HW9 - 8'1' calmlmé Ixand 1" for cross sections...

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Unformatted text preview: 8'1' calmlmé Ixand 1" for cross sections (a) through (I) in Figure 8.19. Given: 1'; = 4000 psi (27.6 MPa). normal-weight concrete 1;. = 60.000 psi (414 MPa) = 29 x 10“ psi (200.000 MPa) . ld—SOIn. I762 mml——'-l Sin. e .5' 5 L L s ' an .— e -0. 1' h In g m ” N g a No. 10 '- eg - 4 No.9 1 3 l a T—l /..\ ‘ 14 in ' ' 3 in. (355.6 mm) (d) {H Figure 8.19 Beam cross-sections for deflection calculations. In E.._-.-. 51000517 = 5100050006 = am It)" “fab ‘ "5.0L I0“ ‘5 1‘3: 1L1 L113 I'2. 22,5104 ' in” 121 V ‘ i din. d . 2019.8 3 5 s. 4No.B o 0 EL Sin. 1m. 21. In. [533.4 mm] 21 In. In cf + m mm, .. 15.x woman: 0 , 0. = 3.3 m. 2 I“: M3533 + m Lama-$531 = mac in“ 3 a 13: mm? = 415m 3n“ . \7. 50.‘ + m Lune. - m (.3950an =0 ) 0,: 1.0 m. 1“: wow + m Lambs-1th 3m» in“ 3 3 on :6: 10003 = mum m" ‘2. 5e} 43.x US$03 + 1.\ koml c. - m Hiatus? -'I.\ comm): o 0.: 5.1 \n Ie“: + N L\$1\1.8~5:D‘ +13 Lemma .. 2531 = 2) 51-1 mu 5 ' . d3 13-.- 12utf = qlzm m“ l2 ugh [3.x mm +1.1 US$31 o. - m mum mu. Lnsfitznth: b C: '3 bug . 9. . 1;m = \2 ms? +$.\L$J®L\%-US\Z+‘L\ (mm [9.8 «15.03 = mum m“ 3 c) ‘5: mJLQLwfi 4: mum?) = ‘M ‘m. 50 Lab + LIQU‘D . 1'22 8.2. Calculate the maximum immediate and longoterm deflection for a 6-in.-thick slab on simple sup- ports spanning over 13 ft. The service dead and live loads are 70 psi (33.5 kPa) and 120 135‘ (57-45 kPn}. respectively. The reinforcement consists of No. 5 bars (Ié-mm diameter) at 6 in. center to center ([54 mm center to center). Also check which limitations. if any. need to be placed on its usage. Assume that 60% of the live load is sustained over a 30-month period. Given: f;'=-4500 psi (31 MPa) f} = 60.000 psi (414 MPa) E. = 29 >5 10“ psi (200.000 MPa) Glitleg 1 h: (m (math 9. = 15 9t Dead Load 2 10 1%!» (mi including set? migh’h LNG. Load = 12.0 19639, A$3 = 110.5 bats @ tom e-e. .___ otuz‘m‘ ea: 51,000 W = ammo" cat = 2‘? no" test. ' n:- 1.33 a; mm = 503 we. 03 mini mum Reggth mm -, hm“ = 3302) =- 156 in > hag: L0 in 2.0 b3 Etheiive. Moment oi? Ineriio. L19: I3: M3 2 21b in“ la 3+; =-. .919. = ‘35 m 2 . Hm: 505a munch = 6L0,225 tn—tb 3.0 124 “To “\nd devi'h OV'neu‘rfm 0.1%: d: 5.0 .A = 0.192 m2 (0&2: 1.3% Lawns-c3 ‘3 0.: ms in 3 . In: MUM-Q +1.3?) Lem) Um «ma: = 10.1 in“ 3 Mn: “mu/.2 gmucmz m: 50,183 ‘m~\b ‘5 ML: 19.00531 1 :2 2 some 'm—vo Hg! 2. BL? = Ha, Big 15% Id: 10.1 +quszm-1ofl =. 2m in“ FDY "f (.9070 : fiwM— = om HoL 33,153+ on £30,141») 3;: ._- 10.1 + Low? Lsz 10.13 s “2.6 m“ Fm “D.L. -\- L.L : __Hu '4 SUIZZB : 0.5% H“; 3b,]53+ 60,420 _. Le: 10.1 1- £0510" Cum—10.13:- qmp in“ 125 ‘isno‘m' TERM beam-rams: -. A: 5H0} sg I21.me = 0.0007 _H_ m '43 51c uttsmmo") L It IHHEDiakc L.L. 'beHec’Hon: AL: 0.0007 (9141?, — 00001 $188 =05?) in 45.0 211.1 In». median 13.1.. bemecfion: AB: 0000': L at» 13% = o.I2 m 21H Immedtak [9070 L.L. Semecfimz AR: 0.0001 55mg)-on : 0.22 m. 112.5 L000 -TE 2H “02‘: FLEGTIDID : ALT: 0.85 0940.12)» IJSLOJfi: \.0\ in. benecfion Regain men’r‘é : 123: I502.) = 0551 in > AL B0 .YBD 3; = 0343 ‘m > 5' .300 W) [26 k = om: in '4 ALT 240 The use. 0? ‘Yh‘xe. 3MB is Limi’fid #0 flows or «>on rm mepomnfl or afiqcnec; +0 nun—Mg 0mm e\emen‘ca. mom (1% amid-Hem. I2? ...
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HW9 - 8'1' calmlmé Ixand 1" for cross sections...

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