as at room temperature

This relation is summarized in Table 6–4



Temperature Factor kd If ultimate strength is known for operating temperature, then just use

that strength. Let kd = 1 and proceed as usual.

If ultimate strength is known only at room temperature, then use Table

6–4 to estimate ultimate strength at operating temperature. With that

strength, let kd = 1 and proceed as usual.

Alternatively, use ultimate strength at room temperature and apply

temperature factor from Table 6–4 to the endurance limit.

A fourth-order polynomial curve fit of the underlying data of Table 6–4

can be used in place of the table, if desired.




Example 6-5

Reliability Factor ke From Fig. 6–17, S'e = 0.5 Sut is typical of the data and represents 50%

reliability.

Reliability factor adjusts to other reliabilities.


Fig. 6–17


A reliability

of R = 0.90 means

that there is a 90%

chance that the

part will perform

its proper function

without failure.

Reliability Factor ke


Table 6–5


Miscellaneous-Effects Factor kf

Consider other possible factors.

Residual stresses

Directional characteristics from cold working

Case hardening

Corrosion

Surface conditioning, e.g. electrolytic plating and metal

spraying

Cyclic Frequency

Frettage Corrosion

Limited data is available.

May require research or testing. Dr. Mohammad Suliman Abuhaiba, PE


Stress Concentration and Notch Sensitivity

For dynamic loading, stress concentration effects must be applied.

Obtain Kt as usual (Appendix A–15)

For fatigue, some materials are not fully sensitive to Kt so a reduced

value can be used.

Define Kf as the fatigue stress-concentration factor.

Define q as notch sensitivity, ranging from 0 (not sensitive) to 1 (fully

sensitive).

For q = 0, Kf = 1

For q = 1, Kf = Kt



Notch Sensitivity Obtain q for bending or axial loading from Fig. 6–20.

Then get Kf from Eq. (6–32): Kf = 1 + q( Kt – 1)


Fig. 6–20


Notch Sensitivity Obtain qs for torsional loading from Fig. 6–21.

Then get Kfs from Eq. (6–32): Kfs = 1 + qs( Kts – 1)


Fig. 6–21


Notch Sensitivity Alternatively, can use curve fit equations for Figs. 6–20 and 6–21 to get

notch sensitivity, or go directly to Kf .


Bending or axial:

Torsion:


Notch Sensitivity for Cast Irons

Cast irons are already full of discontinuities, which are

included in the strengths.

Additional notches do not add much additional harm.

Recommended to use q = 0.2 for cast irons.



Example 6-6


Application of Fatigue Stress

Concentration Factor

Use Kf as a multiplier to increase the nominal stress.

Some designers sometimes applied 1/ Kf as a Marin factor to reduce Se .

For infinite life, either method is equivalent, since

For finite life, increasing stress is more conservative. Decreasing Se

applies more to high cycle than low cycle.


1/ f eef

f

K SSn

K s s



Example 6-8

Fluctuating

Stresses Define sm as midrange

steady component of stress

(mean stress) and sa as

amplitude of alternating

component of stress


stress ratio

amplitude ratio


Application of Kf for Fluctuating Stresses For fluctuating loads at points with stress concentration, the best

approach is to design to avoid all localized plastic strain.

In this case, Kf should be applied to both alternating and midrange

stress components.

When localized strain does occur, some methods (nominal mean

stress method and residual stress method) recommend only

applying Kf to the alternating stress.

Dowling method recommends applying Kf to alternating stress and

Kfm to mid-range stress, where Kfm is



Modified Goodman

Diagram

Midrange stress is plotted

on abscissa

All other components of

stress are plotted on the

ordinate


Fig. 6–24


Master Fatigue Diagram Displays four stress components as well as two stress ratios


Fig. 6–26


Plot of Alternating vs Midrange Stress Modified Goodman line from Se to Sut is one simple representation of

the limiting boundary for infinite life



Plot of Alternating vs Midrange Stress Experimental data on normalized plot of sa vs sm

Demonstrates little effect of negative midrange stress


Fig. 6–25


Commonly Used Failure Criteria Gerber passes through the data

ASME-elliptic passes through data and incorporates rough yielding

check


Fig. 6–27


Modified Goodman is linear, so simple

to use for design. It is more

conservative than Gerber.

Soderberg provides a very conservative

single check of both fatigue and

yielding.

Langer line

represents standard

yield check.

It is equivalent to

comparing max

stress to yield

strength.

Equations for Commonly Used Failure Criteria

Intersecting a constant slope load line with each failure criteria

produces design equations

n is the design factor or factor of safety for infinite fatigue life



Summarizing Tables for Failure

Criteria Tables 6–6 to 6–8 summarize the pertinent equations for Modified

Goodman, Gerber, ASME-elliptic, and Langer failure criteria

The first row gives fatigue criterion

The second row gives yield criterion

The third row gives the intersection of static and fatigue criteria

The fourth row gives the equation for fatigue factor of safety

The first column gives the intersecting equations

The second column gives the coordinates of the intersection



Summarizing Table for Modified Goodman


Summarizing Table for Gerber


Summarizing Table for ASME-Elliptic


Example 6-10


Example 6-10


Fig. 6–28

Example 6-10


Example 6-10


Fig. 6–29

Example 6-10


Fatigue Criteria for Brittle Materials For many brittle materials, the first quadrant fatigue failure criteria

follows a concave upward Smith-Dolan locus,

Or as a design equation,

For a radial load line of slope r, the intersection point is

In the second quadrant,



Torsional Fatigue Strength Testing has found that steady-stress component has no effect on

endurance limit for torsional loading if the material is ductile,

polished, notch-free, and cylindrical.

However, for less than perfect surfaces, the modified Goodman line

is more reasonable.

For pure torsion cases, use kc = 0.59 to convert normal endurance

strength to shear endurance strength.

For shear ultimate strength, recommended to use



Combinations of Loading Modes When more than one type of loading (bending, axial, torsion) exists,

use the Distortion Energy theory to combine them.

Obtain von Mises stresses for both midrange and alternating

components.

Apply appropriate Kf to each type of stress.

For load factor, use kc = 1. The torsional load factor (kc = 0.59) is

inherently included in the von Mises equations.

If needed, axial load factor can be divided into the axial stress.



Static Check for Combination

Loading Distortion Energy theory still applies for check of static yielding

Obtain von Mises stress for maximum stresses (sum of midrange

and alternating)

Stress concentration factors are not necessary to check for yielding

at first cycle

Alternate simple check is to obtain conservative estimate of s'max

by summing s'a and s'm Dr. Mohammad Suliman Abuhaiba, PE

1/2

2 2

max

max

3a m a m

y

y

Sn

s s s t t

s

max a ms s s


Chapter 3 Load and Stress Analysis -...

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