Question

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time Q1:- Calculate the instantaneous and Long- deflection (time limit of 5 years of a no ft span simply supported beams (n=8). fr=uksi, fy=boksi (50%) w =) кі 10= 12к Mp3 2 Kik Po=104 2.5 $ 2#jo Зор Jolt 22.5 Олония ## VoITAllMAXB MARTIN 3.5"

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 on Instantan cous Sinst due to udl. w z 2+1 3+lr Sinst 5 wi" 5x/3x1000) (Dia.)": nu (udy 38u Ele 384x 3604996.53178038.775 2.66in Sinst due to Point load W- 12+10 .220. Maximum deflection occurs at 'C' where 5 b (La) C- * 30x (40+10) = 22:36ofti Sirst Wach (22x103) 16 x (10X12_)\n X (20-36 x2 13:03 (PL) BLEle. 3x (10x12_)in X3604996.53 Psx1803829.99 0.5uin. Total Instantaneous / short term Neflection Siner Sinst + = 2.66 +0.54 3.2011 Singt (PL) (udl) Longterm deflection Singterm Sinst A Sint t multiplier from &=& where §= 2.0 for 5 years lhig of code Complementary 1750e' 2.54 1:20 é: As' Td 17 0.00 705. 1767x000905) lung 16, 22.5 ® Stotala 3:20 + (1-448x3-20) = 7.quin Slong term -7.quin