HW4_Soln - AE 322 HW #4 Solutions Q 1. The components of a...

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Unformatted text preview: AE 322 HW #4 Solutions Q 1. The components of a displacement field are (in meters): ux = ( x2 + 20) 10-4 , u y = 2 yz 10-3 , uz = ( z 2 - xy ) 10-3 , (a) Consider two points (2, 5, 7) and (3, 8, 9) in the undeformed configuration. Find the change in distance between these points. (b) Compute the components of the infinitesimal strain tensor. (c) Compute the components of the rotation tensor. (d) Does this displacement field satisfy compatibility? Soln: Distance between points in undeformed configuration is, = = ( + - ) + ( - ) +( - ) = 3.74165 Coordinates of points in the deformed configuration can be evaluated using the displacement field as, = + = 5 + (2 5 7) 10 = 5.07 = 2 + (2 + 20) 10 = 2.0024 = = = + + + = 7 + (7 - 2 5) 10 = 7.039 = 8 + (2 8 9) 10 = 8.144 = 3 + (3 + 20) 10 = 3.0029 = + - = 9 + (9 - 3 8) 10 = 9.057 ) + ( - ) +( - ) = 3.8108 Distance between points in deformed configuration is, D= ( Hence change in distance between points is given by D-d = 0.069222 Infinitesimal strain tensor, The linear strain displacement relations for small displacement scenario are, = = 2 10 =2 =2 = = =2 = = + + = = 2 10 + = 2 10 = 2 10 =0- - 10 =0+0 10 =- = (2 - ) 10 10 Hence, the infinitesimal strain tensor for the given displacement field 0.2 0 - /2 0 2 (2 - )/2 10 = - /2 (2 - )/2 2 =- =- = = 1 2 - - Rotation tensor: =- = Hence the infinitesimal strain tensor for the given displacement field is, 1 2 1 2 - = 1 = (2 + ) 10 2 1 (- - 0) 10 2 1 = (0 - 0) = 0 2 = 0 - 10 2 0 0 1 - (2 + ) 10 2 0 1 (2 + ) 10 2 0 2 10 Compatibility condition + = 0+0= 0 + + - - + - + + = = 0+0=0 =2 0+0 =0 (0 + 0 - 0) = 0 (0 - 0 + 0) = 0 (-0 + 0 + 0) = 0 =2 =2 + Q2 . For steel E = 207 GPa and = 0.3. Assume that the strain in a ( x1 , x2 , x3 ) frame at a given point is 0 0.004 0.001 ij = 0.006 0.004 sym. -0.001 . r Find the normal and shear components of the traction Tn acting at that point on a surface with unit normal (1/3, 1/3, -1/3). Solution: Give E =207 GPa and = 0.3 = = 2(1 + ) + = 79.615 e= = (1 + )(1 - 2 ) + = 0.009 = 119.42 = = = +2 +2 +2 = 1.7117 = 2.030192 =0.915577 = = = 0.159231 = =0 =0.636923 0.159231 2.030192 0.636923 0 0.636923 0.915577 1.7117 [ ] = 0.159231 0 1.7117 = 0.159231 0 = Normal stress Shear stress = 0.159231 2.030192 0.636923 1/3 1.08018 0 0.636923 1/3 = 0.89633 -0.16088 0.915577 -1/3 =1.08018 + 0.89633 - 0.16088 = 1.23402 | | - = 0.687946 Q3. Find the axial force diagram N(x) vs. x for the system shown. Gravity is the only load acting. The weight density of the bar is . Cross-sectional areas are given by the Ai , i = 1, 2 ,3 . Soln: 4. Draw the shear and moment diagrams, Vz vs. x and M y vs. x , using the sign conventions specified in class for both cantilevered (shown) and simply-supported boundary conditions. Neglect gravity. a. b. Solution: Cantilevered boundary condition Hence shear force and bending moment diagram can be constructed as, Simply supported end conditions Reactions at the support From the free body diagram in Fig. (b), the reactions at the support can be calculated: - Force Balance + - Moment Balance + = - = + = + - + 2 =0 2 + 2 3 - 3 =0 = 1- - 6 R1 R1 (a) 0<x<a From the free body diagram (a), imposing force and momentum balance. + = + 2 - =0 + ( )= = = 2 (b) a<x <L 6 2 + + 2 - 3 1- 1- + - 6 - 6 =0 From the free body diagram (a), imposing force and momentum balance. + = - 2 + 2 - =0 ( )= = 2 + - + 2 - 3 6 + - =0 = SFD and BMD- SS-BC 6 - + 6 + 2 + 1- - 6 6 - s 1- - 6 a L 6 - 2 1- Q4b Uniformly distributed load Cantilevered boundary condition = Hence BMD and SFD are constructed as below. + + Q4 b. Simply supported end conditions Reactions at the support From the free body diagram in Fig. (b), the reactions at the support can be calculated: - Force Balance + = + - 2 - Moment Balance + = + + 3 3 + 2 4 - 8 + =0 = - 8 = - (a) 0<x<L/4 From the free body diagram (a), imposing force and momentum balance. + + = =- = - = - 8 - =0 + =0 8 + (b) L/4<x<L/2 From the free body diagram (a), imposing force and momentum balance. + + = = =- - - 8 =0 + + =0 (c) L/2<x<L = - 8 - + = =- = - 8 + + + + + + - - - 5 - 2 - =0 2 8 2 + 4 + - 2 =0 + = = - 2 + SFD and BMD + L/4 L/2 L x 8 0 - 8 1 4 - 8 5 - 8 - 1 3 4 + 8 ...
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This note was uploaded on 04/12/2010 for the course AE 353 taught by Professor Voulgaris,p during the Spring '08 term at University of Illinois, Urbana Champaign.

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