Question

Prob. 9-63 094- 19-64. The shown. Determi that act perpendic The grains at The frame supports the triangular distributed load Determine the normal and shear stresses at point D berpendicular and parallel, respectively, to the grains. ains at this point make an angle of 350 with the horizontal as shown. 9-67. in-plan orienta 9-65. The to shown. Determi The frame supports the triangular distributed load n. Determine the normal and shear stresses at point E tact perpendicular and parallel, respectively, to the grains. e grains at this point make an angle of 45° with the Пе horizontal as shown. 900 N/m 507 75 mm 1200 mm 2.4 m 0.6 m 100 mm 121 3 m inte tors E 45° – 50 mm 50 mm poin 30 mm 1.5 m 100 mm Probs. 9-64/65

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pro.9-64

Answer 1

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free Body Diagrams goo N/m L/90 NE

forces horizontally (for PRO O Resolving EF=0 - VE=0 living forces vertrelly C for frp BY-NE =0 ->> 0 Now let us consider moment about point RRD a 0, & take the 2 Me-o X3%900 X2-By (3)=0 1350-BBy=0 - By=40N Put this value in egr

--0-N80 - NE=450N - Area A (100x103) XC 50x103). 0-00 m2 A Here 6x = okog stress in a doreaften 1. NE - A 690 0.005 --golpg @ Normal stress (Gal) Gala Gatsy) & (64-by) eseo Thysn20 = o-got of (2x35t o bel= -29.6090 kPa

7 Sheer stress - Tely1 = ( 6x-sy) sin 20 - Dreyes 20 . 20-(-90) sin(9x29) - 0 Txly = 42.2861 kpq