INFORMATIO

# 3 3 3 3 mpa f a a 16 5 12 mpa f a b 166 4 666 16 5 12

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3 3 3 3 MPa f a A 166 . 29 666 . 16 5 . 12 MPa f a B 166 . 4 666 . 16 5 . 12 B A mm 200 mm 250 mm 450 mm 100 kN 500 4 3 3 4 B A mm 750 mm 150 kN 500 4 3 3 4 C 3 R 2 R 1 R 1301 . 53

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9 Stresses at a Point General State of Stress Plane Stress 2 sin 2 cos ) 2 ( ) 2 ( xy y x y x x ……………………..(1) 2 sin 2 cos ) 2 ( ) 2 ( xy y x y x y …………………….(2) 2 cos 2 sin ) 2 ( xy y x y x …………………….(3) The planes defining maximum or minimum normal stresses are found from: y x xy p 2 2 tan ………………(4) The planes of maximum shearing stresses are defined by : xy y x s 2 2 tan ………………(5) The planes of zero shearing stresses may be determined by setting τ equal to zero. y x xy 2 2 tan ……………….(6) Equation 6 and 4 show that maximum and minimum normal stresses occur on planes of zero shearing stresses. The maximum and minimum normal stresses are called the principal stresses.
0 Equation 5 is the negative reciprocal of equation 4. This means that the values of 2θ s from equation 5 and equation 4 differ by 90 ο . This means that the planes of maximum shearing stress are at 45 o with the planes of principal stress. 2 2 . min . max ) 2 ( ) 2 ( xy y x y x …………………(7) 2 2 max ) 2 ( xy y x …………………..(8) 2 y x avg ………………….(9) Example: The state of plane stress at a point is represented by the element shown. Determine the state of stress at the point on another element oriented 30 o clockwise from the position shown. θ=-30 o 2 sin 2 cos ) 2 ( ) 2 ( xy y x y x x ) 60 sin( 25 ) 60 cos( ) 2 50 80 ( ) 2 50 80 ( x MPa x 849 . 25 2 sin 2 cos ) 2 ( ) 2 ( xy y x y x y ) 60 sin( 25 ) 60 cos( ) 2 50 80 ( ) 2 50 80 ( y MPa y 15 . 4 2 cos 2 sin ) 2 ( xy y x y x ) 60 cos( 25 ) 60 sin( ) 2 50 80 ( y x MPa y x 791 . 68 MPa 50 MPa 80 MPa 25 x y MPa 791 . 68 MPa 849 . 25 MPa 15 . 4 MPa x 80 MPa y 50 MPa xy 25

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Example: When the torsional lo pure shear stress in the material associated average normal stress 0 x 0 y MPa xy 60 a) 2 2 max ) 2 ( xy y x MPa 60 ) 60 ( ) 2 0 0 ( 2 2 max 0 2 0 0 2 y x avg xy y x s 2 2 tan 0 60 2 0 2 tan s θ s =0 b) . min . max 2 ( ) 2 ( x y x ) 60 ( ) 2 0 0 ( ) 2 0 0 ( 2 2 . min . max σ max =60 MPa σ min =-60 MPa y x xy p 2 2 tan 0 60 2 2 tan p 45 4 2 2  p p or 135 o 2 cos ) 2 ( ) 2 ( y x y x x x ) 90 sin( 60 ) 90 cos( ) 2 0 ( ) 2 0 ( 1 oading T is applied to the bar shown l. Determine a) the maximum in plane s. b) the principal stresses.
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• Winter '15
• MAhmoudali

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