Question

Long-time Q1:- Calculate the instantaneous and deflection (time limit of 5 years) of a no pt span Simply supported beam (n=8), fé-u ksi, fyz boksi (50%) 205 W2 = 1x18t WO = 2k/ft 12 = 12 * Po=104 2 #10 22-5 3oft post 6#) 3.5 t 16"

Answer 1

bh² Gross second (19) moment of Area 12 b = 16" h= 26" 1 16x263 Ig Ĉ 23434.667in 12 Cracking moment (Mer] frig _(Aci eq.9.8) yr fro 7.55 fcl 7.5*4ooo 470-34 psi.; Yra Mara 47u3ux 23434-667 = 13in 3 855079.89 Abris - 855.079 Krin = 71.256 kft. area method Depth of neutral anis (n) by transformed n As Cd-x) = (n-1) As'Ce-d') + bxxxx As: 6x1.567 9.36 in. 5 6 #11 As': C2#107 2X 1.27 = 2.5uinr. no Es -8.; Ecz 57000f> 3604996.53 psi Ec 8x9.36 (22--15-1) = 7x2:5ux (2-2,5) + 16xxxn 쓸 1684-8-74.581 > 17.78x - 44.45+ 8nr. En +92.66% 1729.25-0. By solving abore. quadratic equation so, depth of neutral anis Calla 10.0lin x= 10.olin. Area for- br3 Cracking Second moment of bux + (n-1) A' CX-d') z nas (d-wit E 16x 10-013 + 16x10.01%(19.01) + 7x2-54R (10601 -2-05) 7 8x 936x 12 12 (225-10.01)

=18033.43 inu Actual moment of the beam (Ma) PL=124 'Polloc 3klft tom manchan B. 30' tota UDARA - (3 xuorud 2)+(22830) RA 676.50 k. RA - (3xuo) + 22] -76.50 RB = 65.506 where sf is Zero maximum moment O clur nr. 65.50 3 21.833 ft RB -(3xx) =0. M max RBX21.833 -( 3x 21.833 21.823" = 715-0u16 Koft Effective second moment of Area Wes 3 [.- (mer fer le - Montag big isourc) 171.256 - 3 X23434.667 7 71.256 [n- en ons 122* 715.0u16 715.0u16 18033.43 23.1915 t 18015.583 7 Je = 18038-775 inu.

Short term deflection corn Instantaneous deflection . Since udl. sustained load due to ws: Wouw ws 2+1, 3+++ Sirst . الوں 5x13x1000) * (40x12)"inu ludy) 38u Ele 384x 3604996.53x18038.775 2.66in Sinst due to Point load wo PO+PL WE 12+10=221. Maximum deflection occurs at c- ☺ where ca 35(ltag Şx 30* (u0+10) - 22.-36#*. posts : (22x103) Jbx (10x12_)\n X (22.36 x2) in 40 - Sirst (PL) Wach 3 LEIE. 3x(U0x12)in x3604996.53 PS)x18038-73 225 0.5uin. Instantaneous / short term Total deflection Sinar Sinsrt Ludl) = 2:66 +0.54 3.2017 Sinst (PL) Longterm deflection Singterm Sinst 1 Sinist + 5 years [mul tiplier from 2= & 17500 e: As' bd where §= 2.0 for where p'e steel ratio in compression 2.54 l=20 code Complementary . 0.00 705 THEO XUODAS) El.vas 16, 22.5 Stotal a 3.20 + (1-448x3-20) - 7.quin Slong term = 7.quin