1. Another Type of Stress and Strain (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results) Or Shear Bliss

2. Parallel Reading Chapter 2 Section 2.7 (Do Reading Assignment Chapter 2B) Section 2.11 Section 2.9 (Do Reading Assignment Chapter 2C)

3. Shear Forces Try to Cut the Material in Two Across its Width

4. One Common Location Is at Pin Connections

5. Applying a Shear Load Causes Our Material to Twist

6. That Twist Angle is Called Shear Strain Note the units of shear strain are radians.

7. We’re In Heaven Because We Find Another Linear Relationship τ ϒ We call the Slope G Modulus of Rigidity Or Shear Modulus

8. Assignment 6 Do Problems 2.8-4 and 2.7-14 As you do the problems explain step by step what you are doing in each calculation until you reach the answer. (Just putting down an answer with scribbles above and expecting the grader to figure out your steps will get you marked Wrong even if your result is correct).

9. Even More Delightful Young’s Modulus, Shear Modulus and Poisson’s Ratio are Related If you know 2, You can always Get the 3 rd.

10. But Shear Can Show Up in Some Unexpected Places We met our Tensile test

11. What If I Reversed That And Tried to Mash my Specimen in Compression? Comment on Engineer’s tricks. Do you notice that I am doing very simple Tests on small specimens and then using mathematical models to extend Those results to very large structures.

12. Its Crush Time! Wait a Cotton Pickin Minute Here! Its as if the material is failing by pushing out wedges on a 45 degree Angle rather than mashing flat like a pancake. What is going on here!

13. Back to the Drawing Board Something very interesting is happening when we put an axial load on things. Ya it stretches, just like Hooke’s Law said. Yes it thins down as we would Expect from Poisson’s Ratio. But we are also producing shear strains in the bar. The presence of a shear strain means we are also producing shear stresses.

14. It Turns Out If you put an axial load on a specimen You will get a shear stress maximum that Is ½ the axial stress on a 45 degree plane (see Bear’s and Johnson 1.3 for the big Ugly proof)

15. What Happens If You Hit a Shear Stress Limit Sooner than a Crush Limit? The specimen will fail on the shear plane! The mystery of the wedgee failed compression Test is solved.

16. A Handy Device for Seeing Where Normal and Shear Stresses Are At It’s called Moore’s Circle and it shows us What kinds of stress are on a plane of any orientation

17. Draw a Graph of Shear and Normal Stress σ τ

18. Consider Our Bar Subject to Tension of Magnitude P τ σ Plot our normal stress in the X direction τ Tension this side Compression This side

19. Now Plot Any Y Axis Compression (of course there is none) τ σ Similarly there is no need To move the points off the τ = 0 line since there is No shear being applied directly

20. Now Draw a Circle That Hits the Points on the Outer Edges τ σ This is Moore’s Circle. We Can use it to find the stress On planes of different Orientation.

21. Where Is the Shear Stress Maximum? Moore’s circle doubles the angle So 90 degrees on the circle means 45 degrees from our pull direction. The shear stress is ½ the magnitude of our tensile stress.

22. Assignment 7 Do problems 2.9-2, 2.9-3, 2.9-4 using Mohr’s Circle Explain how you drew the circle and how you read the answers off of Mohr’s Circle. Obviously show the answer too. Do not just put down a circle with A few points labeled and write down an answer – you’ll get marked wrong if You do.