hw6_s04_sln

hw6_s04_sln - EMA 405, Spring 2004 Solutions for HW #6 1.1....

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EMA 405, Spring 2004 Solutions for HW #6 1.1. Introduction In general, the deflection of a thin beam with an off-center load is a combination of bending in two directions and twist. The relatively small dimension of the thickness of each web in this problem makes it challenging for modeling via 3D finite elements, but this approach provides the maximum amount of information possible if one can obtained converged results. 1.2. Preliminary Analysis The approach for an off-center load acting on a straight beam is described in Section 8.14 of Roark. Since the applied load in this problem is directed parallel to a principal axis, we can consider the behavior to be a linear combination of the same (vertical) force passing through the flexural center and torque applied about the flexural center. The magnitude the torque is the magnitude of the force multiplied by the moment arm, which is 2.5 cm because the pressure distribution is uniform over a region from 0 cm to 5 cm from a line that is directly above the flexural center. Thus, T =(20 kN)(0.025 m)=500 Nm. The location of the cross-section centroid and the area moment of inertia for bending about the x -axis (vertical bending when the beam is aligned with the z axis) can be obtained from the information provided in Roark’s Table A.1, entry 4. for a Tee section with t = t w =0.01 m and b = d =0.1 m. m 0325 . 0 ) ( 2 ) 2 ( 2 = + + + = d t tb d t td bt y w c (vertical distance from the centroid to the top surface) 4 6 2 3 3 m 10 3542 . 2 ) ( ) ( 3 ) ( 3 × = + + = c w x y t d A t b d t d b I Although this cantilevered beam has a distributed load, it is over a sufficiently small region that we lose little accuracy by considering the vertical bending as if the entire load were concentrated at the end of the beam. The predicted vertical deflection for the beam is then m 0142 . 0 ) m 10 Pa)(2.3542 10 2 ( 3 m) kN)(1 20 ( 3 4 6 - 11 3 3 = × × = x a EI L F v y We can also expect the peak bending stress to occur at the bottom of the vertical web, since that location is furthest from the centroid. At z =0 (the wall end), we have MPa -658.4 ) ( ) ( = = = h y I L F y I M c x a x zz y σ 1/10
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To predict the angle of twist at the end of the beam, we will need the effective polar moment of inertia, K , from Roark’s Table 10.1, entry 23 with a = c =0.1 m, b = d = t = t 1 =0.01 m, and r =0. b d b D cd K ab K D K K K 4 15 . 0 0105 . 0 3 1 021 . 0 3 1 2 2 3 2 3 1 2 1 + = = + + = α Some of the above relations have been simplified based on the parameters for this problem. The net K is 6.7179 × 10 -8 m 4 . With G = E/ 2(1+ ν )=76.92 GPa, the expected angle of twist is rad 0968 . 0 = = KG Tl θ considering the torque to be applied at the end of the beam, l =1 m. Finally, since the force is directed vertically, there is no horizontal deflection of the centroid. 1.3.
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hw6_s04_sln - EMA 405, Spring 2004 Solutions for HW #6 1.1....

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