Materials Methodology Stress Factors

8/3/2019 Materials Methodology Stress Factors

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2008 ANSYS, Inc. All rights reserved. 1 ANSYS, Inc. Proprietary

2008 International

ANSYS Conference

Using FEA Results To DetermineStress Concentration Factors

W.D. GrowneyEngineering Specialist

Eaton Aerospace, Conveyance Systems Division


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Agenda

Overview of Eaton Corporation

Background Stress Concentration Factors from FEA results

Example: Finite Plate with a Central Hole

Example: Square Shoulder With Fillet (PureBending)

Summary


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Vocabulary

A Area

F Force FEA Finite Element Analysis Kt Stress Concentration Factor

P Pressure max Maximum Stress nom Nominal Stress


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Eaton Worldwide

Founded in 1911 by J.O. Eaton

World Headquarters inCleveland, Ohio USA

Customers in more than 150

countries More than 70,000 employees

worldwide

Chairman & CEO Alexander M. Cutler


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A premier diversified industrial manufacturer

A global leader in: Electrical systems and components for power quality, distribution

and control

Fluid power systems and services for industrial, mobile and aircraftequipment

Intelligent truck drivetrain systems for safety and fuel economy

Automotive engine air management systems, powertrain solutions

and specialty controls for performance, fuel economy and safety

Fluid Power AutomotiveTruckElectrical

Eaton Business Groups


6/21 2008 ANSYS, Inc. All rights reserved. 6 ANSYS, Inc. Proprietary

Background

Stress Concentration factors (Kts) for numerous simplegeometries have been determined by researchers

(analytical equations) Roark and Peterson have compiled these into easy to use

tables

Using simple Kts for complex geometries can induce error

Determining stress concentration factors (Kt) for complexgeometries can be difficult and expensive

If strain gages cannot be applied to the maximum stresslocation, remote stresses must be used to determine the

peak stress value (induces error)



Using FEA Results To Determine Kts

Kt = max/nom

FEA results can be used to easily determine themaximum stress (max)

Determining the nominal stress (nom) can bemore difficult

Knowledge of the stress gradient provides ameans of determining the nominal stress (nom)



Finite Plate With A Central Hole

Quarter-section Finite Element (FE) model.

Example (from Roarks Formulas for Stress and Strain, 7TH edition):

FEA Geometry (see Figure, quarter model shown):

Length = 1.500 in (Length (Roark) = infinite)Width = 0.750 in (D (Roark) = 1.500 in)r = 0.250 in (r (Roark) = 0.250 in)

t = 0.125 in

A1 = 0.750 in * 0.125 in = 0.09375 in2

A2 = 0.500 in * 0.125 in = 0.06250 in2

Inputs (Area for applied load (pressure):P1 = 320 lbf/in

2

F = 320 lbf/in2 * 0.09375 in2 = 30 lbf

A1 = 0.09375 in2A2 = 0.06250 in

2

Length = 1.500 in

Width = 0.750in

r = 0.250 in



Uni-Axial Stress Gradient

The graph depicts the stress gradient for a finite plate witha central hole and a uniform stress gradient

The uniform stress gradient is the nominal stress (

nom) That would exist if the stress concentration were not present

nom for a finite plate with a central hole is simply:

nom = Pend * A1/ A2 or

nom = Fend/ A2 nom = 320 lbf/in

2 * 0.09375 in2/ 0.0625 in2 = 480 lbf/in2

Finite Plate w/hole Stress Gradient

0

200

400

600

800

1000

1200

0 0.1 0.2 0.3 0.4 0.5 0.6

Distance (in), from hole free surface

(lbf/

in2)

Stress Gradient

Uniform Stress Field



Area Under Stress Gradient Curves

The area under the two stress gradient curves (stressconcentration and uniform stress field) must be equal(equivalent energy)

The area under the uniform stress field curve is:

AKt = nom * Distance (from hole free surface)

Finite Plate w/hole Stress Gradient

0

200

400

600

800

1000

1200

0 0.1 0.2 0.3 0.4 0.5 0.6

Distance (in), from hole free surface

(lbf/in2)

Stress Gradient

Uniform Stress Field



Area Under Stress Gradient Curve

ANSYS Post1 (post processor) provides integrationcalculations via path operations

AKt (ANSYS integration) = 240 lbf/in

Solve for nom: nom = AKt/ Distance = 240 lbf/in / 0.5 in = 480 lbf/in

2

Table 1 ANSYS Integration Operation Results (AKt ), Plate w/hole

S SX AKt S SX AKt S SX AKt

0.000 1115.5 0 0.167 485.27 113.93 0.333 378.81 184.66

0.008 1030.3 8.9405 0.175 476.69 117.93 0.342 374.85 187.8

0.017 957.51 17.223 0.183 468.74 121.87 0.350 370.85 190.91

0.025 894.93 24.941 0.192 461.35 125.75 0.358 366.81 193.98

0.033 840.82 32.174 0.200 454.46 129.56 0.367 362.7 197.020.042 793.79 38.985 0.208 448.02 133.33 0.375 358.5 200.03

0.050 752.72 45.428 0.217 441.96 137.03 0.383 354.2 203

0.058 716.68 51.551 0.225 436.25 140.69 0.392 349.77 205.93

0.067 684.92 57.391 0.233 430.84 144.31 0.400 345.2 208.83

0.075 656.82 62.981 0.242 425.7 147.87 0.408 340.47 211.68

0.083 631.86 68.351 0.250 420.79 151.4 0.417 335.56 214.5

0.092 609.6 73.524 0.258 416.09 154.89 0.425 330.44 217.27

0.100 589.67 78.521 0.267 411.56 158.34 0.433 325.1 220.01

0.108 571.77 83.36 0.275 407.18 161.75 0.442 319.5 222.69

0.117 555.63 88.058 0.283 402.92 165.12 0.450 313.63 225.33

0.125 541.03 92.627 0.292 398.77 168.46 0.458 307.44 227.92

0.133 527.78 97.08 0.300 394.7 171.77 0.467 300.93 230.45

0.142 515.71 101.43 0.308 390.68 175.04 0.475 294.05 232.93

0.150 504.68 105.68 0.317 386.71 178.28 0.483 286.77 235.35

0.158 494.57 109.84 0.325 382.76 181.49 0.492 279.05 237.71

0.500 270.86 240



Calculating The Kt (Finite Plate w/hole)

The Example from Roarks Formulas for Stress and Strain, 7TH edition(pgs. 773-774), provided photo-elastic stress analysis results with

max = 1130 lbf/in2

The Roark (photo-elastic) Kt for a finite plate with a central hole is: Kt (photo-elastic) = max/nom = 1130 lbf/in

2/ 480 lbf/in2 = 2.35

The Roark (formula) Kt for a finite plate with a central hole is:

Kt (formula) = 3.00 3.13(2r/D) + 3.66(2r/D)2 1.53(2r/D)3 = 2.31

The FEA Kt for a finite plate with a central hole is:

Kt (FEA) = max/nom = 1115 lbf/in2/ 480 lbf/in2 = 2.32

Error (photo-elastic to FEA): % error = {(Kt(photo-elastic) Kt(FEA)) / Kt(photo-elastic)} * 100 % error = {(2.35 2.32) / 2.35} * 100 = 1.3%

Error (formula to FEA):

% error = {(Kt(formula) Kt(FEA)) / Kt(formula)} * 100 % error = {(2.31 2.32) / 2.31} * 100 = -0.43%



Conclusion (uniform stress Kt from FEA)

FEA results (path operations and integration to determinethe area under the stress gradient curve) provides an

accurate method of determining the nominal stress (nom) The Kt from FEA results provide a value with less than 2%

error for a uniform stress field

Reference documents (Kt analytical equations) are notrequired

Kt

s are for actual geometry

Further verification/validation of the methodology is required



Square Shoulder With Fillet (Pure Bending)

Roarks Analytical model Square shoulder with fillet in a member of rectangular cross-section

Example (from Roarks Formulas for Stress and Strain, 7th edition)




Half-section Finite Element (FE) model. Square shoulder with fillet in a member of rectangular cross-section

Example (from Roarks Formulas for Stress and Strain, 7TH edition):

Geometry (see Figure, quarter model shown):

D = 3.000 in (Roark D = 3.000 in)r = 0.500 in (Roark r = 0.500 in)h = 1.000 in (Roark h = 1.000 in)L/2 = 1.500 in (Roark L = 3.000 in)

Inputs (applied load (force):Fy = -1000 lbfD

r

L/2 (half model)

h(3 places)




Roarks Analytical model The h/r ratio = 2.0, therefore, the Kt was calculated using the two forms of

the analytical equation 0.1 < h/r < 2.0 and 2.0 < h/r < 20.0

Square shoulder example (from Roarks Formulas for Stress and Strain,7th edition)

Both forms of the equation provide a Kt of 1.222

Kt Theoretical Calculations (from Roark's "Formulasfor Stress and Strain", 7th edition)

0.1h/r 2.0 2.0h/r 20.0h 1.0 L/D 1.0 C1 2.359214 C1 2.358758

D 3.0 2h/D 0.6667 C2 -2.171769 C2 -2.171388

r 0.5 h/r 2.0 C3 0.471304 C3 0.470505

L 3.0 L/D 0.951366 C4 0.341252 C4 0.342126

Kt 1.221947 Kt 1.221649




FEA model

The max stress (12,242 lbf/in2) occurs adjacent to the beginning of the fillet

For ease of calculation (known cross-section and moment arm) the Kt is determinedat the beginning of the fillet where the max stress is 10,900 lbf/in2

For pure bending max = MY/I = nom (stress if Kt did not exist)

nom = MY/(bh/12) = 1.500 in * (1000 lbf) * (0.500 in) / ((1.000 in)*(1.000 in) / 12))

nom = 9000 lbf/in2

Fy = -1000 lbf




Stress Gradients The area under the pure bending stress gradient is:

AKt = max * Distance to neutral axis

max

= 2 * AKt

/ Distance

Using path operations and integration, the area under theRect_Bar_Gradient curve is:

AKt = 2166.30 lbf/in

max = nom = 2 * AKt/ Distance = 2 * ( 2166.3 lbf/in) / (0.5 in) = 8665.2 lbf/in2

Rectangular Bar Stress Gradient

0

2000

4000

6000

8000

10000

12000

0 0.1 0.2 0.3 0.4 0.5 0.6Distance to Neutral Axis (in)

Bending

Stres

s(lbf/in2)

Rect_Bar_GradientRect_Bar (IF Kt=1)

S x AKt S x AKt S x AKt0.0000 10900 0.00 0.1667 5548 1313.60 0.3333 2461 1965.200.0083 10507 89.20 0.1750 5363 1359.10 0.3417 2335 1985.200.0167 10115 175.12 0.1833 5178 1403.00 0.3500 2191 2004.100.0250 9722 257.77 0.1917 5003 1445.40 0.3583 2070 2021.800.0333 9454 337.67 0.2000 4833 1486.40 0.3667 1949 2038.600.0417 9125 415.09 0.2083 4663 1525.90 0.3750 1807 2054.200.0500 8796 489.76 0.2167 4493 1564.10 0.3833 1690 2068.800.0583 8547 562.02 0.2250 4336 1600.90 0.3917 1573 2082.400.0667 8258 632.04 0.2333 4178 1636.30 0.4000 1432 2094.900.0750 7969 699.65 0.2417 4014 1670.50 0.4083 1318 2106.400.0833 7734 765.08 0.2500 3867 1703.30 0.4167 1204 2116.900.0917 7480 828.47 0.2583 3720 1734.90 0.4250 1064 2126.300.1000 7225 889.75 0.2667 3560 1765.20 0.4333 952 2134.700.1083 7004 949.03 0.2750 3421 1794.30 0.4417 840 2142.200.1167 6777 1006.50 0.2833 3283 1822.30 0.4500 701 2148.60

0.1250 6551 1062.00 0.2917 3127 1849.00 0.4583 591 2154.000.1333 6341 1115.70 0.3000 2995 1874.50 0.4667 480 2158.500.1417 6137 1167.70 0.3083 2864 1898.90 0.4750 342 2161.900.1500 5933 1218.00 0.3167 2712 1922.10 0.4833 232 2164.300.1583 5733 1266.60 0.3250 2586 1944.20 0.4917 122 2165.80

0.5000 12 2166.30

Table 1 ANSYS Integration Operation Results (AKt ), Rectangular Bar




FEA Kt

Using the FEA smax (10,900 lbf/in2) at the beginning of the

fillet and snom from the area under the curve, the FEA Ktfor the square shoulder with a fillet geometry is:

Kt (FEA) = maxFEA/nom = 10900 lbf/in

2/ 8665.2 lbf/in2 = 1.258

Analytical Kt (Roark)

The Kt using the analytical formula from Roark is:

Kt = 1.222 Error (formula to FEA):

% error = {(Kt(formula) Kt(FEA)) / Kt(formula)} * 100

% error = {(1.222 1.258) / 1.222} * 100 = -2.95%



Conclusion (Pure Bending stress Kt from FEA)

The Kt obtained from the equivalent energymethod has an error of less than 3% to the

analytical Kt

Analytical Kts are obtained from empirical data

Reference documents (Kt analytical equations)are not required

Kts are for actual geometry

Further verification/validation of the methodologyis required


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Summary

The preliminary work presented for determining Kts fromFEA results is very promising

Obtaining actual geometry Kts from FEA results eliminatesthe need to find, and the error associated with, usingempirically based analytical Kts for simple geometry

Minimal error (less than 3%) between empirically basedanalytical Kts and Kts obtained from FEA results isdemonstrated for uni-axial and pure bending stress fields

Additional verification/validation of the methodology isrequired

Expansion of the methodology to include multi-axial states ofstress will be investigated

Materials Methodology Stress Factors

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