Question

time - Calculate the instantaneous and Long- deflection (time limit of 5 years of a no ft span Simply supported beam (n=8). fc = Hksi, fy= boksi W2 = 1x18t Wo=2k1ft 12 = 12k 2.5 # Po=16* 2#10 3oft (22.5 Klopp 6#) 13.5 th

urgent

Answer 1

8.1 9.36 in b = 16 in d = 3.5 in d = 22.5 in de = 2.5 in As 6 x 1.56 A = 2 x 1.27 = 2.54 in fi'= UKS toksi n=8 W, = 1 klft P = 12K Wo ty = 2 k 2 klft Po = 10k

Instant aneous dead load deflection (Ap). Ig 1/2 bh² 3 x 16 x (22.5 +3.5) 12 = 28 435 in fo Ig Mes = y f = 7.5ndf Assume aal ya h = 13 in 2 3 10 Mapa х 7.541 X 4000 x 2 3435 13 = 71.26 Koft 12 Maz No [m a=30 ft b = 10 ft Po xarb 8 L 2x Gou 10x30x10 + 8 40 475 koft = Now using transformed aree method calentate 16 in (2n-1)AS 2.5in I NA d=22.5 in 13.sin. nas

4 + 5 Wp L PL (32_45) 384 & te 48 & Te 10600 x 40X12 (31408123-4%(104123] 38483-605 X 106x1912858 48 x 3.605 x 106 x 1912 8.8 2000 x (40X1232 12 peof fo.918 = 2.588 in. Instantaneous deflection for dead of live loads (foth). (W, +W₂ ) { (P + PL) x axb M 8. L 3 x 400 22 x 30x10 40 8 = 765 koff ana> 7 Mer I 19093.4 1 2x 71.26 (n-1 su se 19093.4 234 35 3:4) 34 765 19107

ADAL 5(000+602) (Pot PL) L ( 32% - 46) 48 € Ie 384 Ę Ie 5x 3000 x(40X1) 384 x 3.605 x 106 x 19107 12 22000 x 40x12(3x480481209) 48x3.605x 186 x 19107 = 2.508 + 2.024 = 4.532 in mitial deflection for full live load (4) A = ADAL = 4.532 - 2.588 = 1.944 in Long ferm deflection for dead & live load 5 years time limit. (ALT). & n 2. 1 + 508 9 1 + 508 2. -3 1 +50 +7.06 x 10 bo 2.54 16x22.5 = 7.06x 10 3 = 1148 2 1 + 50x7.06 X103 1.48 sun ALT A + 24 + a syg. Ase = 1.944+ 1.48x2.588 to =deflection for sustained load = 5.274 in.