hw2_s04_sln

# hw2_s04_sln - EMA 405 Spring 2004 Solutions for HW#2 1.1...

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EMA 405, Spring 2004 Solutions for HW #2 1.1. Introduction Here we consider a simply supported beam with a linearly varying transverse body force. From the notes, we know that the second derivative of moment is equivalent to the transverse force per unit length, so we would expect transverse displacement (still two more integrals) to be a fifth-order polynomial. The problem therefore helps us see the behavior of beam elements in an application where they do not reproduce the exact solution at all locations along the beam. 1.2 Preliminary Analysis Although we can use tabulated information to determine the deflection of each end of the beam, it is straightforward to use basic methods and integrating the beam equation. Summing the moments at the right side, gives the reaction force at the left side () = + L l dx x q L F 0 0 Writing with q >0 meaning an upward force (here, q’ =-2 × 10 () x q x q = 4 N/m 2 ), we have () 0 6 2 V L q F l = = where V is the shear that satisfies x q dx M d dx dV = = 2 2 . Integrating once and using V (0) gives the distribution of shear. () = 2 6 2 2 x L q x V Integrating again and using the fact that neither end has an applied moment gives the distribution of moment. () () ( ) x L x L x x q x M 2 3 2 2 .) 3333 ( 6 = Nm. We can then integrate the beam equation, M dx v d EI = 2 2 , to get the transverse displacement. The two constants of integration are determined by the boundary conditions associated with the simply supported ends, v (0)= v ( L )=0. 1/16

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() () + = = + = 360 7 12 24 therefore and, 360 7 36 120 4 2 2 4 4 3 2 5 L x L x EI q x dx dv x L x L x EI q x v θ At the two ends, angle of deflection evaluates to (0)=-5.732 × 10 -3 rad and (5 m)=6.551 × 10 -3 rad. For beam elements, the work-equivalent load of a transverse body force amounts to a force and a moment on each node. For a single beam element, the work-equivalent forces will be counteracted by the reaction forces from the supports, so we only need the work-equivalent moments. Using information from the nodes and from Cook, Chapter 2, ()() ()() 20 30 2 3 0 2 4 3 0 3 0 2 4 3 2 0 L q L x L x dx q x q x dxN M L q L x L x x dx q x q x dxN M L L J L L I J I = + = = = + = = For the given magnitude of q’ , the work-equivalent moments are M I = -8.333 × 10 4 Nm and M J = 1.25 × 10 5 Nm. 1.3 Computation Description The ANSYS computation can be run with a single BEAM3 element with the material properties and real constants specified in the problem. Vertical displacement is constrained to be zero at both ends, representing the simple supports, and horizontal displacement is constrained at one end so that rigid-body translation is avoided. The linearly varying transverse body force is
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## This note was uploaded on 07/29/2008 for the course EMA 405 taught by Professor Witt during the Spring '04 term at University of Wisconsin.

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hw2_s04_sln - EMA 405 Spring 2004 Solutions for HW#2 1.1...

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