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Published byMerry Chambers Modified over 2 years ago

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Purpose of Mohr’s Circle Visual tool used to determine the stresses that exist at a given point in relation to the angle of orientation of the stress element. There are 4 possible variations in Mohr’s Circle depending on the positive directions are defined.

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Sample Problem x = 6 ksi y = -2 ksi xy = 3 ksi Some Part A particular point on the part x y x & y orientation

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Mohr’s Circle (CW) x-axis y-axis x = 6 ksi y = -2 ksi xy = 3 ksi (6 ksi, 3 ksi) 6 3 (-2 ksi, -3 ksi) 2 3 of Mohr’s Circle Center of Mohr’s Circle

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Mohr’s Circle (CW) avg = 2 ksi x-face y-face (6 ksi, 3ksi) (-2 ksi, -3ksi) ( avg, max ) x = 6 ksi y = -2 ksi xy = 3 ksi ( avg, min )

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Mohr’s Circle (CW) x-face y-face (6 ksi, 3ksi) x = 6 ksi y = -2 ksi xy = 3 ksi 4 ksi avg + R 7 ksi avg – R ksi ( avg, max ) (2 ksi, 5 ksi) ( avg, min ) (2 ksi, -5 ksi) 3 ksi R

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Mohr’s Circle (CW) x-face y-face (6 ksi, 3ksi) x = 6 ksi y = -2 ksi xy = 3 ksi 22 4 ksi ( avg, max ) (2 ksi, 5 ksi) ( avg, min ) (2 ksi, -5 ksi) 3 ksi

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Principle Stress (CW) x-face (6 ksi, 3ksi) 1 = 7 ksi 2 = -3 ksi 22 4 ksi ( avg, max ) (2 ksi, 5 ksi) ( avg, min ) (2 ksi, -5 ksi) 3 ksi = 18.435° Principle Stress Element Rotation on element is half of the rotation from the circle in same direction from x-axis

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Shear Stress (CW) x-face y-face (6 ksi, 3ksi) avg = 2 ksi max = 5 ksi 22 4 ksi ( avg, max ) (2 ksi, 5 ksi) ( avg, min ) (2 ksi, -5 ksi) 3 ksi 22 Maximum Shear Stress Element = 26.565°

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Relationship Between Elements avg = 2 ksi max = 5 ksi = 26.565° 1 = 7 ksi 2 = -3 ksi x = 6 ksi y = -2 ksi xy = 3 ksi = 18.435° + = 18.435 ° + 26.565 ° = 45 °

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What’s the stress at angle of 15° CCW from the x-axis? = ? ksi = ? ksi = ? ksi Some Part A particular point on the part x y U & V new axes @ 15 ° from x-axis 15° U x V

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Rotation on Mohr’s Circle (CW) avg = 2 ksi x-face y-face ( avg, max ) ( avg, min ) 30° 15° on part and element is 30° on Mohr’s Circle ( U, U ) ( V, V )

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U = avg + R*cos(66.869°) U = 3.96 ksi V = avg – R*cos(66.869°) V = 0.036 ksi UV = R*sin(66.869°) UV = 4.60 ksi Rotation on Mohr’s Circle (CW) avg = 2 ksi x-face y-face ( avg, max ) ( avg, min ) 66.869° R ( U, U ) ( V, V )

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What’s the stress at angle of 15° CCW from the x-axis? U = 3.96 ksi V =.036 ksi = 4.60 ksi Some Part A particular point on the part x y 15° U x V

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Questions? Next: Special Cases

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Special Case – Both Principle Stresses Have the Same Sign Cylindrical Pressure Vessel X Y Z

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Mohr’s Circle xx (CW) yy xx yy Mohr’s Circle for X-Y Planes This isn’t the whole story however… x = 1 and y = 2

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Mohr’s Circle 11 (CW) yy xx zz z = 0 since it is perpendicular to the free face of the element. z = 3 and x = 1 33 Mohr’s Circle for X-Z Planes xx z = 0 max xz

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Mohr’s Circle 11 (CW) 22 yy xx zz z = 0 since it is perpendicular to the free face of the element. 33 max xz 1 > 2 > 3

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Pure Uniaxial Tension y = 0 x = P/A Ductile Materials Tend to Fail in SHEAR 1 = x 2 = 0 Note when x = S y, S ys = S y /2

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Pure Uniaxial Compression y = 0 x = P/A 1 = 0 2 = x

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Pure Torsion T T 1 = xy 2 = - xy CHALK 11 Brittle materials tend to fail in TENSION.

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Uniaxial Tension & Torsional Shear Stresses Rotating shaft with axial load. Basis for design of shafts. x = P/A xy = Tc/J 1 = x /2+R x /2 x, xy ) 2 = x /2-R 0, yx )

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