HW_CH4_set2

HW_CH4_set2 - Shear function V(x = R 1< x 0> w< x...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 4-23a Statement: A beam is supported and loaded as shown in Figure P4-11a. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data giv in row a from Table P4-2. 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 4-23A Distance to extreme fiber c 2.0010 2- ⋅ m ⋅ = Free Body Diagram for Problem 4-23 Solution: See Figures 4-23 and Mathcad file P0423a. 1. The reactions, maximum shear and maximum moment were all found in Problem 3-23a. 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 4-23a FIGURE 4-23aB 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 ....
View Full Document

This note was uploaded on 07/23/2011 for the course EML 4501 taught by Professor Staff during the Spring '11 term at UNF.

Page1 / 12

HW_CH4_set2 - Shear function V(x = R 1< x 0> w< x...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online