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Stress Mohr's circle
- 1. 1 Figure Q1 shows the stress condition on a piece of wooden structure at a critical point. Determine:- a) The principal stresses and sketch the stress element; b) The maximum in-plane shearing stress and show the stress element in this condition; c) The normal and shear stresses in the wood grain direction and sketch the stress element. Figure Q1 x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain
- 2. 2 Firstly, it is suggested to place your graph paper in the landscape orientation.
- 3. σ 3 Draw σ–axis along the mid-width of the graph paper.
- 4. σ σavg 4 • Calculate σavg = (σx + σy)/2. • Locate σavg on σ–axis. It should be somewhere at the middle of the graph. It is because you are going to draw a circle with the centre (σavg,0).
- 5. σ τ σavg 5 Draw τ–axis that passes through σ = 0, with positive τ downward.
- 6. σ τ σavg 6 Also label +ve τ as CCW and –ve τ as CW. Note: the scale for σ:τ axes must be 1:1.
- 7. x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain σ τ σavg σx 7 • Locate (σx, τxy). • Note that σx has a shear stress component τxy that tends to rotate the element clockwise. τxy Therefore, (σx, τxy) must be located at the upper half of the circle.
- 8. x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain σ τ σavg σx 8 τxy By using σavg as the centre, draw the circle.
- 9. σ τ σavg σx, τxy σy, τxy 9 Draw a straight line that connects (σx, τxy) and σavg. You will get (σy, τxy) at the other end of the line. x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain
- 10. x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain σ τ σavg σx, τxy σy, τxy 10 Note that (σy, τxy) is located at the lower half of the circle. It means that σy has a shear stress component τxy that tends to rotate the element counter clockwise.
- 11. σ τ σavg σx, τxy σy, τxy 2θp 11 To reach the principal stresses (σ1 and σ2), you need to rotate an angle of 2θp in the CCW direction.
- 12. σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 12 Label σ1 and σ2 on the Mohr’s circle.
- 13. σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 13 θp x To draw the stress element, by using x–axis as reference, rotate the plane by an angle of θp in the CCW direction.
- 14. σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 14 θp x x’ y’ Label it as x’–y’plane.
- 15. σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 15 θp x x’/2 y’/1 At the same time, since rotating 2θp brings σx to σ2 and σy to σ1, label x’=2 and y’=1.
- 16. σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 16 θp x x’/2 y’/1 Sketch the element in x’–y’ plane.
- 17. σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 17 θp x x’/2 y’/1 σ1=164MPa σ2= -24MPa Label σ1 and σ2 along their respective axis. τ=0 Note that on the principal plane, τxy = 0.
- 18. 18 σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs To reach the maximum in-plane shear stress τm,ip, you need to rotate an angle of 2θs in the CW direction.
- 19. 19 σ τ σavg σ1σ2 σy’, τm,ip σx’, τm,ip σx, τxy σy, τxy 2θp 2θs • Label (σx’, τm,ip) and (σy’, τm,ip) on the Mohr’s circle. • Note that σx’is on the top, and σy’is at the bottom of the circle.
- 20. 20 θs x σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs To draw the stress element, by using x–axis as reference, rotate the plane by an angle of θs in the CW direction. σy’, τm,ip σx’, τm,ip
- 21. 21 θs x y’ x’ σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs Label it as x’–y’plane. σy’, τm,ip σx’, τm,ip
- 22. 22 θs x y’ x’ σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs Sketch the element in x’–y’ plane. σy’, τm,ip σx’, τm,ip
- 23. 23 θs x y’ x’ σavg= 70MPa σavg=70MPa σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs Then, label σx’ = σy’ = σavg. σy’, τm,ip σx’, τm,ip
- 24. 24 θs x y’ x’ σavg= 70MPa σavg=70MPa σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs Note that x’–axis is located on the top. σy’, τm,ip σx’, τm,ip Hence, it means that x’–axis is having the τm,ip component that tends to rotate the element in CW direction.
- 25. 25 θs x y’ x’ σavg= 70MPa σavg=70MPa σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs At the same time, you could also notice that y’–axis is located at the bottom, … σy’, τm,ip σx’, τm,ip and thus having τm,ip component that tends to rotate the element in CCW direction.
- 26. 26 θs x y’ x’ σavg= 70MPa τm,ip= -94MPa σavg=70MPa σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs Complete τm,ip accordingly on the element. σy’, τm,ip σx’, τm,ip
- 27. 27 θs x y’ x’ σavg= 70MPa τm,ip= -94MPa σavg=70MPa σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs Note that this is a –ve τm,ip. σy’, τm,ip σx’, τm,ip +ve τ means that the shear stress is in the +ve x’–y’ quadrant.
- 28. 28 σ τ σavg σ1σ2 σx, τxy σy, τxy 2θp 2θs x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain σy’, τm,ip σx’, τm,ipAlong the wood grain (θ=50o CCW) means that to rotate by an angle of 2θ=100o CCW on the Mohr’s circle.
- 29. 29 σ τ σavg σ1σ2 σx, τxy σy, τxy 2θ 2θp 2θs x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain σy’, τm,ip σx’, τm,ip Measure the angle and draw the line that passes through σavg.
- 30. 30 σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs • Label (σx’, τx’y’) and (σy’, τx’y’). • Note that y’ is attained after rotating 2θ from y. Similarly, x’ comes from x. x y σx=20MPa τxy= 80MPa σy=120MPa 50o Wood grain σy’, τm,ip σx’, τm,ip
- 31. 31 θ x σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs • To draw the stress element, by using x–axis as reference, rotate the plane by an angle of θin the CCW direction. • Note that rotating y by 50o in CCW is the same as rotating x by 50o in CCW. σy’, τm,ip σx’, τm,ip
- 32. 32 θ x x'y' σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs Label it as x’–y’plane. σy’, τm,ip σx’, τm,ip
- 33. 33 θ x x'y' σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs Sketch the element in x’–y’ plane. σy’, τm,ip σx’, τm,ip
- 34. 34 θ x x'y' σy'=140MPa σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs • Then, label σx’ and σy’. • Note that σx’ = 0 for this case. σy’, τm,ip σx’, τm,ip
- 35. 35 θ x x'y' σy'=140MPa σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs Note that σx’ is located at the bottom. Hence, it means that x’–axis is having the τx’y’ component that tends to rotate the element in CCW direction. σy’, τm,ip σx’, τm,ip
- 36. 36 θ x x'y' σy'=140MPa σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs At the same time, you could also notice that y’– axis is located at the top, and thus having τx’y’ component that tends to rotate the element in CW direction. σy’, τm,ip σx’, τm,ip
- 37. 37 θ x x'y' σy'=140MPa τx‘y’=63MPa σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs Complete τx’y’ accordingly on the element. σy’, τm,ip σx’, τm,ip
- 38. 38 θ x x'y' σy'=140MPa τx‘y’=63MPa σ τ σavg σ1σ2 σx, τxy σy, τxy σx', τx’y’ σy', τx’y’ 2θ 2θp 2θs Note that this is a +ve τm,ip. σy’, τm,ip σx’, τm,ip +ve τ means that the shear stress is in the +ve x’–y’ quadrant.