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Meeting Our First Loads. Or May the Force Be With You (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results). Parallel Reading. Chapter 2 Section 2.2 Normal Stress Section 2.3 Extensional Strain; Thermal Strain
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Meeting Our First Loads Or May the Force Be With You (Credit for many illustrations is given to McGraw Hill publishers and an array ofinternet search results)
Parallel Reading • Chapter 2 • Section 2.2 Normal Stress • Section 2.3 Extensional Strain; Thermal Strain • Section 2.4 Stress-Strain Diagrams • Section 2.6 Linear Elasticity • Section 2.10 St. Venant’s Principle • (Do Reading Assignment Problems 2A)
A Force P Pulling on a Bar Since the bar is not moving forces P and P’ must be equal and opposite The forces are trying to pull the bar in two We call such this situation Tension When we are doing our Statics calculations we represent such forces to be positive We also note these forces are pulling down the length of the bar and that The forces are aligned and pass through the “centroid” of the bar We call this Axial Loading
A Definition to Stress We can take our happy tensional force and divide it by the area of our bar P/A (or more commonly Force is represented with an F F/A) The force per unit of area has its own name - Stress σ = Force/Area Obviously this is an average stress
Units and Stress • Force • In God’s chosen unit it is the Lb (after all our system was designed by Kings! What commoner designed the metric system?) • In Metric it is the Newton (Not to be confused with Figg Newton) • Area • In the English System the area is usually in square inches • In metric the area is in square meters • English System Stress • Psi (lbs/in^2) or Kips (1,000s of lbs/in^2) • Metric • Newton/Square Meter gets its own name first Pascal • Since a Pascal is a small number we measure in KPa (1,000s of pascals)
But What Good is Average Stress?Couldn’t the Actual Stress Distribution be Anything? The Saint’s to the Rescue! Stress will spread out evenly down The length of the bar very quickly. Only near the load points will our Stress distribution be funky St Venant’s Principle
Getting Some Joy Breaking ThingsUsing St. Venant’s Principle A set up and test Specimen to break!! Note that the test area is Located well away from the Loading points so we will See average stress.
An Example of Stress 1.0 in 0.5 in Tommy Towing decides he will use his SUV to pull his disabled semi- Truck to the garage using a steel tow bar with dimensions as above. The pull to overcome the rolling resistance of the truck and accelerate The load is 4,000 lbs. We will ignore any questions of traction. What is the stress in the steel tow bar?
Apply the Definition of Stress The force in the tow bar is 4,000 lbs 1 in 0.5 in Area for a rectangle is base*height 1 * 0.5 = 0.5 in^2 Note the lower case sigma symbol is commonly used for a“normal” stress – ie stress directly into the surface
A Few Bell And Whistle Observations Suppose the steel can stand a load of 20,000 psi without any permanent damage. What is our Factor of Safety?
Technique of Dividing into Segments Our Acting Force is 60 kN 60 kN 60 kN 0.03 Meters Check out Stress Here Get our cross sectional area
Apply the Definition of Stress 60 kN A side note on your books naming conventions. It likes to use P for a load or force applied at a specific point.
Now for the Next Segment We know the sum of these Forces has to be zero or the rod would take off moving. 125 kN ? 60 kN 125 kN Here is our Mystery Load
Continuing Our Exciting Problem! 190 kN 190 kN Oh Really – Can We!! 0.05 Meters And using our stress definition.
A Note on Conventions Note that the convention is that tensile stress (pulling apart) is positive, while compressive stress (scrunching together) is negative. Our answer to stress here Why? Our answer to stress here
We Are Considering Alternatives if You Don’t Learn What Stress Is This is an important concept for this class and for many of the classes you will be taking that build on Mechanics of Materials.
Will Stress Always Refer to Average Stress? No – But even if the stress distribution gets funky it is still a force per unit area.
So What Happens We Put a Tensile Stress on a Body? Our bar will stretch Now we need a new term - Strain
Pollock Question of the Day!What are the units of strain? Note that both L and the stretch δ are In units of Length (inches, ft, meters) Length ---------- Length Units cancel Strain is Unit less!
Lets Do Some Problems with Strain Gage marks Are now 250.28 mm apart Gage marks are placed 250 mm apart on an Aluminum rod. What is the strain? A load is applied
Looking at the Definition of Strain Strain is the change in length per unit of length. We need to find the change in length δ Was 250 mm Is 250.28 mm
To Finish the Calculation We Need the Undeformed Length Let me see – What was that? Oh yes – 250 mm Time to plug in
Most Strains are Small – Lets Try One that’s not A person jumps from a bungee platform with an 18 foot bungee chord. At the end of the fall the chord has stretched To 56 ft. What is the strain in the bungee chord?
Back to the Definition of Strain We know the original length Was 18 feet We need that change in length Plug In
Some Items to Note About Strain Our book and many others uses a lower case Delta for change in length. Our book (and a lot of others) uses a lower case Epsilon as the symbol for strain. Did we have to know anything about forces acting to determine the strain?
Strain is another Must Learn Go Ahead Make My Day Ways considered to force you to learn.
Your First Assignment Problem 2.2-1 Problem 2.3-3 When you do the problems, first put the formulas you will use to solve them. Then explain step by step how you are using the formulas to reach a solution. (Warning – if it looks like just a jumble of chicken scratches with no explanation Of what is going on it can be marked as wrong – even if the answer is right).
Lets Plot Stress Over StrainWe have to get our jollies some how. Oh Manny! It’s the Engineers Favorite – A straight line – A simple linear equation! (That might even explain why someone a long time ago Decided to build strength of materials analysis around Stress and strain plots) Stress σ Strain ε
We Are So Happy We Decide to Name the Slope of the Line Stress σ The name of the slope of the Line is E Modulus of Elasticity Or Young’s Modulus Strain ε
Lets Return to Our Previous Problems Our rod The load is 12,000 Newtons Lets throw in that the Aluminum rod was 140 mm in diameter
Lets See if We Can Calculate Young’s Modulus for Aluminum We already know the strain from our previous problem.
Now Lets Get the Stress The force is 12,000 Newtons For a circle the area is 0.07 M (70mm = 140mm/2) 140 mm Area = 0.01539 M^2
Going After Young’s Modulus σ Plot our stress and strain values (σ=779,500 / ε = 1.11643X10-4 ) We know we have a straight line so ε Got it!
Young’s Modulus is Another Must Learn Do You Get It! ?
Lets See Some Ways the FE May Explore Your Understanding of Hooke’s Law Oh No! We are forced to Do some Statics to get at The Strength of Materials Problem
Skipping the Statics • The cable is being stretched by a force of 1667 Newtons • The definition of Stress is • Force/Area • Force is 1667 Newtons • Area is 2 cm^2 (given in problem) • 1667/2 = 833.5 Newtons/cm^2
Moving on to Hooke’s Law 833.5 1.5X10^6 (given) 5.555X10^-4
Applying the Definition of Strain 5 meters (Pythagorean Theorem and given Sides of the right triangle are 3 and 4) 5.555X10^-4 5.555X10^-4 * 5 = 0.002777
Choosing an Answer Our Answer 0.0027777 Pick (A) If we know the definition of stress, the definition of strain, and Hooke’s Law – that problem’s going DOWN!
We Are So Excited Now We Have to Try Again! We have one of those Tensile test bolts
We Remember Hooke’s Law Or with Algebra σ/ε = E 38/0.17 = 223.59 KN/cm^2 Looks like B
Assignment #2 • Problem 2.6-1
But More Things Happen When We Stretch Materials There is a εy and a εz Obviously these strains are Negative compared to the Positive strain of stretching We get the skinny on the material
It Turns Out that the amount of Shrinkage is proportional to the amount of stretching
Of Course the FE considers testing our knowledge of Strain ratios to be Fair Game A pull test specimen is loaded to 40,000 Newtons/cm^2 and the axial strain is 0.015. If Poisson’s ratio is 0.3 what is the approximate change in diameter of The specimen?
Working the Solution • The axial strain is 0.01 given • Poisson’s ratio is 0.3 given • Therefore the strain in the y and z directions is • 0.015 * 0.3 = 0.0045 (definition of Poisson’s ratio)
Remember Strain is change in length per unit length • The specimen diameter is 0.25 cm (0.0025 meters) • The strain is 0.0045 per unit • Therefore 0.0025*0.0045 = 0.00001125 meters We pick A
Modulus of Elasticity • Modulus of Elasticity and Poisson’s ratio are properties of the material • Different materials have different Youngs Modulus • As engineers we can pick materials that give us the flexing that we need • Obviously we don’t want our buildings to distort too much