Example 3: Full Strength W30 x 99 Beam to W14 x 193 Column

Transcription

1 Example 3: Full Strength W30 x 99 Beam to W4 x 93 Column Design a full strength T-stub connection between a W30 x 99 beam and a W4 x 257 column. Use A steel for the beam, column, and all connection components. Assume a service moment of 7,500 k-in. The beam length is 36-0 cc of supporting columns. Use the alternative capacity formulation for the tension bolts with a resistance factor of Step : Find the Required Flange Force Designing the connection as full strength according to the 997 LRFD Seismic provisions; P reqd.f ye Z x d beam F ye R y F y where R y. for A steel P reqd.2f y Z x d beam P (.2) ( 50 ksi) ( 32 in 3 ) reqd ( in) P reqd kip Step 2: Select a Bolt Size and Grade Tension Bolts: The force that the T-stub needs to resist for this connection is relatively large. As a result, the tension bolts will have to be diameter. Using the alternative design strength for a diameter A490 tension bolt with a resistance factor of 0.90, (Section 9..2.) 398

2 φb n φ π bf db n F u th φb n ( 0.90) π ( 50 ksi) φb n 8.8 kip bolt Shear Bolts Find the required slip resistance: P slip,reqd M service d beam P slip,reqd ,500 k-in kip in The slip resistance of a single diameter, A490 bolt is 6.5 kip. Step 3: Layout the Shear Bolts kip 5.3 bolts 6.5 kip bolt use 6, diameter, A490 shear bolts Use 6, diameter, A490-X shear bolts. B f 0.45 in for the beam. The required edge distance.25 for diameter bolts. The maximum gage, g s, that is permitted by the beam flange can be written as g s 0.45 ( 2) (.25 ) 8.0 A shear bolt gage of 7 will be used instead of the maximum of 8 to avoid stress concentrations between the edge of the T-stem and bolt line and the edge of the beam flange and bolt line. 399

3 use g s 7.0 Use a bolt spacing of 3d b 3. The stem layout will be similar to that shown below. /6" dia. (typ) 7" 0 /2" 3" /2" Assuming a beam set-back of 2, a beam with a depth of roughly 28 will be required for the T-stub section. Step 4: Layout the Tension Bolts Find the minimum center distance for a W4 x 257 column. s center For a W4 x 257, k 3/6 400

4 min s center 2k (,col + d b ) min S center ( + ) Check wrench clearance: min s center 2clearance ( ) + t w,col Assume a wrench diameter of 3.5 for larger tools. (LRFD Table 8-5) min s center 2.75 ( ) O.K. Use 5 center spacing to allow extra clearance: W T-stub 4.0 4" /2" 3" 5" 3" /2" /6" dia. (typ) B f g t Step 5: Find the Minimum Stem Thickness φr n,stem φ f F u A stem,net t s,min P reqd 2d h,eff φ f F u [ W eff ] Find the effective stem width, W eff W eff s sb ( n sb 2) tan( θ eff ) + g s W T-stub 40

5 W eff ( 3 )( 6 2) tan( 30 ) W eff Solve for t s,min : W eff 4.0 t s,min kip ( 0.75) ( 65 ksi) [( 4.0 ) ( 2) (.25 )] t s,min.2 Check for t s,min based on bearing failure: t s,min P reqd φ f 2.4n sb F u d b t s,min kip ( 0.75) ( 2.4) ( 6) ( 65 ksi) ( ) t s,min 0.34 O.K. Step 6: Find the Minimum B f and t f for the T-stub Flange Find the required flange width, B f,reqd φ b f B n T reqd reqd δ db T reqd δ p 2W T-stub δ d h n tb p p ( 2) ( 4.0 ) δ Find T reqd 402

6 T reqd P reqd n tb T reqd kip 79.6 kip 8 Solve for b reqd : b 8.8 kip 79.6 kip reqd ( 2) ( ) 79.6 kip b reqd 0.35 Since this is obviously too small to be practical, use b reqd.5 d b.5 b reqd.5 g t,reqd 2b ( reqd + t s,min ) 2.5 ( +.2 ) 5.24 B f,reqd g t,reqd + 4d b ( 4) ( ) 9.24 Find the range for t f t f,max 4T reqd b reqd φ b pf y ( + δ ) t ( 4) ( 79.6 kip) (.5 ) f,max ( 0.90) ( 3.50 )( 50 ksi) (.696) t f,max.337 t f,min ( T reqd )[ 8d b b reqd d b ( 2d b + b reqd )] φ b 3.75d b pf y 403

7 t f,min ( 79.6 kip) {( 8) ( )(.5 ) ( )[( 2) ( ) + (.5 )] } ( 0.90) ( 3.75) ( )( 3.50 )( 50 ksi) t f,min.070 Step 7: Select a Trial Section for the T-stub An examination of the available W sections shows that there are none that satisfy the stem and flange requirements. This leaves two options. We can () use a wider T-stub or (2) design the flange for zero prying. The first option is the most attractive but while increasing the T-stub width decreases the required stem thickness, it also decreases the maximum flange thickness. As a result, this option will not work in this case. To design for zero prying, determine the minimum gage of the tension bolts for tightening. Use a conservative estimate of g t 6.0 g b trial ---- t + t 2 s,min b trial From EQ 9-33, t f,reqd 4φB n b reqd φ b pf y t ( 4) ( 8.8 kip) (.88 ) f,reqd ( 0.90) ( 3.5 )( 50 ksi) t f,reqd.976 Try a W30 x 326 B f 5.37 t s.4 t f 2.05 k -5/6 d

8 Check the flange capacity: Check minimum tension bolt gage, g t, based on bolt and wrench clearance. min g t 2k ( + d b ) min g t ( + ) Check wrench clearance assuming that the tension bolts will be installed before the shear bolts: min s center 2clearance ( ) + t s Assume a wrench diameter of 3.5 for larger tools. (LRFD Table 8-5) min s center 2.75 ( ) use g t 5.0 t, k s t seff, t seff b gt -- ( t 2 seff, ) a Bf -- ( g 2 t ).25b b -- ( ) a -- ( ).25b 2 2 b.559 a b b d b a a d b

9 b a φr n,flange n tb φt n φt φ b 2a d b 2 pfy t 4 f a b d b ( a + b ) ( 0.90) ( 2) ( ) ( 3.5 )( 50 ksi) ( 2.05 ) 2 4 φt ( 4) ( )(.059 ) ( )( ) φt kip/bolt 2 φ φt f B n a φ b pf y t f a + b + 4a ( + b ) φt 2 ( kip) ( ) ( 0.90) ( 3.5 )( 50ksi) ( 2.05 ) 2 ( ) ( 4) ( ) φt kip/bolt φt 3 φ f B n φt kip/bolt φt 3 governs since it s smallest, 406

10 φr n,flange n tb φt 2 φr n,flange ( 8 bolts) ( 8.8 kip/bolt) φr n,flange kip O.K. The figure below shows the solution space for the flange capacity of the W30 x 326 with a tension bolt gage of 5.0 The thickness of the W30 x 326 is denoted by a dashed vertical line. Note that the intersection of the dashed line with the design curve falls within no prying range Nominal Capacity Curve 600 T-stub Capacity (kip) Design Capacity Curve Ductile Range Flange Thickness (in) Solution Space for a W30 x 326 T-stub with g t 5.0 Check the stem capacity: Net section fracture, φr n φ f F u A stem,net A stem,net ( W eff 2d h,eff )t s A stem,net [ 4.0 ( 2) (.25 )] (.4 ) 3.40 in 2 407

11 φr n,stem ( 0.75) ( 65 ksi) ( 3.40 in 2 ) kip O.K. Bearing failure, φr n,bearing φ br n sb 2.4F u d b t From EQ 9-9 φr n,bearing ( 0.75) ( 6) ( 2.4) ( 65 ksi) ( )(.4 ) φr n,bearing 234. kip O.K. Check t s,max to insure a balanced T-stub failure. If the capacity of the stem is less than the capacity of the flange, then a more desirable stem fracture will precede a tension bolt fracture. In this design, φt 3 governed the capacity of the flange, which indicates a tension bolt fracture without the development of flange yielding. As a result, the T-stub should be proportioned such that yielding in the stem precedes a tension bolt fracture. φ y F y A stem,net < φr n,flange ( 0.90) ( 50 ksi) [ 4.0 ( 2) (.25 )] (.4 ) < φr n,flange kip< kip O.K. Check Block Shear Failure: From EQ 9-20, φr n,stem φ f [ F u ( g s d h,eff )t s +.5F eff ( L sb + L e )t s ] where F eff ( C L )F y + C L F u C L ( L sb + L e ) and 408

12 n L sb s sb 2 sb L sb ( 3 ) L e.5d b (.5) ( 3 ).5 Solving, C L ( ) 0.08 use C L 0.0 (conservative) F eff ( 0.0) ( 50 ksi) + ( 0.0) ( 65 ksi) 50.0 ksi Substituting into the capacity equation, φr n,stem ( 0.75) [ ( 65 ksi) ( )(.4 ) + + (.5) ( 50.0 ksi) ( )(.4 )] φr n,stem,432 kip O.K. Check the beam capacity: Bearing failure of the beam flange, φr n,bearing φ br n sb 2.4F u d b t From EQ 9-9 φr n,bearing ( 0.75) ( 6) ( 2.4) ( 65 ksi) ( )( 0.67 ) φr n,bearing,254 kip O.K. The provisions of Section B.0 of the LRFD code will not be checked. Step 8: Design the Shear Connection Section 9.2c of the Seismic Provisions states that the shear connection from the beam to the column must be able to withstand the application of. R y F y Z x in 409

13 the opposite sense at each end of the beam. From this, the required shear force can be written as, For our beam, ( 2) (.R V y F y Z x ) reqd L beam V ( 2) (.2) ( 50 ksi) ( 32 in 3 ) reqd ( 432 ) V reqd 87.4 kip The longest single shear plate that will fit between the flanges of the top and bottom T-stubs can be approximated by, L plate,max d beam B f,t-stub L plate,max Select a single plate shear connection from Table 9-0 of Volume II of the LRFD. Using 4, diameter A490-N bolts, a plate 9/6 thick will be sufficient with a capacity, φr n, of 99.3 kip. Note that the beam set-back of the connection will be 3. This leads to a shear tab that is much wider than the design tables account for. As a result, a more rigorous design check should be included for the shear connection. Figures of the connection and T-stub details are shown below. 40

14 Continuity Plates as Required " dia A490 Tension Bolts (typ) " dia A490-X Shear Bolts (typ) 2" x 6" x 9/6" Plate with " dia A490-N Shear Bolts 7/6" W30 x 99 T-stubs cut from W30 x 326 (see detail) W4 x 257 Connection Detail -/6" dia (typ) 4" -/2" 3" 5" 3" -/2" -/6" dia (typ) 5-3/8" 6" -/8" 7" 0-/2" 4-/2" 3" -/2" 27" 2-/6" T-stub Detail 4

15 The bolt sizes required are shown in the table below. They were determined using a design guide in the Structural Bolting Handbook (SSTC, 996). Bolt Location Grip Size Column Flange to T-stub Flange 3.94" "-8 x 5-/2" T-stem to Beam Flange.8" "-8 x 3-/2" Shear Tab to Beam Web.08" "-8 x 2-3/4" Step 9: Compute the Initial Stiffness Compute the flange stiffness: where, 2n K tb EIa β ( a + 3b β b ) i,flange b 3 β b ( 4a β a + 3b β b ) From EQ 9-38 I pt f 2 I ( )( 2.05 ) in 4 2 β a EI Gpt f a 2 β a ( 2) ( ksi) ( 2.53 in 4 ) ( 200 ksi) ( 3.5 )( 2.05 )( ) 2 β a β b EI Gpt f b 2 42

16 Solving for K i,flange, β b ( 2) ( ksi) ( 2.53 in 4 ) ( 200 ksi) ( 3.5 )( 2.05 )(.059 ) 2 β b.276 K ( 2) ( 8) ( ksi) ( 2.53 in 4 )[( )( 5.445) + ( 3) (.059 )(.276) ] i,flange (.059 ) 3 (.276) [( 4) ( )( 5.445) + ( 3) (.059 )(.276) ] K i,flange , 885, 308 k-in 288, kip/in , in Compute the stem stiffness: t s EW ( T-stub W beam ) 2, L stem ( W T-stub W beam ) + W W beam ln beam W T-stub K istem where, L stem ( L T-stub k) L stem ( ) (.4 )( ksi) ( ) 2, ( ) ( ) + ( 0.5 ) ln K istem Computer the slip stiffness: K istem, 34,930.6 kip/in K i,slip P slip slip From EQ

17 where, P slip the connection slip load, P slip n sb α( 0.70F u )( 0.75A b )µ slip (determined statistically) α.0 for A325 bolts and 0.88 for A490 bolts µ coefficient of friction on the faying surface P slip ( 6) ( 0.88) [( 0.70) ( 50 ksi) ][( 0.75) ( in 2 )] ( 0.33) P slip kip Solving for K i,slip, K i,slip kip 37, kip/in Combine the component stiffnesses: K i,t-stub,ten K i,flange K i,stem K i,slip From EQ 9-36 K i,t-stub,ten , kip/in ,930.6 kip/in , kip/in K i,t-stub,ten , kip/in K it-stub,comp, K i,stem K i,slip K i,t-stub,comp ,930.6 kip/in , kip/in K i-t-stub,comp 85.9, kip/in Compute the connection rotational stiffness: 44

18 K i,conn 2 d beam K i,t-stub,ten K i,t-stub,ten K i,t-stub,comp + K i,t-stub,comp K ( ) 2 ( , kip/in) ( 7,895. kip/in) i,conn ( , kip/in + 7,895. kip/in) K i,conn 7735,, 54 k-in/rad The T-stub force-deformation response and connection moment-rotation response are shown in the figures below. The initial stiffnesses calculated above are plotted and compared to the full monotonic responses as predicted by a computer implementation of the stiffness model outlined in Chapter 7. The ultimate rotation is radians. Note, however, that since the connection is full strength, beam plasticity can be expected to contribute to the connection s rotational capacity Load (kip) Full Monotonic Stiffness Model Initial Stiffness Estimate Deformation (in) T-stub Force-Deformation Response 45

19 Moment (k-in) Full Monotonic Stiffness Model Initial Stiffness Estimate Rotation (rad x 0 3 ) Connection Moment-Rotation Curve 46