Solution6 - AE 321 Solution to Practice Problems Chapter 6:...

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AE 321 – Solution to Practice Problems Chapter 6: Extension, Bending, and Torsion 1. (a) Tractions on the cylindrical surface . Since b a >> , we can use the St. Venant’s principle, which states that stresses and strains far from the point of application of an external force are not significantly altered if the applied force is replaced by a statically equivalent load (i.e. same net resultant). Therefore an equivalent uniform distributed load can substitute the linearly distributed load at the ends of the bar. Graphically The solution of the statically equivalent problem is obtained by assuming no body forces and taking the components of the stress tensor as follows ! yy = P 2 b , and xx = zz = xy = xz = yz = 0 It can be easily showed that the applied tractions (see solution of problem 1 in homework set 4), equilibrium equations, stress-strain law and compatibility equations (nonzero stress component is constant) are satisfied by the assumed components of stress tensor. The traction vector on the cylindrical surface is obtained from T i ! n ( ) = ij n j . The unit normal vector of the cylindrical surface (see figure above) is given by ! n = cos ˆ e x + sin ˆ e y . Therefore the traction on the cylindrical surface is

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! T " n ( ) = 0 0 0 0 P 2 b 0 0 0 0 ! " # # # # \$ % & & & & cos sin 0 ! " # # # \$ % & & & = P 2 b sin ˆ e y (b) 2D displacement field in the bar. In this case we are dealing with a 2D plane-stress problem, then the stress is specified only by σ xx , yy , and xy and are functions of x and y only. The displacements u and v , along the x - and y -direction, respectively, are determined from the strains components obtained after using the constitutive equations for a linearly elastic homogeneous isotropic material. Then ! xx = " u " x = 1 E # xx \$ %# yy [ ] = \$ % P 2 Eb & u x , y ( ) = \$ P 2 Eb x + u 0 y ( ) yy = " v " y = 1 E yy \$ xx [ ] = P 2 Eb & v x , y ( ) = P 2 Eb y + v 0 x ( ) (5) and xy = 1 2 " u " y + " v " x # \$ % % & ( ( = 2 μ ) xy = 0 (6) substituting (5) in (6) ! u ! y + ! v ! x = du 0 y ( ) dy f y ( ) ! " # + dv 0 x ( ) dx g x ( ) ! " # = 0 " du 0 dy = # dv 0 dx = C constant ( ) \$ u 0 = Cy and v 0 = # Cx substituting this in Equations (5) we obtain u = ! " P 2 Eb x + Cy and v = P 2 Eb y ! Cx Now we need to determine the value of C . That can be done by noticing that, because of the loading of our equivalent problem, there is not vertical displacement at plane 0 = y , i.e.
v x ,0 ( ) = 0 = P 2 Eb 0 ( ) ! Cx " C = 0 Therefore u = ! " P 2 Eb x and v = P 2 Eb y The shape of the cylindrical surface after deformation is obtained by remembering that the position of any point on the cylindrical surface after deformation can be obtained by X = x + u = 1 ! vP 2 Eb " # \$ % & x , (7) Y = y + v = 1 + P 2 Eb ! " # \$ % & y , (8) Before deformation, the equation of the cylindrical surface was x 2 + y 2 = b 2 ! " # \$ % & 2 (9) solving Equations (7) and (8) for x and y , and then substituting them in Equation (9) we obtain the equation for the deformed shape of the initially cylindrical surface X 2 1 ! P 2 Eb # \$ % & ( 2 + Y 2 1 + P 2 Eb # \$ % & ( 2 = b 2 # \$ % & ( 2 or X 2 b 2 ! P 4 E # \$ % & ( 2 + Y 2 b 2 + P 4 E # \$ % & ( 2 = 1 (10) Equation (10) represents an ellipse with axes having the following dimensions: b !

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This note was uploaded on 09/16/2009 for the course AE AE321 taught by Professor Lambros during the Spring '09 term at University of Illinois at Urbana–Champaign.

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Solution6 - AE 321 Solution to Practice Problems Chapter 6:...

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