1. Mock Course – Plane Stress Transformation (ref MCHT 213, Hibbeler text) 1 hr 15 min format: –Present Theory/Overview with Power Point, 15 min. –Problem solving on board (example problems), 45 min –Industry Examples, 5 min –Answer HW questions Chapter 9 – Plane Stress Transformations

2. Stress Analysis - Review 1.Determine critical point 2.Solve for internal forces at that point (or reduce to cantilever) 3.Solve for stresses at that point 4.Add “like stresses, i.e.:  total =  1 +  2 +  3 + ….………..  total =  1 +  2 +  3 + …………… 5.Summarize stresses at that point on a stress element. 6.May be necessary to use Stress Transformation or Mohr’s circle to get max stresses! Chapter 8 Chapter 9

3. Recall from Chapter 8, already did steps 1 – 5:

4. Drill Bit Isolator ( Thrust Load = 8,000 to 10,000 lb Bending Load= 125 lbs)= K*  Torsion Load= 300 lb-ft Drill Rod Chuck Isolator Drill Bit HOT SPOT! Recall from Chapter 8, already did steps 1 – 5:

5. Now What???? Solve for stress at a point using standard SoM Equations. Summarize these stresses on an initial stress element or aligned stress element. Must find MAXIMUM stresses at that point, may be different then the applied stresses can occur at some other orientation plane or angle. Compare max stresses to material allowables to determine: Is it safe, will it fail??? step 6

6. Step 6. Stresses on other planes? Really nothing new, recall Chapter 1: Average normal stress and shear stress (see example 1-10):

7. a) In general, can have 6 independent stresses (3 normal and 3 shear) acting at a point. a) Many practical engineering problems involve only three independent stresses – called plane stress. a) Stress state for plane stress can be summarized on a 2D element. 9.1 Stress Elements: Now, instead of stresses on “planes” transform stress at a point:

8. 9.2 Plane Stress Transformation: In this course –force equilibrium in 2D) Graduate course –force equilubruim in 3D but math solved with matrices and tensors!!/ solve eigenvalue problem) Derivation of the plane-stress transformation equations:

9. A. Given Plane Stress State: B. What are new stresses at element rotation of  ??: Note, positive stress directions shown. Note, positive angle (ccw) shown.

10. Easy, cut element, sum forces in the x’ and y’ directions:

12. (9.1) (9.2) (9.3)

15. 9.3 Principal Stresses and Maximum In-Plane Shear Stresses What are the maximum stresses at a point? Will they be different than what’s shown on the initial stress element?

16. (9.1) Recall equation for normal stress at new angle  : Want max and min stresses so what do we do? (9.4) Solving for  : This is the principal plane

17. Next, plug 9.4 into 9.1 and get: (9.5) Max principal Min principal

18. So, to summarize: (9.5) (9.4) Principal Stress/Princip al plane – Note-shear stresses are zero on principal plane!

19. Next, do the same thing for shear stress: (9.2) Solve for angle, then plug into eqn 9.2 to get:

20. (9.6) (9.7) (9.8) 1. On planes of max shear, normal stress is not zero but  avg as shown in 9.8. 2. The planes for max shear stress can be determined by orienting an element 45deg from the position of an element of max principal stress!

21. Max principal stress Max shear stress Summary:

22. Important Points: The principal stresses represent the maximum and minimum normal stress at the point. These stresses are shown on the principal stress element. When the state of stress is represented by the principal stresses, no shear stress will act on the element. The state of stress at the point can also be represented in terms of the maximum in-plane shear stress. In this case an average normal stress will also act on the element. This is called the maximum in-plane shear stress element. The maximum in-plane shear stress element is oriented 45 degrees from the principal stress element.

32. 210/418/470 – Capstone Design Projects!!