Chapter 10 - CHAPTER 11} ~ PLATE GERBERS 10.4"]: Check...

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Unformatted text preview: CHAPTER 11} ~ PLATE GERBERS 10.4"]: Check web width-thickness ratio: wflmii_ _ MKM— ” ______29=000 : a- Iw _ 3/8 420, flip—saéfiy W3.!6 50 90.55 _ _E_ x 290‘0‘0 fl 1,—5.701Fy 5.70J—50 137.3 Since AP < it < Ar, web is noncompact. AISC Section F4 appiies, but Section F5 may be used. Compute the section moduius: 2 2 I. m Jun/a3 +2.44%? ) = {Ea/8x45? +2(10)(-452ftl ) = 13,430 111.4 Ix 13,430 a (11/2 + 2})" " "(45/2 +1) From AISC Equation F 5-10, the tension flange strength based on yielding is M» :- Fnyt 2- 50671.5) = 2. 858 x 10“ in.«kips m 2382 ft-kips = 571.51n.3 The compression flange strength is given by AISC Equation F557: Mn : Rngchxc where the criticai stress Fer is based on either flange local buckling or yielding. For flange locai buckling, the relevant sienderness parameters are film: 10 m = L: 129000 5 J1. th 2mm 5, 1p 0.38 Fy 0.38 50 9.152 Since A < 13,, there is no flange local buckling. The compression flange strength is therefore based on yielding, and F c, =- F y = 50 ksi. To compute the bending strength reduction factor Rpg, the value of aw will be needed: 3 at... a 45(3/8) bfil‘fi 10(l.0) From AISC Equation F5-6, __ m aw“ £1“ i R” "‘1 1200+300a.. (a, 57,1113. )51'0 ... _..m1_.£8_8_w-m _ [29000 = "1 1200+300(1.688)(120 5'7 50 l 1‘0” Since 1.017 >11), use Rpg 1.0 Mn = Rngchxc m 1.0(50)(571.5) = 2. 858 x 10“ in.«kips m 2382 ft-kips 21.688 <10 aw no-1] Because the girder has continuous lateral support, lateral-“torsional buckling does not have to be checked M” = 2300 fi—kips 10.4«3 Check web width-thickness ratio: u 12 W 60 2. m E‘_ 29,000 w 2 _ may _ _3/8 -160, 2., — 3.76 lmFy __ 3.76 IWSO _ 90. 55 _ {L m [M a 2, m 5.70 Fy 5.70 50 137. 3 Since A > A” web is slender and AISC Section F5 applies. Compute the section modulus: 2 2 n = 712—20123 +2.44%) m “fie/exec? +2(—’§;~ x12)(60:w8) = 2‘621X104lflf' alsmmLm 26210 = .3 Sx C (II/2+9") (60/2+7/8) 848.9m. From AISC Equation F540, the tension flange strength based on yielding is Mn m Fnyr m 50(848.9) : 4. 245 x 104 in.-kips = 3538 ft-kips The compression flange strength is given by AISC Equation F5—7: M,1 = 16ng “SM where the critical stress Far is based on either flange local buckling or yielding. For flange local buckling, the relevant slenderness parameters are a 5 gamma 2 “25;, = 29000 2 2 2” 20/8) 6.857, 2p 0.38 Fy 0.38/ 50 9.752 Since 3. < M, there is no flange local buckling. The compression flange strength is therefore based on yieiding, and Far = F), = 50 ksi. To compute the bending strength reduction factor R p , the value of aw will be needed: awzflméwnzd43<10 bfctfi- 12018) From AISC Equation F543, _. “may. __._ mile“ "E" RPVI 1200+300aw(fw 5‘7 ijsl‘o e ——_~2¢‘$3—n_ _ 29000 g I i200+300(2.143)(16O 5-7") 50 j 09736 Mn = Rngchxc = 0.9736(50)(848.9) = 4. 132 x 104 in.-kips m 3443 ft—kips [10mm Check lateral—torsional buckling. 7/8” —W12"~——+ ..L 3/8"»H*- (not to scalc) if. ; £9. = ‘ g _1_ 3 “L 3 = - 4 6 6 10:11., I 12(10)(3/8) «3 12(7/8x32) 125.0111. ... _ - 2 m 2: x 126 = - A .~ 10(3/8)+12(7/8) _ 14. 2s 1n. , r; a /—«-«~----~--~——14n25 2. 974 m. Lb = 40/2 :. 20 ft LP: 1.1:“; (—135— =1.1(2.974) 21%!)- $78. 79 in.=6.566ft }’ L. a m, 0%? = “(2.974) 6% = 268. 9 £11.: 22.40 ft Since L1,, < Lb < L, the girder is subject to inelastic lateral—torsional buckling. From AISC Equation F5-3, ' F” = C5[Fy—~O.3Fy(?—:LL—E~)] s Fy 7' P :1.3o[50«(0.3 x 50)(%)] = 43. 46 ksi s 50 ksi where Cg, a l. 30 is from Figure 5.15 in the textbook. LTB has the lowest critical stress and controls. M... 2 Rngchxc : 0.9736(48.46)(848.9) : 4. 005 X 104 in.—kips : 3338 fi—kips ' Mn 3 3340 ft—kips [104] 10.4»5 Check web Width~thickness ratio: 2: ’1 mfl=156, 9113:3715 Exam w :9055 “:17 1/2 F, 50 2 i I 29000 g 2. 5.70/5; 5.70/ 50 137.3 Since it > L, web is slender, and AISC Section F5 applies. Compute the section modulus: 2 2 1.. a itwh?’ +2.4,(WL) = ll—Zn/zx'zsw +2(2.5 >< 22)(m—EZ¥§—) = l. 980 x 10501.4 I 1 1. 980 x :05 . x w —x— m min—mm = m— 2: 4 ‘3 S c (0/2 + 5:) (78/2 + 2. 5) m m From AISC Equatiou F540, the tension flange strength based on yielding is Mn = Fny, 2 50(4771) u 2. 386 x 105 in.~kips = 19,880 ft—kips The compression flange strength is given by AISC Equation F 5W7: Mn = 1?ng as)“: where the critical stress F c, is based on either flange local buckling or yielding. For flange local buckling, the relevant slendemess parameters are 25322: a _E_.= 29000: l 21? 2&5) 4.4, 2,, 0.381% 0.381 50 9.152 Since 11 < zip, there is no flange local buckling. The compression flange strength is therefore based on yielding, and F N = F), = 50 ksi. To compute the bending strength reduction factor Rpg, the value of aw will be needed: _ her... 78(1/2) bfirf. 2205) From AISC Equation F5—6, _ fl aw 11;, .E. RPS ‘1 1200+300aw(1w 5'71153, )5 1'0 = _._ 0. 091 ( _ 29000): 1 1200+300(0.7091) 156 5-7 50 0-9906 Mn m Rngchxc = 0.9906(50)(4771) m 2. 363 x 105 in.-kips = 19,690 ft~kips = 0.7091 <10 Because the girder has continuous lateral support, lateral-torsional buckling does not have to be checked. Compression flange buckling controls, and Mn = 19, 690 ftwkips. (a) LRFD solution 051%,, = 0.9009690) = 17,700 ft—kips {1M1 wt,=1,2wD 1- 116m, $1.2(1.0)+1.6(2): 4. 4kips/ft PM a 1.613;, a 1.6(475) 2 760.0 kips —i LEW}. 2 Mu m SwuL 1' 4 8(4.4)(30) + m 18,700 ftukips > 17,700 ft—kips (N.G.) Flexural stmngth is not adequate 760(80) 4 (b) ASD solution A431, m 19690 2 a - Qb — L67 11,800ftkips Wu : w0+wL =1.0+2.0 : 3kips/f‘t Pa 2 PL 2 475 kips : 11,900 ft~kips > 11,800 ft~kips CNS.) Fiexurai strength is not adequate kvE m 10(29000) _ 1.37/ Fy -1.37/W50 _104.3 Since i > CV _ z = ir‘w H (h/tw)2Fy (140)2(50) Tension field action cannot be used in an end panel: V,2 m 0.6FyAva r»— 0.6(50)(0.5 >< 70)(0.4468) = 469. 1 kips Vfl x469 kips 2 (b) % 7.. 2%1 22.857 < and<3 (am)2 {2. 857)? [10-5] H WA” 5,6135%9009). £62.76 y 0 Since {3” > 62. ’76, tension field action can be used, and 1_C Vn = Aw CV + Mummm V ' y [ 1.15 1+(am)2 ) ComputoC'v. 1.37 fig 21.37/W =78.17 F), 50 Since ll. > 7&11’ C” x m 2 0,2503 Iw (Mr...) F). (140) (50) 1— 0.2508 - V. a 0.6(50)(o.5 x 70) 0.2508 + ——-——~—~.--a__-::- = 489. 3 klps [ 1.1s.l1+(2.857)2 ) 17,. = 4.89 kips (o) If no intermediate stiffeners are used, % > 3, and [CV = 5 (no tension field permitted) 9. {on _ /5<29.000) _ 1.10 Fy —l.10 50 —5 24 Since il— > 59.24, V“ : 0.6FyAva. Compute C1,. tw L828. _ /5<29,000) __ 1.37 Fy ——l.37 50 ~73.78 mom .. I-5109000X5 =0.2234 Since it > 73.78, C. = 2 m 2 w (2/3...) F), (140) (50) V” = 0.6824va : 0.6(50)(0.5 x 70)(0.2234) = 234. (Skips V” m 235 kips 10.5-3 Before developing the LRFD and ASD solutions, compute the nominal shear strength of each panel. 11. 2 W66 :— Iw 5/16 211. 2 Endpanel: a = 6(12)+2 = 78 111., "g = gag— =1.121< 3 2 2 260 3 260 m .2 (film) (211.2) “16> h [10-6] 1665+ 5 =5+--—5— =3.979 (6/6)2 (1.121)2 [161,13 = 8979129000) w 1.10 Fy 1.10 50 ~79.38 I kvE _ 8.979(290002 _ 1.37 Fy m 1.37 50 — 98. 87 Since it- > 93.37, Cv = = = (11763 w (MW) 63. (211.2) (50) Tension field action cannot be used in an end panel: V" : 0.653,}!va m 0.6(50)(5/16 x 66)(0.1763) m 109. I Rips Second panel: a =12(12)+9—74 = 79111. 2 6:12,: 260 h 66 1.197<[Mw) and<3 k.=5+ 5 as m~i———~=8.490 (6/6)2 + (1.197)'~’ [£6111 _ 1849909000) _ 1.10 Fy WIJO ——-————-—««50 —77.19 Since —h— > 77. 19, tension field action can be used, and I‘w V” 2 0.51.361“, Cv+.__......_l_“:_C_v.___ 1.15 1+(a/h)2 Compute C... 1.37 ’“E =1.37/W39-90—)— =96.14 1/ 1:3. 50 Since J1— > 96.14, C. — M - 151(29000) 8'490) = 0.1667 1.. '" (lg/mfg _ (211.3260) ,. = 0.6(50)(6/16 x 66) 0.1667 + ——1—"9-—-—1657——-~ a 390. 6 kips 1.15 1~+~(1.197)2 Middle panel: a = 55(12) — 2(12 x 12 —1— 9) = 354111., wfi- a 3—53? = 5. 364 > 3 16,, = 5 and tension—fiefd action cannot be used. 66 1.10 I‘VE 61.10 599000 =59.26 1/13, 1/ so [10-7] 1.37 I‘VE $1.37 “29000 273.78 F. 50 Compute C... 1.511%. z W 2 9.817x10“2 : "Ki/MW. (211.3260) Vn m 0.6FyAva : 0.6(50)(5/16 x 66)(0.09817) = 60. ’74 kips (a) LRFD soiution End panel: Design strength 245.14. m 0.90(109. l) = 98.2 kips w“ = 1.21423 +1.6w; =1.2(0.225)+1.6(2.0): 3. 470 kips/ft Since 74— > 73.78, C. W Left reaction: V], m “720 55) = 95. 43 Rips < 93.21am (OK) Second panel: Design strength = $1.17,. = 0.90690. 6) m 352 kips V1. at beginning ofpanel : VL ~— wux a 95.43 —~ 3. 470(6. 167) = 74.0 kips < 352 kips (OK) Middle panel: Design strength a reel/n 2 0.90(60.74) m 54.7 kips V“ at beginning ofpanei = V1, — wax ~—"" 95.43 - 3.470(12. '75) a 51.2 kips < 54.7 kips (OK) Girder has enough shear strength (b) ASD solution End panel: Allowable strength: V" 2 109 1 = 65.3 kips .0. 1.6"? wa = 1429 +wL = 0.225 +2.0 2 2. 225 kips/f’t Left reaction —.= V; = a 61.19 kips < 65.3 kips (OK) Second panel: Allowable strength = g" = = 234 kips Va at beginning ofpanel = P]; — wax = 61.19 — 2. 225(6. 167) = 47.5 kips < 234 kips (OK) Middle panel: Aiiowable strength = g" z = 36.4 kips Va at beginning ofpanel : V1, — wax = 61.19 — 2. 22502.75) = 32. 8 kips < 36.4 kips (OK) Girder has enough shear strength [10»3] 10.6—1 Bearing strength: Apb = (6 — 0.5)(1/2) x 2 = 5. 5 in? Rn = 1.8FyApg, m 1.8(36)(5.5) m 356. 4kips Compressive strength: Use a length of web equal to 25m : 25(5/16) 2 7. 813 in. Compute the radius of gyration about an axis along the middle ofthe web: I = —1-1+2-(7.813)(5/16)3 + 2[1—12-(1/2)(6)3 + some + 5732?] = 77. 79 in.4 A = 7.813(5/16) +2(6)(l/2) = 8. 442 in.2 _ “L = [77.79 m - r— J; “mu-H.442 3.036 m. - .. flwm_0.75(56)_ The slenderncss rat1013 r —— r — W3‘036 - 13. 83 <25 Pn = FyAg = 36(8442) w 303. 9 kips (a) LRFD solutiOn Bearing strength 2 M6,, = 0. 75856.4) : 267 kips Compressive strength = 6P" 2 0. 90(303.9) = 274 kips Bearing controls: Maximum factored concentrated load = 267 kips (b) ASD solution - _ Rn m 356.4 _ - Bearmgstrength— Q m 2.00 M178k1ps - _ "131 __ 303.9 = - Compresswe strength— Q m 1.67 1821:1105 Bearing controls: Maximum service concentrated load m 178 kips [10-9] ...
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This note was uploaded on 10/06/2010 for the course CE 406 taught by Professor Drake during the Spring '10 term at Cal Poly Pomona.

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Chapter 10 - CHAPTER 11} ~ PLATE GERBERS 10.4"]: Check...

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