1. 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)

2. 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)

3. 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

4. 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

5. 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)

6. 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

7. Getting Some Joy Breaking Things Using 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.

8. 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?

9. 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

10. 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?

11. Lets Try Another

12. Technique of Dividing into Segments Check out Stress Here 60 kN Our Acting Force is 60 kN 0.03 Meters Get our cross sectional area

13. 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.

14. Now for the Next Segment 125 kN 60 kN ? We know the sum of these Forces has to be zero or the rod would take off moving. Here is our Mystery Load

15. Continuing Our Exciting Problem! 190 kN 0.05 Meters And using our stress definition. Oh Really – Can We!!

16. A Note on Conventions Our answer to stress here Note that the convention is that tensile stress (pulling apart) is positive, while compressive stress (scrunching together) is negative. Why?

17. 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.

18. Will Stress Always Refer to Average Stress? No – But even if the stress distribution gets funky it is still a force per unit area.

19. So What Happens We Put a Tensile Stress on a Body? Our bar will stretch Now we need a new term - Strain

20. 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!

21. Lets Do Some Problems with Strain Gage marks are placed 250 mm apart on an Aluminum rod. A load is applied Gage marks Are now 250.28 mm apart What is the strain?

22. 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

23. To Finish the Calculation We Need the Undeformed Length Let me see – What was that? Oh yes – 250 mm Time to plug in

24. 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?

25. Back to the Definition of Strain We know the original length Was 18 feet We need that change in length Plug In

26. Some Items to Note About Strain Our book (and a lot of others) uses a lower case Epsilon as the symbol for strain. Our book and many others uses a lower case Delta for change in length. Did we have to know anything about forces acting to determine the strain?

27. Strain is another Must Learn Go Ahead Make My Day Ways considered to force you to learn.

28. 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).

29. Lets Plot Stress Over Strain We have to get our jollies some how. Strain ε 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

30. We Are So Happy We Decide to Name the Slope of the Line Stress σ Strain ε The name of the slope of the Line is E Modulus of Elasticity Or Young’s Modulus

31. Lets Return to Our Previous Problems Lets throw in that the Aluminum rod was 140 mm in diameter Our rod The load is 12,000 Newtons

32. Lets See if We Can Calculate Young’s Modulus for Aluminum We already know the strain from our previous problem.

33. 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

34. Going After Young’s Modulus σ ε Plot our stress and strain values (σ=779,500 / ε = 1.11643X10 -4 ) We know we have a straight line so Got it!

35. Young’s Modulus is Another Must Learn Do You Get It! ?

36. Hooke’s Law Is that Linear Stress – Strain Relationship

37. 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

38. 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

39. Moving on to Hooke’s Law 833.5 1.5X10^6 (given) 5.555X10^-4

40. Applying the Definition of Strain 5.555X10^-4 5 meters (Pythagorean Theorem and given Sides of the right triangle are 3 and 4) 5.555X10^-4 * 5 = 0.002777

41. 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!

42. We Are So Excited Now We Have to Try Again! We have one of those Tensile test bolts

43. We Remember Hooke’s Law Or with Algebra σ/ε = E 38/0.17 = 223.59 KN/cm^2 Looks like B

44. Assignment #2 Problem 2.6-1

45. But More Things Happen When We Stretch Materials We get the skinny on the material There is a ε y and a ε z Obviously these strains are Negative compared to the Positive strain of stretching

46. It Turns Out that the amount of Shrinkage is proportional to the amount of stretching

47. 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?

48. 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)

49. 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

50. 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

51. What if Our Force Tries a Squeeze Play? The sign on our Force will be represented as Negative to indicate Compression Compression Squeezes material Young’s Modulus is the same in compression And tension Poisson’s ratio is the same in tension or compression St Venant’s Principle still applies The definitions of stress and strain are unchanged

52. Assignment #3 Problem 2.6-6

53. There is One More Thing that Can Stretch that Metal Bar Heating things up makes them Expand!

54. Another Beloved Linear Relationship The amount of expansion is proportional to the change in temperature

55. Visiting the Subject on the FE At one end as is heated 60 degrees C above the neutral temperature. What will be the elongation? 6 X 10^-6 cm/(cm*C) 60 30 0.0108 cm We will pick C you see.

56. Thermal Expansion Can Be Stressful What happens if our heated member is held In place by other members and can’t expand? Obviously this can put stresses into our Structure.

57. The Resisting Member Has to Push Back Enough to Cancel the Expansion Substitute the Definitions for Thermal and Physical deformation Solve for P Apply the Definition of stress

58. So Lets Try That With an FE Question 3X10^7 6X10^-6 60 -10,800 N/cm^2 (the sign reflects that it is a compressive stress, but the FE only wants To know about the magnitude in this case. Pick B).

59. Practical Application I’m going to lay some railroad track. My rails are steel and 10 meters long. Young’s Modulus for steel E=200 GPa Coefficient of Thermal Expansion α=11.7X10 -6 / ̊ C It is a cool 6 degrees when I lay my track and I weld the sections together to form A smooth continuous track One day the sun comes out a blisters down on the Landscape. My rails sitting on open rock in the Hot sun reach 48 ̊

60. What Happens to My Rails? The rails will attempt to expand Each 10 meter length will gain 4.9 mm Of course my rails are welded together and pinned down so there will be push Back and stress in the rails.

61. Fact of the Matter I do not know the stress or the load on The rails What I do know is that the deformation From the induced compressive load will Cancel the thermal expansion. I’ll get my delta P and just Admit that I have an Unknown in the value.

62. Substituting Back and Solving One equation – one unknown – this is a piece of cake. 98 Million Pascals! Holy Crud (over 14,000 psi)

63. And More Holy Crud! What makes me think That nothing good can Come of this?

64. What can we do about this? We use expansion joints. Suppose I took my previous story on this time Instead of welding the rails together I used Expansion joints with 3 mm gaps every 10 meters.

65. Lets Try It Now This time we know we have 3mm to play with Plug in Solve for the stress

66. Expansion Joint Issues We could try to pick the gap to make sure we kept the stress in the rails tolerable –Most of us have heard the clank-clank of trains going by over expansion joints For some higher speed applications we really need the welded rails –Now what do I do about expansion

67. How About This?

68. There Are Other Places We See Expansion Joints

70. Another Place

71. Lets Check Out Another Issue Concrete posts are regularly reinforced with steel Our concrete has E= 3.6X10 6 psi and α= 5.5X10 -6 / ̊ F Our steel bars are 7/8 th inch diameter E= 29X10 6 psi α= 6.5X10 -6 / ̊ F What happens if I have a 65 ̊ temperature rise?

72. No Load on the Post Means No Load? Coefficient of Thermal Expansion Steel Coefficient of Thermal Expansion Concrete The steel is going to expand more and The concrete is going to have to restrain That expansion (or the system comes apart).

73. If Both Were Free to Expand But we must constrain the system so Our Strain Equalizer comes from physical Stress developed in the Concrete (Hooke’s Law)

74. Add in Our Hooke’s Law Term There will be a restraining force developed in the concrete This tensile force will make the concrete expand more That opposing force will make the Steel expand less. Notice the load term is the same for Both but E and area are different for The concrete and steel.

75. Now Imposing the Expansions are Equal Since the bars and the concrete post are both the Same length – obviously the strain is the same Substituting our terms for The strain in the steel and The concrete And doing a little algebra to have One equation with the load in the Concrete being the only unknown This one will be a piece of Chocolate cake.

76. Of Course We Have to Solve for Our Concrete and Steel Areas For 6 steel bars, 7/8 th inch in diameter Our concrete area is 10 by 10 minus the Space occupied by the steel

77. Now Plugging into Our Set Up Now solving for the pressure in the concrete Holding back the expansion

78. Now Getting the Stresses is Easy You can actually put some pretty nice stresses on composite members Without putting any external load at all. One issue you can look forward to in your Civil Engineering Materials is what kind Of bond strength you can really get to hold the steel and concrete together.

79. Assignment #4 Problem 2.3-16