Question

Qli- Calculate the instantaneous and long-time deflection (time limit of 5 years of a no ft span Simply supported beam (n=8), fc-uksi, fyz boksi (50%) W2 = 1 x 1 lt 12 = 12k 2.5 WD2K/ft Po=104 2 #10 3oft 22.5 posta 6#1 3.5" t 16 +

Answer 1

Solulion Given data, m=8; ہ = 4 ksi fy= 60 = boksi Gross second moment of area (Ig) bh² here, b= 16" 12 ; h=26" Ig 16 x 262 = 23434.667 in 12 Cracking moment (mer) :- Mera for Ig -lespor Ac eq YE eq9-8) fo= 75ffc = 7.5.14000 = 474.34 psi Y = hla 26 وا 13 in. Mar 474.34 X23434.667 = 13 855079.84 ab in = 855.079 k-in Mora 71.256 k-ft

D Pepth of neutral axis by transformed corea method n As(d-x) = (n-1) As' (x-d') + b*x* & here As=6#1 =) 6x1.56 = 9-36 in Asia &#107 2 x 1.27 gruin no Es =8 2 & Ec=5700 offel E B = 3604996.53 pi i. 8x9.36 *(22.5-x) = 7x2.54(X-25 ) + 16? 16x ~ 1684.8 -74.88x = 17.78% -44.45 + 8x' 8+ 920668.-17 29.25 =0 by Edving above can ; we get a value as 2= 10:01in; i depth of neutral axis (a) = 10.olin Gorror ti

3 Cracking, Second moment of Aurea; I cr bx² + bx ($)+(2-) As' (2-d') +nas(d-2) 12 3 16 x(10.01) +1666-01) (1009)* +7 (2:54) (10-01-25) 2 + 8x9.36 ( 22.5 -10-01) Isor 18033.43 iny Actual moment of the beam (mali. & 3k/ft PL=12k Po= lok n ma to BIA 30' - >k 10' let us Calculde moment at A YO (RA) 2 / 3 x 40 x 40 ) + 11 x49 ) + (22x30) RA 76.50 K * * a Rog Efy=0; RB+ RA (3440) + 22 RB = 120 +22-76.50

RB= = 65.50k. * maximum moment occurs which shear force 30; RB-(3 xx)=0 65.so =121.833 ft 3 :. m max = R8 (21-833) – (3*21.823") Mmar - 715.0416 kaft Effective Second moment of area (Ie):- 3 Te = mer Ig + [1 photo ] ma Ico 71.256 3 715.0416 ca ) x 23434.667 + 171.256 3 715.0416) X 18033.43 = 23.195 + 18015-583 Ie = 18038.775 in 4

( ? Instantaneous a) Shrew term deflection:-(Sinst) ; Einst due to uniformy distributed load, w=2+1=3k/ft Sintludl) wdy 5. 384 Х Ete 5x 5x(3x 1000 12 (40X 12)“Ciny X 384 x 3604 996.53X180 38.775 2.66 in Sinst due to point load: W=10+12= 22 k; maximum deflection occurs at c ; b (La) where X 30 ( 40+10) C= 22.36 ft.

6 Sinst War² (22) 3LEIC (22x103) 1b * (10x 12)10 x * (lox 12)in x (22-36 X12) 3x(40X12 Kin)x 3604996-53(psi) x 18038.795 in) 3 en V 0.54 in 2. Total instantaneous / shat time deflection:- Sinst finst finst (PL) Cudej + 9. 66 + 0.54 * Siost 3.20 in Long term deflection Slong term Sinst + X Sinst . है 1 +50 el . where &=2 for syears [Multiplica from fig of code complementary e'- As' bd 2.54 16x22.5 0:00705

G 2 = 1.478 1+50*0.00705) ng term = 3.20 + (1-478 X 3.20) Siong Siong term = 794in * *