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MFGT 290MFGT Certification Class

8: Strength of Materials

Chapter 11: Material Properties

Professor Joe Greene

CSU, CHICO

MFGT 290

• Stress and Strain

• Practice Problems

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

• 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

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

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

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

• 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

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%

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

• 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 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

• 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

• 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.

• 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.

• 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

Transverse

Strain

Axial

Strain

• 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

• Structure of Matter

• Material Testing Agencies

• Physical Properties

• Mechanical Properties and Test Methods

• Stress and Strain

• Fatigue Properties

• Hardness

• Practice Problems

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

• 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

• 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 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 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

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:

• 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

• 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)

• 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)

• 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)

• Rockwell hardness includes the value and the scale letter

• 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

• 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

• For a Rockwell C values, HRC, values greater than 40,

• Example,

• Convert the Rockwell hardness number HRc 60 to BHN

• Review Questions