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1 AISC Live Webinars Thank you for joining our live webinar today. We will begin shortly. Please stand by. Thank you. Need Help? Call ReadyTalk Support: There s always a solution in Steel AISC Live Webinars Today s audio will be broadcast through the internet. Alternatively, to hear the audio through the phone, dial: Access Code:

2 AISC Live Webinars Today s live webinar will begin shortly. Please stand by. As a reminder, all lines have been muted. Please type any questions or comments through the Chat feature on the left portion of your screen. Today s audio will be broadcast through the internet. Alternatively, to hear the audio through the phone, dial Access Code: AISC Live Webinars AISC is a Registered Provider with The American Institute of Architects Continuing Education Systems (AIA/CES). Credit(s) earned on completion of this program will be reported to AIA/CES for AIA members. Certificates of Completion for both AIA members and non-aia members are available upon request. This program is registered with AIA/CES for continuing professional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner of handling, using, distributing, or dealing in any material or product. Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation.

3 AISC Live Webinars Copyright Materials This presentation is protected by US and International Copyright laws. Reproduction, distribution, display and use of the presentation without written permission of AISC is prohibited. The 017 The information presented herein is based on recognized engineering principles and is for general information only. While it is believed to be accurate, this information should not be applied to any specific application without competent professional examination and verification by a licensed professional engineer. Anyone making use of this information assumes all liability arising from such use. Course Description This presentation will highlight efficient ways to take advantage of the various design aids in the AISC Steel Construction Manual (14 th Edition). The speaker will walk through the different sections of the Manual and highlight useful resources. The speaker will then focus on shortcuts for connection design, including: stiffeners, web yielding/crippling, welds, and more. Shortcuts for member design including composite members will also be reviewed. Learning these shortcuts are a must for those designing steel structures and checking designs. 3

4 Learning Objectives Locate important design shortcuts in the Manual Identify appropriate uses for eccentrically loaded bolt group tables Apply table shortcuts for beam bearing and column stiffener checks Design beams using composite beam tables in the Manual Based on the 14 th Edition Manual and AISC Presented by Carol Drucker, PE, SE Drucker Zajdel Structural Engineers Chicago, IL 8 4

5 Manual Organization- General Parts Part 1: Dimension and Properties d Part : General Design Considerations 9 Manual Organization Main Member Design Part 3: Flexural Members Part 4: Compression Members Part 5: Tension Members Part 6: Members Subject to Combined Forces 10 5

6 Manual Organization-Connections Part 7: Design of Bolts Part 8: Design of Welds Part 9: Design of Elements Part 10: Simple Shear Connections Parts 11,1: Moment Connections Part 13: Bracing and Truss Connections Part 14: Base Plates, Anchor Rods, Column Splices Part 15: Hanger Connections, Brackets 11 Manual Organization-Main Member/Connections Part 16: Specification, Commentary, and Codes AISC » Chapter A- N: Main Member, Stability, Connections, Fabrication, etc» Appendix 1- Appendix 8: Ponding, Fatigue, Fire, Existing, Stability, Analysis, etc» Commentary for Chapter A-N» Commentary for Appendix 1-8 RCSC: Specification for Structural Joints Using High-Strength Bolts AISC: : Code of Standard Practice 1 6

7 Manual Organization-Main Member/Connections Part 17: Miscellaneous Data and Mathematical Information Metric Shapes Gage metal thickness Coefficient of thermal expansion Material Weights Properties of Shapes General Nomenclature Index 13 Part 1: Structural Properties A, d, t w, b f, t f, etc =INDEX('Database v14.0.1'!$1:$ ,match("w1x96",'database v14.0.1'!$c:$c,0),match("bf",'database v14.0.1'!$:$,0)) Link to shapes database:

8 Part 1: Structural Properties Angle Gage in Table 1-7A Welded Flat Widths HSS g g 1 g 15 Part : General Design Considerations Bearing Connections Table -4: Preferred Material Specification- Shapes Table -6: ASTM Fasteners (Bolts, Rods) ASD to LRFD 1.5 Ω = φ 16 8

9

10 Part 3: Main Member Table 3-: Selection by Z x Table 3-6: Maximum Total Uniform Load Table 3-10: Available Moment vs Unbraced Length Table 3-3: Beam Shear, Moment, and Deflection 19 Part 4, Part 5, Part 6: Main Member Table 4-1: Compression of W-Sections Table 4-8 to 4-10: Axial Compression Double Angles Table 4-1: Eccentrically Loaded Single Angles Table 4-: kl/r Table 0 10

11 Part 4, Part 5, Part 6: Main Member Table 4-8 to 4-10: Axial Compression Double Angles 1 Part 4-6: Main Member Manual Organization Table 4-1: Eccentrically Loaded Single Angles The effects of eccentricity can be neglected using an effective slenderness ratio if (Spec E5): Member are loaded at the ends in compression through same on leg Member attached by welding or by a minimum of two bolts No transfer loading 11

12 Part 4, Part 5, Part 6: Main Member Table 4-: KL/r Table 3 Table 4- Use for Whitmore Check for Gussets KL KL 1 = r tgusset Bo Dowswell (01) Technical Note: Effective Length Factors for Gusset Plates in Chevron Braced Frames in AISC Engineering Journal 3 rd Quarter. Free download for AISC members! 4 1

13 Part 7: Bolts Table 7-1, 7-, and 7-3: Bolt Shear and Tension Values Table 7-4: Bolt Dimensions Table 7-6: Coefficients C for Eccentrically loaded bolt groups Table 7-15 and 7-16: Bolt Installing Tolerance s s s s e x 5 Hex Head Bolt (ASTM A35) TC Bolt (ASTM F185) 6 13

14 Bolt Installation-TC Bolts Example: gwyinc.com 7 BOLT TENSION rt = n ' M c b dm CG (tension group) r t d m NA CG (compression group) See Manual Fig

15 See Manual Fig. 7-6 Case I 9 M = Σn ( rt ) d n i i See Manual Fig. 7-7 Expanded (similar to DG4* and DG16*) *free download for AISC members! 30 15

16 Part 8: Welds Figure 8-16: Susceptible and Improved Details Table 8-: Prequalified Welds (AWS) Table 8-3: Electrode Strength Coefficient, C1 Table 8-4 to Table 8-11: Coefficients C for Eccentrically Loaded Weld Groups (0.9)(80 ksi) C = 1 70 ksi 31 Part 8: Welds: Table 8-1 Cost of Welds Fillet and Groove Welds 3 16

17 Part 9: Connections Cope Beams: Single Coped and Double Coped Prying Action Rotational Ductility Filler and Shims Table 9-4: Beam Bearing Prying 33 Part 9: Connections Cope Beams: Single Coped and Double Coped Prying Action Existing Rotational Ductility Filler and Shims Table 9-4: Beam Bearing WT Connection End Plate Connection 34 17

18 Part 9: Shear Rupture at Welds t min to develop weld: Shear Rupture = Fillet Weld Strength φw(0.6)( Fexx )()(0.707)( w) tmin = Manual Eq.(9 ) φ(0.6)( Fu ) D 0.75(0.60)(70 ksi)()(0.707)( ) = (0.6)( Fu ) (1.39)( D) = 0.75(0.6)( Fu ) 6.19D = Manual Eq.(9 3) (-Sided Connections) F u 35 Part 10: Connections Figure 10-3: Encroachment Table 10-1 to Table 10-9: Connection Tables Figures: Skewed Plate Welds Fig Fillet encroachment (riding the fillet)

19 Skewed Plate Welds 37 Skewed Plate Welds 38 19

20 Skewed Plate Welds PLs Over ½ Support 3/16 Max Gap for Fillet Weld Support WO WA S(E) WA 45 BTC-P4 39 Skewed Plate Welds 1-Sided Welds CJP S (E) W 45 CJP PJP+REINFORCING FILLET (AWS Annex A) 40 0

21 Alternate Bent Plate Detail Large Skews Plate Welds Typically 1/ Max Plate Thickness Design from Bend Line Design of Skewed Connections by Larry Kloiber and William A. Thornton, EJ 3 rd Quarter Examples when rotational ductility and a fillet weld size of 5/8t is not always needed Plate Welds Bracing Connections Moment Connections 4 1

22 Welded Angle Legs 5/8 5/8 5/16 43 Part 11-Part 1: Partially Retrained and Fully Restrained Moment Connections Extended End-Plate Moment Connections Design assumptions 1-11 End-plate effective width = b f + 1 in. CJP welds are treated as PJP welds between beam flange-to-web fillets. No weld access holes Also see DG4* and DG16* *free download for AISC members! 44

23 Part 13: Bracing See DG9 and Design Examples Link to design examples: 45 Part 14: Beam Bearing Plates, Column Base Plates, Anchor Rods, and Column Spices Table 14-: Recommended Maximum Sizes for Anchor-Rod Holes in Base Plates (also see DG1*) Tables 14-3: Cases I-XII *free download for AISC members! 46 3

24 Part 15: Hanger Connections, Bracket Plates, and Crane Rail Connections Figure 15-: Bracket Plates Table 15-3: Z net plates Table 15-4 to 15-6: Clevis, Pins, and Turnbuckles 47 Part 15: Brackets Cont. 48 4

25 Part 15: Clevises and Pins For design: Specification Section D5 and J7 Example: clevelandcityforge.com 49 Part 16: Specification for Structural Steel Buildings (360-10), User Notes Commentary 50 5

26 Part 16: Specification for Structural Steel Buildings (360-10), 51 RCSC Masking Requirements: Fig C-3.1 Section 4:PT and SC Joint Applications Section 6: Washer Requirements Section 8: Bolt Installation Methods 5 6

27 Part 16: Code of Standard Practice Section 3.1.: Delegate Design Fabricator shall submit representative samples of connections Shop and Erection drawing review Section 6.4 Fabrication Tolerance Section 10: AESS Steel 53 Connection Details for Delegated Design Show all information to allow for checking 7/8"Ø A35-N BOLTS, STD HOLES, U.N.O. 7-1/4" MAX GRADE 50 PLATE 3" 3" 1-3/4" TYP AISC , LRFD TYPE A Member forces and sizes 7/16 7/16 6-7/8" MIN 4-1/" MIN (3x3) 3" O.C. W14x V = 35 KIPS (LOAD GIVEN IN RFI 5R) Slopes ASD or LRFD with Specification Edition 1-3/4" TYP 1-1/" TYP SHAPED PL5/8 w/ SSL HOLES Plate thickness, edge distances, etc W7x84 EXTENDED SHEAR PLATE CONNECTIONS AT A-17 SLAB ELEVATION CHANGE, S11-0C NOTES 54 7

28 Bearing Example Given: AISC Specification AISC 14 th Edition, LRFD W Shapes ASTM A99, F y = 50 ksi 1 cap plate, Grade 50, F y = 50 ksi 7/8 dia. A35-N, STD holes, φr v =4.3 kip Load P u = 00 kips d e d/ W14x90 COL W36x135 BEAM P u = 00 k e=4 55 Web Local Yielding AISC Specification Eq. J10-3 φ R = φ F t.5k+ l ( ) n yw w b ( φ=1.00) W14x90 COL Determine l b 4 W36x135 BEAM 56 8

29 Length of bearing l = t + 5t b wc pl ( ) = 0.440in in. = 5.44in. W14x90 COL 4 t pl 1 l b W36x135 BEAM.5 57 Web Local Yielding l b = 5.44in. AISC Specification Eq. J10-3 φ R =φ F t.5k+ l ( ) ( )( )( ( ) ( )) n yw w b = ksi 0.600in in in. = 79kips P = 00kips o.k. u ( φ=1.00) W14x90 COL 4 W36x135 BEAM 58 9

30 Web Local Crippling ( φ=0.75) AISC Specification Eq. J Web Local Crippling l b = 5.44in. AISC Specification Eq. J l 1 3 b t EFywt w f φ R n =φ tw + d tf tw in in. 9000ksi 50ksi 0.790in. = 0.75(0.80) ( 0.600in. ) in in. ( 0.600in. ) = 389kips P = 00kips o.k. u ( φ=0.75) ( )( )( ) 60 30

31 Web Local Yielding l b = 5.44in. AISC 14 th Edition Eq. 9-45a ( φ=1.00) n ( ) φ R =φ R + l φr 1 b kips = 116kips + ( 5.44in. ) 30.0 in. = 79kips P = 00kips o.k. u 61 Web Local Crippling l b = 5.44in. AISC 14 th Edition Eq. 9-49a ( φ=0.75) ( ) φ Rn = φ R + l φr 3 b 4 kips = 149kips + ( 5.44in. ) 8.3 in. = 389kips P = 00kips o.k. u 6 31

32 Column Stiffener Checks Given: AISC Specification AISC 14 th Edition, LRFD W Shapes ASTM A99, F y = 50 ksi 3/4 x9 flange plates, grade 50, F y = 50 ksi 7/8 dia. A35-N, STD holes, φr v =4.3 kip Load M u = 300 kips-ft W4x55 BEAM W14x90 COL 63 Flange Force M u Pfl = d 300kips-ft = 3.6in. = 153kips W4x55 BEAM W14x90 COL 64 3

33 ( 5 ) φ R =φ F t k + t n yw w des pl ( )( )( ( ) ( )) = ksi 0.440in in in. = 161 kips P = 153kips o.k. fl ( φ=1.00) Web Local Yielding AISC Specification Eq. J10- W4x55 BEAM W14x90 COL 65 Web Local Crippling ( φ=0.75) AISC Specification Eq. J tpl t EFywt w f φ Rn =φ 0.80t w 1+ 3 d t f tw in in. ( 9000ksi)( 50ksi)( 0.710in. ) = 0.75(0.80) ( 0.440in. ) in in. ( 0.440in. ) = 19kips P = 153kips o.k. u 66 33

34 Web Compression Buckling AISC Specification Eq. J tw EFyw φ Rn =φ h 3 ( ) ( )( ) 14in. ( 1.31in. ) in. 9000ksi 50ksi = (0.9) = 194kips P = 153kips o.k. fl ( φ=0.9) W4x55 BEAM W14x90 COL W4x55 BEAM 67 Flange Local Bending ( φ=0.9) AISC Specification Eq. J10-1 φ Rn =φ6.5fyftf = 0.9( 6.5)( 50 ksi)( 0.710in. ) = 14 kips Pfl = 153kips n.g. Stiffeners are needed See Blodgett

35 Web Local Yielding ( φ=1.00) AISC 14 th Edition Eq. 4-a φ R = P + P l n wo wi b kips = 144kips in. in. = 161kips P = 153kips o.k. fl ( ) 69 Web Local Crippling ( φ=0.75) AISC 14 th Edition Eq. 9-49a ( ) φ Rn = φ R + l φr 3 b 4 kips = 88.8kips + ( 0.75in. ) 9.9 in. = 19kips P = 153kips o.k. u 70 35

36 Web Compression Buckling AISC 14 th Edition Eq. 4-3a ( φ=0.9) φ R = P n wb = 194 kips P = 153kips o.k. fl 71 Flange Local Bending ( φ=0.9) AISC 14 th Edition Eq. 4-4a φ R = P n fb = 14 kips P = 153kips n.g. Stiffeners are needed fl 7 36

37 Polling Question 73 Understanding Tables Determine End Reaction from UDL Given: AISC Specification AISC 14 th Edition, LRFD W14x ASTM A99, F y = 50 ksi L= 0 ft End Reaction = 50% UDL V+ w l M

38 Understanding Tables Shear Based on Flexural Yielding Strength AISC Specification Eq. F-1 φ M = φf Z n y x 3 ( )( ) = ksi 33.in. = 1494 kip-in (1 ft/1 in.) = 15 kip-ft ( φ=0.9) wl flexure φm n(8) = l 15kip-ft 8 = (0ft) = 49.8 kips ( ) M max () = wl 8 75 Shear Based on Shear Strength AISC Specification Eq. G-1 φ Vn = φ(0.6)( Fy)( Aw)( Cv) = 1( 0.6)( 50ksi)( 13.7 in. )( 0.3 in. )(1) = 94.5 kips wl shear = φv () n ( ) = 94.5 kips = 189 kips Understanding Tables ( φ = 1) V max () = wl 76 38

39 Understanding Tables Determine End Reaction Based on Percent UDL (%UDL) wlshear Vu = min(( wlflexure ), ) 100% (50%) 189 kips = min(49.8 kip, ) 100% = min(4.9 kip, 94.5 kip) = 4.9 kip wl= 49.8 Vu Vu 77 Understanding Tables UDL Cautions: For 100% composite, φm n 97 kip-ft. This gives: φm n(8) Vu = l() ( ) 97 kip-ft 8 = 0ft () = 59.4 kips 119% UDL 78 39

40 Understanding Tables UDL Cautions: For Point Loads φ Vu_max = φ(0.6)( Fy)( Aw)( Cv) ( )( )( )( ) = ksi 13.7 in. 0.3 in. (1) = 94.5 kips 190% UDL 79 Understanding Tables Working backwards: Given: Connection strength, Find: Minimum beam length, l min φv n _ connection DOUBLE ANGLES l min_ flexure l min_ shear φm n(8)(%udl) = 100%( φv ) φm n(8) = φ V () n n_ connection Lweb BEAM n 3 O.C. COLUMN l max( l, l ) min = min_ flexure min_ shear UNCOPED 80 40

41 Understanding Tables Working backwards: Given: Connection strength, φ V n _ connection = 71 kip Find: Minimum beam length, l min l l min_ flexure min_ shear φm n(8)(%udl) (15 kip-ft)(8)(50%) = = = 7.04 ft 100%( φv ) 100%(71 kip) n_ connection φm n(8) (15 kip-ft)(8) = = = 5.9 ft φv () (94.5 kip)() n V u = 14k/=71k l = max( l, l ) = 7.04 ft min min_ flexure min_ shear 81 Understanding Tables 8 41

42 Instantaneous Center of Rotation Method Spandrel example Given: AISC Specification AISC 14 th Edition, LRFD W Shapes ASTM A99, F y = 50 ksi 3/8 plates, grade 50, F y = 50 ksi 7/8 dia. A35-N, STD holes, φr v =4.3 kip Bolt spacing: 3 vertical, 3 horizontal Load: Façade moment M f = 100 kip-in W4x55 BEAM W14x48 BEAM 83 3 in. I I x y 6 i= 1 6 i= 1 i i ( ) = x = 61.5in. = 13.5in ( ) = y = 4 3in. = 36in I = I + I p x y 3 in. 3 in. x 1 y 1 W14x48 BEAM = 49.5in W4x55 BEAM 84 4

43 R R M max f i= 1..6 x = = I p = 3.03kip M f i= 1..6 y = = I p = 6.06kip ( yi) ( 100kip-in. )( 1.5in. ) 49.5in. ( xi) ( )( ) max 100kip-in. 3.0in. ( 3.03kip) ( 6.06kip) = x + y = + R R R = 6.78kip 49.5in. R x R y R y R y R x R x R y R y R x W14x48 BEAM R y W4x55 BEAM 85 nv Mmax_1 M f R ( 4.3kips) ( φr ) = = 100 kip-in 6.78 kips = 360 kips-in. Use the Instantaneous Center of Rotation Method with C' C ' = 15.8 (From Table 7-7) max_ ( )( ) M = C'( φ r ) = 15.8in. 4.3kips nv = 384 kip-in. M R f 100kip-in. = = 14.8in. C ' = 15.8 in 6.78in

44 l i max Δ lmax C' = l i 1 e, in. Manual Eq. (7-1) 87 Façade and Gravity Moments Should Act Opposite M f M f V g V g M g + FACADE GRAVITY 88 44

45 V g V g M g M g M f M f V g V g + = e e =e + M f / V g FACADE GRAVITY TOTAL 89 NON EXTENDED SPANDREL CONNECTIONS 90 45

46 Pick a clevis: rod, F y = 36 ksi, full strength π( drod ) π( in. ) Arod = = 4 4 = 3.14in. Tmax =φ Arod Fy _ rod = 0.9( 3.14in. )( 36 ksi) = 10 kips Select Clevis Number 6, Design strength = 135 kips 10 kips 91 Secrets of Manual Check Double-Angle Tolerance using Fig 10-3 W4x6 t w = 7/16 L3x3x3/8 W W8x8, T= 6-1/8 k det = 15/16 t f =7/16 Fig Fillet encroachment (riding the fillet) kdet tf = in. in.= in Encr. = in

47 Secrets of Manual Check Double-Angle Tolerance using Fig Wmax = T + () Encr. = 6 in.+() in. = 6 in W = tw + () bangle = in. + ()3 in. = 6 in W W o.k. max W4x6 t w = 7/16 L3x3x3/8 W W8x8, T= 6-1/8 k det = 15/16 t f =7/16 93 Determine Strength of Column at Grid B/1 A B C A B C A B C 1 3 nd Floor, TOS=10-0 3rd Floor, TOS=5-0 Roof, TOS=

48 1 Determine Strength of Column at Grid B/1 Given: AISC Specification AISC 14 th Edition, LRFD W14x90 ASTM A99, F y = 50 ksi r x = 6.14 in. r y = 3.70 in. L x = 30 ft, L y = 15 ft K = Determine Unbraced Length Since unbraced length differs in the two axes, select member based on y-y axis then check equivalent y-y axis length for x-x axis. KL r x x KL KL x y KL = r rx = r y y y KL y_ EQ KL x = rx r y 96 48

49 1 Determine Unbraced Length rx 6.14 in r = 3.70 in. = y (See Table T4-1a) 15-0 KL y_ EQ KLx 30.0 ft = = = 18.1 ft r 1.66 x r y (Controls) 15-0 KL = 15.0 ft y 97 Determine Strength Using KL y_eq = 18.1 ft and interpolation: φ P n = 96 kips 98 49

50 Check Using Specification KL r x KL r x y y 1(30.0 ft)(1 in./ft) = = in. 1(15 ft)(1 in./ft) = = in. (Controls) E 9,000 ksi F = 50 ksi = y 99 Check Using Specification π E π (9,000 ksi) Fe = = = 83.3 kl ( 58.6) r F cr Fy 50 ksi Fe 83.3 ksi = Fy = ksi=38.9 ksi (E3-4) (E3-) φ = (0.9)38.9 ksi = 35.0 ksi F cr φ P = F A = o.k. n 0.9 cr g (35.0ksi)(6.5 in. ) =97 kips 96 kips

51 Determine Strength of Beam Between Grids A/ to B/ Given: AISC Specification AISC 14 th Edition, LRFD W18x35 ASTM A99, F y = 50 ksi L= 30 ft Braced points only at beam-to-girder connections Equally spaced bays C b = 1.0, conservative 1 A B C From Table 3- L p = 4.31 ft L r = 1.3 ft y y y y = x x x x ( ) L L p Mnx M px Mrx M px L r L p φ =φ + φ φ ( ) Solving for y: y = y + y y x x0 x x ft 4.31 ft = 49 kip-ft + ( 151 kip-ft 49 kip-ft ) = 179 kip-ft 1.3 ft 4.31 ft 10 51

52 Check if Compact for Flexure 103 From Manual Table 3-, C b = 1 (See Table 3-1 for other C b ) 104 5

53 From Specification: 105 Deflection

54 From Specification: 107 From Specification:

55 Guide to Stability Design Criteria for Metal Structures, Galambos 109 Manual Fig 3-1 (also see Commentary Figure C-F1.)

56 Determine Strength of Composite Beam (shown) Given: AISC Specification & AISC 14 th Edition, LRFD W16x6 ASTM A99, F y = 50 ksi, L = 30 ft 3 NLWT concrete (f c = 4 ksi) on 3 metal deck ¾ dia. stud 111 Determine Effective Width of Concrete Slab (Specification I3.1a) Each side of beam centerline, minimum of: ( 30 ft)( 1 in/ft) Lb 1) 1/8 of span = = = 45 in., controls 8 8 bo 1 ) 1/ c.c. beam spacing = = = 60 in. 3 spaces 3) distance to edge of slab = N/A ( 30 ft)( 1 in/ft) ( ) b eff For this example, b eff =( sides of beam centerline)(45 in) = 90 in

57 Load Transfer Between Steel Beam and Concrete Slab (Specification I3.d(1), Positive Flexural Strength where concrete slab is in compression) (a) Concrete crushing (concrete fully in compression) V ' c = 0.85 f ' c Ac = 0.85(4 ksi)(3 in)(90 in.) = 918 kips ( Note : only concrete above deck considered.) (b) Tensile yielding of steel section (steel fully in tension) V ' = A F = (50 ksi)(7.68 in ) = 384 kips < V ' PNA in concrete s s y c (c) Shear strength of steel studs V ' = Q = min( V ', V ' ) for 100% composite q n c s 113 Stud Strength (Specification Eq. I8-1) Q = 0.5 A f ' E R R A F n sa c c g p sa u A sa area of stud, in E = concrete modulus of elasticity = w f ', ksi 1.5 c c c F = specified minimum tensile strength of stud = 65 ksi u = 1.5 ( ) ( ) ( ) Qn = in (4 ksi) 145 pcf 4 ksi = 6.1 kips/stud RRA F = g p sa u 1.0(0.6)( in )(65 ksi) = 17. kips/stud, controls

58 Use Table 3-1 for Stud Strength Deck Perpendicular Weak Studs per rib (R p = 0.60) 115 Determine Number of Studs For 100% composite, studs required between points of maximum and zero bending moment (half span of beam): V ' = min( V ', V ' ) q c s 384 kips # of studs = = =.3 46 studs for beam total length Q 17. kips/stud However, for 46 studs, there will be () studs per rib. For () studs per rib, R g = 0.85 Q = 14.6 kips/stud n bending moment (half span of beam): n V ' = min( V ', V ' ) q c s 384 kips # of studs = = = studs for beam total length Q 14.6 kips/stud n *See Design Example I.1 on page I-15 for more information

59 Determine 100% Composite Strength Qn 384 kips a = 1.5 in f ' b = ksi 90 in. = c eff ( )( ) 1.5 in. Y = ( 3 in. conc. + 3 in. deck) = 5.38 in. Y = Distance top of steel to beam to center line of the concrete flange force - Sum moments about a/: dbm 384 kips 15.7 in. φ bmn =φ AsFyY + = in. 381 kip-ft + = 1 in./ft 117 a/ a For Use with Table 3-19 (see also Figure 3-3 in Manual) Y1 = distance from top of steel flange to any of the PNA locations listed PNA locations: TFL = Top of Flange (beam top flange), also 1 BFL = Bottom of Flange (beam top flange), also 5,3,4 = 4 equal spaces within flange 6 a Y = Ycon Q n = 7= 0.5FA y s at BFL + Q at 7 Qn a = 0.85 f ' b and Y con = top of steel to top of concrete c eff n Y con Y 1 Y

60 Table 3-19 From above, V ' c V ' s = 918 kips = 384 kips Enter Table with PNA at TFL and Qn 384 kips Qn = 384 kips (100% composite) a = 1.5 in f ' b = 0.85( 4 ksi)( 90 in. ) = c eff 1.5 in. Y = ( 3 in. conc. + 3 in. deck) = 5.38 in. 119 Moment Capacity based on Table 3-19 (100% composite) 5.5 in in. φ bm n = 384 kft ( 384 kft 370 kft) = 381 kip-ft 5.5 in. 5 in. Compare to dbm 384 kips 15.7 in. φ bmn =φ AsFyY + = in. 381 kip-ft + = 1 in./ft 10 60

61 Using Table 3-19 for 63% composite For 63% composite, enter Table with kips 4 kips a ( )( ) Q n _63% = = Q 4 kips = n _63% 0.85 f ' cb = eff 0.85( 4 ksi)( 90 in. ) = in in. Y = 6 in. = 5.60 in. 11 Using Table 3-19 for 63% composite For 63% composite, enter Table with Q n _63% ( )( ) = kips = 4 kips Note : PNA at 4. 6 in. 5.6 in. φ bm n_63% = 333 kft ( 333 kft 34 kft) = 36 kip-ft 6 in. 5.5 in. 1 61

62 Determine Moment Capacity using 63% Composite From AISC Commentary Section C-I3.a and Equation C-I3-10, Sum Fx = 0 C+ A sc F y = P y -A sc F y A sc F y = (P y C)/() For partial composite action, there is a neutral axis in the concrete and a neutral axis in the steel. M = Cd ( + d) + P( d d) AISC Commentary Eq. C-I3-10 n 1 y 3 Sum moments about C s / 13 Determine Moment Capacity using 63% Composite (cont d) where: C = P d d 1 Q = A F n y s y = distance from centroid of compression force, C, in concrete to top of steel, in. = distance from centroid of the compression force in steel section to top of steel section, in. ( d d3 = distance from P y = 0 for no compression in steel) to top of steel section, in. 14 6

63 Determine Moment Capacity using 63% Composite (cont d) P = 384 kips y C63% = Q n _ 63% = 0.63(384 kips) = 4 kips C63% 4 kips a = in f ' b = 0.85(4 ksi)(90 in.) = c eff d d 1 a in. = tslab = 6 in. = 5.60 in. ( 384 kips 4 kips) ( ) 15.7 in. d 3 = = = 7.85 in. x 1 AF s y C63% 1 = = = = 0.19 in. bf f y (5.50 in.)(50 ksi) d bm 15 Determine Moment Capacity using 63% Composite (cont d) M = C ( d + d ) + P ( d d ) n_ 63% 63% 1 y 3 ( 4 kips)( 5.60 in in. ) ( 384 kips)( 7.85 in in. ) = + + = 4351 k-in. ( ) kip-in φ M n _63% = = 36 kip-ft

64 Determine Number of Studs For 63% composite, 0.63( 384 kips) # of studs = = studs for beam total length 17. kips/stud Note : Portion of steel beam will be compression. 17 Summary Be familiar with the information contained in the Manual Use the Database when possible Read descriptions of the tables Include all design information on details to facilitate review and checking Manual can facilitate efficient design

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