Of linearly elastic material this is what

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Unformatted text preview: s distributed uniformly within a prisma\c bar of area (A) and length (L), the strain energy density can is expressed as: ✓2◆ 1 PL w= AL 2EA U w ( x) = w = AL 1/17/13 P2 w= 2EA2 2 w= 2E w= ✓ ◆ 1 EA 2 L w= AL 2L E2 w= 2L M. Mello/Georgia Tech Aerospace ✏ 2 E ✏2 w= 2 17 EXAMPLE: STRAIN ENERGY OF A SUSPENDED PRISMATIC BAR N (y ) = A(L y) + P y y d⇠ d⇠ weight of lower sec\on exer\ng Tensile load on cross- sec\on A (y ) = (L y) + P A stress follows from (y ) = (L y ) P ✏(y ) = + E EA strain follows from Hooke’s law [N (y )]2 [ A(L y ) + P ]2 w (y ) = = 2EA2 2EA2 U (y ) = Z y Aw(⇠ )d⇠ = 0 1/17/13 N (y ) A Z y 0 [ A(L SUSPENDED PRISMATIC BAR IN RESPONSE TO IT’S WEIGHT DENSITY (γ) AND APPLIED LOAD (P) Elas\c strain energy density func\on ⇠ ) + P ] 2 d⇠ 2EA Elas\c strain energy func\on M. Mello/Georgia Tech Aerospace 18 SUSPENDED PRISMATIC BAR PROBLEM (cont.) U (y ) = 2 A2 L2 + 2P LA + P 2 y 2EA 2 22 A2 L + P A 2 A3 y+ y 2EA 6EA Poten\al energy increases as a cubic func\on of posi\on (y) At y=L : U (L) = W Where, = AL 1/17/13 2 AL3 P L2 P 2L + + 6E 2E 2EA Total poten\al energy stored within bar of length (L) M. Mello/Georgia Tech Aerospace = y W AL A L g P 19 IMPACT LOADING: MAXIMUM ELONGATION OF AN IMPACTED BAR g Impact load on a horizontal bar Impact load on a prisma\c bar AB due to a falling object of mass M. ☞  INVOKE CONSERVATION OF ENERGY 2 EA max M g (h + max ) = 2L 1/17/13 Kine\c Energy = strain energy of the bar under maximum elonga\on HIGH SPEED VIDEOS OF ELASTIC AND INELASTIC IMPACTS: Gravita\onal P.E = MAX. strain energy of the bar under maximum elonga\on MAXIMUM ELONGATION = (δmax) ☞  INVOKE CONSERVATION OF ENERGY 2 1 EA max 2 Mv = 2 2L hvp://www.youtube.com/watch?v=wrpgq7WHtgU hvp://www.youtube.com/watch?v=2Y57pw_iWlk hvp://www.youtube.com/watch?v=aMqM13EUSKw M. Mello/Georgia Tech Aerospace 20 g 2 max 1 EA M v2 = 2 2L 2 EA max M g (h + max ) = 2L Solve quadra\c eqn. for δmax (posi\ve root): ✓ ◆2 ✓ ◆ 1/2 WL WL WL + + 2h max = EA EA EA ²་  NOTE: (WL/EA) is sta\c elonga\on resul\ng from a slowly applied sta\c load WL i.e., static = and so, EA max = static 1/17/13 +[ 2 static + 2h static ] max = r M v2 L EA ²་  MAXIMUM ELONGATION (δmax) of the elas\c bar in response to impact by rigid block of mass (M). ²་  Block impact speed = (v) ²་  Bar length = (L) ²་  Axial rigidity of bar = (EA) 1/2 M. Mello/Georgia...
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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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