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by: Jaylan Rath


Jaylan Rath
Texas A&M
GPA 3.77

Lee Lowery

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Lee Lowery
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This 183 page Class Notes was uploaded by Jaylan Rath on Wednesday October 21, 2015. The Class Notes belongs to CVEN 446 at Texas A&M University taught by Lee Lowery in Fall. Since its upload, it has received 92 views. For similar materials see /class/226110/cven-446-texas-a-m-university in Civil Engineering at Texas A&M University.

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Date Created: 10/21/15
CLACE S 33gt LSfQEO 71 1 E Aw vwwwma FIGURE 742 422 Chapter 7 FIGURE 744 Solution LRFD Solution Answer l 4 5 Simple Cnnectio s PL12x4 ifl WW was VA an E A36 steel M The base metal in this connection is A36 steel so E7 OXX electrodes will be used N o restrictions on connection length have been stipulated so weld length will not be lim ited and the smallest permissible size will be used 2 Del 47 4 Minimum size 3 in 16 AISC Table J24 Try a 316in fillet weld using E70XX electrodes The design strength per inch is 1392 x 3 sixteenths 4176 kipsin The shear yield strength of the base metal is ER 1 Milli 0613 0636 81 kipsin V and the shear rupture strength of the base metal is 04511 seeingj 9788 kipsin V The weld strength of 4176 kips in governs The factored load is P 12D 16L 126 1618 36 kips Vquot and Required length 3 6 2 862 m Mquot V 4176 3 1 6 075 in lt 862 in OK Minimum length 4w 2 4 Use two 45inlong side welds for a total length of 2 X 45 9 in For this type of connection the length of the side welds must be at least equal to the transverse dis efwwjj tance between them or 4 inches in this case The provided length of 45 inches will it i4 therefore be adequate Emit Use a 316 inch fillet weld with E70XX electrodes with a total length of 9 inches as shown in Figure 745 444 Chapter 8 Eccentric Connections FIGURE 83 Pm p b replacement is made the load will be concentric and each fastener can be assumed to resist an equal share of the load given by pc Pn where n is the number of fas teners The fastener forces resulting from the couple can be found by considering the shearing stress in the fasteners to be the result of torsion of a cross section made up of the cross sectional areas of the fasteners If such an assumption is made the shear ing stress in each fastener can be found from thetorsion formula mi Sszfvgirl fz T where d 2 distance from the centroid of the area to the point where the stress is beingg A computed 39 J polar moment of inertia of the area about the centroid 445 82 Eccentric Bolted Connections Shear Only and the shear force in each fastener caused by the couple is pmAfvA Mdz 2 A2d 2d The two components of shear force thus determined can be added vectorially to ob tain the resultant force p as shown in Figure 83b where the lower right hand fastener is used as an example When the largest resultant is determined the fastener size is selected so as to resist this force The critical fastener cannot always be found by in spection and sgyerglwfgrcewg a lcrtlations may be necessary 39 quot more BonkE i trtts work with rectangular components of forces For each fastener the horizontal and vertical components of force resulting from di rect shear are 6 t mt where Px and Py are the x and ycomponents of the total connection load P as shown in Figure 84 The horizontal and vertical components caused by the eccentricity can be found as follows In terms of the x and ycoordinates of the centers of the fastener areas 6 2 7 2 z Mai 2612 2062 y2 can 392 gas 39 d we 93 where the origin of the coordinate system is at the centroid o the total fastener shear area The x component of pm is 13 dz gm 3 P a X a itrtateatepax quot5 and Pay 53 x i Mfrtmg a ee 11 E Md 2139 d 202 yz 4Q Mmy mmiwjjgg Pm 2 3 Pnt 71 Similarly p Mx my 2x2y2 V l and the total fastener force is plzarcmr V FlGURE 84 446 Solution FIGURE 85 Determine the critical fastener force in the bracket connection shown in Figure 85 The centroid of the fastener group can be found by using a horizontal axis through the lower row and applying the principal of moments 20 7 y 25 28 211 6 in v 8 The horizontal and vertical components of the load are 1 V 2 P502236kislt and 504472kis x 5 p P J5 p Referring to Figure 86a we can compute the moment of the load about the centroid M 447212 275 223604 6 4807 in kips 39 clockwise y t ital 3 e Z Z 32 3 3H 3 5 W RM 3 512quot 82 Eccentric Bolted Connections Shear Only 447 FIGURE 86 P 22 336quot 14 r M a W gem i is at a Figure 86b shows the directions of all component bolt forces and the relativ fmagni tudes of the components caused by the couple Using these directions and relative mag nitudes as a guide and bearing in mind that forces add by the parallelogram law we can conclude that the lower righthand fastener will have the largest resultant force The horizontal and vertical components of force in each bolt resulting from the concentric load are 2236 4472 pcx 2795 kips and p6 T 5590 kips t For the couple 2amp2 y2 82752 262 12 22 52 1925 in My 48076 4 was 2 1498 kips lt r Pm m a n a 202 yz 1925 s 57 a Mx 4807275 ire e 68671a t 9 p 208 yz 1925 1 3 3 W Ma a gs 0 3 2px 2 2795 1498 2 1778 kips lt 05 lyeW 5 2p 2 5590 6867 1246 kips l 51 a g fl yf C pl778212462 217 kips see Figure 860 t mmmWgwaxwymrmaMWwwpmmww v M 5 M f 63 MOMENT AMPLIFICATION 05 is The foregoing approach to the analysis of members subjected to both bending and axial load is satisfactory so long as the axial load is not too large The presence of the axial load produces secondary moments and unless the axial load is relatively small these ad ditional moments must be accounted for For an explanation refer to Figure 63 which Z 63 Moment Amplification FIGURE 63 nglwm g V w VAVKV Figure 64 shows a simply supported member with an axial load and an initial out of straightness This initial crookedness can be approximated by sin y e 39 0 L where e is the maximum initial displacement occurring at midspan For the coordi nate system shown the moment curvature relationship can be written as 5 31 M ii Wt d362 E M FIGURE 64 y Z 83amp 63 MomentAmplification M P ew iw ai max ue PePu1 t l puew M l Pam 1 a l A QP quot Wquot s WlK 53 l V ij K Pg 1 5 M0 5 1 4mm where M0 is the unamplified maximum moment In this case it results from initial crookedness but in general it can be the result of transverse loads or end moments The moment amplification factor is therefore 1 laPg 057 As we describe later the exact form of the AISC moment amplification factor can be slightly different from that shown in Expression 64 284 Solution As we describe later the exact form of the AISC moment ampli cation factor can be slightly different from that shown in Expression 67 i E 39 1 Fans Em What Use Expression 67 to compute the LRFD amplification factor for the beam column of Example 61 fCD EU Since the Euler load Pg is part of an amplification factor for a moment it must be com puted for the axis of bending which in this case is the x axis In terms of effective length the Euler load can be written as 72EI 752EIx n229000272 Pa X KL2 1fo 10gtlt17x122 1871 kips From the LRFD solution to Example 61 P 2004 kips and 1 1 l PuPe 1 20041871 l12 285 64 Braced versus Unbraced Frames which represents a 12 increase in bending moment The amplified primary LRFD moment is 112 gtltMu1121071z 120 ft kips Amplification factor 112 Answer 64 BRACED VERSUS UNBRACED FRAMES AISC Equation C2 1a 23 15 Zquot Mr Ble BZMU where 39 M required moment strength M for LRFD Ma for ASD FIGURE 65 r b 33 3 WWW WM 2 1 mt Ev Qt W W 286 Chapter 6 BeamColumns big 5 EMquot maximum moment assuming that no sidesway occurs whether the frame is 6 actually braced or not the subsoiipt nr is for no translation Mquot will be a factored load moment for LRFD and a service load moment for ASD W36 M3 maximum moment caused by sidesway the subscript Z is for lateral translation This moment can be caused by lateral loads or by unba lanced gravity loads Gravity load can produce sidesway if the frame is unsymmetrical or if the gravity loads are unsymmetrically placed M 6 will be zero if the frame is actually braced For LRFD M F Will be a factored load moment and for ASD it will be a service load moment B1 amplification factor for the moments occurring in the member when it is braced against sidesway Ba amplification factor for the moments resulting from sidesway We cover the evaluation of B 1 and 82 in the following sections t B use it gm as 1 Pitp MEMBERS IN BRACED FR ES 3 9 254 Z The amplification factor given by Expression 67 was derived for a member braced against sidesway that is one whose ends cannot translate with respect to each other Figure 66 shows a member of this type subjected to equal end moments producing singlemcurvamre bending bending that produces tension or compression on one side throughout the length of the member Maximum moment ampli cation occurs at the center where the de ection is largest For equal end moments the moment is con stant throughout the length of the member so the maximum primary moment also OC curs at the center Thus the maximum secondary moment and maximum primary moment are additive Even if the end moments are not equal as long as one is clock wise and the other is counterclockwise there will be single curvature bending and the maximum primary and secondary moments will occur near each other l W l Etta FIGURE 66 wk ModP5 65 Members in Braced Frames 287 HGUREBJ p 996 Mz WWII P M 2 M max M max g or M1 M0 P5 MmaxM2 MmaxgtM2 A B 2 1 AISC Equation 222 l a PrRel Bug Where N Pug Z P required axial compressive strength Pu for LRFD k qt PC for ASD Eglg lrwegfa 288 Chapter 6 Beam Columns ac LOO for LRFD 160 for ASD Pd EZEI W u g AISCEquationCZ Srlwz g K1102 The required compressive strength P has a contribution from the P A effect and is given by P Pm Bsz Where Put 2 axial load corresponding to the braced condition P4 axial load corresponding to the sidesway condition AISC Equation C2 1b No I 2 As an approximation P can be taken as PrPmPZt The moment of inertia I and the effective length factor K1 are for the axis of bending and KJ 10 unless a more accurate value is computed AISC C21b Note that Kati 5 i throughout Chapter C of the Specification the subscript 1 corresponds to the braced condition and the subscript 2 corresponds to the unbraced condition 65 Members in Braced Frames 289 FlGURE 69 N KN KN OI 01 v c u v Negative M 3 M2 Positive M 1M 2 BM W369 1 If there are no transverse loads acting on the member C cm W Cm 06 04 TJIL AISC Equation C24 2 M1 M2 is a ratio of the bending moments at the ends of the member M J is the end moment that is smaller in absolute value M2 is the larger and the ratio is pos itive for members bent in reverse curvature and negative for singlecurvature bending Figure 69 Reverse curvature a positive ratio occurs when M1 and M2 are both clockwise or both counterclockwise 8 2 For transversely loaded members Cm can be tahen as 10 more refined pro C cedure for transversely loaded members 13 provrded in Section C2 of the Corn mentary to the Specification The reduction factor is Cm 1 114051quot AISC Equation CC2 2 sg image ei For simply supported members 726091 11 2 MOL 1 where 50 is the maximum de ection resulting from transverse loading and M0 is the maximum moment between supports resulting from the transverse loads The factor I has been evaluated for several common situations and is given in Commentary Table cc21 its l w 33 3 STEEL COMPRESSION MEMBER SELECTION TABLES 2 0 06V 4917 Table 4 1 continued F 50 k Available Strength 1n I I I Y AXIaI Compressmn klpS W Shapes W12 Shapev W12gtlt thft 37 79 72 65 Design I 4an 0P 0Pll Fir90 I ban an f 1351 1353 1133 11163 L333 3 LRFD 0 1270 1150 quot 394 1040 951 kh E a E D a u g 12 930 337 307 E 13 953 362 734 13 14 924 336 761 3 3 15 395 309 736 E 6 3 2 5 a 1 E 5 E Pwkips 157 79013 133 Pwkips n 335 143 213 wakips 278 213 Lp 103 107 Lcft 466 430 399 374 Ag inZ 232 256 232 211 611104 333 740 662 597 Iy 104 270 241 216 195 300 309 307 305 304 atlo rzsjr 176 175 175 175 PexKL 1y04k1n2 23300 21200 13900 17100 f 1Lquot 104kin2 7730 6900 6180 5530 LRFD c 167 60 090 AMERICAN INSTITUTE OF STEEL CONSTRUCTION INC 65 Membersin Braced Frames E291 I r lame SQ WM 15 FF if C 06 04 06 04 m 09437 M2 824 c gal c 09437 Wm B 1023 Am 1 lamePal 1 000PuPel 174295405 Fat ne 39 as ex From the Beam Design Charts with Cb 10 and Lb 14 feet the moment strength is 0an 345 ftkips 3 l 21 For the actual value of Cb refer to the moment diagram of Figure 611 C 125me 25Mmax 3MA 4MB 3MC 125824 1 060 25824 3737 4766 3795 Q 4an C1 345 1060345 366 ftkips E 4 R116 But bMp 356 ftkips from the charts lt366 ftkips use an 356 ft kips 4 Mm 4 Since a W12 X 65 is noncompact for F y 50 ksi 356 ft kips is the design strength based g 6 on FLB rather than full yielding of the cross section The factored load moments are gig tame 5 Mn ft kiPS M91 0 I t From AISC Equation C21a the required moment strength is Mr Mquot B Mquot BZME 1023824 0 8430 ft kips Mm From Equation 63 AISC Equation Hl la Jig M My 06131 843OOO824lt10 OK CPII 9 anx ley 9 Answer The member is satisfactory K OZg MAXIMUM TOTAL UNIFORM LOAD TABLES q 1 0V 3 121 F 2m 39 Table 3 10 continued m 39 W Shapes Available Moment vs Unbraced Length IPII9II Available Momenthle 2 kip increments 42Mquot 3 kip Increments 16 18 20 increments 10 12 Unbraced Length AMERICAN INSTITUTE OF STEEL CONSTRUCTION INC FIGURE 53 CLASS 14 52 Bending Stress and the Plastic Moment 75 l 1 W Moment L I amp a k samp b l i ampamp R 39376 Chapter 5 Beams the same as the elastic neutral axis From equilibrium of forces CT ACFy AFy ACA WA 5 l Thus the plastic neutral axis divides the cross section into two equal areas FOE apes that are symmetrical about the axis of bending the elastic and plastic neutral axes are the same The plastic moment Mp is the resisting couple formed by the two equal and opposite forces or f i E M a FyltA a Fyi jai Fyz where A total crosssectional area A 2 W ytww F V 2 8 39E a 2 distance between the centroids of the two half areas Z a plastic section modulus At L EP b Jig bemm FlGURE 55 Plastic neutral axis f CV quot 4 KW x it Zzt lnwaii t iitza m 3 178 Chapter 5 Beams Rm 7 1M TABLE 52 Component A y Ay if Flange 8 6 5 52 q 2 5 5 4 Web 3 9 j Sum 11 61 FIGURE 57 I 1 81 4 39 i f g Centroid of 39 quot quot CD 37 half area y A E if 7 M y 2 gt L Answer f W1 Figure 57 shows that the moment arm of the internal resisting couple is a 237 26545 2 1109 in and that the plastic section modulus is m G 110109 12221113 The plastic moment is Mp FyZ 50122 6100 in kips 508 ft kips Z 122 in3 and Mp 508 ftkips Example 52 Solution Compute the plastic moment Mp for a W10 X 60 of A992 steel g 4 From the dimensions and properties tables in Part 1 of the Manual W 74 A 176 in D 2 5 W W A 88 in2 2 2 The centroid of the half area can be found in the tables for WT shapes which are cut from Wshapes The relevant shape here is the WTS x 30 and the distance from the outside face of the ange to the centroid is 0884 inch as shown in Figure 58 278a CVEN 446 CLASS NUMBER DATE PAGE was 39 r PM LCM V ri mw a og 6 WW Q 21 2 FT 3 lg L WEE m E AM AWEfng A Agmm m lt9 N3Z 5OKiz 02quot S 1235 amp4gt cg Qp OO MK 1 508i ltxgifd w nggiax I 99 i g A 390 g 55 7 WEE g x g 1 W z W 53 Stability 179 FIGURE 58 0884quot 39quot 39 w Amie h jg m d102quot I I i I w 55 lt2 i it usuacwtfzitw 2 H m Foot a d 20884 102 20884 8432 in Z 533 13 88432 7420 in3 This result compares favorably with the value oiven in the dimensions and properties tables the difference results from rounding of the tabular values Answer Mp 1532 500420 2 3710 in kips 309 ftkips 53 STABILITY T AL 1 1 A 1 6 54 Classification of Shapes l8 l 54 CLASSIFICATION or SHAPES Eat A l AISC classifies cross sectional shapes as Compact noncompact or slender depend gx ing on the values of the width thickness ratios For I shapes the ratio for the pro jecting ange an unsti ened element is pf25c and the ratio for the web a sti ened 1 element is hrw The classification of shapes is found in Section B4 of the Specifica My 39 gag i tion Local Buckling in Table B4l It can be summarized as follows Let IEquot Bi 3 width thickness ratio 5 Q a xlp upper limit or ompact category m quotW 2 A 2 upper limit for moncompact category WW a cngQArquia L lt9 o The s if ft S ft and the ange is continuously connected to the web the shape is compact if 2 lt ft S if the shape is noncompact and a Q if 7t gt 1 the shape is slender a mirw f mg w gM VTheacategory is based on the worstsvhlth thickness ratio of the cross section For ex i he web is compact and the ange is noncompact the shape is classified as pm W33 noncompact Table 53 has been extracted from AISC Table 341 and is specialized EM a was for hotrolled l shaped cross sections W Table 53 also applies to channels except that xi for the ange is bftf 353 i a 5 TABLE 53 Element VG u Q A ftp 2n g va Width Thickness 1 Parameters Flange bf 038 E 10 E I F 39 p 7 quotg i 22f y EQWaaagaxas a f it h E nigmajm 5 w I y i Vt a Dams Eff Eff i I A For hot rolige pzi isiiapes in flexure at lls gii wwe iiw CUEN HG SWZOO39 ip NW 1297Mquot TOPS PRWTING BRYAN TEXAS W 36 CM 1 g E 6 thu an MA Vu Wth 4 4 395 EELS 540 Fy39fa p 1 VP l 006Fy A l0065ol lxo35 lgg lu VL 06 13an 046G 9 gt 750 NU MW 9 Bo U86 R325 LSSUML walk ALMA M Shad Pl alignA H MD V 21549 12569 abo i 391 5 TV 54 Aazsmz 5T1 cm WK 5 CUP WMU L EMMA Use 9 bow 51mm c Wed 5 7 2254 224343 2 am 5 Le W1 7 Legtkea quot quotquotTZL6 REL 014 t L 5Nb4h2 L5 9M 2ejquot L 61quot 4 T3 72quot Ana1L Use 91 34 A iZB39boffs 3 wfarng quotO39 xf kph 13 quot0 Column l 0 TLK i CUM4 5 1 m7 7 D QYWIM AnaL Thiang 1 0 MI fu 5 VAgt 93405157410 75 s 07506Y5gt 534 egtt2gt EZ 2252 2230 Pg4W 5139 v 5 75 VA 0 0er AW 05434fwaz VP 7514 4 W am I 60W ESE AW 20260 1quotz316mL A quotquot75 2m ami VgWoz WW W W 2Wz Izltaxvayazsob 0555th 05 2 Miami Am Hum 071069 A30 Hi M3 4 01106mmzmmgww a a 424 W s 016on MM 376 4L 6 1 0536312ng quot w quot239 DEM 87 39quot 7 Vngq 04 0 39 4 11 air Wtwrsa llu 03185 Mm 15v mo wisme Wilma 0340 LC 64w 50 Wm Mquot kwmadan Law 4 0759H 01gt 0761L65 my 54 0346 07503 65x74 345 5159 MJQKIbaH wubok 172M 6 0213 MW gtu gt422 001 TOPS PRINTING BRYAN TEXAS TUL I cum445 5 3mm 2007 a Pm A39 0 L6 1 by 512gt39l z3 i lbgt 4061 WA 0715 I 711 59x mam 10250 4amp5 k4 WW4 07501 5 xzxogso akML W LC 248quot 1336 OJSUM ngL 2 8871210250 5 Sit60 q 1 57 l WM g 3 1 Tom M 25255 2m 6 Tow mum gtvw7aou om 1 Use Z L 4W4 x Vix 0 4quot EN Tsl hmrh ps TOPS PRINTING BRYAN TEXAS TLK i ClEIU446 KAS AampM ERSFI Y 223quot 31 Ea mz 29 83932 1 59 04 2007 Wm th Wk Design 0 Wireng WWW M1 Mu 990mm 03639 2y 050W23 5039sz 2 Na 2 0mm mom 752 Dzkwmw ak 1ng ff MU 352 34g t d 39 le Harm W46 Wz39cH k 79913 6439 I 7044 25049 H22 3014mm hmquot 2 104 304 Eddy Hum Wale Wickmg 172 g HA mg 52 Dime NSF H 47quot 7 52quot 04 Eeftm 523 was Memos 4 0532 a2 Ni 6 0756Y07g1593 K rk 439 3 am A3 13 4 0 taeb gt Assam 7035 104 cm H 175 Ae 0754ram A6 H 9156QOJSK ex 34 47K Calling 0 3 V b TLK 1 CUEU 4 1 3mm 2007 5g W 0 PM We WWW9 t mum 0555 34quot 65 0555 k L MMMIPS Mm V4quot 45 2 arsesquot ma 324 4mm 5499 06626quot To mm a max 9156534 Q5025 5 Bow me Ems A cM a w 7 t N 5amp9 WWW 5126 MM M one Pass l 4mm V4quot lt L 509 4 Max911 Lem H l3 8 9971019 stmebYa 0160e70c7o76 LQEGF H8 7 20quot Of 12 Far kad U86 5amp9quot 1939 Mr WeH 10 om eacks39fda 01C bo vm 1 9 PINE i M CWUL 6a WM Shaw le 152 MMELOCWRtL answea MmDOWN50m566i20036u16 6610565sz lt15sz mum 33036550 GWA 330 gt 1 6 8 01 1 1 PMxLe Dimksz ms a 1194 WM Q xo wsquot Bo vm 51w 8quot K L H quot ab wx s in Q41 befwum MM 344 pCMK M a artI144 dagwk TOPS PRINTE IG BRYANT TEYAS i5 1K marl4 395142007 Warm W46 49 obWt Flkuzuz WWWQI JD MM 0340quot34quot 7quot 5 HM ZZXtoe m W19 15 58 mu 3243 N0 4m 115 636 AMUS D jr mmz WeH S ze U ff39gtaiiae 10w LT0rF QM QJ F 84434 EWHM39 me 9 2 56 r725quot Lem 7 H i mam stmeOYYLw deer 39gt 4 2 04512 397504Y7o0707 5 5 Uee Qquot HM WM No Lg basermw uf ofmm flan13 MAM ag WeMs s Lof M sham jg lt10 m CWCK We WM SWMPMW TOPS PRINTENG BRYAN TEXAS TLK CVEuHH A 169mg 2007 W m FEXASMCM i l f2 L4 X4 f quotx Okqu Umvmzsm39 2 39 34 WIbM S M423 WVWW lquot t Fr L 39 rwmmbqu FL 55 my x 01 W All was EYOXX dedmdas mg 3mm CHE M3915 n m mb 393 m 39 77 z I a E Z Mvv a gt r 3 F35 A I 51 m aquot I at f 39 Th 4quot New dyau f same IQN 50 fears THAN t CVEN4449 l Spam 2007 1 i CEMZ Urk157QPFW F M 39W I w 3 7QW Wm 2a 4123quot Jaw 0200quot I 403 ff10425 C a r 9 ng 1717124253 MW 4w 2 0725 Team a 31 1 0575quot CELQMN gi imjtma Fquot DEED 142 w z w g TEXAS313534quot 4396 VVAquot h ll27541 UNIVERSITY m 5 J W vaA 600FY 1442 00616ot23lt92lt403 g ngw a CoMEUWLQNVmQ gmiiga I TQPS PRINTING BRYAN TEXAS VW T QJ WW 04561574 Vu 432quot Ddamm BOW SR6 ZEQLTE We 335 432 Mme NboH 2 K PTZZEJgdl AMONI 3 57D Vn2 gt Wm Ivesam 3 52quot 839 a g l wen LO 251 N3 QWch 1025mz6 35o02eo g iiij EL2 9239 7 W 432 00 Maui WULTWU NAz 3amp5 0240 t 5 Eiwaf 37065 W Engt9150qowzl 4N4 0 2W4f1 015IQHOWZEJOZHamp02 0260 5 quoxSox39o125 34 0amp25 Hi 200 421 UK I CVEN 4 i 3PM 2007 3 KAT42JK lt V 2K go QQE quot N 0 I W N 0 9 quot W A W Dan wmm hfacmg f Vammm 0 b i I kac 6 2 V21 NZ 7Q x M M lt2 6 21 2 2 225 mnwwwge e e 226 1376 0m 41562 7 4X45Zx076 WWW O x 39 Kb 561 7 938m 53 78 dose 0050 oi 84 L G 4 Lquot i 3783 4 2 Chan M 54am E WQ U Wc m sawg 5075 EA 9530515 2 3732 4 mm 20 39016 li lc5 076M39663976a575 39 En TWP 25 Xlto WK 7 V 4556 L NU LZ m W l iwg 9 75 MEANS is rw invt IabQ Fo evbo is A 7quotan MaulWM 39 Zaizg TUL l man4 I SW6 2007 Beam 3fo Wetdc f1 2 76 E mm 0 425u fagEn 3939 0426 am 6mquot wamzsquot mum 7ng g NO gradW you39re WJ S My H MW W M M 3 34 W42 39 L Sthlize 1 9 an MNWDZSI n I Nl 39 dls ned 36 I MM w W23 Ln Wm M612 s 4 2 59 73 q5 A325 Mrs 25 JNJ Le Hivequot gtlt 0 42 Aw TDPS PRINTING BRYAN TEX 3x8 L L S aA1W6 a 1 V at 2y H J 2 die444 43 AESC Requirements 107 With Equation 49 a direct solution for inelastic columns can be obtained avoiding the trialand error approach inherent in the use of the tangent modulus equation At the boundary between inelastic and elastic columns Equations 48 and 49 give the same value of F This occurs when KL r is approximately quOO 3 47147 56 ll winMM To summarize When 12 471 3 Er 0658FquotF Fy 410 r Fy When 55 gt 471 5 087717 M4 411 r Fy The AISC Speci cation provides for separating inelastic and elastic behavior based on either the value of KL r as in equations 410 and 411 or the value of Fe The limiting value of Fe can be derived as follows From AISC Equation 1334 KL zzE r F e L I WEBV39Z Forjjb gs47l 971 o ZZJBLLEquot witWMMEWUmeaZr 39 t 39 108 FIGURE 48 Chapter 4 Compression Members W O A Er L 3 392 MN 521512 Cu RV 5 KLr jet 240 Solution 6 A7W14 x 74 of A992 steel has a length of 0 feet and ned ends Compute the design compressive strength for LRFD and the Iowa compressive strength for ASD Slenderness ratio Maximum KL 1 20x12 248 r Ty 471 47129 5300 2113 7 V y 9 Since 9677 lt 113 use AISC Equation E3 2 r 9677 lt 200 OK 2 2 zwzmm 9677 9 KLr2 658Fquot Fy z 06158503056 50 2521 ksi mi W4 Z W14 XML 920 08 a ez 02fjg 61 2YES 305 M KL 6077 4200 ah f I i Y M ML jg 2 10 VW47I 4 4 35 I Saw KLMM q 77 4 54 3 Fm Oh i W pa Eb a EEO3amp5quot 50 2252 M 239 t Z 8l125 1 54g gquot PM 00615461493 MIMDK uv Mgquot 2 quot 5n IIIn39 39yH i My i lll H WH v a N v V thle II WMJ1F39 L 39 w 5 I I MW u Iquot H quot W 39 v H I m Jinan H M m w HM m n 39U W l I 39ll IlllH llllal ll m I 39 39 I l k H 1 39i M I 4 u m 7 In 3 1quot I 1 Jlul V WW1 Mil l39 u af lil m quot4quot 39 us H H quotr1 L Ll IV m n w In M lxl 0 211 J 1 Mn M W A quot2 IH mm auM 19 IL H39u39 339 x w m m H M v I quotWm uun mm II Wm I II I 39Im IMIIJIIIIWII um W W I W H j39 I III IIII n MN I I Iu I Li M w IIur IIII I J Ml I I h a H Iv H 1 x I M 4 III 39 m I 39 H Iw r H 3 U x 39l HMquot Hllf l39 quot 1 w W H I M39 1 H W H quot39l H u um II III u 4 II NH Ill MI h I J H39 39W I I I H CHI I I JI r I39 II 39 391 H I mM II I39 IIMIHHI I m Iquot I II I x m I H m HHH Ixquot II I II IIM I H I r quotI HI HII IIIIIH w quot 39 I39 39uquot m I quotNINTH Hn H HI HM H x lml I NHIWH Illhwlrl Irllir wr H 39 H N P If M w II I39IIIIMIIIIIIIHIIIIIIIIIIJI HmIIw III III I 39h u H H I I 3 1 I h 2 va 1 391 i I IIIin IIg I 39II I I I 39 M39I IIII L 5 IIIIIuI I I r 39H l 1 Mr I 39 II39m 39I I c h Jquot I J W 39a Wei1 II 4 I jHI39 I Hm I ImeI 1 I 2 f mwrudu h 55 AUE VJW T LPN aw 4mm MmAaa I llm 11qu 110 Chapter 4 Compression Members and H shapes is either the ratio bt or htw bth of whic defined presently If 2 is greater than the specified limit denoted if the shape is slender and the potential for local buckling must be accounted for We postpone a discussion of the compact and noncompact categories until Chapter 5 Beams For I and Hshapes the pro jecting ange is considered to be an unstiffened element and its width can be taken as half the full nominal width Using AISC notation gives EelQ kgzbmfzzg w w C 5 where bf and If are the width and thickness of the ange The upper limit is M 1410 Lassa1 rgwl 39x 205Q g39 ft 5q The webs of I and H shapes are stiffened elements and the stiffened width is the dis tance between the roots of the flanges The width thickness parameter is WW mm w WW Turf W am 39 39 FIGURE 49 t t b Jt T 39 f b b CZ d tw i t J D I 4 7g bt 30564797 Insosa E 4075475 1 g erS 4 717 TU l l I w n HOG 43 AIEC Requirements 7 quotquot Z it W 939 A L w 1 s 056 s 0454 b s 056 MWSI49W V m w FD N E if 1 s 14M 1513 1 s 149 lEFy DA 3 011EFy Vtw S 149V EFy 124 DIMENSIONS AND PROPERTIES I OEI d20 Table 1 1 continued w W Shapes Dimensions i Web ViFlange 7 7 Vipi tance Area Depth v 39 Work shape A d Thickness amp Mdth Thickness I k 7 k1 T able in 2 b 1 kdes km Gage in2 in in in in in in In in in in 50 124 112 1510 914 5V2 916 118 1 516 914 512 W12x58 170 122 1214 0360 33 39315 100 10 0040 x53 155 121 12 0345 35 315 100 10 0575 DIMENSIONS AND PROPERTIES 1725 Table 1 1 continued W Shapes Properties W12 W10 Compact Torsional quot 39 39 section Axis xx Axis vv Properties Inal rt h Criteria J 0 A A I s r z I s r z W lbllt 2 Tu in in in in3 in in3 in in I in in in in 58 782 270 475 780 528 864 107 214 251 325 282 116 210 3570 3953 869 281 425 706 523 I 779 958 192 248 291 279 115 158 3160 CVEN446501 Date 9 z 3106 L CDOL I Lied 44 Local Stability 111 4Z aa O I IN 1quot Investigate the column of Example 42 for local stability 0 7 8 5 For a W14X74 bf 111 tf 0785 in and b 7 f 49 643 th 2O785 r 21 056 5 0561 lw 135 gt 643 OK ew wa 5 50 h d 2kd 142 2138 as 254 W rw r 0450 Example 43 Solution mMW r tailing value is the largest 1r 2149 E5 149 29800 359gt254 OK 3A Mp Wk y Answer Local instability is not a problem Kg f 16 CLASSIFICATION OF SECTIONS FOR LOCAL BUCKLING Sect B4 Limiting WidthThickness Ratios for Compression Elements Limiting Width V Thickness Ratios 0 Description of ness AP 39beOlESLENDER Element Ratio F compact noncompact Example 1 Flexure in flanges of bf 038 E F 10 E F rolled Ishaped y y sections and R M N 5 channels 7 LP 5TIC of M BEA M GYR Al l o 2 Flexure in flanges of bt 038 EF 0 r k F alibi doubly and singly y 3995 CE L symmetric Ishaped builtup sections 3 Uniform bt NA 056 E F compression in y 9 flanges of rolled W 7 V 5 I shaped sections A E plates projecting 2 from rolled Ishaped quot sections 398 outstanding legs of 5 pairs of angles in E continuous contact 17 and flanges of g channels 4 Uniform bt NA a compression in 0 64 kCE F Y mam mam flanges of builtup WK I shaped sections and plates or angle legs projecting from builtup I shaped sections 5 Uniform bt NA 045 E F compression in legs y of single angles W t legs of double angles with separators and all other unstiffened elements 6 Flexure in legs of W o54E T o91 E F r rl single angles y y t WC Sect B4 CLASSIFICATION OF SECTIONS FOR LOCAL BUCKLING f 17 f k d TABLE B41 cont Limiting WidthThickness Ratios for Compression Elements Limiting Width 8 With Thickness Ratios 8 Thick Description of ness p r Element Ratio compact noncompact Example 7 Flexure in flanges of bt 038 f EFy 10 EFy b tees i gem A 8 Uniform dl NA 075 E F m compression in W y L stems of tees 7 V t 9 Flexure in webs Of htw 376 E F 570 E F J doubly symmetric y y I shaped sections and channels t 21 quot w 551mm DES A 10 Uniform h t NA 149E F J r compression in W y webs of doubly Li a symmetric Ishaped Co U m N 5 sections 1 E 2 m 0 11 Elexureln webs of hctW he E 570 EFy cu smglysymmetrlc B F 5 I shaped sections P Y 2 lt A E Mp 5 054m 009 12 Uniform bt 112E F 14orE F compression in y y anges Of CopW rectangular box and hollow structural sections of uniform thickness subject to M bending or compression flange cover plates and diaphragm plates between lines of fasteners or welds 13 Flexure in webs of Mt 2427575 570 9 rectangular HSS 515 Biaxial Bending 249 cannot buckle in the other direction so lateral torsional buckling is not a limit state If the shape is compact then M 2 Mp 1732y S 16FySy AISC Equation F613Z gvg sausrieo When a a t r h i Z x jagawed w 0 Elgl M 3MQ QK F r was y If the shape is nowact because of the ange width thickness ratio the strength will be given by it 21 MW Mp Mp 07115 A M A AISC Equation F6 2 r P This is the same as AIS C Equation F3l for ange local buckling except for the axis of bending f him w ii gt t if 315 Example 5 18 A W21 X 68 is used as a simply supported beam with a span length of 12 feet Lateral support of the compression ange is provided only at the ends Loads act through the shear center producing moments about the x and y axes The service load mo ments about the x axis are M Ox 2 48 ftkips and MM 2 144 ft kips Service load moments about the y axis are MD 6 ftkips and M Ly 18 ftkips If A992 steel is used does this beam satisfy the provisions of the AISC Specification Assume that is all moments are uniform over the length of the beam N W 7quot Cb 2 I563 W i WES D My 2 Mai Solution First compute the nominal exural strength for x axis bending The following data 5 g Warsaw for 21 W21 X 68 are obtained from the Zx table The shape is compact no footnote to a W E indicate otherwise and i 92496 3 193 L1 636 nL187 ft Egg gigging gags mt itaisg 2 31357 gh tgg i i The unbraced length Lb 12 ft so LP lt Ll lt L and the controlling limit state is inelastic lateraltorsional buckling Then Lb Z jf Le 4 E p g a g Elias welwa z Cb Mm Mm 07135 250 Chapter 5 Beams Because the bending moment is uniform Cb 10 M M Fa sic f 55 M 2108000 8000 07 X50X140 H nx 39 39 6583 in kips 548 6 ft kips life a For the y axis since the shape is compact there is no ange local buckling and M MP 2 132 5amp244 1220 in kips 1017 ft kips 2 L gt E eigt 53 l ZSEWK r 30 a Check the upper limit E3 0 g4 244 r F 1393955ltquotV 16 9 if at W age 439 LRFD Solution For x axis bending MM 2 12M Dx 16M 1248 16044 2 2880 ftkips For y axis bending M 2 12M 16M 126 1618 360 ftkips Check interaction equation 522 EH i quotquot i b M Vim Mux w 2880 360 0977lt10 OK 2b Mm 114 0906486 0900017 Note that QM can be obtained from the beam design charts 3 J 3 150a Answer The W21 x 68 is satisfactory 252 Chapter 5 Beams FIGURE 549 4 K a quot km 36 K3 czar gt Pcl 32145 Peh b Design of Roof Puriins FIGURE 550 515 Biaxiai Bending 253 Example 5 19 A roof system consists of trusses of the type shown in Figure 551 spaced 15 feet apart Purlins are to be placed at the joints and at the midpoint of each top chord member Sag rods will be located at the center of each purlin The total gravity load including an estimated purlin weight 5 42 psf of roof surface with a live load W 39 to dead load ratio of 10 A u g this is I go itica loading condition use A36 x steel and select a channel shape fo t 3 FIGURE 551 x rj quot wv 39 W WW AM rquot quot y 55 A i a 39 3 4 15 0 60390quot l r i P LRFD Solution For the given loading condition dead load plus a roof live load with no wind or snow load combination 3 will control w 12wD 16Lr 1221 1621 5880 psf The width of roof surface tributary to each purlin is W Then Purlin load 58807906 4649 lbft P 3 N a1 t 4649 4410 lbft 39 01m componen m x W P 1 s g Parallel component 4649 1470 lb ft sigma Q 10 f and l 39 V Mux glt04410152 1240 ft klps gt x WW a With sag rods placed at the midpoint of each purlin the purlins are two span continu ous beams with reSpect to weak axis bending From Table 322c Continuous Beams 3 20 the maximum moment in a twospan continuous beam with equal spans is at the interior 3 support and is given by 5 M 012514212 where v H M w uniform load intensity l t7 span length 515 Biaxial Bending 1 5 FIGURE 556 600333 quots 4 150 60 0quot mmwmu M a Pkg 1 Q5 1 i w if 2 MW 0441015 1240 ftklpS w C 8 A3007 7 farmM 39 a R Met m 5125 QN f W Net 254 Chapter 5 Beams The maximum moment about the y axis is therefore May O125014701522 1034 ft kips To select a trial shape use the beam design charts and choose a shape with a relatively large margin of strength with respect to major axis bending For an unbraced length of 152 75 ft try a C10 X 15394 i 424 start BBK CWiCb 5 O For C1 10 pme 330 ft kips From Figure 51513 Cb is 130 for the load and lateral support conditions of this beam Therefore 98214 130330 4290 ft kips OQCE was M From the uniform load table for C shapes 431114 430 ft kips gt 4290 ftkips 3 age a 39 use 9me 429 ftkips 39 This shape is compact no footnote in the uniform load tables so 6an be gbbF Z O9036234 7582 in kips 6318 ft kips J P y y 4 3 OR Rat A Fit t as 44 m gym 51 W a 4 EL 23934 203 gt 16 424 m gm 5y 6it 4 g g m M4 b16a5y 090l636115 4 at 5962 inkips 4968 ft kips 3f Because the load is applied to the top flange use only half this capacity to account for the torsional effects From Equation 522 Mm Mm 1240 1034 QM 951Mquot 429 49682 0705 lt 10 OK The shear is V 2 0442005 M 331 kips 29m 4 2 From the uniform load tables awn 4467 kips gt 331 kips 0K CVEN 446 CLASS NUMBER DATE PAGE POE U a 378 LRFD Solution Try Viainch bolts The nominal bolt area is 2 A 7E 5 03068 1112 The shear strength is 5 Rn 2 5111va 07548Ab 0754803068 1104 kipsbolt assuming that the threads are in the shear plane No slip is permitted so this connection is slip critical We will assume class A sur faces and for a 58 inch diameter A325 bolt the minimum tension is T 12kips FIGURE 719 abO3 39 77 Design Examples 379 I o 1 l o from AISC Table I 31 From AISC Equati n J34 the slipcritical strength for one bolt is Rn uDuhSCTbNS 10O35l13l019l 7515 kipsbolt 4 l I 0 K433i De 1 t oil The shear strength is Rn 07548Ab 0754806013 2165 kips bolt assuming that the threads are in the shear plane The minimum tension for a 78inch A325 bolt is Tb 39 kips so the slip critical strength is w Eb m g Q 3 Rn bQLlDuhschNs 1003511310391 quotg 1542 kipsbolt controls E fQ S The number of 7sinch bolts required is p 4 9 31 bolts 1542 5 w 2 380 Chapter7 Simple39Connections if Sic eg g esse FIGURE 721 112 4 3 r 31 3 gt11 le I x E Le il75 6 gns E K e 1 3g 5 a 2 g g z rgse Eseif e v MW 0 O O O a g as v a I 391 64961 Four 7s inch diameter A325 bolts will be used From AISC 133 the minimum 1 spacing is quot s 266703 2667 g233 in 0rpreferably 3d 3 263 in From AISC Table J 34 the minimum edge distance is E J w C57 Le 15 in assuming sheared edges 77 Design Examples 381 Compute U with Equation 31 7 3 E 3599 U 7 l 2 EL 1 M O9326 will 9 l t a e W Where f 2 0607 inch for the long leg vertical The effective net area is AG AnU 119O9326 111 in2 gt 0985 in2 OK Now check the bearing strength The edge distance for the angle is the same as the edge distance for the gusset plate and the angle is thinner than the gusset plate so the angle thickness of 14 inch will be used For bearing strength computation use a hole diameter of hzd17 115 1n 16 8 16 16 quot G For the hole nearest the edge of the gusset plate 3 a h 1516 I WWWW u m L L 15 1031in N E c e 2 2 x L C Rn 12LCtE 07512103 1 65 1508 kips Vi 7 1 Upper limit 2 24th 07524 65 2559 kips gt1508 kips We use 45Rquot 2 1508 kips for this bolt For the other holes 15 m Lcsh3 1 6 2063in l2Lchu 075122063 65 3017 kips WW 24dr12 2559 ldps lt 3017 kips use 12 2559 kips for this bolt V The total bearing strength for the connection is V I bRn 1508 32559 919 kips gt Pu Now check block shear With the bolts placed in the long leg at the usual gage dis tance seeGhapter igure zt the failure block is as shown in Figure 722 The shear areas are NM 7 l 382 Chapter 7 Simple Connectidns FIGURE723 112 3 3 H 3 gt gt1I2quot T l 1 ab2quot vd W 33 j O o o o 7T 112 I 386 78 HIGHSTRENGT H BOLTS IN TENSION The application of force F will increase the bolt tension and cause it to elongate by an amount 5b Compression in the ange of the structural tee will be reduced result ing in a distortion E in the same sense and amount as 61 A relationship between the applied force and the change in bolt tension can be approximated as follows fa 39 FIGURE 724 78 HighStrength Bolts in Tension 387 FIGURE 725 millilitl 3 w 5 y k6 tranaaaiymggi 539 Am g x 2 f kg 1 F b C From elementary mechanics of materials the axial deformation of an axially loaded uniform member is PL 671 M 73 where P applied axial force L original undeformed length A cross sectional area E modulus of elasticity Equation 73 can be solved for the load AE5 r L P 74 The change in bolt force corresponding to a given axial displacement 5b therefore is 14171312 512 Lb AT 75 Where the subscript indicates a property or dimension of the bolt Application of Equa tion 74 to the compression ange requires a somewhat more liberalinterpretation of f z 1 a V 388 Chapter 7 Simple Connections i O f the load distribution in that N must be treated as if ityvere unifoimly applied over a sur face area A The change in the force N is then obtainedwfrom Equation 75463 I A E 6 f 7 ft ft f1 ANf MMEMW Mg 1 Mnmw Q Where L is the ange thickness As long as the connected parts the two flanges remain in contact the bolt deformation 6 and the ange deformation 6 will be equal Be cause Efg approximately equals Eb Bickford 1981 and A is much larger than Ab f 4quot r c t A 5 a r aired nx digg h V pg A41 7gb Lfe Lb quot 7 quot W i 39 45 if L and therefore 39 72 Azy gt AT g The ratio of AT to AN is in the range of 005 to 01 Kulak Fisher and Struik 1987 Consequently AT will be no greater than 01AN or equivalently maximum ATAN 01 demonstrating that most of the applied load is devoted to relieving the com pression of the connected parts T 0 estimate the magnitude of the load required to overcome completely the clamping effect and cause the parts to separate consider the free body diagram in Figure 726 When the parts have separated T F or To AT F 77 At the point of impending separation the bolt elongation and plate decompression are the same and AbEb 6b Lb Lb AbEb 5 78 AT where E is the elongation corresponding to the release of the initial compression force NO From Equation 73 Nd 5fe AfEEfg z26 HGURE7 T i t j E Q i i l g L777 1 u mmmmwww a 34 i 79 Combined Shear and Tension in Fasteners 399 FIGURE 732 f gquot n ma segue weeFtner fv r i t Mm tiliwiv igg The AISC Specification approximates the elliptical curve with three straight line segments as shown in Figure 732 The equation of the sloping line is given by g new E H To avoid going above the line we 2 If Equation 724 is solved for the tensile strength f we obtain for a given fv A fv AlaWM quot WWWMW WWW f 131 EE ww 725 Let available strength 2 gx nominal strength or nominal strength avalable Ength where i I I for LRFD 1 39 ugma for ASD W n W lt3 f If f is viewed as the available tensile strength in the presence of shear then from Equation 725 the corresponding nominal strength is memwwmwm y g F A yumm 73965 Nd N WAQAWKQ n wa g Wag i 285 K K O C UAY gD lt 64 Braced versus Unbraced Frames F moment The amplified primary LRFD i 8 i L quot P 7f9ezxx 15214 64 BRACED VERSUS UNBRACED FRAMES AISC Equation 32121 EA 397quot M 37 31M BzMw where M required moment strength 2 M L for LRFD P A FIGURE 65 1 3 x I WWMMWX 44 a 96 4 b M Am a i xwamm wjf iv gg ggl g E W awrmw wgm P P 236 Chapter 6 BeamColumns 6 Cf EM maximum moment assuming that no sidesway occurs Whether the frame is AIL 3 actually braced or not the subscript m is for no translation Mquot will be a factored load moment for LRFD and a service load moment for ASD 4 M5 2 maximum moment caused by sidesway the subscript 2 is for lateral translation This moment can be caused by lateral loads or by unba lanced gravity loads Gravity load can produce sidesway if the frame is unsymmetrical or if the gravity loads are unsymmetrically placed M 6 Will be zero if the frame is actually braced For LRFD M 6 will be a factored load moment and for ASD it will be a service load moment B1 amplification factor for the moments occurring in the member when it is braced against sidesway Ba 2 amplification factor for the moments resulting from sidesway i We cover the evaluation of B1 and 32 in the following sections J W i Rea mii es parP MEMBERS IN BRAC FR ES s The amplification factor given by Expression 67 was derived for a member braced against sideswaywthat is one whose ends cannot translate with respect to each other m K a MAW HGUREas P 4 an if P M0P5 FIGURE 67 287 65 Members in Braced Frames M0 Maps Mmax M2 That is not the case if applied end moments produce reversecurvature bending as shown in Figure 67 Here the maximum primary moment is at one of the ends and max imum moment amplification occurs between the ends Depending on the value of the axial load P the amplified moment can be either larger or smaller than the end moment Vi The maximum moment in a beamcolumn therefore depends on the distributi mapEa of bending moment within the member This distribution is accounte r by a fac quot tor Cm applied to the amplification factor given by Expression 67 The amplifica tion factor given by Expression 67 was derived for the worst case so Cm will never be greater than 10 The final form of the amplification factor is W L I AISC Equation c2 2 I 61 8amp4 L gt where w Ef cjf quot P required axial compressive strength P l quotW P1 for LRFD enigma t 2 Pa for ASD ma w m WWt vmh Sect C2 1a 1b CALCULATION OF REQUIRED STRENGTHS 28 76 16121 M Methods of SecondOrder Analysis Second order analysis shall conform to the requirements in this Section 3122 Z 5 f Z 3 Z Any second order elastic analysis method that considers both P A anAd PB e ects may be used BL LO W General SecondOrder Elastic Analysis The Ampli ed First Order Elastic Analysis Method de ned in Section C2lb is M an accepted method for second order elastic analysis of braced moment and combined framing systems SecondOrder Analysis by Ampli ed FirstOrder Elastic Analysis e User Note is proVided section toVaCcOunt I effeCtsin frames by amplifying the axial forces and moments in members 39connections from a rst y39Ql daranaly sisf I L39 iquot A 39 39 39 A 1 i The following is an approximate secondorder analysis procedure for calculating the required exural and axial strengths in members of lateral load resisting systems The required second order exural strength Mr and axial strength Pr shall be determined as follows Mw o M 2 81Mquot 32M CZ1a Pr Pm 3213 Mm 02 l l 4 CM 24 We Where BAZZ C DLi 2 1 322 1 aP Pe1 For members subjected to axial compression B1 may be calculated based on the rst order estimate P Pm P 1 65 Members in Braced Frames 239 FIGURE 69 KN N KN OF 0139 DUU L Negative M1 M2 Positive M1 M2 W 1 If there are no transverse loads acting on the member 39 X a C we v Cm 06 041 AISC Equation C24 2 M 1 M2 is a ratio of the bending moments at the ends of the member M1 is the end moment that is smaller in absolute value M2 is the larger and the ratio is pos itive for members bent in reverse curvature and negative for singlecurvature bending Figure 69 Reverse curvature a positive ratio occurs when M and M2 are both clockwise or both counterclockwise 2 For transversely loaded members Cm can be taken as 10 A more refined pro cedure for transversely loaded members is provided in Section C2 of the Com mentary to the Specification The reduction factor is dMLRFD Cm 1 my Pr 51 AISC Equation 002 2 e a 3n weaeazs For simply supported members 2 T W 1 M01 where 50 is the maximum de ection resulting from transverse loading and M0 is the maximum moment between supports resulting from the transverse loads The factor 1 has been evaluated for several common situations and is given in Commentary Table cc21 5 3 g m f 4 flquot f 9Q GAE Memhpre in Rrsmnrl Fr moe fa r le 64 The horizontal beam column shown in Figure 613 is subjected to the service live loads shown This member is laterally braced at its ends and bending is about the x axis Check for compliance with the AISC Specification Q N e was i Sitamw w N4 FIGURE 613 W8 X 35 l A992 tang ma 4 w M P1 Examp 28k gt Y lt 28quot Q Q gm b ca a wax a 539 I 539 k 90 A we 7 1039 The factored axial load is LRFD Solution PM 1628 448 kips The factored transverse loads and bending moment are 7 Q1 1628 448 kips a V L 4 wquot 120035 0042 kips ft w 2 A4 Liam WO39OQSGO 1125 ft kips This member is braced against sidesway so M e 0 294 Chapter6 Beam Columns Xa Elf Compute the moment amplification factor For a member braced against side sway and transversely loaded C can be taken as 10 A more accurate value can be up P53 cm 1W r r w up 142 aha2 6quot From Commentary Table C CZ l I 02 for the support and loading conditions of this beam column For the axis of bendingX 933 weigh r kxg L 2524 kips 21 quotP 71 mix 75229000127 5W P 2 2 2 z 2 KIL 161 10 x 12 53 0113 10013 448 x g M 21 4 Cquot 1 PPd l 02 Pd 022524 09965 M m l 0 3 Hang Z 3 Cam gt The amplification factor is C c 09965 g z B1 W 1 an131 1 100311 1 448 2524 The amplified bending moment is MU A S M Ble 32114 10151125 0 1142 ft kips R quotJ W MFromrn the design charts for 1110 ft I 11M 123 ftkjps quot w Because the beam weight is very small in relation to the concentrated live load C1 may be taken from Figure 5150 as 132 This value results in a design moment of g I 4415 r an 132123 1624 ft klps r 39 J This moment is greater than bMp 130 ftkips so the design strength must be lim ited to this value Therefore 41M13qrtkips 3 i izegig l Check the interaction formula From the column load tables for KL 10 ft 413 358 kips 9g 4 il Wag21 Wis Lu W I ywv 0 a f e quot w Pquot 448 p 4 01251 lt 02 use Equation 64 AISC Equation Hllb C53 413 358 W Mm M 01251 1142 0 benx 12M My 2 1 0941 lt 10 OK w Answer A W8 X 35 is adequate 295 Chapters BeamColumns 39y mple 65 The member shown in Figure 614 is a W12 X 65 of A242 steel and must support the service loads and moments shown One end is pinned and the other is subjected to moments about both the strong and weak axes Use Kx K 10 and investigate for compliance with the AISC Specification Exa P 50k Mx 2253 My 5ftk FIGURE 614 Dead loads Live loads P 150k Mx 675ka 15 W12 x 65 My 15 A242 steel quot gt MOmeMQ l Solution First determine the yield stress F y From Table 2 3 Part 2 of the Manual we see that A242 steel is available in three different versions From the dimensions and proper ties table in Part 1 of the Manual a W12 X 65 has a ange thickness of tf 0605 in XE 51 em 50 65 Membersin Braced Frames 297 A M 5 This matches the thickness range corresponding to footnote 1 in Table 23 therefore F 50 ksi LRFD Solution Compute the factored axial load and moments r P 1250 16050 300 kips Mm 12225 16675 1350 ft kips v Mm 125 1615 300 ftkips v 3 my g Compute the strong axis bending moments A 5 gigs k S 2 A me 0396 39 wi 06 39quot 04 0 06 4P V FWngg M2 1 Em Oug i 2 2 2 M 75 EI 76 EIx TE 29000533 6L 2 2 z 2 4708 laps XL XL 10x15x12 mewwr f 5 5 5 goggle B Crux me MMWW M M 1 1 anPm 1 1002I1n 1 gowsw ff g 0641 lt 10 use Bix 10 5 l o Z lt3 7 The required moment strength is O 3 0 7 M 2 M BIXMW ngMm 10135 O 135watkips M From the Beam Design Charts with Cb 10 and Lb 15 feet the moment stre oanx 340 ftkips and DbMpx 356 ft kips tagg i2 I From Figure 515g Cb 167 and V 39 C x qwa for Cl 2 10 167640 5678 ft kips This result is larger than obMpx therefore use prm obMpx 356 ftkips MMMCompute the weak axis bending mognents TMTWWW M WWWW a W 3 2 i Cmy 06 04941 06 040 06 M2 2 2 1 7 E1 Pely E19 2 n 3 klpS KILT KyL 10 X 15 gtlt12 B Cmy Cmy y 1 an1231 1 Loon121 1 3001537 j 2 0 6 lt 10 3 use 393 1 i i i 74 3 O The required moment strength is MM my Blmey BzyMefy O 298 Chapter 6 Beam Columns YWW Because the ange of this shape is noncompact see footnote in the dimensions and properties table the weak axis bending strength is limited by FLB sic Section 515 of this book and Chapter F of the AISC Specification 1 5 b t f992 f 47K 5 ing 1 038 E 038J29 000 9152 559 39i 0 it x if E 29 000 83 I ti0 10 2408 we 230 a 50 m i V Sincekpltlltkp r E V 7 Pgm Sg ti M M M 07F s J AISC Equation F6 2 p p y 3 A A r p M y Zy g24838 kms f f Mgznawleis asaS OJxsoxzahd f 17a7hams prW 0900787 2 1608 ft kips r 1 96 3 23 quot The value of prny is also given in the Zy table listed as bMW For 21 W12 X 65 it is a given as 161 ft kips Determine the compressive strength For KL 1015 15 feet the axial com pressive design strength from the column load tables is Mawquot quotquot MmMDVWWMWMW quot W39tmmmw Mwm39ww w quot39 wmequotquot mwltwquot 39 g M g MMTW E Mit Determine which interaction formula to use W Pquot 2 04532 gt 02 use Equation 63 AISC Equation Hlla 0613 662 M Pquot M quoty 04532 135 3O 61 9 anx beny 9 0956 lt 10 OK Answer The W12 X 65 is satisfactory E 34 CLASS 13 Example 4 72 The rigid frame shown in Figure 415 is unbraced Each member is oriented so that its web is in the plane of the frame Determine the effective length factor K for columns AB and BC Solution Column AB ForjointA 210156 2 83312107012 1586094m 62 we ZIgLg 1350 20183018 1692 E Forjoint B g ZICLC 107012107015 1605 0 95 3 366mm 218 L3 1692 1692 39 Answer From the alignment chart for sidesway uninhibited AISC Figure CC24 with GA 094 and GB 2 095 Kx 13 for column AB For column BC I 158 3 33 Wee L 2 For joint B as before L G095 I 1 W FIGURE 415 g A L a 12 g 3 L939 Wu w24 55 w24 x 684 W24gtlt 55 3 WM x 68 mg a B 16929 3 Wgt 15 E l C JL u C L465 13 47 More on Effective Length 135 W As pointed out in Example 412 for a pinned support G should be taken as 100 W for a fixed support G should be taken as 10 The latter support condition corresponds to an in nitely stiff girder and a exible column corresponding to a theoretical value of G 0 The discussion accompanying the alignment chart in the Commentary rec ommends a value of G 10 because true fixity will rarely be achieved 376 7 DATEPAGEL MHV CVEN 446 CLASS NUMBER coMPOSITE COMPRESSION MEMBER SELECTION TABLES 4 317 Table 4 21 Stiffness Reduction Factor a I ksi column given by Equa tion 46b Dividing it by the cross sectional a a the buckling stress 7213 mag The rotational stiffness of a column in this state would be proportional to EJCLc and the appropriate value of G for use in the alignment chart is zan ra gw g g m a Ginelastic 2 E18 Lg Ev eli StiC A175 0quot Because E is less than E Ginelastic is less than Gelastic and the effective length factor K will be reduced resulting in a more economical design To evaluate E E called the sti izess reduction factor denoted by 1 consider the following relationship for a col umn with pinned ends Hr nelastic zErLr2 WE Fcrelastic 7E2ELr2 E ix AISC uses an approximation for the inelastic portion of the column strength curve so Equation 413 is an approximation when AISC Equations E3 2 ande33 are used for For We can approximate F a by the compressive strength a 30 e F 5 forLRFD Week a Ken 5 8 7 it 5 am 12 1 a W are Then in the elastic range Femindastm is approximately V 1y FF 5 quot 3 E W W 4in 96143 0658 for LRFD l a o e Wm gw 51 5 A yin foriA39SE xx xxx x a We can solve for Fe then compute Fcrelastic e The stiffness reduction factor 1 can then be computed Qa Egg 138 Chapter 4 Compression Members Example 4 13 Compute the stiffness reduction factor Ta for an axial compressiv stress of 25 ksi and F y 50 ksi LRFD Solution l 25 ksi a In 2 r nelastic Z 3 0658 393Fy be1g 090 W m Wm or 2778 7 0658lt59Fegt50 Fe 2 3561 ksi FCQ WJJZ 0877FQ 08773561 3123 ksi xt 59963 B Z ED i 35 The stiffness reduction factor is therefore M W Answer crinelastic a 0890 F 3123 W 559 C T cr elastic Solution Compute elastic G factors For joint A M 2ICLc 17112 1425 15239 Jet 3 EggLg quot 8862088618 quot 9352 FIGURE 417 4 LRFD Solution 0 Effective L hgth 39 W12gtlt 14 ArW12gtltl4 j W l l l W14gtlt22 B W14gtlt 22 W10 x 33 W10 x 33 l 3 a I l 2039 2039 J 18 394 A For joint B ZUcLC 217112 285 EggLg 19920 19918 quot 2101 quot 136 7 t 39 l f 162 From the alignment chart for unbraced frames Kx 145 based on elastic behavior Determine whether the column behavior is elastic or inelastic I I K L 14512x12 3 7quot 4983 Kquot g 419 z 4714 29 000 113 50 A Since E h L t iwkpwwwmmv Fy I I gt4 471 rx behavior is inelastic and the inelastic K factor can be used 3911 5 39 gm Ww The factored load is b 9 P z 12D 16L 1235516142 2698 kips Enter Table 4 21 in Part 4 of the Manual with 39 31 269398 2779 ksi Ag 971 and obtain the stiffness reduction factor 1 2 08105 by interpolation E 6 lt3 3 x 1 149 Chapter 4 Compression Members For joint A g Gingham a gtlt Gm 08105152gt123 mg i 22 J 3 5 For joint B 3 ii Gindastic From the alignment chart XX 2 515 Because of the support conditions normal to the frame Kv can be taken as 10 M Answer Class 10 13938 Chapter 4 Compression Members Example 45 Compute the available strength of the compression member of Example 42 with the aid of a Table 422 and b the column load tables W 171 X74 R5 5Q LRFD Solution 21 From Example 42 KL r 9677 and F y 50 ksi Values of pol in Table 4 22 are given only for integer values of r for ldecimal values KL r may be rounded up or linear interpolation may be used For uniformity we use in terpolation in this book for all tables unless otherwise indicated For KLr a m Wngjme 9677 and F 501m wt 3 C l 34 9561252267 ksi W papa LFcAg 267913 igmkips YE 2amp4 I L e l b The column load tables in Part 4 of the Manual give the available strength for L selected W HP singleangle WT HSS pipe double angle and composite shapes We cover composite construction in Chapter 9 The tabular values for the symmetrical shapes W HP H88 and pipe were calculated by using the minimum radius of gyration for each shape From Example 42 K 10 so KL 1020 20 ft For 21 W14 X 74 17 50 ksi and KL 20 ft Eel r7 gtCP494 kips vmfm1devmmeu M Wnnxewquotquot w V 4318 quot M O 1 DESIGN OF COMPRESSION MEMBERS a O 2 vwiagi V SLEHBEE MESS quot at Table 4 22 M Available Critical Stress for K Compression Members ml lWlm m K 4420 44 m ksi ksi ksi ksi ksi ksi ksi ksi ksi m mm an mum 1 210 315 1 216 324 1 2517 378 1 275 414 1 299 450 2 210quot 315 2 216 324 2 2515 378 2 127551 414 2 299 450 3 2092 315 3 2215 324 3 251 378 3 275 414 3 299 450 4 2097 315 4 21539 324 4 951 47 l m u W33 it I ASD O 167 be 090 AMERICAN INSTITUTE OF STEEL CONSTRUCTION INC 4 12 DESIGN OF COMPRESSION MEMBERS NOTE Q2 23 L mmggg 0220000 Table 41 continued i Available Strength in VWM a 2 r F50k39 AXIaI Compressmn klps y 339 er W14 W Shapes Imxc gm wm WWVEQWMjiy 623 Shape W14gtlt Wtft 257 233 21 1 193 1 76 159 I llQC CPII clgt12PI1 PnQc 0PI1 0Pn P QG 0Pn PnQc DEF Design ASD39 LRFD ASD LRFD VASD39 LRFD ASD LRFD Asa LRFD ASD LRFD Oz 0 2260 3400 2050 3080 quot1j85Q 2790 1700 2550 155 2330 1400 2100 L nnqn rmrm nnrin nrun 401m nmn zen ncnn an nnnn 4nIn nnx n 337W Nuva 39 iASD LRFD Q 20167 00090 aim 2 O 5 AMERICAN INSTITUTE OF STEEL CONSTRUCTION INC 46 Design 121 Example 48 LRFD Solution EM Select 21 W18 shape of A992 steel that can resist a service dead load of 100 kips and a service live load of 300 kips The effective length KL is g feet Pl 121 16L 12100 16300 600 kips r N V 35 Try Fa 33 ksi an arbitrary choice of twothirds F y E g W l5 2 Bi Required Ag 5 600 202 in2 EEQIM gchcr 09033 pk QEL AG TryaW18x7l 029 A8 2081112 gt 202 1112 OK he 5 3E 26 1835 lt 200 OK wig rmjn 170 sz 7z229000 e KLm2 18352 471 471129300 113 We y 85 ksi 122 Yet 065 Chapter 4 Compression Members Since 511 gt 471 E AISC Equation E3 3 applies r F A 5 Q g r 0877 087785 7455 ksi 00131 617 A O907455208 140 kips lt 600 kips NG Cr g Because the initial estimate of F or was so far off assume a value about halfway between 33 and 7455 ksi Try Fm 20 ksi P RequiredA 99 O 333 in g ltch 09020 CI Try a W18 X 119 Ag 351 in2 gt 333 in2 OK KL 26X12 1160 lt 200 OK 269 51 772E 2 29000 e KLr2 11602 r min 2127 ksi Since lg gt 471 7 113 AISC Equation E33 applies r y F6 0877173 08772127 1865 ksi ban QF A 0901865 351 589 kips lt 600 kips NG cr g This is very Close so try the next larger size 240 Eratell l germ Chapter 5 Beams If the beam is not laterally braced at the load point in such a way as to prevent relative lateral displacement between the loaded compression ange and the tension ange the Specification requires that web sidesway buckling be investigated AISC J104i When loads are applied to both anges web compression budding must be checked AISC J 105 39 Ceumn Base Plates As with beam bearing plates the design of column base plates requires consideration of bearing pressure on the supporting material and bending of the plate A major dif ference is that bending in beam bearing plates is in one direction whereas column base plates are subjected to two way bending Moreover web crippling and web yield ing are not factors in column base plate design The background and development of the plate thickness equation is presented here in LRFD terms After some simple modifications the corresponding ASD equation will be given Column base plates can be categorized as large or small where small plates are those whose dimensions are approximately the same as the column dimensions Fur thermore small plates behave differently when lightly loaded than when they are more heavily loaded 514 Beam Bearing Plates and Column Base Plates 241 FIGURE 541 in warms Pug 9f Pct gmw 035 Ni 53 a The thickness of large plates is determined from consideration of bending of the portions of the plate that extend beyond the column outline Bending is assumed to take place about axes at middepth of the plate near the edges of the column anges Two of the axes are parallel to the web and 0 Obf apart and two axes are parallel to the anges and 09501 apart Of the two l inchgan 1 Ever strips labeled m and n in Figure 541 the larger is used in place of n in Equationwg kl to compute the plate thickness or 213422 E t 2 0903ng t 2 g LL 09OBNFV or 515 514 Beam Bearing Plates and Column Base Plates 243 Where n39 dbf 517 M These three approaches were combined by Thornton 1990b and a summary of the 39 39 resulting unified procedure follows The required plate thickness is 2 g i 090miy where 0 393 wg ixjk iiw E maxmn39Ln N 095d m 2 8 08 n f 2 ZxE i39 1 Has QQC 8 GAS 4dbf 31 23 Q gr I X 2 C3 5327 dbf chW mo lt v g n7i gfv J L Mg gbc 060 m i r P nominal bearing strength from AISC Equation 181 or J82 gquot There is no need to determine whether the plate is large or small lightly loaded or heavily loaded As a simplification 2 can always be conservatively taken as 10 Thornton 1990b This procedure is the same as that given in Part 14 of the Manual Design of Beam Bearing Plates Column Base Plates Ameling and Column Splices 244 Chapter 5 Beams Example 5 17 A W10gtlt49 is used as a column and is supported by a concrete pier as shown in Figure 544 The top surface of the pier is 18 inches by 18 inches Design an A36 base plate for a column dead load of 98 kips and a live load of 145 kips The concrete strength is fc 3000 psi w I use FIGURE 544 1 V 18quot LRFD Solution The factored load is PM 12D 16L 1r298 16145 3496 kips Compute the required bearing area From AISC Equation J 8 2 W E with JVQ Wgw quot I f I a 4 39 1631 i Q Pp 085fc A1 32 317 Al 3 ta WW Q i 1 CNN kissing For 11th 2 Pquot g 0600853A1 2 3496 f 39 1 A 2 2 940 1 A21611n i 1 M m Checkupperhmm as was quot 017fc A1 060173161 493 kips gt 3496 kips OK Also the plate must be at least as large as the column so M bfd 100100 100 in2 lt 161 in2 OK F0 N 13 inwAi provided 1303 169 in2 5e Answer 514 Beam Bearing Pfates and Column Base Plates 245 W The dimensions of the cantilever strips are N m NO95d z 13 09510 1175 in 26 2 2 o 8 08 r From Equation 517 1 I1 quot 4 As a conservative simplification let 1 10 giving 70 3Q maxmnIn39 maX175 25 25 25 in db 2 100100 392 25 in From Equation 518 the required plate thickness is r E I 2Pquot 25 2349 6 0893 in V 09OBNFy l O90131336 UseaPL1x13X13 WW amiamwwmv 245 CVEN 446 CLASS NUMBER DATE PAGE H K V K 1 Ea m A 1 mx lt15 34 a immWWWWWWW x 2 g i M A 1 3 3 K bf A EQWEE P 58 J Fr g w 42 3 a XE COCO Gangquot w f l WV mymmamu WWWMmM K 3358 x 2 0979 p Biaxial Bending 247 HGURESAG HGUH 111 Puntmmys WWW Wm W HGURE735 Backing bar F Corner Butt a Complete penetration groove welds 3 b Partial penetration groove welds FIGURE 736 412 39Chapter 7 Simple Connections FIGURE 737 Throat w x cos 45 W X 0707 Failure P W M f 2W where w is the weld size Ifthe weld ultimate shearing stress FW is used in this equation the nominal load capacity of the weld can be written as 9 E a t we a 9 E w Wag w Rquot 0707wLFW W V FIGURE 738 711 Fillet Welds 413 f f i V ml Axis of weld quot 26 w lt The design strengths of welds are given in AISC Table I 25 The ultimate shear ing stress F W in a fillet weld is 06 times the tensile strength of the weld metal de noted FEXX The nominal stress is therefore FW 060FEXX I Drag Maj AISC Section J 24a presents an alternative fillet weld strength that accounts for the direction of the load If the angle between the direction of the load and the axis of the weld is denoted 9 see Figure 738 the nominal fillet weld strength is FW 060FEXX10 050 sin 9 AISC Equation J25 Table 72 shows the strength for several values of 6 As Table 72 shows if the weld axis is parallel to the load the basic strength given by FW 2 06017 EXX is correct but when the weld is perpendicular to the load the true strength is 50 higher For simple that is concentrically loaded welded connections with both longi tudinglkand transverse welds AISC J24c specifies that the larger nominal strength great the following two options be used 1 1 Use the basic weld strength F W 06FEXX for both the longitudinal and the transverse welds L I a X wk 3 B R RM R AISC Equation 32 where RM strength of the longitudinal welds 06FEXX R strength of the transverse weld 06FEXX 2 Use the 50 increase for the transverse weld but reduce the basic strength by 15 for the longitudinal welds That is use F W O85O6FEXX for the longi tudinal welds and F W 15O6FEXX for the transverse welds R 085ng 1513 AISC Equation 12913 414 FlGURE 739 Chapter 7 Simple Connections WW4 mrMampstgziawz ampWP f A B Gusset plate thickness r a b L KW l 39i we 2 t W T Shear area 0 along line AB Figure 73 and subject an area of IL to shear Figure 739c The shear strength of the weld AB cannot exceed the shear strength of the base metal corresponding to an area IL Arnhem 1 The nominal shear strength of the base metal will be based on either the limit state of yielding or the limit state of rupture fracture As in previously discussed pro visions the shear yield and ultimate stresses are taken as 06 times the tensile yield and ultimate stresses AISC J 42 gives the shear yield strength as 3 s12 R 06FyAg AISC Equation 13143 Where Ag is the gross area of the shear surface and is equal to tL For LRFD the re sistance factor is q 100 and for ASD the safety factor is Q 150 The shear rup ture strength is Rll O396FltAlll AISC Equation 144 415 Chapter 7 FIGURE 740 Simple Connections 316quot fillet welds E70 electrodes 419 738 Example 713 A connection of the type used in Example 711 must resist a service dead load of 9 kips and a service live load of 18 kips What total length of 14inch fillet weld E70XX electrode is required Assume that both connected parts are ff inch thick PM 12D 16L 129 1618 396 kips M LRFD Solution F E1 The shear st ength of the 1d per inch of length is WE EUC 3 MM it sw a 1392 x 4 srxteenths 5568 lapsin will The shear yield strength of the base metal is Xm 06F z 0636 81 kipsin 7 g V I gsaesequot y 8 Wm 35 t if Pa 1 N and the shear rug ture strength of the base metal is I i IIII 0453 04568 9788 kipsin W C The weld strength of 5568 kips in governs 51 The total length required is x 396 kips if 711 in 5568 klpsin Answer Use 8 inches total 4 inches on each side k Ifquot 420 gm seine er Eth r tie FIGURE 741 4 5 Practical design of welded connections requires a consideration of such details f as maximum and minimum weld sizes and lengths The requirements for fillet wax X are found in AISC J 22b and are summarized here 5 afk a Wwwmmwm 3 WWJWVAWW Minimum Size 1 The minimum Size permitth is a function of the thickness of the thinner connected part and is given in AISC Table 124 This requirement is taken directly from the 7 America Welding Society Structural Welding Code AWS 2004 f f A i J Lj i f 2 n I I f Aquot Maxnmum Size W 56 W Along the edge of a part less than 11 inch thick the maximum fillet weld size is equal to the thickness of the part For parts Mi inch or more in thickness the maximum size 5 is t 116 inch where t is the thickness of the part For fillet welds other than those along edges as in Figure 741 there is no max imum size Specified In these and all other cases the maximum size to be used in strength computation would be that limited by the base metal shear strength Minimum Length The minimum permissible length of a fillet weld is four times its size This limitation is certainly not severe but if this length is not available a shorter length can be used if the effective size of the weld is taken as one fourth its length Flat bar tension member connections of the type shown in Figure 742 similar to those in the preceding exam ples are in the special case category for shear lag in welded connections covered in Chapter 3 The length of the welds in this case may not be less than the distance between them that is L 2 W 44 CLASS 4 Chapter 3 Tension Members Example 33 A double angle shape is shown in Figure 35 The steel is A36 and the holes are for FIGURE 35 Solution Answer 12 inch cliameter bolts Assume that A6 075A k a Determine the design tensile strength for LRFD 3659 M75 aw bTaDete nineatheallewableastrengthwforwASDtw 3 g ltgt gt i 2L5 X 3 X 516 LLBB strength based on the gross area is my g aw v pquot FAg 36241 8676 kips I S Y at pawaeea Jame The gross area has been taken from the table form angle shapeslt There are two holes in each angle so the net area of one angle is A241 5 11 gtlt22019m2 16 2 8 The effective net area is A3 0752019 1514 in The nominal strength based on the net area is Pquot 5A6 58l514 8781 kips a The design strength based on yielding of the gross area is 613 0908676 7808 kips Q38 The design strength based on fracture of the net area is 6P 0758781 6586 kips M35 was Because 6586 kips lt7 808 kips fracture of the net section controls and the design strength for the two angles is 2 X 6586 132 kips We CVEN 446 CLASS NUMBER DATE PAGE 39 P varies as it progresses back into the plate 900k 800k 600k 300k K T iw ii 33 EFFECTIVE AREA Ag AU AISC Equation D31 Pgm gw g For welded connections we refer to this reduced area as the effective area rather than the effective net area and it is given by 4anfmwmz gm Ae Ag MW 5 i FiGURE 36 x OOO Section CVEN 446501 W wwwmwi z mAvwmww 45 Chapter3 Tension Members a 1 For any type of tensio lembei ekpt plates and round HSS with E V 30 x W 39 5 U 1 where rm f distance from centroid of connected area to the plane of the connection 3 length of the connection 31 2 Plates In general U 10 for plates since the cross section has only one element and it is connected There are two special cases for welded plates x WWW 21 Connected with longitudinal welds on each side and no transverse weld see 5 W6 iii W Figure 39 vmwtui 9 For E 22w U 10 a For 15w S E lt2w U 087 I V W e For w S E lt15w U 075 W W39WW WWW v W lt m Q t b Connected Wit transverse welds only U 10 and Aquot 2 area of connected element Figure 310 illustrates the difference between transverse and 10ngi tudinal welds Connections by transverse welds alone are uncommon 3 Round HSS with 6 2 13D see Figure 37d U 10 33 Effective Area 49 FlGURE 311 000 gt 0000 WTS x 225 gt e O O O O bf 2 W shape 7 0794 gt 3 for parent shape U 070 U 090 Figure 311 illustrates the alternative values of U for various connections AISC 1933 mandates that for shapes such as angles double angles and WT Wagme Wapes the value of U should not be less than 060 g L Q g 50 Chapter3 Tension Members g M Z f 4 299 FIGURE 312 L6 x 6 x 12 1 lt 1 67 v iwwwmmm mm 523 W f 53 in bolts Section Exam MW MWMmemmWIMWWM MW WWth M 5 gig W W w 34 Staggered Fasteners 51 FIGURE 313 gt1 W 167 L6gtlt6gtlt12 34 STAGGEREB FASTENER F My 39 FIGURE314 X g g j amp E gig QM R PM 2 Q EL sf 52 Chapter 3 Tension Members WII1Vg Ed 2 9 5 quotquot EP wgzd W x w Wmwg Zd22 Egt mg N 4 a Example 36 Compute the smallest net area for the plate shown in Figure 315 The holes are for 1inchdiameter bolts D W FIGURE315 f 3 ug1o2i V8 339 ll 1 o 3 539 giW m 3 3 3H 4 W m M 3 G O 76 SlipCritical and Bearing Type Connections 355 r 5 quotif Example 74 Solution The connection shown in Figure 713a uses inch diameter A325 bolts with the threads in the shear plane No slip is permitted Both the tension member and the gus set plate are of A36 steel Determine the strength of the connection Both the design strength LRFD and the allowable strength ASD will be computed For efficiency the nominal strength for each limit state will be computed before spe cializing the solution for LRFD and ASD Shear strength For one bolt 766402 4 44 04418 in V Mjizj 39 V a 5 R 1va 4804418 2121 kipsbolt 3 350 gigs e t 0 l x For four bolts 13 42121 8484 kips 366 Chapter 7 Simple Connections 9ng 59 MW 13 1 the minimum bolt tension is Tb 28 kips From AISC Equation J 3 4 Eb b 19 f 39 1 1631 1 03 13 itDuhschNs 035l13102810 1107 kipsbolt For four bolts quot Cm K13 E m 1 a a Ora Rn 41107 4428 kips Bearing strength Since both edge distances are the same and the gusset plate is thinner than the tension member the gusset plate thickness of 38 inch will be used For b 39 g st th utation use a hole diamete of t earm reng comp 1 I 1 L a i W h d 1 3 11 31n new 164161639 e For the holes nearest the f the gusset plate amp L I ffa LCLe El5 1094in w 2 2 insomnia t R 12LCtFl 121094 582855 kiij 1 Wimgv s 3 3 w Upper hmit 24thl 58 W 3915 kips gt 2855 kips use R 2855 kips for this bolt ML 153mg For the other holes LC s h31 321881n 16 3 M R 12L IF 12 2188 58 5711k n c u 8 Ips W W V g 5 Upper limit 241th M 3915 ldps lt 5711 kips use R1 3915 kips for this bolt M The nominal bearing strength for the connection is Rn 22855 23915 1354 kips Check the strength of the tension member Tension on the gross area P Eng 366x 1080 kips Gig Y 77 Design Examples 369 77 DESIGN EXAMPLES Although an elementary bolt design was illustrated in Example 73 most examples so far have been review or analysis Examples 75 77 demonstrate more realistic de sign situations l lSo39l H We Example 75 A 5sinch thick tension member is connected to two 14lnCh splice plates as shown in Figure 714 The loads shown are service loads A36 steel and 5As inchdiameter A325 bolts will be used If slip is permissible how many bolts are required Each bolt centerline shown represents a row of bolts in the direction of the width of the plates WW r3 FIGURE 714 a C C 3 15quot 3 15quot 39 t IA H 4 quotwe l Kl l l I ll 53 a W a w mm Pu2 Pu2 lt a 9 Solution Shear For shear the nominal bolt area is 9 2 7t 5 8 Ab illL 03068 in2 Assume that the bolt threads are in the planes of shear Then the nominal strength for one bolt is l21 anAb X 2 planes of shear 48030682 2945 kips g l I o It 350 a W E 3 WWW F m 3 g H E Wig M Maw E R f a 8 I q d m rm a N n wasquot if FL 1 a f I W fge iv r kgk g E X gs 3 3333 quz J yw WQWWJDHQ a a 5 W g 4 3 Z 5wa M ng g L 3 fl g Q g a 2W A a 39 a a if M W 3 i g Ma a M gt G M VIM9quot Lg r jag 751 r r M M 3 W g w w amg mw 4 i 370 Chapter 7 Simple Connectiohs V For the holes nearest the39 dge of the plate h 1 116 LCZL815 1156 in 4 t W 2 2 Mquot LC 39 L c R 12LctF 121156G 211 58 4023 kips x 5quot a W e 5 1 1 653 Upper limit 124sz 24 Z 249 58 1 31 kips gt4093 kips f use Rn kips for this bolt e m For the other holes 7 WWWWWWWWMWWW 11 LC sh3 2313 in 16 Rquot 121412 1293136 58 8049 kips Upper limit 24th 4351 kips lt 8049 kips use R 2 4351 kips for this bolt The shearing strength per bolt is smaller than both of these bearing values and there fore controls LRFD Solution The design strength per bolt based on shear is 9512 0752945 2209 kips 8591 T The factored load is Pu 12D 16L 2 1225 1625 70 kips total load load per bolt 70 317 bolts 2209 Number of bolts required Answer Use four bolts two per line on each side of the splice A total of eight bolts will be required for the connection 77 DesignEaamples 37 1 Exa ple 76 The C8 X 187 shown in Figure 715 has been selected to resist a service dead load of 18 kips and a service live load of 54 kips It is to be attached to a 3s inch gusset plate with 7sinch diameter A325 bolts Assume that the threads are in the plane of shear and that slip of the connection is permissible Determine the number and required lay out of bolts such that the length of connection L is a minimum A36 steel is used FIGURE 715 3squotPL C8 X187 C O m 13 0 c r L I L lt a 43 Le L b Solution Determine the nominal capacity of a single bolt This ill e used in both the LRFD and the ASD solutions Shear 5W m 31 39 2 W W A1 Lg82 06013 in2 s E 1 g p Li R 17va 4806013 2886 kipsbolt Em 033 Bearing The gusset plate is thinner than the web of the channel and will control As A SSW PT QEA sume that along a line parallel to the force the length LC is large enough so that the upper limit will control Then 7 3 39 lt7k R 24th 2394 58 4568 klps Be 1 g The bearing strength will need to be verified once the actual bolt 0 ayout is determined LRFD Solution The factored load is 12D 16L 1218 1654 1080 kips M e LC w 53 371 0L K 5 L MWWWWM mmWm ggmwwmmw LZLCXNRV 2 ZAdit 39 W M W 1594 g vggpgg i r 5 LC C3 xM w an E w w a 973quot w m 2 man 2 44 m gu uw f1 55 g QMQ W39faW fiwgg E L gt 2 n nwW 8 gig M MDquot W 9 4255 w 73 E m 33quot 6 5 372 Chapter 7 Simple Connections FIGURE716 nggfiz M 7Zb gasegesi 000 000 Pn FAg 36651 1984 kips G gY The design strength is btPz 0901984 179 kips Tension on the effective net area f m 7 1 2 U a 3 A 551 4g g0487 4536 m seems The exact length of the connection 18 not yet known so Equation 31 for U cannot be used Assume a conservative value of U 060 A a 2 eeeee mmm gag mansu 1 2 Y Li kn21 a As AnU 4536060 2722 m 372 amp W552 x r gng x Pn FuAe 582722 1579 kips QM TE 45121 O751579 118 kips We A S y a i 5 e 21 we BS 1 7 e LW The member capacity is thiarefore adequate with two gage lines of bolts 3 E Check the spacing and edge distance transverse to the load From AISC J 33 0 I 0g 7 g Mlnlmum spacmg 26678 233 111 W From AISC Table J34 2 o t e i Q Minimum edge distance 118 in 77 Design Examples 373 A spacing of 3 inches and edge distances of 2V2 inches will be used transverse to the load 50252 9 A The minimum length of the connection can be established by using the minimum per T WWW C l missible spacing and edge distances in the longitudinal direction The minimum spacing A I 03911 in any direction is 223d 233 in Try 212 in The minimum edge distance in any di rection is 1 in Try 1 in These distances will now be used to check the bearing strength of the connection For the bearing strength computation use a hole diameter of 1 7 115 gdgg et g hd m 16 8 16 16 V For the holes nearest the edge of the gusset plate 54 3 LC Le 11251 16 06563 me 3 39 W MR 12LctFu 12O6563 58 1713 laps 0517 7 3 WM Cg l kueUpper limit 245Mquot 24 58 4568 kips gt 1713 kips use RR 1713 kips for this bolt For the other holes W if g F chs h225 1563in Rn 12ch13 l21563 58 4079 kips Upper limit 240311 4568 kips gt 4079 kips use R 4079 kips for this bolt The total nominal bearing strength for the connection is Rn 21713 44079 1974 kips The design bearing strength is 13 O751974 148 kips gt PM 108 hips OK The tentative connection design is shown in Figure 717 and will now be checked for E g block shear in the gusset plate the geometry of the failure block in the channel is identical but the gusset plate is thinner Shear areas Agv 2gtlt 252511254594 in2 V A V 2 x goizs 2500 2719 in V Tension area Aquot 32 10 07500 1112 2 g i g a filtl7ggjwdw R Ef e e t e 1 he 7 374 Chapter 7 Simple Connections FIGURE 717 212quot 311 2127 j 39 18n 21211212H 1quot WWW gt r 4 For this type of block shear U by 10 From Rn O396F11Anv UbsFuAnt 06582719 105807500 1381 kips with an upper limit of 06F A UbsFuAm 06lt36lt4594 105807500 1427 kips ygv The nominal block shear strength is therefore 1381 kips and the design strength is DRM 0751381 104 kips lt 108 kips NG The simplest way to increase the block shear strength for this connection is to in crease the shear areas by increasing the bolt spacing or the edge distance we will increase the spacing Although the required spacing can be determined by trial and error it can be solved for directly which we do here If we assume that the upper limit in AISC Equation 145 does not control the required design strength is th UbsElAnr 0750658Aw 105807500 108 kips RequiredAm 2888 in2 235 5 a AW gm s 1125 25102 2888 in H4 Requireds 261 in use 3 3 in ZW Compute the actual block shear strength 3 2 A ZX 331125 5344 m gv 8 Equot Am 5344 125 x102 3469 ill2 95Rquot 2 07506FAW U bsFuAnt O7506583469105807500 123 kipsgt 108 kips OK w ag LE 7 law 5 9 2m 3 f Q 7 3 m MM 51 g a kw E W 3w 33 W f 3m K QQQQ 2 Va 281 63 Moment mpli cation MOMENT AMPLIFICATION The foregoing approach to the analysis of members subjected to both bending and axial load is satisfactory so long as the axial load is not too large The presence of the axial load produces secondary moments and unless the axial load is relatively small these ad ditional moments must be accounted for For an explanation refer to Figure 63 which shows a beam column with an axial load and a transverse uniform load At an arbitrary point 0 there is a bending moment caused by the uniform load and an additional mo ment P3 caused by the axial load acting at an eccentricity from the longitudinal axis of the member This secondary moment is largest where the de ection is largest in this case at the centerline where the total moment is wLZ 8 P6 Of course the additional moment causes an additional de ection over and above that resulting from the transverse load Because the total de ection cannot be found directly this problem is nonlinear and Without knowing the de ection we cannot compute the moment 252 FIGURE 63 63 MomentAmplification KW y sin m e 0 L where e is the maximum initial displacement occurring at midspan For the coordi nate system shown the moment curvature relationship can be written as d5 Fi Ewww rwmg 239 v El hwmwmmwmw HGUREGA y A yo e Pquot r j 4 Xgt39 ymax 3 83amp 63 Moment Amplification The maximum moment occurs at x L2 M ffz 393 M P a e will 63 g up Wi w w it quot m v quotr H max u 39 FePu 1 W L i efiiix we 3 2 Pug g pm We ma apg i PP 1 we x r e 2 Pg 45 Wm I 1 l 3 wiv if M0 a W waitPg where M0 is the unamplified maximum moment In this case it results from initial crookedness but in general it can be the result of transverse loads or end moments The moment amplification factor is therefore 1 1 Pue 7 As we describe later the exact form of the AISC moment amplification factor can be slightly different from that shown in Expression 64 WWWWM team column of went it must be arms of effective ZOCgt BL 827 W 0145 which represents a 12 increase in bending moment The ampli ed primary moment is 112gtltMu 1120063 119 ft kips I BLAlf we 7 meow Answer Amplification factor 112 WWWM Solution can be slightly different from that shown in Expression 67 l E l EM ttie Em 9ampf wg gifgyw m Use Expression 67 to compute the LRFD amplification factor for the beam column of Example 61 taste g 5 Since the Euler load P3 is part of an amplification factor for a moment it must be c0 puted for the axis of bending which in this case is the x axis In terms of effectigie W length the Euler load can be written as quot anI nZEIx n229000272 P X KL2 1fo 10gtlt17gtlt122 e 1871kips 3 From the LRFD solution to Example 61 P 2004 kips and 1 1 1 Pu11 1 20041871 56 bf mg 9 szmy WW K4 a g4 5 E g ng fi r 5 gt 3 54 r 5 g W f s L w 352 w W f4 V quot 1 M if a EAR xxx3mm 1 W 05204quot 7 53 5 W 2 2 0578KJ 2m 3 510 6 499 45mg A 55 X igx i y wg f 813 EM If U K w K y f K 39 351i 43 2n l i m 5 WW 3 BQWKEQQ WM 394w quot gamw sAYfgw razxwwmmm 0 o fHJ K v41 50x 2 u f r 3 q ag7quot MC 1 J 2 Jugi WET f at 3 1 M Kl m W C Cg g W M 01914 M X M 1 E 5quot g f Cs 55 W I W V MM 2 6 3 39 2 A I a 2 Mi g V 1055 g 3amp3 fa i 5 H CO Qggaa Amplification Factor Input Shape W12x50 about axis of bending weak or strong 394 lnquot4 r Loading condition Uniform 195kft K 1 ratio M L 20 feet quotquotquot Axial load 102 kips we Mmax1 1356 kip feet quotm Delta 1 no axial effects bending only 0982 inches Using LRFD Method Peuler about axis of bending Final Moment Amplification Factor Using Iteration Method Delta 1 no axial effects Madded due to Delta 1 Delta 2 with Delta effect Madded due to Delta 2 Delta 3 with Delta effect Madded due to Delta 3 Delta 4 with Delta effect Madded due to Delta 4 Delta 5 with Delta effect Madded due to Delta 5 Delta 6 with Delta effect Madded due to Delta 6 Delta 7 with Delta effect Final Moment Amplification Factor 1 0982 835 1064 904 1071 910 1071 911 1071 911 1071 911 Get from moment tables page 5 162 LRFD Ta es Get from de ection tables page 5162 LRFD Tables kips using equation inches kip feet inches kip feet inches kip feet inches kip feet inches kip feet inches kip feet nches Without Cm using iteration 3 g 7 x k g W 2 Tk P Squot 802 45 i 1 W 49 wwwsz a E 5 a x EM W mm WA Ew W 3 k r W APE 51 W is ack W If W5 I A 2 ig m 43982 3 7W 3 w if g 23 W H W 5 mi wz E s g 7 mg f 2 Z 095825Mywwf T s5 ltP T Amwggv I u g 4W2 t o 55 I g 2 It w E QS f 3 3 gt lo M m Ema g m Amplification Factor input Shape W12x50 I about axis of bending weak or strong 394 Loading condition lnquot4 Uniform 195kft K 1 ratio L 20 feet Axial load 1632 kips Mmax 1 975 kip feet Delta 1 no axial effects 0614 inches Using LRFD Method Peuler about axis of bending 1957814 kips Final Moment Amplification Factor Using Iteration Method Delta 1 no axial effects 0614 Madded due to Delta 1 835 Delta 2 with Delta effect 0665 Madded due to Delta 2 905 Delta 3 with Delta effect 0670 Madded due to Delta 3 911 Delta 4 with Delta effect 0670 Madded due to Delta 4 911 Delta 5 with Delta effect 0670 Madded due to Delta 5 911 Delta 6 with Delta effect 0670 Madded due to Delta 6 911 Delta 7 with Delta effect 0670 Final Moment Amplification Factor using equation inches kip feet inches kip feet inches kip feet inches kip feet inches kip feet inches kip feet inches Get from moment tables page 5162 LRFD Tables Get from de ection tables page 5162 LRFD Tables Amplification Factor input Shape W12x50 about axis of bending weak or strong 394 lnquot4 Loading condition Uniform K 1 ratio L 20 feet Axial load 102 kips Mmax 1 156 kip feet Delta 1 no axial effects bending only 0 inches Using LRFD Method Peuler about axis of bending Final Moment Amplification Factor Using Iteration Method Delta 1 no axial effects 0982 inches Madded due to Delta 1 835 kip feet Delta 2 with Delta effect 1064 inches Madded due to Delta 2 904 kip feet Delta 3 with Delta effect 1071 inches Madded due to Delta 3 910 kip feet Delta 4 with Delta effect 1071 inches Madded due to Delta 4 911 kip feet Delta 5 with Delta effect 1071 inches Madded due to Delta 5 911 kip feet Delta 6 with Delta effect 1071 inches Madded due to Delta 6 911 kip feet Delta 7 with Delta effect 39nches Final Moment Amplification Factor Without Cm using iteration Get from moment tables page 5162 LRFD Ta es Get from deflection tables page 5162 LRFD Tables kips using equation 6 Mm M if Wsw 39 1Q ag i XE A Em g f A I pal5L A 3amp2 2 WCLltC LXL 5J 1 ETEL QEEL L A MKLE m M Li M23 i z A fmg aim 5 2 AL 2 Z 30 Wm 49 7 Sawsz r2f2log z 055 W 0 23000 KmzNE ZNE 5 A 1 m M Emab M9 cfi z a Zquot ZO3 MMM OZKZ 3MmZw gt 6 a7 5 bigExaf mm 1 QZSf H27 MUEL 3 K wy m gt55 R x I JQAWE k w W 16 it it Amplification Factor Rut Shape W1 2x50 l about axis of bending weak or strong 394 lnquot4 Loading condition Equal end moments of 975 K 1 ratio L 20 feet Axial load 102 kips Mmax 1 975 kip feet Delta 1 no axial effects bending only 0738 inches Using LRFD Method Peuler about axis of bending kips Final Moment Amplification Factor using equation Using Iteration Method Delta 1 no axial effects 0738 inches Madded due to Delta 1 627 kip feet Delta 2 with Delta effect 0784 inches Madded due to Delta 2 667 kip feet Delta 3 with Delta effect 0787 inches Madded due to Delta 3 669 kip feet Delta 4 with Delta effect 0787 inches Madded due to Delta 4 669 kip feet Delta 5 with Delta effect 0787 inches Madded due to Delta 5 669 kip feet Delta 6 with Delta effect 0787 inches Madded clue to Delta 6 669 kip feet Delta 7 with Delta effect 39nches Final Moment Amplification Factor Without Cm using iteration Get from moment tables page 5162 LRFD Tabl Get from deflection tables page 5162 LRFD Tables Vm x i X 5 M W W M a xwxzx Ex I g y E H N E 206 LRFL ASE 59 CLASS 8 DEFLECTION In addition to being safe a structure must be serviceable A serviceable structure is one that performs satisfactorily not causing any discomfort or perceptions of unsafety for the occupants or users of the structure For a beam being serviceable usually means that the deformations primarily the vertical sag or de ection must be limited Excessive de ection is usually an indication of a very exible beam which can lead to problems with vibrations The de ection itself can cause problems if elements attached to the beam can be damaged by small distortions In addition users of the structure may View large deflections negatively and wrongly assume that the struc ture is unsafe For the common case of a simply supported uniformly loaded beam such as that in Figure 522 the maximum vertical de ection is 5 wL4 384 EI 3 2 I g 208 Chapter 5 FIGURE 523 Answer Beams WD 500 Ibft WL 5501bft HHiHH amp W18gtlt35 3039 U egg l 7 The live load de ection is 3341144io5501230x124 L 384 E1 384 29000510 W 0678 in The maximum permissible live load de ection is L 3012 0K 360 360 10 in gt 0678 in The beam satisfies the de ection criterion y 6Q 3419 In Example 510 it was first assumed that a compact shape would be used and then the assumption was verified However if the search is made based on available strength glibMp or MpQb rather than section modulus it is irrelevant whether the shape is compact or noncompact This is because for noncompact shapes the tablulated values of gile and MpQl are based on ange local buckling and not the plastic moment see Section 56 This means that for laterally supported beams the Z table can be used for design without regard to whether the shape is compact or noncompact Beam Design Charts by Many graphs charts and tables are available for the practicing engineer and these aids can greatly simplify the design process For the sake of efficiency they are widely used in design offices but you should approach their use with caution and not allow basic principles to become obscured It is not our purpose to describe in this book all available design aids in detail but some are worthy of note particularly the curves of moment strength versus unbraced length given in Part 3 of the Manual These curves will be described with reference to Figure 525 which shows a graph of nominal moment strength as a function of unbraced length Lb for a particu lar compact shape Such a graph can be constructed for any crosssectional shape and specific values of F and C b by using the appropriate equations for moment strength The design charts in the Manual comprise a family of graphs similar to the one shown in Figure 525 Two sets of curves are available one for W shapes with F y 50 ksi and one for C and MC shapes with Fy 36 ksi Each graph gives the exural strength of a standard hot rolled shape Instead of giving the nominal strength M however both the allowable moment strength MQ and the design moment strength 1141 are given Two scales are shown on the vertical axis one for MnQb and one for bM All curves were generated with Cb 10 For other values of C1 simply mul tiply the moment from the chart by Cb However the strength can never exceed the value represented by the horizontal line at the left side of the graph For a compact shape this represents the strength corresponding to yielding reaching the plastic LY L4 FIGURE 525 510 Design 213 W2 X 14613 50 ksi compact L LP 20 L 1 lt2 9 39 3 3 7196 moment Mp If the curve is for a noncompact shape the horizontal line represents the ange local buckling strength 214 Chapter 5 Beams D FIGURE 527 M W WI 4 CM Somme i Name M39v 4 m 1 WW 2 6 49 7 EMA iquot O39 lt1wa zz40 31 L3 50 r13 72 5 M nominal strength based on FLB L 7 L Lquot M L L I J Form of the strength curve in the charts Crag El 400 Chapter 7 Simple Connections gigsemi quot i or M r a e x K6 E l I f F Equot 13Fm I v 31 Egan s M V AA 043 Q refer WEM 1 r mi or M l w as if A w x s 39 F f 1quot tter 431 s a W 2 1 31quot Lfv 4 ill quot5 E t Etii g ats 39 will n gab 726 WWW W where WM F f nominal tensile stressgi the presence of shear 2 W quotU Fm nominal tensile stress in the absence of shear 7 A Fm nominal shear stress in the absenCe of tension 553 T3 39 a wwwnvfr39 3 A i i fl required shear stress W w was qng Wi kevsWeeie ii Gitesii W that Fft must not exceed F and fv must not exceed Fm The nominal tensile TM 32 strength is then Rut FfA AISC Equation 13 2 Equation 726 will now be presented in the two design formats LRFD and I E21 Eu 1315quot m V S En AISC Equation J 3 3a 14731 r its 1 to if b 075 ASD l 1 f I Q ffX and k KIQEU z 13Fny 7 fv AISC Equation 13 31 nv 5 5 Where Q 200 The Commentary to AISC J 37 gives alternative interaction equations based on the elliptical solution Either these alternative equations or the elliptical equation it self Equation 723 may be used in lieu of AISC Equations J 33a and J 33b In this book we use AISC Equations 13 321 and J3 3b In slip critical connections subject to both shear and tension interaction of We me shear and tension need not be investigated However the effect of the applied ten 1 sile force is to relieve some of the clamping force thereby reducing the available g Lw wg wg friction force The AISC Specr cation reduces the slipcritical strength for this case E is 514 gratifw I Q e 393 w Example 79 79 Combined Shear and Tension in Fasteners 401 This reduction is not made for certain types of eccentric connections that will be covered in Chapter 8 The reduction is made by multiplying the slip criticaiftrength by a factor ks as follows Pi Wa g ggjwng new For LRFD TM gift T dd k k 1 I a W swat wreak s w Duan QQMCQf g 2 AISC Equation 13 521 AISC Equation J3 5b Where Tu total factored tensile load on the connection D 2 ratio of mean bolt pretension to specified minimum pretension default value is 113 Tb prescribed initial bolt tension from AISC Table I 31 Nb number of bolts in the connection The AISC Specification approach to the analysis of bolted connections loaded in both shear and tension can be summarized as follows Bearingtype connections 1 Check shear and bearing against the usual strengths 2 Check tension against the reduced tensile strength using AISC Equation J33a LRFD or J3 3b ASD I Egg07 Slipcritical connections 1 Check tension shear and bearing against the usual strengths 2 Check the slipcritical load against the reduced slipcritical strength A WT105 X 31 is used as a bracket to transmit a 60kip service load to 21 W14 x 90 35 g column as previously shown in Figure 730 The load consists of 15 kips dead load and 45 kips live load Four 7s inch diarneter A325 bolts are used The column is of A992 steel and the bracket is A36 Assume all spacing and edge distance require ments are satisfied including those necessary for the use of the maximum nominal strength in bearing ie 24 szu and determine the adequacy of the bolts for the fol lowing types of connections a bearing type connection with the threads in shear and b slipcritical connection with the threads in shear The following values are used in both the LRFD and ASP solutions 13 1 foT toea 31 ea lasts a as am M Shaw A sea at 5 g a vegan Vi Compute the nominal bearing strength ange of tee controls 7 24 558 39 R 24th gm 61 7491k1psE k W v V i 402 Chapter Simple Connections Nominal shear strength 2 Ab 7018 06013 in 1 ieqqz Maximal R EwAb 18506013 289 upsgemquot 7101 1ol N x LRFD Solution Pu 12D 16L 1215 1645 90 kips 1 S a The total shearbearing load 18 V 90 54 kips quot The shear bearing force per bolt is 14 54 a a Kbol 135kips Bert x w ggy 4 galTS W The design bearing strength is a x V a Wm Qa M 05Rn 0757491 562 kips gt 135 kips D OK x LT The design shear strength is L A Fig 9 seventies We 11Rz 075089 217 klps gt 135 Mpg 0K 0LT TEWS O N The total tension load is Tu 90 72 kips The tensile force per bolt is a 72 I w wnyw Wea a i Rho T 18 klps Aid h j f h agmiw s To determine thejgavallable tensile strength use AISC Equation J33a ECO lQLi39l M F j w www 6 13 13E quot E was figw 4 ml 3 3 a K 5 r j e as w m Fm nominal tens e stress 1n the absence of shear 90 1311 Fm nominal shear stress in the absence of tension 48 ksi 39 m CM Vilbolt f 2245 ksi Ab 06013 Then e j g rs E r 1390 0 7048 2245 6088 ksi lt 90 ksi fng B Exit r l f is R e 79 Combined Shear and Tension in Fasteners 403 The nominal tensile strength is Wm quot 90 RM FAb 2 6088O6013 3661 kips 80 Wk and the available tensile strength is R 0753661 275 kips gt 18 kip OK our Answer The connection is adequate as a bearing type connection In order not to obscure the combined loading features of this example prying action has not been included in the M 1 Avg 392 C A a a Liv 5 14 s Mm mmma Mis t W b From Part a the shear bearing and tensrle strengths are satisfactory From AISC Equation J 34 the slip critical strength is 1 s We my X 121 2 LLDuhschNS U Av 39 Ma r k I 1 Egg E13524 L g 075 MW WW1 3 E a i A 1 3291 From AISC Table J 31 the prescribed tension for a 78inch diameter A 25 bolt is Tb 39 kips If we assume Class A surfaces the slip coefficient is u 2 035 and for four bolts R 1 thsz X 4 03511310391 x 4 6170 kweea 39 i I k gmquot 106170 6170 kips pityg5 54 5 Since there is a tensile load on the bolts the slipcritical strength must be reduced by a factor of 0t was T 72 Ismscase ng sing u k 1quot 05916 DuTbNb 113394 WW vi we veth The reduced strength is therefore ks6l70 O59166l70 365 kips lt 54 kips NG Answer The connection is inadequate as a slip critical connection ft x a w i i M 405 79 Combined Shear and Tension in Fasteners Connections with fasteners in shear and tension can be designed by trial but a more direct procedure can be used if one assumes that the design is controlled by the strength that is reduced If the assumption turns out to be correct no iteration is re quired This technique will be illustrated in the following example N Example 710 2d LRFD Solution A concentrically loaded connection is subjected to a service load shear forCe of 50 hips and a serviceload tensile force of 100 kips The loads are 25 dead load and 75 live load The fasteners will be in single shear and bearing strength will be controlled by a 516inchthick connected part of A36 steel Assume that all spacing and edge dis tances are satisfactory including those that permit the maximum nominal bearing strength of 24th to be used Determine the required number of 3 iinchdiameter A325 bolts for the following cases a a bearing type connection with threads in the plane of shear and b a slip critical connection with threads in the plane of shear All contact surfaces have clean mill scale 7 Consider this to be a preliminary design so that no consideration of prying action is necessary 3 page as L i s 6 Factored load shear l202550 l607550 75 kips Factored load tension l2025100 l6075100 150 kips Lem e L 1591 406 Chapter 7 Simple Connections L V 7 a For the bearingtype connection with threads in the shear plane assume that ten sion controls 7 gt hi 39 xxk x k K 5 I H a 117 25fv 390 BMW 48 W 513 075017 25 fv 3 07590 Low 8775 1875 fv s 675 Let 17713 150 and 75 2A1 2A where ZAb is the total bolt areajgubstituting and solving for ZAb we get A x it 150 75 8775 1875 j e 9 i a 2 a ZAb EA M tr 150 87752Ab 187575 2A1 3312 1712 5 The area of one bolt is 2 A1 73 04418 in w and the number of bolts required is Nb 2 341128 HOEOCf SV a b V 6 if 4 Try eight bolts First check the upper limit on 132 M Eig p 12sz 3 L M i fV mb 8014418 2122kS1 4 Qt75ltlg 563 V F 2117 25fv117 252122640 ksi lt 90 ksi OK m r WWWWWWM 1 k m ngaa MWCheck shear w R11 I WAb be 07548044188 n 1271ltips gt 75 kips OK 5 3 g t t g 79 Combined Shear and Tension in Fasteners 407 Check bearing we M M g wwm mx 13 24th x 8 bolts 07524 588 196 kips gt 75 kips might WWW La g o I Answer Use eight bolts L h Slip critical connection Assume that the slip load controls The reduced slip critical strength is we39vemm W 5 Where l C 1 150 4741 g s D1111le Nb h i l where Tb 28 kips from AISC Table J 3 1 E i For one bolt Rn uDlh5CTNY l 0035113l02810 1107 klps Eon l g l Setting the total factoredload shear load to the reduced slip critical strength for Nb bOItS E 7 E 4 0 75Nb1 411107 Mb AP 7L H g I a 1107Nb 4 5248 g N1 115 t l Since eight bolts are adequate for shear bearing and tension with a reduced tensile strength these limit states will not have to be checked 5 l Answer Use twelve inch diameter A325 bolts l l Rmt Q5 lt M 2 37 Steel heali e igg gag 3 lea et gt t 9 5r Z 393 ME 55quot 33 W Ff if eke Alleges 409 710 WELDED CONNECTIONS Structural welding is a process whereby the parts to be connected are heated and fused with supplementary molten metal added to the joint For example the tension member lap joint shown in Figure 733 can be constructed by welding across the ends of both connected parts A relatively small depth of material will become molten and upon cooling the structural steel and the weld metal will act as one continuous part where they are joined T he additional metal sometimes referred to as ller metal is deposited from a special electrode which is part of an electrical circuit that includes the connected part or base metal In the shielded metal arc welding SMAW process shown schematically in Figure 734 current arcs across a gap between the electrode and base metal heating the connected parts and depositing part of the electrode into the molten base metal A special coating on the electrode vaporizes and forms a protective gaseous FIGURE 733 lt gt W53 3 WWWWMWMWWM E f g 410 FIGURE 734 Chapter 7 Simple Connections Direction of travel Coated electrode Arc Slag Gaseous shield Bead ten J 1 Base metal shield preventing the molten weld metal from oxidizing before it solidifies The elec trode is moved across the joint and a Weld head is deposited its size depending on the rate of travel of the electrode As the weld cools impurities rise to the surface forming a coating called slag that must be removed before the member is painted or another pass is made with the electrode 111 FHthlLe We W AWW HGURE735 Bu Ike MW WHMM f v Corner 1 Complete penetration groove welds WW gall W m 2 b Partial penetration groove welds M FIGURE 736 connected parts can be treated as completely continuous at the joint The strength of a partial penetration groove weld will depend on the amount of penetration once that has been determined the design procedure will be essentially the same as that for a fillet weld CVEN 446 CLASS NUMBER DATE ff PAGE W 6 1 20aw Qim Er W gm 2 r V E3 3 3 ieffod g 5 gm 21 gw fe g f z a gm 2 ag i m O o 348 M W W m A X 90 k s gilt N 39 g 3 W gm 2ng mm Qxigimm m 102 5 Fi gilig f 3 MM f quotFEW g X r W A a g as I m ngEaEQ w B g 39 a quot quot W i gig lt2 CVEN 446 CLASS NUMBER DATE PAGE 91 In rmdum inn 553 Section 91 Introduction 555 We first require that the strain in the concrete at any point be equal to the strain in any replacement steel at that point 8683 or f c f EC ES and Es s cnc f ch f where E modulus of elasticity of concrete E n 5 modular rat10 C 91 Introduction 557 FIGURE 95 REF m 1 A1 131412 Z a 7617 TABLE 91 Component A y Ay 8quot Concrete 5440 250 1360 2 75 W16X36 106 1295 1373 a f E 6500 2733 TABLE 92 Component A y l d 7 Ad2 Concrete 5440 250 1133 1705 2714 W16 X 36 106 1295 448 8745 1259 550 Chapter 9 Composite Construction FIGURE 97 91 Introduction 551 Example 92 Compute the available strength of the composite beam of Example 91 Assume that sufficient shear connectors are provided for full composite behavior Soiution Determine the compressive force C in the concrete horizontal shear force at the in terface between the concrete and steel Because there will be full composite action K this force will be the smaller of ASP and 0853940 W 5570 AsFy 10650 530 kips 08572710 08545 x 87 1479 kips MW 353 at The steel controls C 530 kips This means that the full depth of the slab is not needed to develop the required compression force The stress distribution shown in Figure 98 will result The resultant compressive force can also be expressed as C 085 fcfab from which we obtain ti C 530 a l792in 085 fc b 08b54 37 39 1 0852 FIGURE 98 b gi g a 1 Vz K L i 5amp0 ti it a 6 C I w weequot d2 y d 5 TQ 12 530 amp CVEN 446 Structural Steglra s lfg 0L 1 Prepared by Z Date gZ lr b a Page g39 quotof 39Q H g whim m Am n mgg h efogmmfc 466 39oux 2m 0 139 6Lffd I n 1 nt d 39cizacj Wl 2 quotM3 e MW 39 g 539 17 MA 0M3 Coy 2676 L h DW kOVlMCI applicmm II Me 6 14 g K m MM 2 L19an E WM 1 12 quotmaze 243 1an ever S c fw but 413 TB 1 We 49 mm Wa v i fp nm 1L m cws Li LIN 0kg Mo l mwh as Ear lie quotWW 0 WMW quotsz fa 0dr 2 0 CL res H o quotMum w raw 4 quotQM qua llv W as 07M 111M S Pvt83M JUL WM u I x a Texas AampM University College Station Texas 77843 CVEN39446 Prepared by 2 2 lPate 26 39 age r Structural Steel Design L 5 150 425 a Legal Atogum f mam Mammals MW quot gawok 2S wayfum MV FW P umbm 4ath w 39 39 0 WMNSIeeJUB 39a 99 9 AR Ci e I W WMLFQS K We 0 as l S o Mos Man a Magel bu hafcede famed 61 W WAFGCZJ f0 Mags 7quot 21W Mrzr dg w 7 13 A lam mJJ A KIN vi iiluv v l vat 3 uv wnr 51 051 bug a h ak Usedm ch S U 25 41 Malayan Raw c 5002m EHar Code quot144 Sou39 em 5Ciie quot Con Mai 39 I I Uazd PmmaIV n SowWm skies 9 BM H44 Bug 12A ampMmm 39 39 USLA Acp j k Texas AampM University College Station Texas 77843 CVEN 446 Structural Stee Design Prepared b 1 K Date Page a of am ajmct malm Infiu WW 4 WK QwA 0 Coda Coma Na192 s W t 069 58C ALFtale r a 5064 r 6657 ninirrqu an f mu 5 0M Mix U WW but it L4 j 7 41m SW 211 6amp5 ms rmJ 12sz 31 Wu it Was i 1242 Una Eh M ML W6 M g i as Mai CI d g n svec A L MMCW 0V1 dim 0ng Le aidi w ai r Mtg PW 81 Mimi Texas AampM University College Station Texas 77843 CVEN 446 Prepared by Date 2 06 Structural Steel Design Page of 3 S bf l bui EULA Canmgfwkong U tr Tr My mimic quot MVSH TO i Newly bn Ajzsami newt I S mkS 554145035 Cancee4e3 Kailwau EN 34gquot 5 dialed 5chth T 5460 W294i CommCe I P 7 g Hemefem Paihumfr Emahw wt mmwtmce 94 Way gg k a 395 COLA 143m fight e dd Head 33 7 Hma awg L6 SWJWI 3965 Shuzr wd pmM es M We bum 1 12 4000 52 u r Pffx g 4 K36 Mae ka Warmeel 61712er g lea 1A lye A rem 5 IT D39T k Udallaw bd AL W M baf e MC 6amp0 Texas AampM University College Station Texas 77843 CVEN 446 Structural Steel Design Prepared by Date Page f 9 gtz oe VM REMJail Unti d amysec nmf Mea Q 39 T n 5M PrE E z I I z ny A T at r P Lu p H v P 3 L 4391 AL A L L quot 9 Ir I H 196 SWSg fr 2 g 1 I C 5262151 F 39th 1Q 53 L I Y E I 3 F t quot VMWWW E Ei39a hifquot mmquot harm quotT Macaw w 71 FLU e Texas AampM University College Station Texas 77843 Design 3 24425 M mea Vega Hes x Aaglam FV Fa I i I L L 9066 va e 64 j emmp 2 F s Sieei i l3 PiaOx Uibovx i eai 2 Low 40 Siee W 04 Speci iw y S39i eei i 05439 Cammm Seed g lli quot quotigr Duigpwl m quot g452 A35 3 5 37ng 2070 A572 6e 50 50 6 5 897 he AWZ 60 9 M T0347 WLUS E COMMMKV U eci Ska I Texas AampM University College Station Texas 77843 484 Chapter 8 FIGURE 835 Moment FEM w z 090 FEM w Eccentric Connections 39 399 lt4 W3 F E9 9 LD X V Qlt 13 9v Lg M39 amp 8Q 070 Rotation 86 Moment Resisting Connections 485 FIGUREBgs HHHMMMWMM 792 7 L L J I l R if 13 WW a 11 w g 3 11224 A g g 95 2 353 M 12 I 8 b wL216 1me gy w may Wm Conventional shim En WE Finger shim d 86 Moment Resisting Connections 487 Example 8 10 Design a three plate moment connection of the type shown in Figure 838 for the connection of 21 W21 X 50 beam to the ange of a W14 X 99 column Assume a beam setback of V2 inch The connection must transfer the following service load effects a dead load moment of 35 ftkips a live load moment of 105 ft kips a dead load shear of 55 kips and a live load shear of 165 kips All plates are to be shop welded to the column with E70XX electrodes and field bolted to the beam with A325 bearing type bolts A36 steel is used for the plates and A992 steel is used for the beam and column E PL FIGURE 838 W14 x 99 LRFD Solution M 12 MD 16 ML 1235 16105 2100 ftkips p Vu 12 V 16 V 1255 l6165 330 kips V W515 1919176 D L Tag For the web plate try 34inchdiameter bolts Neglect eccentricity and assume that EC L IV the threads are in the plane of shear The nominal shear strength of one bolt is Rquot 1 va 4804418 2121 kjps V o m 488 Chapter 8 Eccentric Connections quot 2 1 3 1 13 gr Me hdwg WE 16 4 16 16 m EQXR QC For the hole nearest the edge LC Le 15 13 1 1094 in V The nominal bearing strength is Rn 12LctFu 121094t58 7614 X subject to an upper limit of 24 24gt58 1044 1quot Use RR 7614 and 1 gm 07506140 5711 kipsbolt 9 For the other holes LC s h3 E 2188 in 1 16 R 12ch F l22188t58 1523 V 24thquot 1044t controls 1i RM 0751044t 7830t kipsbolt V 489 86 MomentResisting Connections it we eat R060FyAg0y603 6 39 R1001836t For shear rupture the net area is gigabmt From AISC Equation J44 9375 Z Lit Rquot 06FuAW 0658 r875z 2045 WA qugiima 43R 0752045z lief CST9 git R i wm xiweiaaa t 53 Shear rugture controls Let E e 0 tea Wt 33 r 021 5in W Wt A3E9ta2 ig g For the connection breather platetat aiumn ange liai aatt a stt ngm 5 R 3 per inch is Y 33 3667 kipsin 5 tease eet 18 3 39 r t a From Equation 735 the base metal shear yield strength per unit length is at 06143 0636 5400 kipsin VA 0 an 44 From Equation 736 the WtaLshea rggture strength per unit length is WWm4w 7 90Rquot 2 04513 045ss 625 Ripsin t A an tent 5 1 The base metal shear strength is theil ig iig ggnkip 111 which is greater tm WWWWM quired strength of For welds onMBBt m taestmattut plate the quotrequired strength per weld is 3667 ill8 34 kips To determine the required weld size let 13921 1834 D 132 sixteenths of an inch 0 The minimum weld size from AISC Table J24 is 18 inch based on the thinner con W R 0 5003 tit 3 We i3 f i at nected part the shear plate There is no maximum size requirement for this type of 39 connection one where the weld is not along an edge Use a 18inch llet weld on each side of the plat f w w 39 yg WIhe largest required thickness ispfgorthelimit state ofshearruwpture Try t 14 M 93 M 490 Chapter 8 Eccentric Connections Wee I The minimum width of the plate can be determined from a consideration of edge dis tances The load being resisted the beam reaction is vertical so the horizontal edge 1o EQYE distance need only conform to the clearance requirements of AISC Table 134 If we assume a sheared edge the minimum edge distance is 1 inch With a beam setback of 12 inch and edge distances of 112 inches as shown in Figure 83 the width of the plate is Q g 05 215 35 in 397 Q Try a plate 14 X 312 Check block shear The shear areas arena j v3 42 1 fr V AgVZ33151875 in E Sheila AWi 3315 25 El 1328 in M 4 4 8 3 The tension area is f 1 3 1 2 l t Aquot Z15 05Z 02656 in t 83 GE 3 iii For this type of block shear U 10 From AISC Equation J45 Rn 06FAL UbsFuAn to l 3 5 Z 06581328 105802656 6162 kips with an upper limit of 06FyAgv UbsFuAm 0636l875 105802656 5590 kips lt 6162 kips pRu 2 0756590 419 kips gt 33 kips OK 81 M El eaplateigm FLA 1 6g For the ange connection first select the bolts From Figure 840 the force at W the interface between the beam ange and the plate is a r i N etamm M 21002 1212 kips A En H M E in d 208 W 39 mwi lxre d is the depth of the beam Although the moment arm of the couple is actually the distance from center of ange plate to center of ange plate the plate thickness is not yet known and use of the beam depth is conservative w i 1 2 m a FIGURE 840 f M 86 Moment Resisting Connections 491 M Try inch A325 bolts Since 3At irichdiameter bolts were selected for the shear connection try the same size here If bolt shear controls the number of bolts required is 12 762 use 8 bolts 4 pairs ERM 4 Use edge distances of 1 inches spacings of 3 inches and determine the minimum FLAG it 65 i Keg plate thickness required for bearing Use a hole diameter of m 1 3 1 13 wt u 5 3 E h W a d 16 4 16 16 m For the hole nearest the edge LC Lew 15 1 1094 in 2 2 The nominal bearing strength is R 21 425 7614t gymw g 12LCrF 1098 W subject to an upper limit of 241th 24 58 1044 Use Ru 2 7614t and 25Rn 07576l4t 5711t kipsbolt For the other holes g t on m Lcsh3 1 32188 in 16 4 Rn 12Lchu 24mm 10442 controls f i if 3n g 2 R 07500449 78302 kipsbolt WM Welshman MW v the required thickness equate the total bearing strength to the applied load ONLY M 4453 25711z67830r1212 r0208 in The ange plate will be designed as a tension connecting element the top w P E A W KEN plate then checked for compression the bottom plate The minimum cross section t required for tension on the gross and net areas will now be determined From AISC a X 2 Equation J4 1 WWWWYW Rn FyAg 451 090F A 3 R y g H 1212 Required Ag quot 39 3741 in2 1901 2 09013 09086 492 Chapter 8 Eccentric Connections From AISC Equation J4 2 El en l 5quot l l 393 RIZ EIAE M E l pa 07512146 M Required Ag R H 1212 2786 in2 075Fu 07527 07558 Try a plate width of wg 2 65 in equal to the beam ange width Compute the thickness needed to satisfy the gross area requirement r f 3 3 L A8 65t 3741 in or t 0576 in Compute the thickness needed to satisfy the net area requirement if 7 A3 A twquot twg Zdhole 65 4750t Let 47501 2 2786 in2 or t 0587 in controls K t 33 as This thickness is also greater than that required for bearing so it will be the miltimum a 3 W 7 acceptable thickness Try a plate gtlt 612 04225 3 a WWW Wwwnam WM WWW a a m quot 3WWChec bompreSSion Assume that the plate acts as a fixedend compression memV ber between fasteners with L 3 in and K 065 was a 7 3 E22 l e a a if m 3 W i r 1 639558 2 01804 in V W m g A 6558 h t I lt3 i C735 73 quot 53 5 Ii9 6 1081 K ana r 01804 Mei 3 DQ t l l 79 From AISC J44 for compression elements with KLr lt 25 the nominal strength is PM FyAg M a WWW CW AISC Equation J4 6 and for LRFD q 09 9le 09FyAg 0936 x 65 132 kips gt 1212 kips OK 39 8 was mas it Check block shear in the plate In the trartsVE fseilirection use bolt edge distances of quot 112 inches and a spacing of 312 inches quot his places the bolts at the workable gage F336 i location in the beam ange see Part 1 of the Manual Figure 841 shows the bolt39lay 4 M MW out and two possible block shear failure modes The shear areas for both cases are I sea aquot Agv 33315gtlt21313in2 An 33315 35 JX2 9297 in2 86 MomentResisting Connections 493 FIGURE 841 T swxa tester w e h it W A A ii CB4 ij eegwb ig I M 5M I V V 41 lt 3 lt 3 3UH g Reggae 1 1 1 n 5 2 12 I W J a a V W L V2 9 CRtTlCAL N f amp k E e at T a r 57 th I 3 112 by 999 gig M v J 312 l l l l I l l I I I I I l I I I l I l I l I l VO f gm 1 V2 29 i l 12 112quot 3 3 3 112 M 1 If a transverse tension area between the bolts is considered Figure 84121 the Width is 35 in If two outer blocks are considered Figure 84119 the total tension Width is 215 30 in This will result in the smallest block shear strength Am 15 05 Hx2 13281n 8 Rn O 6FHAIZV UbsFuAnt I O6589279 10 581328 3999 kips if Upper limit 06143919 UbsFuAquot O6361313 10581328 3606 kips arequot 0756606 271 kips gt 121 kips OK sis WCh eclE Block shear in the beam ange The bolt spacing and edge distance are the same as for the plate i 1 Aggy i 3 3 494 Chapter 8 Eccentric Connections 24 egg 4q A M in 352 f f g 06657958 10651137 3843 kips 06FA UbsFuA 06501124 10651137 4111 kips gt 3843 kips gtgV Since 3843 kips gt 3606 kips block shear in the plate controls Wmaw MN Use a plate 58 x 612 Part of the beam ange area will be lost because 01 thewb olt holes zk N wmw WM W Use the provisions of AISC F131 to determineWhether we neEEi to account for this loss The gross area of one ange is E 9 l gt 1 Afg 2129 0535653 34941112 The effective hole diameter is dh 2 l Z in 4 8 8 1 and the net ange area IS 5 E 9 2 A Afg deh 3494 05352 gtlt 2558 in2 EA 652558 1663 kips 4 Determine K For A992 steel the maximum F y F I ratio is 085 Since this is greater than 08 use Y 11 1131492 KFyAfg 11503494 1922 kips Since FuAfn lt YtFyAfg the holes must be accounted for Frorm AISQ Equation FIB 1 H97 lt 012 were i F A WWW i q Mn fquot 5x l633945 4498 inkips quot 3 Afg 3494 QMquot 0904498 4048 in kips 337 ftkips gt 210 ft kips OK V Answer Use the connection shown in Figure 842 column stiffener requirements will be con sidered in Section 87V 40740 Calculation of Flange Ant CLASS 35 455 l to per t ll o39l Example 83 Use the tables in Part 7 of the Manual to determine the available strength based on bolt FIGURE 811 Solution shear for the connection shown in Figure 811 The bolts are 3At inch A325 bearing type with the threads in the plane of shear The bolts are in single shear 7 38 V W z i Ii Sn L OO nquot 46 This connection corresponds to the connections in Table 7 8 for Angle 2 0 The ec centricity is ex8l595 in The number of bolts per vertical row is n 3 From Table 78 M 739 3 a C 153 by interpolation The nominal strength of a 3 t inchdiameter bolt in single shear is rquot FWAb 48O4418 2 2121 kips Dist i quot l 0 439 Here we use r for the nominal strength of a single bolt and R for the strength of the connection The nominal strength of the connection is Rn Cr 1532121 3245 kips Answer For LRFD the available strength of the connection is 12 O753245 243 kips 7 58 Table 7 8 MiBouts DEE 455a Coefficients C for Eccentrically Loaded Bolt Groups LOAD Angle 0 B39Qmm U Available Strength of a bolt group where of or RnQ is determined with P required force Pu or Pa kips ltxgt19 6 R 0X r nominal strength per bolt kips g V 7 quot quot 2 00 e eccentricity owaith respect quot O 4 Pl 4 039 5 1 Q 39 t0 centroid of bolt group in quot O V or not tabulated may be M A O LRFD ASD determined by geometry f f mmk e horizontal component ofe in V P 9P X cmin I cmmTa s bolt spacmg 1n 3 n n 0 coe ICIent tabulated below 41 Number of Bolts in One Vertical Row n s1n em quot 1 2 C3 4 5 e 7 8 l 9 I 10 11 12 9 mm 954 442 659 R79 me 190 15n17nl10n91n 930 a 0 023 337 534 50 355 10 123 142 1031 10 026 353 490 542 810 991 118 138 159 C3 12 022 301 419 551 701 863 104 122 142 I VERTILALSACIN G Homework b Verify the Table 7 value for a bolted connection consisting of 3 bolts horizontal and 2 bolts deep The bolts are spaced 6quot horizontally and 3quot vertically The load is directed downwards at an angle of 30 degrees clockwise from vertical as shown on all Table 7 gures and is 9 inches to the right and level with the elastic centroid of the bolts Use 1quot diameter A325 bolts with the threads in the shear plane with slip permitted llt Pm K Q Cr 267674 Q7528 m N o Fm quotas A325 W 134i fig48M 37 LP m i K 3 2 7lt377 OO39Q K M PRM 075000390 37 You may put the instantaneous center of rotation at any point you like The moments and forces resisting and applied below just won39t equal each other Calculations are good for vertical and horizontal loads and are based on Ru 74k and deltamax 034quot deformation for A325 34quot bolts Adjust these nal answers to account for 1 Your bolt39s area and grade of steel and 2 The reduced bolt capacity allowed by the LRFD code See page 378 Segui for example except that these calculations use all X and Y referenced to bottom left bolt Number of bolts in connection ote that any rows below this bolt number in the table below will be ignored X location Y location quotX location Y location Angle of lt Angle measured from vertically down and counterclockwise of lC of IC of l 01 219395 37 Elastic Xbar Ybar Centroid 6 15 Data entry only permitted in colored cells Use Excel39s quotSolverquot for solution and set TARGET cell J33 equal to max by changing cells B8C8 subject to constraints l33l29 amp K33K29 Fastener Xref Yref ch from lC Lcy from lC r from IC Delta Resistance rR Rsuby Rsubx to bolt to bolt to bolt to bolt to bolt Bolt Deform 1 20840 28336698 3517 01160 60166 211631 35646 48470 2 39160 28336698 4834 01594 65309 315689 52910 38286 3 99160 28336698 10313 03400 72631 749046 69836 19957 4 20840 01663302 2091 00689 50434 105437 50274 4013 5 39160 01663302 3920 01292 62020 243090 61964 2632 6 99160 01663302 9917 03270 72439 718410 72429 1215 7 20840 61663302 6509 02146 69107 449812 22126 65469 8 09160 61663302 6234 02055 68628 427824 10084 67883 9 20840 91663302 9400 03099 72146 678188 15994 70351 10 09160 91663302 9212 03037 72026 663500 7162 71669 Cell Typical Equation Due to bolt xampy force components Nifty Function J5 SUMB16B25COUNTB16B25 Moment Sum y Sum x To get only certain cells added K5 SUMC16025COUNTC162025 Sum gt 234330 17122 9885 lt SUMOFFSETK1600D51 D16 B8B16 E16 C8C16 Due to location of applied forces F16 D16quot2E16quot2quot05 Moment Sum y Sum x G16 034F16MAXOFFSETF1600D51 Sum gt 234330 17122 9885 H16 741 EXP10G16quot055 I16 F16H16 Final force on connection 19771 kips J16 D16F16H16 Your bolt strength 100 kips K16 E13F13H13 Permitted force on connection 267 kips 0413576 2417933 l29 SUMI16l25 a 39 1 J29 SUMJ162J25 Im l il i mlnwlbmm K29 SUMOFFSETK1600D51 4 I33 J29D8B8K29E8CB J33 J29 K33 J33TANPI180F8 P9917 r 453 Chapter 8 Eccentric Connections For the hole nearest the edge use LE 2 15 in Then LC Le 15 1 3 161094 in BM 1a 12Lctli39 121094056065 4779 kips A Upper limit 24th 2480560x65 6552 kips gt 4779 kips 7t34 2 Ab 04418 1n 4 qomimaQ a LMu 4amp Rn anAb 4804418 2121 kips 39 39 Since the nominal shear strength is less than the nominal bearing strength of any bolt the shear strength controls LRFD Solution The factored load is PM 12D 16L 2 1220 1640 88 kips and the shearbearing load per bolt is 88 11 kips The shear design strength per bolt is 45R 0752121 159 kips gt 11 kips OK Compute the tensile force per bolt and then check the tension shear interaction Be cause of symmetry the centroid of the connection is at middepth Figure 815 shows l I o the bolt areas and the distribution of bolt tens e forces FIGURE 815 0000 0000 83 Eccentric Bolted Connections Shear Plus Tension 459 The moment of the resisting couple is found by summing moments about the neutral axis ZMNA 2r45 15 15 45 24r The applied moment is M Pue 88275 242 inkips l Equating the resisting and applied moments we get 24rr 242 or r1 1008 kips The factored load shear stress is h M to M 1 1 fv 04418 2490 k31 200x904 MAJ J uf M and from AISC Equation 133a the nominal tensile stress 2 Fm lt PM Fm 13Equot fv Fm m 411v at L 1 439 90 T 1390 2490 5475 ks1 lt 90 k51 WW 017548 24 W The design tensile strength is bRn O75FfA 075547504418 181 kips gt 1008 kips OK Answer The connection is satisfactory


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