Physics 6B Stress, Strain and Elastic Deformations Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

Slides:



Advertisements
Similar presentations
Ch. 18 Solids. Characteristics are due to its structure, or arrangement of its atoms.
Advertisements

Physics 6B Oscillations Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Stress and Strain TUTORIAL 6 to answer just click on the button or image related to the answer.
Materials for Automobiles CEEN0903 Summer 2011 Classes of Materials Metals Made of metallic elements: Sn, Cu, Fe Ceramics Compounds between metallic.
Simple Harmonic Motion and Elasticity
Properties of solid materials
Fisica Generale - Alan Giambattista, Betty McCarty Richardson Copyright © 2008 – The McGraw-Hill Companies s.r.l. 1 Chapter 10: Elasticity and Oscillations.
The Circle Introduction to circles Let’s investigate… Circumference
PH0101 UNIT 1 LECTURE 1 Elasticity and Plasticity Stress and Strain
Chapter 13 - Elasticity A PowerPoint Presentation by
Springs and Elasticity ClassAct SRS enabled. In this presentation you will: Explore the concept of elasticity as exhibited by springs.
Elasticity by Ibrhim AlMohimeed
Particle movement in matter What happens when a particle moves in another matter?
Stress, Strain, and elastic moduli
Gravity Physics 6B Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Physics 2 Chapter 11 problems Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
A 10-m long steel wire (cross – section 1cm 2. Young's modulus 2 x N/m 2 ) is subjected to a load of N. How much will the wire stretch under.
Deformation of Solids Stress is proportional to Strain stress = elastic modulus * strain The SI unit for stress is the Newton per meter squared (N/m 2.
Stress and Strain. Deforming Solids  Solids deform when they are subject to forces. Compressed, stretched, bent, twistedCompressed, stretched, bent,
Copyright © 2009 Pearson Education, Inc. Lecture 8a – States of Matter Solids.
Copyright © 2009 Pearson Education, Inc. Chapter 12 Elasticity.
Physics. Properties of Matter Session Session Objectives.
Deforming Solids.
Fluids - Hydrostatics Physics 6B Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Fluids - Hydrostatics Physics 6B Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Objectives  Understand how elasticity is related to Hooke’s Law for springs.  Know that the change in length of an object is proportional to the force.
1.3.4 Behaviour of Springs and Materials
Springs. Hooke’s Law Any spring has a natural length at which it exerts no net force on the mass m. This length, or position, is called the equilibrium.
Physics 6A Stress, Strain and Elastic Deformations Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Elastic Stress-Strain Relationships
Strong forces are needed to stretch a solid object. The force needed depends on several factors.
PHASES OF MATTER There are 3 basic phases of matter: solid, liquid, gas (plasma). Plasma is a gas that contains ions and conduct electricity. 99% of the.
Physics 121.
Equilibrium and Elasticity
FYI: All three types of stress are measured in newtons / meter2 but all have different effects on solids. Materials Solids are often placed under stress.
10.7 Elastic Deformation.
R. Field 10/31/2013 University of Florida PHY 2053Page 1 Definition of Strain System Response – Linear Deformation: System Response – Volume Deformation:
LECTURE 9.1. LECTURE OUTLINE Weekly Deadlines Weekly Deadlines Stress and Strain Stress and Strain.
In this lecture you will learn about pressure(PSI), different types of force, and be introduced into a stress strain diagram.
The ratio of stress and strain, called modulus of elasticity. Mechanical Properties of Solids Modulus of Elasticity.
Statics Activities. Stress  Force per unit area (  ) Typical engineering units – psi (lb f /in 2 ) – N/m 2 Stress = Force/Area – Applied by external.
Strengths Chapter 10 Strains. 1-1 Intro Structural materials deform under the action of forces Three kinds of deformation Increase in length called an.
Axial Members AXIAL MEMBERS, which support load only along their primary axis, are the most basic of structural members. Equilibrium requires that forces.
STRUCTURES Outcome 3 Gary Plimer 2008 MUSSELBURGH GRAMMAR SCHOOL.
Fluids - Hydrostatics Physics 6B Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Copyright © 2009 Pearson Education, Inc. An object with forces acting on it, but with zero net force, is said to be in equilibrium. The Conditions for.
1.To understand the keywords associated with the deformation of different types of solids 2.To be able to calculate stress, strain and hence Young’s modulus.
Chapter 9: Mechanical Properties of Matter
1.To understand the keywords associated with the deformation of different types of solids 2.To be able to calculate stress, strain and hence Young’s modulus.
Unit 1 Key Facts- Materials Hooke’s Law Force extension graph Elastic energy Young’s Modulus Properties of materials.
This section of work is also known as Hookes Law.
-Elastic Properties of Solids AP Physics C Mrs. Coyle.
Inner Product, Length and Orthogonality Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Hooke ’ s Law. Elasticity: The ability of an object to return to its original shape after the deforming force is removed.
Gravity Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Elasticity Yashwantarao Chavan Institute of Science Satara Physics
Robert Hooke Hooke’s Law deals with springs.
Stress and Strain1 Stress Analysis Mechanics: Stress and strain.
Strain Energy Lecture No-5 J P Supale Mechanical Engineering Department SKN SITS LONAVALA Strength of Materials.
FORCES SPRINGS This section of work is also known as Hookes Law.
Stress, Strain and Elastic Deformations
Stress and Strain.
(a) Describe what material the spring is made from;
Tutorial in Mechanical Properties
Elastic properties of materials
Elastic Deformation: Stress, Strain and Hook’s Law
Describing deformation
Forging new generations of engineers
ELASTIC DEFORMATION.
Tutorial.
Presentation transcript:

Physics 6B Stress, Strain and Elastic Deformations Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

When a force is applied to an object, it will deform. If it snaps back to its original shape when the force is removed, then the deformation was ELASTIC. We already know about springs - remember Hookes Law : F spring = -kΔx Hookes Law is a special case of a more general rule involving stress and strain. The constant will depend on the material that the object is made from, and it is called an ELASTIC MODULUS. In the case of tension (stretching) or compression we will call it Youngs Modulus*. So our basic formula will be: *Bonus Question – who is this formula named for? Click here for the answerClick here for the answer Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

To use our formula we need to define what we mean by Stress and Strain. STRESS is the same idea as PRESSURE. In fact it is the same formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

To use our formula we need to define what we mean by Stress and Strain. STRESS is the same idea as PRESSURE. In fact it is the same formula: STRAIN is a measure of how much the object deforms. We divide the change in the length by the original length to get strain: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

To use our formula we need to define what we mean by Stress and Strain. STRESS is the same idea as PRESSURE. In fact it is the same formula: STRAIN is a measure of how much the object deforms. We divide the change in the length by the original length to get strain: Now we can put these together to get our formula for the Youngs Modulus: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Problem 11.6 A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Youngs modulus for this nylon? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

ΔL=1.1m L 0 =45m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.6 A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Youngs modulus for this nylon?

ΔL=1.1m L 0 =45m A couple of quick calculations and we can just plug in to our formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.6 A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Youngs modulus for this nylon?

ΔL=1.1m L 0 =45m 7mm A couple of quick calculations and we can just plug in to our formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.6 A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Youngs modulus for this nylon? Dont forget to cut the diameter in half.

ΔL=1.1m L 0 =45m 7mm A couple of quick calculations and we can just plug in to our formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.6 A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Youngs modulus for this nylon? Dont forget to cut the diameter in half.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

L 0 =2m ΔL=0.25cm 400N diam=? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Youngs modulus for steel in a table: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Youngs modulus for steel in a table: The only piece missing is the area – we can rearrange the formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have? L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Youngs modulus for steel in a table: The only piece missing is the area – we can rearrange the formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Youngs modulus for steel in a table: The only piece missing is the area – we can rearrange the formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Youngs modulus for steel in a table: The only piece missing is the area – we can rearrange the formula One last step – we need the diameter, and we have the area: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Youngs modulus for steel in a table: The only piece missing is the area – we can rearrange the formula One last step – we need the diameter, and we have the area: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Youngs modulus for steel in a table: The only piece missing is the area – we can rearrange the formula One last step – we need the diameter, and we have the area: double the radius to get the diameter: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 11.7 A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB (MC) When a weight is hung from a cylindrical wire of diameter D, it produces a tensile stress X in the wire. If the same weight is hung from a wire having twice the diameter as the first one, the tensile stress in this wire will be

We can do this one just by staring at the formula for stress: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB (MC) When a weight is hung from a cylindrical wire of diameter D, it produces a tensile stress X in the wire. If the same weight is hung from a wire having twice the diameter as the first one, the tensile stress in this wire will be

We can do this one just by staring at the formula for stress: The force is the same in both cases because it says they use the same weight. The area is related to the square of the radius (or diameter), so when the diameter doubles the area goes up by a factor of 4. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB (MC) When a weight is hung from a cylindrical wire of diameter D, it produces a tensile stress X in the wire. If the same weight is hung from a wire having twice the diameter as the first one, the tensile stress in this wire will be

We can do this one just by staring at the formula for stress: The force is the same in both cases because it says they use the same weight. The area is related to the square of the radius (or diameter), so when the diameter doubles the area goes up by a factor of 4. Thus the stress should go down by a factor of 4 (area is in the denominator) Answerc) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB