X y 2r cos 2 xy sin 2 6 23b 6 5a

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Unformatted text preview: ses in the xy plane 13 OUT- OF- PLANE SHEAR STRESSES •  BUT DON’T FORGET TO THINK ABOUT THE MATERIAL ELEMENT IN 3D! •  CONSIDER 45° ROTATION ABOUT THE OTHER TWO PRINCIPAL AXES (x and y axes) y1=y 45° or 45° x1=x z1 z1 Material element under stress in reference configura]on (xyz coordinate system) Normal stress a ver ac]ng on planes of max. pos. shear stress (refer to (6- 5a,b) 3/25/13 Plane of maximum shear (τmax)x1 along plane rotated by 45° about the x axis Principal stresses within plane of rota]on: 2 6= 0, (⌧max )x1 = ± aver = 3 Plane of maximum shear (τmax)y1 along plane rotated by 45° about the y axis =0 2 2 y 2 M. Mello/Georgia Tech Aerospace Principal stresses within plane of rota]on: 1 6= 0, (⌧max )y1 = ± aver = 3 1 2 x 2 14 =0 A final note on absolute maximum shear stress •  If σ1 and 2 have the same sign, then either (⌧ ) = ± 2 or, (⌧ ) = ± 2 gives the numerically largest stress. •  If they have opposite signs, then (⌧ = ± ) 2 is the largest. max x1 max z1 3/25/13 2 1 max y1 1 2 M. Mello/Georgia Tech Aerospace 15 EXAMPLE 6- 3: Element in plane stress subjected to σx = 84MPa, σy = - 30MPa, τxy = - 32MPa DETERMINE: (a)  Principal stresses and show them in a sketch (b)  Maximum shear stresses and show them on a sketch SOLUTION: (a) PRINCIPAL STRESSES 2⌧xy USE tan 2✓p = (6- 11) x y SUBSTITUTE σx = 84MPa, σy = - 30MPa, τxy = - 32MPa tan 2✓p = 3/25/13 2( 32M P a) = 84M P a ( 30M P a) 0.5614 M. Mello/Georgia Tech Aerospace 2✓p = 150.6 ! ✓p = 75.3 and 2✓p = 330.6 ! ✓p = 165.3 16 Example 6- 3 (cont.) •  Next ques]on… so which of these corresponds to the max principal stress? •  Subs]tute the two values of 2Θp we just obtained into: x1 x1 x1 = = ) 2 = x + 2 y + x y 2 cos 2✓ + ⌧xy sin 2✓ (6- 5a) 84 + ( 30) 84 + 2 ( 30) cos(150.6 ) + ( 32) sin(150.6 ) = 2 84 + ( 30) 84 + 2 ( 30) cos(330.6 ) + ( 32) sin(33.6 ) = 92.4M P a 2 38.4M P a AND = 92.4M P a ; ✓p1 = 165.3 and = 38.4M P a ; ✓p2 = 75.3 1 3/25/13 M. Mello/Georgia Tech Aerospace 17 EXAMPLE 6- 3: Element in plane stress subjected to σx = 84MPa, σy = - 30MPa, τxy = - 32MPa (cont.) •  ALTERNATIVE SOLUTION PATH: First, obtain 1 , 2 using (6- 17) 1 ,2 x+ = 2 y ± 1 84 + ( 30) = + 2 1 s✓ s✓ 84 x y 2 ◆2 ( 30) 2 2 +⌧xy ◆2 (6- 17) (this formula leads dire...
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This note was uploaded on 09/19/2013 for the course CEE 3001 taught by Professor Zhu during the Spring '09 term at Georgia Institute of Technology.

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