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Unformatted text preview: Tech Aerospace 21 g 2 1/2 max = static + [ static + 2h static ] ✓ ◆1 / 2 2h max = static 1 + 1 + static p h NOTE: For >> 1 max ⇡ 2h static static p WL Recall: v = 2 gh and static = 2 1 EA max 2 Mv = 2 2L r maximum elas\c deforma\on 2L Mv of the elas\c bar in response to max = EA Impact by rigid block of mass M DEFINE “IMPACT FACTOR”: ImpactFactor = max static = q EAv 2 M g2 L EA max 1/17/13 = r M v2 L EA Not surprisingly we a recover similar forms M. Mello/Georgia Tech Aerospace 22 MAXIMUM STRESS OF AN IMPACTED BAR max Invoke Hooke’s law… max = E ✏max ✓ ◆ max = E L Recall: ✓ ◆2 ✓ ◆ 1/2 WL WL WL + + 2h max = EA EA EA (original form/no approxima\ons) ✓ ◆ 2 1/2 W W 2W hE max = + + A A AL 1/17/13 max max M. Mello/Georgia Tech Aerospace =E E = L r = ✓ r max L ◆ M v2 L EA M v2 E AL 23 MAXIMUM STRESS OF AN IMPACTED BAR (cont.) ✓ ◆2 max 1/2 =E ✓ max L ◆ W W 2W hE r + + max = A A AL E M v2 L max = L EA Define: static = W/A r M v2 E ✓ ◆1 / 2 max = 2hE 2 AL ax = static + m static static + L h >> 1 Consider: static r r recover similar forms 2hE static M v 2 E (as before for max = max = L AL displacement) 1/17/13 M. Mello/Georgia Tech Aerospace 24 SUDDENLY APPLIED LOADS ²་  i.e., collar is slowly lowered un\l it is “just above” the flange. ²་  Collar is then suddenly released Finally, consider the limit as h ⇡ 0 2W L lim max = 2 static = h!0 EA 2W lim = 2 static = max h!0 A What do these results imply?? 1/17/13 ²་  Collar s\ll falls downward as it is resisted by the elas\c force generated within the bar ²་  Collar must con\nue to move downward un\l its velocity is brought to zero. ²་  Bar will oscillate un\l coming to rest at the sta\c elonga\on produced by the weight of the collar. Impact Factor = 2 ! M. Mello/Georgia Tech Aerospace 25...
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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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