Assignment5_Solutions

# Assignment5_Solutions - 252 Solutions Manual S.20.2 From...

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Unformatted text preview: 252 Solutions Manual S.20.2 From Eq. (20.6) qs = - Sy Ixx n Br yr r=1 Sy 7 8 50 mm 50 mm 1 2 150 mm 200 mm 3 150 mm xS 4 S 6 5 40 mm 40 mm 80 mm 80 mm Fig. S.20.2(a) where Ixx = 4 2.0 802 + 2 200 502 + 2 200 402 i.e. Ixx = 8.04 106 mm4 Then qs = -1.86 10-4 from which q12 = -1.86 10-4 200 (-50) = 1.86 N/mm q43 = -1.86 10-4 200 (-40) = 1.49 N/mm q32 = 1.49 - 1.86 10-4 250 (-80) = 5.21 N/mm q27 = 1.86 + 5.21 - 1.86 10-4 250(-80) = 10.79 N/mm. The remaining shear flow distribution follows from symmetry; the complete distribution is shown in Fig. S.20.2(b). Taking moments about the mid-point of web 27 Sy xS = 2(q12 150 80 - q32 200 80 - q43 150 80 - q43 40 200) which gives xS = -122 mm (i.e. to the left of web 27) n B r yr r=1 Solutions to Chapter 20 Problems 1.86 253 7 5.21 6 Sy 1.49 5 8 10.79 xS 1 1.86 4 5.21 3 1.49 All shear flows in N/mm 2 Fig. S.20.2(b) ...
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## This homework help was uploaded on 04/07/2008 for the course MAE 157 taught by Professor Valdevit during the Winter '08 term at UC Irvine.

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