Question

A wood beam supports the loads shown. The cross-sectional dimensions of the beam are shown in the second figure. Assume LAB=2.8 m, LBC=1.1 m, LCD=1.6 m, w=12 kN/m, P=6.8 kN, b1=20 mm, b2=75 mm, d1=100 mm, and dz=240 mm. Determine the magnitude of: (a) the maximum horizontal shear stress Tmax in the beam. (b) the maximum tension bending stress max (and location x) in the beam. - X BI ec LAB I LBCI LCDJ bil b2 bil Answers: kPa. (a) Imax = (b) O max = kPa, at x =

Answer 1

Date Page 12 kN/m P= 6.8kN - 2.8m * lilmk l. for equilibrium * Ema se o og beam EMATD & Efy = 0 2.8+1+1+) - g (3.9) + (123 2. 3) thay (6. 8x5.5+ 12x2.8x loy) y - 2165128 kW Efy o Ay tly - ( 12x2.8) - 608 20 Ay = 12x2, 8 + 6.8 - 21065 128 Ay = 18.74872 IN so shear force in section AB I MA 12 kn/m , Ery zo a le TRX - - Ay 18.74892-122 F = 2.8 m - so v=18.749 at x=0 V= 18,74872 - 12/2.8) = -14.85128 KN. & vo 17 18.74872 = 122 = x= 1.56239 m CV ALSO EM zo & M - Ayx + (122) 20 7 M = -6x² + 18.74872 a. so At a = 1,56239 m - ME - 61.51279)² + 18.7487281.56239) 2 M= 31.12042 knom

Section BC A bibkn & Utly-pe K Em 3 r = 6.8 - 21,65722 22 15 128 IN 3 IV = -14,85728 KN 15.5-24) x 7-M + ly [sis-a-1.6) - Pls.s-x) =0 & M = 21.65128/3,9)- 21.15128 x - 6,8x5.5+ 618x & M = 47.04 - 14.85122x] M=28= 47,04 - 14,85128128) = 5.4564kw-m.

Shear force diagram: زنجلا 18.749 Isom -1485 Bending Mousent AMLkN-m) Diagram 31.12 x=1.56239 m -10.89 So Vmax = 18.749 KN max 31.12842 knom.

Page 1 75 x 2403 20x10o² 2 12 12 X2 2youm Iz = 8 9 7 33 333. 333 minst lot 75 20K max = N 31, 12042 x 10 x 10 x 120 89733333.333 max = 41.672 Mpa Tong = 41617.2 Ha la=156239 m a. section There are two criticy at yoo & y = 50 mm 52 % А ч , 2 k Tmax= VAT I b T18.74872) (2x20x 50 x 25 + 75x120x8o) to 25+ Enax TSX120x8o) tlo = 1174,031fa 89733333.333 x los Al it y=50mm emax2 = 18,74872 | 70x75x8526 89733333.333 x 75 Tmax = 1243. 18 Kla