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Published byLynette Lawrence Modified over 9 years ago
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Content Stress Transformation A Mini Quiz Strain Transformation
Click here Click here Click here Approximate Duration: 20 minutes
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Plane Stress Transformation
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Plane Stress Loading z = 0; xz = 0; zy = 0
~ where all elements of the body are subjected to normal and shear stresses acting along a plane (x-y); none perpendicular to the plane (z-direction) z = 0; xz = 0; zy = 0 x y
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Plane Stress Loading x; y; and xy
Therefore, the state of stress at a point can be defined by the three independent stresses: x; y; and xy A x y xy A x y
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If x, y, and xy are known, …
Objective x y xy A A x y State of Stress at A If x, y, and xy are known, …
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…what would be ’x, ’y, and ’xy?
Objective ’x ’y ’xy A A x y State of Stress at A x’ y’ …what would be ’x, ’y, and ’xy?
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Transformation y xy ’xy=? ’x=? x y x A y’ x’
State of Stress at A
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Transformation Solving equilibrium equations for the wedge…
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Principal Planes & Principal Stresses
~ are the two planes where the normal stress () is the maximum or minimum ~ there are no shear stresses on principal planes ~ these two planes are mutually perpendicular ~ the orientations of the planes (p) are given by: gives two values (p1 and p2)
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Principal Planes & Principal Stresses
Orientation of Principal Planes p2 p1 x 90
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Principal Planes & Principal Stresses
~ are the normal stresses () acting on the principal planes
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Maximum Shear (max) max = R
~ maximum shear stress occurs on two mutually perpendicular planes ~ orientations of the two planes (s) are given by: gives two values (s1 and s2) max = R
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Maximum Shear Orientation of Maximum Shear Planes s1 s2 x 90
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Principal Planes & Maximum Shear Planes
45 x p = s ± 45
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Mohr Circles Equation of a circle, with variables being x’ and xy’
From the stress-transformation equations (slide 7), Equation of a circle, with variables being x’ and xy’
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Mohr Circles (x + y)/2 xy’ R x’
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Mohr Circles A point on the Mohr circle represents the x’ and xy’ values on a specific plane. is measured counterclockwise from the original x-axis. Same sign convention for stresses as before. i.e., on positive planes, pointing positive directions positive, and ….
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Mohr Circles x’ xy’ = 0
= 90 = 0 When we rotate the plane by 180°, we go a full round (i.e., 360°, on the Mohr circle. Therefore….
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…..when we rotate the plane by °, we go 2° on the Mohr circle.
Mohr Circles x’ xy’ 2 …..when we rotate the plane by °, we go 2° on the Mohr circle.
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Mohr Circles xy’ x’ 2 max 1
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From the three Musketeers
Mohr circle represents the state of stress at a point; thus different Mohr circles for different points in the body Mohr circle is a simple but powerful technique Get the sign convention right Quit Continue
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A Mohr Circle Problem A The stresses at a point A are shown on right.
40 kPa 200 kPa 60 kPa The stresses at a point A are shown on right. Find the following: major and minor principal stresses, orientations of principal planes, maximum shear stress, and orientations of maximum shear stress planes.
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A (kPa) (kPa) Drawing Mohr Circle 200 kPa 60 kPa 40 kPa 120
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Principal Stresses R = 100 (kPa) (kPa) 1= 220 2= 20
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Maximum Shear Stresses
(kPa) (kPa) max = 100
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A (kPa) (kPa) Positions of x & y Planes on Mohr Circle
120 60 60 40 tan = 60/80 = 36.87°
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A (kPa) (kPa) Orientations of Principal Planes 36.9° 200 kPa
minor principal plane 71.6° 18.4° major principal plane
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A (kPa) (kPa) Orientations of Max. Shear Stress Planes 53.1° 36.9°
26.6° 116.6°
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Testing Times… Do you want to try a mini quiz? YES Oh, NO!
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The state of stress at a point A is shown.
Question 1: A 30 kPa 90 kPa 40 kPa The state of stress at a point A is shown. What would be the maximum shear stress at this point? Answer 1: 50 kPa Press RETURN for the answer Press RETURN to continue
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At A, what would be the principal stresses?
Question 2: A 30 kPa 90 kPa 40 kPa At A, what would be the principal stresses? Answer 2: 10 kPa, 110 kPa Press RETURN for the answer Press RETURN to continue
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At A, will there be any compressive stresses?
Question 3: A 30 kPa 90 kPa 40 kPa At A, will there be any compressive stresses? Answer 3: No. The minimum normal stress is 10 kPa (tensile). Press RETURN for the answer Press RETURN to continue
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The state of stress at a point B is shown.
Question 4: B 90 kPa 0 kPa The state of stress at a point B is shown. What would be the maximum shear stress at this point? Answer 4: This is hydrostatic state of stress (same in all directions). No shear stresses. Press RETURN for the answer Press RETURN to continue
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Plane Strain Transformation
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Plane Strain Loading z = 0; xz = 0; zy = 0
~ where all elements of the body are subjected to normal and shear strains acting along a plane (x-y); none perpendicular to the plane (z-direction) z = 0; xz = 0; zy = 0 x y
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Plane Strain Transformation
Similar to previous derivations. Just replace by , and by /2
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Plane Strain Transformation
Sign Convention: Normal strains (x and y): extension positive Shear strain ( ): decreasing angle positive x y before x y after e.g., x positive y negative positive
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Plane Strain Transformation
Same format as the stress transformation equations
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Principal Strains ~ maximum (1) and minimum (2) principal strains
~ occur along two mutually perpendicular directions, given by: Gives two values (p1 and p2)
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Maximum Shear Strain (max)
max/2 = R p = s ± 45
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Mohr Circles (x + y)/2 x’ xy’ 2 R
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Strain Gauge ~ measures normal strain (), from the change in electrical resistance during deformation electrical resistance strain gauge
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Strain Rosettes ~ measure normal strain () in three directions; use these to find x, y, and xy e.g., 45° Strain Rosette x 45° 90 x = 0 measured 45 y = 90 xy = 2 45 – (0 + 90) 0
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The End
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