HW6_solution - h3~2 a] A : f 2’ ~——>...

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Unformatted text preview: h3~2 a] A : f 2’ ~——> Afgf‘Juggflisfiiflr E Effijfiaflfi I : f U- ACWAQJI 3; 5;; . “M - h 1 ‘ ) f f ‘ I of; {3% : E“ E “'r'” I “.7, I x g. f?- ‘7 /w :2 if “if; V EA. .33. ‘ 5+ 5}! 7t vi}? : gfiaawfl: h:d+%é:m;+%%efigflg a; “:9 f5: i'm‘ .r y was 9 x: {K2 ‘ :m M *v "N 5‘” s2 I! _ \a ‘ gm umwj CB/gMM} :3. ma WW;ng "N m. x _ ' W: i N: = :1 “$132 A3,»! a“ :<’ Ex. «53;; HULL 9{&a.m.»€f;€, r“ ~; g {f 3 5 fin. h" :r .5 0‘ if} ‘5;- .. f 2:» ~- f (5.. 7 fl ' {1) A =’« 1: 2 F M; {D M) CW“ £3” ” e’ . fifflréfig Hé‘gf‘zgw‘iéi’ffi ‘ FD?“ {gigggi’ 3 .35 Ed- Q 56} I, J; , - PM" [fhmwzfigi ~"'§;§£€{~ ' i” '37:? 553 “ (“Ming-p gem mm; ail/w; :74 Ammféfl?‘ 7c)me S/éazsa-r Switf‘awfifl’gé ":3 é ‘ U 6 QC"; J : fiifl v 1 (“a t m 2 L14 ESE g M E :3; 3"? affix R :3» W 1‘3: C35”? M E 5 £1,» Spam-W g; 2 2/3 05 Z 3 5755“) 11"» IE 59.53 U. .1. £6 24f: ‘* w wax? m" f 5“:- 5522‘. 301%" EfiMf Wréfijf~fi§ _ 115% ' fig r3 2: {5%) {:3 w » (0, (am 2:} :32. L5 if @fii Toffltl fifg'fiirg “2:, :31“ ‘9 ( J7!“ “:1 3H j? j; wf V“ (’9 Mfr-{W ,c flip/{£5 89M. a}! W?!“ij _ fie/gfi W243 MW“ Ef'“ “J a} g .IWJ-H mam m fa;ka gig/arm ix?“ » ‘. i 1. j Nmm Md _ f; I f f9 3:21? ?. Jéalfi 2.25; \. 2.46m = 2.4(7/s)(3xs)(53) = 45. 68 kips) 39.96 kips use R" = 39.96 kips For the other bolts, Lc=s—h=3~~%~g— 22.063in. 11,, = 1.2.6.1161. = 1.2(2.063)(3/3)(53) = 53. 84 kips 2.4M. = 45.68 kips < 53.34 leaps use 12,, =' 45.68 kips ” For the cennectien, the nominal bearing strength is 39.96 1- 4015.68) = 222. 7 kips Bearing controls. RR = 222.7 kips. (a) LRFD Solution 612,, = 0.750227) = 167 kips Pu = 1.2D-1— 1.6L =1.2(40)+1.6(100)= 208 kips >167 kips (N.G.) connection does not have enough capacity. (1:) ASD Solution &_.222.i_ - Q — 2‘00 willklps Pa = n+1. .-= 40+ 100 = l40kips> 111 kips (N.G.) connectiOn does not have enough capacity. 7.6-1/ Bolt shear (assume that the threads are in shear): A6 = mil/4 == 7r(7/8)2/4 = 0.6013 in.2 Nominal shear capacity of one bolt is Rn = EMA), x 2 (for double sheet) 2 48(0. 6013) x 2 a 57. 72 kins/bolt For 8 bolts, 8(57.72) r- 461. 8 kips Check bearing 0n gusset plate (it is thinner than the combined thickness of the angles, the edge distance is the same as for the angles, and the ultimate tensile stress Fu is smaller): *1 _L=_1§_' h" s 7“ 16 16 "1' For the hole nearest the edge, _ J; _ __1§/_1_ .. - Lc—Le 2~2 2. —l.5311n. an = 1.2mm = 1.2(1.S3l)(S/8)(58) m 66. 60 kips [7'7] 5% The upper limit is 2.41th“ = 2.4(7/8)(5/8)(58) = 76. 13 kips > 66.60 kips use RN = 66.60 kips For the other bolts, Lash-M3“?— =2.063in. Rn = 1.2LctFu w 1.2(2.063)(S/8)(58) = 89.74 kips 2.4thu 22‘76. 13 Rips < 89.74 kips use R" = 76.13 klps For the connection, the bearing strength is 2(66.60) + 606.13) = 590. 0 kips Tension on the gross section: Ag m 20.13) m 14. 26 in.2 P” w FyAg = 5004.26) = 713.0 kips Netsection: =2 — z: .- i .1. .1. a ' 2 A. A... Etdh .2[7.13 2 3)(8+8)] 11.76m. .. -1: “3.1.2.: 0—1 a 1 9 0.8089 A; = AgUm 11.76(0. 8089) = 9. 513 in.2 Pa = FHA. = 650.513) == 618. 3 kips (a) LRFD solution. Compute the design strength: For bolt shear, 45R” -—- 0.750161. 8) = 346. 4 kips For bearing, 4512,, m 0.756900) w 442. 5 Rips For tension on the gross section, 915,119.. = 0.90(713. 0) = 641.7 kips For tension on the net section, $119,, = 0.7503183) = 463. 7 kips Bolt shear controls. Pu w 346.4 kips. Since D > 8L, load combination 1 controls. 1.41): 1.48.51.) 2 346.4 2:» L=29.11kips P=D+L=¢=8.5(29.11)+29.11=277kips P=277kigs (b) ASD solution. Compute the allowable strength: EL _ 461.8 = - Forboltshear, Q — 2'00 230.9kips For bearing, «€3.21 = = 295.0kips R. _ M = 426. Qldps For tension on the gross section — ’ Q 1.6 ’ 7 to. -. tts. t to... W mm W m. ms mas-mt . jg; w $541.1 4% as stem“ .0me a ~23: 17-3} éfifslwm MHw—fiébw. .Xr‘fl...1£%«€......Berti-g E w \ y H i ‘ 3 ,g; i z .t"/ § .,,«-/‘ a /%F Zf/m Wm aw W0 {2m 55 £2 an a? ,. $555" flfi - imé" :5 [(15.9 a? Wm Mfi? My; A; bfié I w fingl - = wWWflmE “ fifffiflgié. 3A3. am? 4@ W : {iii i L g3 Rm : g0~5§fiM {/3125/ 1L M53 1F? 72%? jfliéjfifi.iégv f~fijj- Efl‘:%fif f p a'”““ k m K. ? { @g‘>x{gg)figggigjifig i3}- 5525; 5?} “'3? f—é’ig‘b j . 1%!“st -' :.W%m W0 ° (5 W (65%? ii. 35‘ Q?) 4:. e: 3“ Z 2; Y J . 2 5"- .3: WW? “3 W“ K m L3 Lg" W525» M5643 x ! Mgwgm F: , m 63 gas m ""35 E Jr” s; - Egg} fiéé 4 f” y 17) rw- .9} ’ j? ‘ E, % 1: ’fi“ Mfg? fight i‘ g? ‘ ‘ 2% 1 m if awn”; mm a "3‘ Zr“ :3» M mnflvmmwfimwm m “WM” ‘ @‘w‘g E LEM “2* E?" é \ ,- ‘ 5b ‘* «.1 ‘- “1; ,v r 1: 5 1: r n l; "- EM“, ‘ki if ; {w}? é | g ‘" ~ (if if? 53 k E2: 1% (4) Bolt Single-shear Slip-critical design Single-shear Siip-critical allowable Diameter design strength strength, one slip plane ailowable strength strength, one slip plane ‘ (in) (kips) (kips) (Rips) (REES) 1/2 7.069 4.746 4.712 3.164 5/8 11.04 7.515 7.363 5.010 3/4 15.90 11.07 10.60 7.383 7/8 21.65 15.42 14.43 10.28 1 28.27 20.17 18.35 13.45 1 1/8 35.78 22.15 23.86 14.77 1 1/4 44.13 28.08 29.45 18.72 1 373 53.46 33.62 35.64 22.41 1 1/2 63.62 40.74 42.41 27.16 (b) Shear never controls in a slip-criticai connection; the slip-critical strength is always smaller. ...
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This note was uploaded on 12/07/2011 for the course CES 4605 taught by Professor Prevatt during the Fall '11 term at University of Florida.

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HW6_solution - h3~2 a] A : f 2’ ~——&amp;gt;...

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