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Unformatted text preview: Mechanics of Solids Lecture 9 by Dr Emre Erkmen Lecturer, School of Civil and Environmental Engineering, The University of Technology Sydney Office: 2.520 Phone:9514 9769 Email: emre.erkmen@uts.edu.au Lecture hours: Tuesdays 11:0014:00, Wednesdays 15:0018:00 Office hours: Tuesdays 15:0017:00 1 Mechanics of Solids  Lecture 9 Outline Outline Stress transformations Maximum and minimum stresses Mohrs circle 2 Mechanics of Solids  Lecture 9 Body in Static Equilibrium Y X F 1 F 2 3 F 3 Cube in static equilibrium Triangular prism in static equilibrium Mechanics of Solids  Lecture 9 Stresses due to internal forces 4 N My A I = VQ T Ib J = + V M N static equilibrium of a free body cut from the beam Mechanics of Solids  Lecture 9 Transformation of plane stresses Uniaxial stress 5 x P A = F = x Mechanics of Solids  Lecture 9 Area of the surfaces b cos tan 6 Mechanics of Solids  Lecture 9 Transformation of plane stresses Uniaxial stress 7 x P A = F = F = x cos cos x A A = sin cos x A A + = 2 cos x = sin cos x = Eq. ( 1 ) Eq. ( 2 ) Mechanics of Solids  Lecture 9 Transformation of stresses Only y acting y 8 F = F = tan sin cos y A A = tan cos cos y A A = 2 sin y = sin cos y = Eq. ( 3 ) Eq. ( 4 ) Mechanics of Solids  Lecture 9 Transformation of stresses Both x acting 2 2 cos sin x y = + sin cos sin cos x y = + y and From the superposition of the previous equations Eq. ( 1 ) + Eq. ( 3 ) Eq. ( 2 ) + Eq. ( 4 ) 9 From trigonometry ( ) 2 1 cos 1 cos2 2 = + ( ) 2 1 sin 1 cos2 2 = 1 cos sin sin 2 2 = ( ) ( ) 1 1 cos2 2 2 x y x y = + + ( ) 1 sin 2 2 x y = Eq. ( 5 ) Eq. ( 6 ) Mechanics of Solids  Lecture 9 Transformation of stresses Pure shear xy A xy A cos xy A sin 10 F = F = sin tan cos cos xy xy A A A = + cos tan sin cos xy xy A A A + = 2 sin cos xy = ( ) 2 2 cos sin xy = Using trigonometric relations we obtain sin 2 xy = cos2 xy = Eq. ( 7 ) Eq. ( 8 ) Mechanics of Solids  Lecture 9 Transformation of stresses All x acting y and From the superposition of the previous equations ( ) ( ) 1 1 cos2 sin 2 2 2 x y x y xy = + + + xy , ( No stresses on the z surface ) Eq. ( 9 ) 11 ( ) 1 sin 2 cos2 2 x y xy = + Given stresses on 2 perpendicular planes at a point x , y , xy , we can calculate stresses , on another plane (at the point) making an angle, , to one of the perpendicular planes....
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This note was uploaded on 08/22/2011 for the course ENG 48331 taught by Professor Brown during the Three '11 term at University of Technology, Sydney.
 Three '11
 BROWN
 Environmental Engineering

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