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MFGT 290MFGT Certification Class

Professor Joe GreeneCSU, CHICO

MFGT 290

8: Strength of MaterialsChapter 11: Material Properties

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Chap 8: Strength of Materials• Stress and Strain• Axial Loading• Torsional Loading• Beam Loading• Column Loading• Practice Problems

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Mechanical Test Considerations• Normal and Shear Stresses

– Force per unit area• Normal force per unit area

– Forces are normal (in same direction) to the surface• Shear force per unit area

– Forces are perpendicular (right angle) to the surface

• Direct Normal Forces and Primary types of loading– Prismatic Bar: bar of uniform cross section subject to equal and opposite pulling

forces P acting along the axis of the rod.– Axial loads: Forces pulling on the bar– Tension= pulling the bar; Compression= pushing; torsion=twisting; flexure=

bending; shear= sliding forces

tension compressionshear

P

P

PPP

P

AP

A

Normal Forces Shear Forces

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Stress-Strain Diagrams• Equipment

– Tensile Testing machine• UTM- Universal testing machine• Measures

– Load, pounds force or N– Deflection, inches or mm

• Data is recorded at several readings – Results are averaged – e.g., 10 samples per second during the test.

• Calculates– Stress, Normal stress or shear stress– Strain, Linear strain– Modulus, ratio of stress/strain

Test Sample

Forces

Fixed

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Stress-Strain Diagrams• Stress-strain diagrams is a plot of stress with the

corresponding strain produced.• Stress is the y-axis• Strain is the x-axis

Stress

Strain

Linear(Hookean)

Non-Linear(non-Hookean)

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Modulus and Strength• Modulus: Slope of the stress-strain curve

– Can be Initial Modulus, Tangent Modulus or Secant Modulus• Secant Modulus is most common

• Strength– Yield Strength

• Stress that the material starts to yield• Maximum allowable stress

– Proportional Limit• Similar to yield strength and is the point where Hooke’s Law is valid

– If stress is higher than Hooke’s Law is not valid and can’t be used.– Ultimate strength

• Maximum stress that a material can withstand• Important for brittle materials

Stress

Strain

Modulus

Yield StrengthProportional Limit

Ultimate Strength

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Allowable Axial Load• Structural members are usually designed for a limited

stress level called allowable stress, which is the max stress that the material can handle.– Equation 8-2 can be rewritten

• Required Area– The required minimum cross-sectional area A that a structural

member needs to support the allowable stress is from Equation 9-1Example 8-2.1 Hinged Beam

• Statics review– Sum of forces = 0– Sum of Moments = 0. Moment is Force time a distance to solid wall

AP

AP allowallow

allow

allowPA

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Strain• Strain: Physical change in the dimensions of a specimen that results from applying a load to the

test specimen.• Strain calculated by the ratio of the change in length, , and the original length, L. (Deformation)

• Where, = linear strain ( is Greek for epsilon) = total axial deformation (elongation of contraction) = Lfinal –Linitial = Lf - L– L = Original length– Strain units (Dimensionless)

• Units– When units are given they usually are in/in or mm/mm. (Change in dimension divided by original length)

• % Elongation = strain x 100%

LLL

L)f(

L

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Strain• Example

– Tensile Bar is 10in x 1in x 0.1in is mounted vertically in test machine. The bar supports 100 lbs. What is the strain that is developed if the bar grows to 10.2in? What is % Elongation?

=Strain = (Lf - L0)/L0 = (10.2 -10)/(10) = 0.02 in/in• Percent Elongation = 0.02 * 100 = 2%

• What is the strain if the bar grows to 10.5 inches?• What is the percent elongation?

10in1 in

0.1 in

100 lbs

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Tensile Modulus and Yield Strength• Modulus of Elasticity (E) (Note: Multiply psi by 7,000 to get kPa)

– Also called Young’s Modulus is the ratio of stress to corresponding strain – A measure of stiffness

• Yield Strength: (Note: Multiply psi by 7,000 to get kPa)– Measure of how much stress a material can withstand without breaking– Modulus(Table 8-1) Yield Strength

• Stainless Steel E= 28.5 million psi (196.5 GPa) 36,000 psi• Aluminum E= 10 million psi 14,000 psi• Brass E= 16 million psi 15,000 psi• Copper E= 16 million psi• Molybdenum E= 50 million psi• Nickel E= 30 million psi• Titanium E= 15.5 million psi 120,000 psi• Tungsten E= 59 million psi• Carbon fiber E= 40 million psi• Glass E= 10.4 million psi• Composites E= 1 to 3 million psi 15,000 psi• Plastics E= 0.2 to 0.7 million psi 5,000 to 12,000 psi

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Hooke’s Law• Hooke’s Law relates stress to strain by way of modulus

– Hooke’s law says that strain can be calculated as long as the stress is lower than the maximum allowable stress or lower than the proportional limit.

• If the stress is higher than the proportional limit or max allowable stress than the part will fail and you can’t use Hooke’s law to calculate strain.

– Stress = modulus of elasticity, E, times strain– Stress= = load per area, P/A– Strain= = deformation per length, /L – Rearrange Hooke’s law– Solving for deformation is

• With these equations you can find – How much a rod can stretch without breaking.– What the area is needed to handle load without breaking– What diameter is needed to handle load without breaking

• Example 10-1• Example 10-3

E

LLLE

LE

AP )f(

AEPL

Eqn 8-3

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Problem solving techniques• Steps to solve most Statics problems

– Set-up problem• Draw picture and label items (D, L, P, Stress, etc..)• List known values in terms of units.

– Solve problem1. Make a Force balance with Free body diagram

• Identify normal forces• Identify shear forces

2. Write stress as Force per unit area• Calculate area from set-up, or • Calculate force from set-up

3. Write Hooke’s law• Rearrange for deflections• Write deflections balance

4. Solve for problem unknowns

E

LLLE

LE

AP )f(

AEPL

Eqn 8-3

AP

AP allowallow allow

allowPA

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Safety Factor• Allowable Stresses and Factor of Safety

– Provide a margin of safety in design for bridges, cars, buildings, rockets, space shuttles, air planes, etc…

– Structural members and machines are designed so that columns, plates, trusses, bolts, see much less than the stress that will cause failure.

• Ductile materials: If the stress is greater than the yield strength or proportional limit of the material.

• Brittle materials: If the stress is greater than the ultimate strength of the material.Since they do not show any yielding, just fracturing.

StressAllowableStressFailureSafetyFactor_

__

AllowableAllowable

LimitoportionalStrengthYieldStressAllowable

StrengthYieldSafetyFactor

_Pr__

__

Allowable

StrengthUltimateStressAllowable

StrengthYieldSafetyFactor

__

__

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Stress Concentrations• Stresses can be higher near holes, notches, sharp corners in

a part or structural member.– Stress concentration factor, K = stresses near hole

stresses far away from hole

– K is looked up in a table or on a graph– Stress at hole can be calculated to see if part will fail.

• Where b is the net width at hole section and t is the thickness.

averageawayfarStressHoleatStressKionConcentratStress

max

_____

btPK

APKK average max

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Thermal Stresses• Most materials expand when heated as the temperature increases.

– As the temperature goes up, the material expands and results in forces that cause stress in the part. As temperature increases the stresses increase in part.

• Examples,– Cast iron engine block heat up to 500F and expands the cast iron block which causes

stresses at the bolts. The bolts must be large enough to withstand the stress.– Aluminum heats up and expands and then cools off and contracts.

» Sometimes the stresses causes cracks in the aluminum block.– Space shuttle blasts off and heats up, goes into space and cools down (-200F), and

reenters Earths atmosphere and heats up (3000F)» Aluminum melts at 1300F so need ceramic heat shields» Aluminum structure expands and cools.

– The amount the material expands is as follows:• Change in length that is causes by temperature change (hot or cold)

– Where, = change in length» = the CLTE (coefficient of linear thermal expansion T = change in temperature (Thot – Tcold)» L = length of member

– Examples

TL

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Strain and Poisson’s Ratio

• Axial strain is the strain that occurs in the same direction as the applied stress.

• Transverse strain is the strain that occurs perpendicular to the direction of the applied stress.

• Poisson’s ratio is ratio of lateral strain to axial strain. Poisson’s ratio = lateral strain axial strain– Example

• Calculate the Poisson’s ratio of a material with lateral strain of 0.002 and an axial strain of 0.006

• Poisson’s ratio = 0.002/0.006 = 0.333– Example

Note: For most materials, Poisson’s ratio is between 0.25 and 0.5Plastics: Poisson’s ratio 0.3

Table 8-1 Metals: Poisson’s ratio = 0.3 steel, 0.33 Al, 0.35 Mg, 0.34 Ti

AxialStrain

TransverseStrain

P, Loada

t

strainaxialstraintransverse

_

_

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Chapter 11: Material Properties• Structure of Matter• Material Testing Agencies• Physical Properties• Mechanical Properties and Test Methods

– Stress and Strain– Fatigue Properties– Hardness

• Practice Problems

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Fatigue Properties• Fatigue Properties

– All materials that are subjected to a cyclic loading can experience fatigue

– Failure occurs through a maximum stress at any cycle.– Test methods

• Subject the material to stress cycles and counting the number of cycles to failure, then

• Fatigue properties are developed.• Table of properties for each material

– How many cycles a material can experience at a certain stress level before failing.

• S-N diagrams are developed (Stress and Number of cycles)– Specify fatigue as a stress value– Design for less than fatigue stress

N, number of cycles

S, stress

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Fundamentals of Hardness

• Hardness is thought of as the resistance to penetration by an object or the solidity or firmness of an object– Resistance to permanent indentation under static or dynamic loads– Energy absorption under impact loads (rebound hardness)– Resistance toe scratching (scratch hardness)– Resistance to abrasion (abrasion hardness)– Resistance to cutting or drilling (machinability)

• Principles of hardness (resistance to indentation)– indenter: ball or plain or truncated cone or pyramid made of hard steel or diamond– Load measured that yields a given depth– Indentation measured that comes from a specified load– Rebound height measured in rebound test after a dynamic load is dropped onto a surface

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Hardness Mechanical Tests• Brinell Test Method

– One of the oldest tests– Static test that involves pressing a hardened steel ball (10mm) into a test

specimen while under a load of • 3000 kg load for hard metals, • 1500 kg load for intermediate hardness metals• 500 kg load for soft materials

– Various types of Brinell• Method of load application:oil pressure, gear-driven screw, or weights with a lever• Method of operation: hand or electric power• Method of measuring load: piston with weights, bourdon gage, dynamoeter, or

weights with a lever• Size of machine: stationary (large) or portable (hand-held)

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Brinell Test Conditions

• Brinell Test Method (continued)– Method

• Specimen is placed on the anvil and raised to contact the ball• Load is applied by forcing the main piston down and presses the ball into

the specimen• A Bourbon gage is used to indicate the applied load• When the desired load is applied, the balance weight on top of the

machine is lifted to prevent an overload on the ball• The diameter of the ball indentation is measured with a micrometer

microscope, which has a transparent engraved scale in the field of view

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Brinell Test Example• Brinell Test Method (continued)

– Units: pressure per unit area– Brinell Hardness Number (BHN) = applied load divided by area of

the surface indenter

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2

dDDD

LBHN

Where: BHN = Brinell Hardness Number L = applied load (kg) D = diameter of the ball (10 mm) d = diameter of indentation (in mm)

• Example: What is the Brinell hardness for a specimen with an indentation of 5 mm is produced with a 3000 kg applied load.

•Ans:

2

22/6.142

)5()10(10)10(

)3000(2 mmkgmmmmmmmm

kgBHN

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Brinell Test Method (continued)• Range of Brinell Numbers

– 90 to 360 values with higher number indicating higher hardness– The deeper the penetration the higher the number– Brinell numbers greater than 650 should not be trusted because the

diameter of the indentation is too small to be measured accurately and the ball penetrator may flatten out.

– Rules of thumb• 3000 kg load should be used for a BHN of 150 and above• 1500 kg load should be used for a BHN between 75 and 300• 500 kg load should be used for a BHN less than 100• The material’s thickness should not be less than 10 times the depth of the

indentation

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Advantages & Disadvantages of the Brinell Hardness Test

• Advantages– Well known throughout industry with well accepted results– Tests are run quickly (within 2 minutes)– Test inexpensive to run once the machine is purchased– Insensitive to imperfections (hard spot or crater) in the material

• Limitations– Not well adapted for very hard materials, wherein the ball deforms excessively– Not well adapted for thin pieces– Not well adapted for case-hardened materials– Heavy and more expensive than other tests ($5,000)

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Rockwell Test• Hardness is a function of the degree of indentation of the

test piece by action of an indenter under a given static load (similar to the Brinell test)

• Rockwell test has a choice of 3 different loads and three different indenters

• The loads are smaller and the indentation is shallower than the Brinell test

• Rockwell test is applicable to testing materials beyond the scope of the Brinell test

• Rockwell test is faster because it gives readings that do not require calculations and whose values can be compared to tables of results (ASTM E 18)

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Rockwell Test Description• Specially designed machine that applies load through a

system of weights and levers– Indenter can be 1/16 in hardened steel ball, 1/8 in steel ball, or 120°

diamond cone with a somewhat rounded point (brale)– Hardness number is an arbitrary value that is inversely related to

the depth of indentation– Scale used is a function of load applied and the indenter

• Rockwell B- 1/16in ball with a 100 kg load• Rockwell C- Brale is used with the 150 kg load

– Operation• Minor load is applied (10 kg) to set the indenter in material• Dial is set and the major load applied (60 to 100 kg)• Hardness reading is measured• Rockwell hardness includes the value and the scale letter

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Rockwell ValuesScale Indenter Applied Load

(kg)A Brale 60B 1/16 in 100C Brale 150D Brale 100E 1/8 in 100F 1/16 in 60G 1/16 in 150

•B Scale: Materials of medium hardness (0 to 100HRB) Most Common•C Scale: Materials of harder materials (> 100HRB) Most Common•Rockwell scales divided into 100 divisions with each division (point of hardness) equal to 0.002mm in indentation. Thus difference between a HRB51 and HRB54 is 3 x 0.002 mm - 0.006 mm indentation•The higher the number the harder the number

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Rockwell and Brinell Conversion• For a Rockwell C values between -20 and 40, the Brinell

hardness is calculated by

• For HRC values greater than 40, use

• For HRB values between 35 and 100 use

CHRxBHN

1001042.1 6

CHRxBHN

100

105.2 4

BHRxBHN

130

103.7 3

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Rockwell and Brinell Conversion• For a Rockwell C values, HRC, values greater than 40,

• Example,– Convert the Rockwell hardness number HRc 60 to BHN

• Review Questions

CHRxBHN

100

105.2 4

60100105.2 4

xBHN 625BHN