Displacement of a gently tapered bar

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Unformatted text preview: cross- sec>onal area as a func>on of distance (x) (2.7) Apply analy>c geometry and calculus principles Analysis con>nues 1/17/13 M. Mello/Georgia Tech Aerospace 8 y EXAMPLE 2- 4 (CONT.) dA 0, 2 L, r = RL N (x)dx 0 EA(x) dB 2 x (2.7) ²་  Line: ²་  Slope: y = r(x) = mx + b m= dB 2 dA 2 L dA ²་  Intercept: b = 2 ✓ dB ²་  Radius: r(x) = dA 2L ✓ dB ²་  Area: A(x) = ⇡ ◆ dA 2L x+ ◆ dA 2 dA x+ 2 2 NOTE: Lower limit of integra>on is zero in fundamental formula ( x) = Let, and, RL 0 ⇡E u( x) = ⇥ du(x) = 1/17/13 P dx dB dA 2L dB dA 2L dB dA 2L dA x+ 2 x+ dx ⇤2 dA 2 R dB 2 dA 2 = P 2L ⇡ E dB dA = P 2L [u 1 ] d2 A ⇡ E dB dA 2 u du dB 2 = 4P L ⇡ EdA db M. Mello/Georgia Tech Aerospace (2-8) Axial displacement of gently tapered bar (BLUE REGION ABOVE) 9 STATICALLY INDETERMINATE STRUCTURES ²་  StaMcally determinate configuraMon: reac>ons and internal forces can be determined solely from free- body diagrams and equa>ons of equilibrium ²་  Forces are determined without knowledge of material proper>es, such as elas>c modulus (E) P Fvert = 0 (reac>on force easily determined) R = P1 + P2 Rigid constraints 1/17/13 ²་  StaMcally indeterminate configuraMon: reac>ons and internal forces cannot be determined solely from free- body diagrams and equa>ons of equilibrium! RA P + RB = 0 ²་  Equilibrium problem must be complemented by compa>bility equa>ons which describe the the constraint of the structure (support condi>ons). =0 AB ²་  Compa>bility equa>on must then be expressed in terms of Ra and Rb using the linear elas>c force- displacement rela>on Pi Li (calculate for each sec>on and combine together) i= Ei Ai M. Mello/Georgia Tech Aerospace 10 ANALYSIS OF A STATICALLY INDETERMINATE BAR RA GIVEN: ²་  Consider a bar of cross- sec>onal area (A) and modulus (E) ²་  A force P is applied at point C as shown ²་  The bar is constrained and prevented from displacing at the fixed ends A and B. DETERMINE: ²་  Reac>on forces RA and RB ²་  Displacement δc of point C ²་  Stresses within the bar (above and below point C) SOLUTION STRATEGY: ²་  EQUILIBRIUM EQUATION: P Fvert = 0 RA P + RB = 0 (ONE EQUATION AND TWO UNKNOWNS) ²་  COMPATIBILITY CONDITON: AB = AC + C B = 0 ²་  FORCE- DISPLACEMENT RELATION: i 1/17/13 = Pi L i E i Ai RA RB RB...
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