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Unformatted text preview: Shear function V(x) = R 1 < x 0> w < x 0> 1 + w < x a > 1 F < x b > 0 + R 2 < x L > Moment function M(x) = R 1 < x 0> 1 w < x 0> 2 /2 + w < x a > 2 /2  F < x b > 1 + R 2 < x L > 1 Modulus of elasticity E 207 GPa ⋅ = Reactions R 1 264.0 N ⋅ = R 2 316.0 N ⋅ = Maximum shear V max 316 N ⋅ = (negative, from x = b to x = L ) Maximum moment M max 126.4 N ⋅ m ⋅ = (at x = b ) 2. Integrate the moment function, multiplying by 1/ EI , to get the slope. θ ( x ) = [ R 1 < x > 2 /2  w < x > 3 /6 + w < x a > 3 /6  F < x b > 2 /2 + R 2 < x L > 2 /2 + C 3 ]/ EI 3. Integrate again to get the deflection. y ( x ) = [ R 1 < x > 3 /6  w < x > 4 /24 + w < x a > 4 /24  F < x b > 3 /6 + R 2 < x L > 3 /6 + C 3 x + C 4 ]/ EI 4. Evaluate C 3 and C 4 At x = 0 and x = L , y = 0, therefore, C 4 = 0. R 1 6 L 3 ⋅ w 24 L 4 ⋅ w 24 L a ( ) 4 ⋅ + F 6 L b ( ) 3 ⋅ C 3 L ⋅ + C 3 1 L R 1 6 L 3 ⋅ w 24 L 4 ⋅ + w 24 L a ( ) 4 ⋅ F 6 L b ( ) 3 ⋅ + ⋅ = C 3 31.413 N m 2 ⋅ = 5. Define the range for x x m ⋅ 0.005 L ⋅ , L .. = 6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z , and a value of one when it is greater than or equal to z . S x z , ( ) if x z ≥ 1 , , ( ) = PROBLEM 423a Statement: A beam is supported and loaded as shown in Figure P411a. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data giv in row a from Table P42. Units: N newton = MPa 10 6 Pa ⋅ = GPa 10 9 Pa ⋅ = Given: Beam length L 1 m ⋅ = Distance to distributed load a 0.4 m ⋅ = Distance to concentrated load b 0.6 m ⋅ = Distributed load magnitude w 200 N ⋅ m 1 ⋅ = Concentrated load F 500 N ⋅ = Moment of inertia I 2.8510 8 ⋅ m 4 ⋅ = FIGURE 423A Distance to extreme fiber c 2.0010 2 ⋅ m ⋅ = Free Body Diagram for Problem 423 Solution: See Figures 423 and Mathcad file P0423a. 1. The reactions, maximum shear and maximum moment were all found in Problem 323a. Those results are summarized here. Load function q(x) = R 1 < x 0>1 w < x 0> + w < x a > F < x b >1 + R 2 < x L >1 Slope and Deflection Diagrams for Problem 423a FIGURE 423aB 0.2 0.4 0.6 0.8 1 2 1.5 1 0.5 y x ( ) mm x m 0.2 0.4 0.6 0.8 1 0.01 0.005 0.005 0.01 θ x ( ) x m DEFLECTION, mm SLOPE, radians y max 1.82 mm = y max y c ( ) = Substituting c into the deflection equation, c 0.523 m = c B B 2 4 A ⋅ C ⋅ + 2 A ⋅ = C 33.547 N m 2 ⋅ = B 16.000 N m ⋅ = A 92.000 N = C C 3 w 6 a 3 ⋅ = B 3 w 6 ⋅ a 2 ⋅ = A R 1 2 3 w 6 ⋅ a ⋅ = Solving for c , θ 1 E I ⋅ R 1 2 c 2 ⋅ w 6 c 3 ⋅ w 6 c a ( ) 3 ⋅ + C 3 + ⋅ 9. Maximum deflection occurs at x = c , where θ = 0 and c < b ....
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This note was uploaded on 07/23/2011 for the course EML 4501 taught by Professor Staff during the Spring '11 term at UNF.
 Spring '11
 Staff
 Machine Design, Shear, Stress

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