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MFGT 290 MFGT Certification Class. 8: Strength of Materials Chapter 11: Material Properties. Professor Joe Greene CSU, CHICO. MFGT 290. Chap 8: Strength of Materials. Stress and Strain Axial Loading Torsional Loading Beam Loading Column Loading Practice Problems.

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Mfgt 290 mfgt certification class
MFGT 290MFGT Certification Class

8: Strength of Materials

Chapter 11: Material Properties

Professor Joe Greene

CSU, CHICO

MFGT 290


Chap 8 strength of materials
Chap 8: Strength of Materials

  • Stress and Strain

  • Axial Loading

  • Torsional Loading

  • Beam Loading

  • Column Loading

  • Practice Problems


Mechanical test considerations
Mechanical Test Considerations

P

P

A

P

P

P

P

shear

compression

tension

  • 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

Normal Forces

Shear Forces


Stress strain diagrams
Stress-Strain Diagrams

Forces

Test Sample

Fixed

  • 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


Stress strain diagrams1
Stress-Strain Diagrams

Linear

(Hookean)

Stress

Non-Linear

(non-Hookean)

Strain

  • Stress-strain diagrams is a plot of stress with the corresponding strain produced.

  • Stress is the y-axis

  • Strain is the x-axis


Modulus and strength
Modulus and Strength

Modulus

Yield

Strength

Proportional

Limit

Stress

Strain

  • 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

Ultimate Strength


Allowable axial load
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 9-1

      Example 8-2.1 Hinged Beam

      • Statics review

        • Sum of forces = 0

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


Strain
Strain

L

  • 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%


Strain1
Strain

0.1 in

1 in

10in

  • 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?

100 lbs


Tensile modulus and yield strength
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


Hooke s law
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

Eqn 8-3


Problem solving techniques
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 problem

    • Make a Force balance with Free body diagram

      • Identify normal forces

      • Identify shear forces

    • Write stress as Force per unit area

      • Calculate area from set-up, or

      • Calculate force from set-up

    • Write Hooke’s law

      • Rearrange for deflections

      • Write deflections balance

    • Solve for problem unknowns

Eqn 8-3


Safety factor
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.


Stress concentrations
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.


Thermal stresses
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


Strain and poisson s ratio
Strain and Poisson’s Ratio

Transverse

Strain

Axial

Strain

P, Load

  • 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 0.5

  • Plastics: Poisson’s ratio 0.3

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


Chapter 11 material properties
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


Fatigue properties
Fatigue Properties

S, stress

N, number of cycles

  • 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


Fundamentals of hardness
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


Hardness mechanical tests
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)


Brinell test conditions
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


Brinell test example
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

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:


Brinell test method continued
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


Advantages disadvantages of the brinell hardness test
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)


Rockwell test
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)


Rockwell test description
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


Rockwell values
Rockwell Values

  • 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


Rockwell and brinell conversion
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


Rockwell and brinell conversion1
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


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