Question

Problem 6: The beam is made of Douglas fir having an allowable bending stress of allow = 1.1 ksi and an allowable shear stress of Tallow = 0.70 ksi. Determine the width b if the height h = 26. (Show your SFD & BMD) 800 lb 80 lb/ft 800 lb -3 ft -+ -6 ft + 3 ft h = 2b b

Answer 1

Answer :- allowell ksi Zallow h=26 o toksi 800 lb 80e6lft 800 lb B A 3ff 64€ 3ft Ro Re RB+Re=800x2+80x6=2080 eb from symmetry Ro=re so Ro = Re=2080 1040db 2 Section AB :- 800ll Vx AV V shear force My Bending moment -V-800 co Voc =-800 e6 VA= Vo(left) = -800 lb Mx (5) + 800x (O)=0 M+ 800X-O Mx - 800X MA= MB=Mx fx = 3613= -800x3=-2400 16-fl O Section Bcn 18oodb 80lb/ft Vx 3 36€ M - Yx - 800+ Re 80 xx = 0 Vac = 1040-800-80x Vas 240-80X Volvight) = 240-0=240.86 Ro Vx=0 240-80x = 0 x = 3ft fxs foom points Mc (3)+800 (3+x) () + 1040x (C) + 80x2 )=0

Mc = - 40x2+1040x 2400 8oox Mx = -240 +240X - 40x2 Mx Sx= 36t? = -2400t Quox 3 - 40 x 32 =-2040 lb ft (111) Section co: Simelar section to coin 800.26 A of 374 С -80 lbs Figure: SFD A 3 A - 2840.lb-ft -2400db-ff "E-2400 lb-ft Figure: Bra

NA neutral axis INA = 6h3 6x (26)3 264 12 12 [h=26 NA for Bending stress: Mmax Omex max 6 INA Momax = 2400 lb-ft=24000x 12 = 16. in Smax h = 6 max = Fallow = 1.lksi -) 1.1 % 103 -2400x 12 6 2/64 b= 0.2942 in (berding = 0.2942.ch For shear stress in Zmax Vmax Ay 1.5 Vmax 16 th Vmax=80006 Zmax= o. 7oksi 2) 6.7ox 103= 1.5X 800 6826 6=0.9258 in = 0.9258 in 6 = (6) sheer maximum of of (b) bending, 6=0.9258 in Answer 6 6) shear?