Question

A simple beam AB of span length L 6.7 m (Fig. 5-14a) supports a uniform load of intensity q 22 kN/m and a concentrated load P 50 kN. The uni- form load includes an allowance for the weight of the beam. The concen trated load acts at a point 2.5 m from the left-hand end of the beam. The beam is constructed of glued laminated wood and has a cross section of width b 220 mm and a height h 700 mm (Fig. 5-14b) (a) Determine the maximum tensile and compressive stresses in the beam due to bending. (b) If load q is unchanged, find the maximum permissible value of load P if the allowable normal stress in tension and compression is о, 313 MPа. P = 50 kN 2.5 m q=22 kN/m B L 6.7 m (a) h 700 mm b 220 mm (b)

Answer 1

JIJI COO TORNA karmt Ke RA+PB = 50+ 22x617 (RA+ RB = Pt 6.79 D RA+RB - 197.4 - taking moment about A (MY)A=0 RB X 6.7 = 50x215 +22X6.7 X 6.7 JRB = 92.356) TRA= 104.64 KN] RB X 6 7 = Px2.5+ 9 x 6.72 RB = 0.3731 p + 3,35 a wa TRA= P + 6.79 -0.373 p - 2359 = 0,627 P + 3.359 at x (SP)xx = -(0.37310 + 3.359) + ax maximum (BM) (SF) = 0 9 x = 0.373P + 3.35a x = 10.373p+3.359 1

MC for PESOKN, = 22 10.373 XED + 3.35X22) 22 x = x 4.197 4.2 maximum moment is actat. that pour from Riser is left = 6.7-4.2 M = RAX215 - 0x2152 = (25) = 104.64 X2.5 - 22x215? 2 M = 192.85kN-m/ Given 7 350 700mm I = 603 = 12 = 220x7003 I = 62.888108 mm 12 I 2207 maximum tension at top) L=3500 mg & comp at bottom/y=350mm) we know that. M = of my = 192.85 X 106(N-mm) X 350 62.88 X 108 7 = 10.734 Mpa loc= 10.734 Mpa

>6) i H3 M = I= 134 62.88 2100 350 M = 233.55 KN-m we know that. Ma RAX215- M= RAX2I5- 9x2.5² M = (0.627P + 3.35422) X215-22725² a 2 233155= 118.05 = 0.627 PX2.5 + 115.5 0.62782.58 P p= 75.31 TP = 75 KN An.