LECTURE+33+COE-3001-A

# Have maximum shear stress out of plane

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Unformatted text preview: h L pr t (7- 5) pr = 2t (7- 6) = 1 2 = = h (800 ⇥ 103 P a)(1.8m) = = 72M P a 0.020m L (800 ⇥ 103 P a)(1.8m) = = 36M P a (Θ=0°) 2(0.020m) M. Mello/Georgia Tech Aerospace “axial stress” (Θ=90°) 9 EXAMPLE 7- 2(b) Problem 7- 1(b): •  Calculate the maximum in- plane and out- of- plane shear stresses SoluJon 7- 1(b): (⌧max )z = pr 4t (7- 8) (maximum in- plane shear stress) (800 ⇥ 103 P a)(1.8m) (⌧max )z = = 18M P a 4(0.020m) (⌧max )x = ± 1 2 =± pr (7- 9a) (absolute maximum shear stress) 2t (800 ⇥ 103 P a)(1.8m) (⌧max )x = ± = ±36M P a 2(0.020m) ⌧max = 36M P a 4/3/13 M. Mello/Georgia Tech Aerospace 10 EXAMPLE 7- 2(c) Problem 7- 1(c): •  Calculate the circumferenJal and longitudinal strains, ε1 and ε2, respecJvely. SoluJon 7- 1(c): Apply strain- stress form of Hooke’s law: 1 ( E 1 ✏1 = ( E ✏y = ✏1 = ✏2 = 4/3/13 ⌫ x) 1 ⌫ 2) Note: In this instance the x axis is aligned with long axis of cylinder but that is the σ2 axis and so σx = σ2 and σy = σ1 in Hooke’s law… 1 (72M P a 200 ⇥ 103 M P a 1 ( E 1 ✏2 = ( E ✏x = y x ⌫ ⌫ ✏1 = 306µ✏ (0.3)(72M P a)) = 0.000072 ✏2 = 72µ✏ y) 2 (0.3)(36M P a)) = 0.000306 1) 1 (36M P a 200 ⇥ 103 M P a M. Mello/Georgia Tech Aerospace 11 EXAMPLE 7- 2(d) Problem 7- 1(d): •  Calculate the normal stress σw and shear stress τw acJng perpendicular and parallel, respecJvely, to the welded seam. x SoluJon 7- 1(c): Two ways to solve this: 1.  Direct applicaNon of the stress transformaNon equaNons 2.  Mohr’s circle 1st let’s apply the stress transformaNon equaNons: x1 = x y1 = x ⌧x 1 y 1 = 4/3/13 + 2 + 2 y + y y x y 2 (6- 5a) cos 2✓ (6- 5b) 2 2 x cos 2✓ + ⌧xy sin 2✓ y x ⌧xy sin 2✓ sin 2✓ + ⌧xy cos 2✓ (6- 6) M. Mello/Georgia Tech Aerospace 12 EXAMPLE 7- 2(d) cont. y y 1 2 A x x ↵ = 55 x B ✓ = 35 ✓ = 90 x1 = x y1 = x ⌧x 1 y 1 = 4/3/13 + 2 + 2 y + y y x y 2 cos 2✓ 55 = 35 (6- 5a) (6- 5b) 2 2 x cos 2✓ + ⌧xy sin 2✓ y x ↵ = 90 ⌧xy sin 2✓ sin 2✓ + ⌧xy cos 2✓ (6- 6) M. Mello/Georgia Tech Aerospace 13 EXAMPLE 7- 2(d) cont. x1 y1 x = x = + 2 y + 2 y y 2 x y 2 cos 2✓ + ⌧xy sin 2✓ (6- 5a) y 1 y 2 x ⌧x 1 y 1 = + x cos 2✓ ⌧xy sin 2✓ sin 2✓ + ⌧xy cos 2✓ (6- 5b) A x 2 (6- 6) In this case we have: ↵ = 55 ✓ = 35 x = 2 y = 1 B pr = 36M P a 2t pr = = 72M P a t = x ✓ = 35 ⌧xy = 0 ✓ = 90 ↵ = 90 55 = 35 Can do some algebra by subsNtuNng into (6- 5a) and (6- 6) or directly subsNtute numerical values 4/3/13 M. Mello/Georgia Tech Aerospace 14 EXAM...
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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 Tech.

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