1. Column Failures (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results)

2. Parallel Reading 10.1 Introduction 10.2 Euler Buckling Load 10.3 The Affect of End Conditions 10.4 Eccentric Loading 10.7 Design of Centrally Loaded Columns (Do Chapter 10 Reading Assignment Problems) Chapter 10

3. Failed Columns The Rush of the Crush The Shear Joy of It Wait a Minute Back-Up This is not Suppose to happen.

4. There Appears to be Another Mode by Which Columns Fail Buckling Failures

5. We Find Columns in Some Unexpected Places

6. Can You Explain Why Somebody Would Brace a Floor in this Manner

7. One of Our Assumptions Has Been that Deformation Does not Change Geometry B But if the Column ever deforms enough to affect geometry we see our eccentric loading problem forming again.

8. The Stiffness and Inertia of the Column to Spring Back Becomes Critical A solution to this restriction is

9. The Term is Called Eulers Buckling Equation Where

10. And r is the Radius of Gyration

11. So Lets Take Our Formula for a Test Drive If E = 200 GPa How much of a load can I put on this puppy before it buckles The key formula will be

12. Working on Our Formula Looks like the only missing term is I

13. Finishing Up

14. Assignment 17 Do problem 10.2-1 Do problem 10.2-2

15. So What Kind of Load Will it Take to Buckle this Column? For very fat columns the load to shear on a 45 or to crush will come before a buckling load.

16. Lets Try Another Application A circular Brass Column 2 feet in diameter is required to handle the maximum load possible. How tall call the column be before we have to worry about it failing by buckling? Brass Compression Strength 130 Ksi Shear Strength 36 Ksi Young’s Modulus 15,000,000 psi

17. Lets See What it Will Take to Fail it In Compression or Shear 130,000 psi Shear max will be 65,000 psi which is over the 36,000 psi we have available. The column will shear first. I will trick Mohr’s Circle into giving me the answer τ σ σ

18. So What Load Produces Shear Failure? 12 inches 36,000 psi τ σ 72,000 psi The compressive stress that will be occurring at the time shear reaches 36,000 psi will be twice that amount or 72,000 psi. P = σ * A = 72,000 * 452.39 = 32,572,000 lbs 32,572,000

19. Now We Get Ready to Solve Euler’s Buckling Equation for Length Looks like we need to find I

20. I for a Circle 12 inches I = 16,286 in 4 = 272 inches = 22 ft 8 inches

21. The Model Used for Eulers Buckling Equation Assumed Pin Connected Ends This is certainly a worse case scenario because the column is getting no help from its’ surroundings. But columns frequently appear much more confined than a pin connection.

22. Extending Eulers Equation If one end is confined and the other is free to move the column can be only half as long as the equation indicates (I’m looking for Effective Lengths) I imagine there is a reason you don’t see a lot of buildings built this way on purpose.

23. More Cases If we stop the upper end from swinging, even if it is free to pivot then only 70% of the column length is considered in Euler’s Equation. A longer column is possible.

24. Of Course the Best Case is to Confine Both Ends Now we can double the column length. Not surprisingly this is how we try to design structures.

25. Lets Try an Application This column design is fully constrained at the base, but for rotation about the z axis it is pin connected on top and for rotation about the y axis it is unconstrained. The column is aluminum What ratio of a to b will give the maximum bending resistance with the least Material?

26. Looking at Eulers Buckling Equation for critical stress We will be the most efficient with material when the critical stress for column buckling is the same in both directions. Obviously when the slenderness ratio in the two directions is the same the condition is satisfied. Slenderness ratio

27. Lets Pick-Off the Effective Length Terms Around the Y axis we are unconstrained so L e = 2*L But around the Z axis we are pin connected so L e = 0.7*L

28. Finding the Radius of Gyration about the Z axis a b

29. Now for the Radius of Gyration about the Y axis b a

30. Equate the Slenderness Ratios

31. Assignment 18 Problem 10.3-1

32. For Our Next What If What if someone loads the column Off-Center?

33. Answer We can equate a load that is Off-Center by a distance e as the same load at the center plus a bending moment equal to the load times the off-center distance e.

34. Some Ugly Math We’ll take derivatives On that deflection distance English translation – the column just buckled it two.

35. The Equation Allows Us to Get Critical Measurements and Loads Remember. A h b Called the Secant Formula

36. It Provides a Rather Smoothed Curve

37. Lets Try Some Problems and See How this Thing Works Once upon a time there lived a Column of structural tubing with The properties shown below. One day the good witch wanted to know how much of a load could be placed on the column and still allow a factor of safety of 2 against buckling. Along came the good engineering student to help the good witch find the answer.

38. We Observe how the column is fixed Fixed bottom, free movement At top. So

39. Plugging into Eulers Buckling Equation given From effective Length calculation

40. Adjusting for Our Factor of Safety

41. But the Story Continues Next a bad witch came along. The bad witch was very drunk And placed the load 0.75 in From the column center where It was suppose to be placed. Now how much load can the Column take before failure?

42. Now We Use the Secant Formula A A little error in loading made a big difference in stress (I wonder whether this explains why Engineers use safety factors)? Note our P/Pcr comes From 2 safety factor

43. Assignment 19 Do Problem 10.4-2 Do Problem 10.4-4

44. Of Course Real World Loading Scenarios are not perfect Columns are not perfectly fabricated Empirical results tend to Follow more of a smooth Curve.

45. Manufactures Have Empirical Guidelines for Materials This empirical design formula is for Aluminum For Dividing line Is L/r = 66 C 1 = 20.2 for Ksi or 139 for MPa C 2 = 0.126 for Ksi or 0.868 for MPa C 3 = 51,000 Ksi or 351,000 MPa Where L is the column length r is the radius of gyration And L/r is the slenderness ratio

46. Different Materials or Formulations Have Different Guidelines This empirical design formula is for Aluminum For Dividing line Is L/r = 55 C 1 = 30.7 for Ksi or 212 for MPa C 2 = 0.23 for Ksi or 1.585 for MPa C 3 = 54,000 Ksi or 372,000 MPa Where L is the column length r is the radius of gyration And L/r is the slenderness ratio

47. We Can Get Some Interesting Twists Greater than 55 I’m design r I will use a formula based on L/r I don’t know r so I can’t know which formula

48. Try this for a Solution L/r must either be greater than or less than 55 Take a guess – in this case we will try >55 Use that formula to solve for r Plug into L/r and see if your guess about L/r was right. If it was – you are done If it was not – use the other formula

49. So We are off to the races Using the L/r >55 formula 750 mm Radius of Gyration for a circle Check out the Slenderness Ratio

50. Assignment 20 Problem 10.7-14