Ejemplo resuelto 4E ( Ej 5 TP5)-AISC Design Guide 04

SAVE OFFLINE

Chapter 4

Design Examples 4.1

Scope

The following examples illustrate design procedures for the (1) four-bolt extended unstiffened (4E), (2) four-bolt extended stiffened (4ES), and eight-bolt extended stiffened (8ES) end-plate connections. Both beam side and column side calculations are illustrated. Two examples are provided for the 4E connection: one is for cyclic/seismic design and the second for wind/gravity loading. Both beam side and column side calculations are illustrated. The connections are symmetric to accommodate load reversal, which is necessary for the cyclic/seismic designs but may not be necessary for the wind/gravity loading. Shear forces are assumed to have been determined from analysis. 4.2

Bolts: ASTM A490 See Figure 3.1 for definition of connection geometry. Beam Side Design 1.

(3.2)

= 1.1(1.1)(50)(126) = 7623 k-in. Location of Plastic Hinge

4E Example A

L p = min

Using cyclic/seismic loading, a four-bolt extended unstiffened (4E) end-plate connection is to be designed to connect a W21×55 beam to a W14×109 column. The beam and column material are ASTM A992 steel and the end plate is ASTM A572 Gr. 50 steel. ASTM A490 bolts are to be used. The required shear resistance, Vu, is 40 kips.

Column: W14×109 dc = 14.3 in. twc = 0.525 in. bfc = 14.6 in. tfc = 0.860 in. kc = 1.46 in. (h/tw)c = 21.7 Workable Gage = 51/2 in. Zxc = 447 in.3

Connection Design Moment Mpe = 1.1 RyFyZx

Four-Bolt Unstiffened Extended (4E) End-Plate Connection

Beam: W21×55 db = 20.8 in. twb = 0.375 in. bfb = 8.22 in. tfb = 0.522 in. kb = 1.02 in. Workable Gage = 51/2 in. Zxb = 126 in.3 Fyb = 50 ksi (ASTM A922 steel) Fub = 65 ksi

= 50 ksi (ASTM A992 steel) = 65 ksi

Fyc Fuc

d / 2 = 20.8 / 2 = 10.4 in. 3b fb = 3 (8.22 ) = 24.66 in.

(3.3)

= 10.4 in. Moment at the Face of Column (Connection Design Moment) Muc

= Mpe + VuLp

(3.1)

= 7623 + 40 (10.4) = 8039 k-in. 2.

Select Connection Configuration: Extended Unstiffened

Four-Bolt

Assumed Geometric Design Data bp ≈ bf + 1 in. = 8.22 + 1= 9.22 in.⇒ Use bp = 9.0 in. g = 5½ in. (same as beam and column “workable gage”) pfi = 2 in. pfo = 2 in. de = 15/8 in. Fyp = 50 ksi Fup = 65 ksi (ASTM A572 Gr. 50 steel) Ft = 113 ksi (ASTM A490 bolts) Using assumed dimensions,

DESIGN GUIDE 4 / EXTENDED END-PLATE MOMENT CONNECTIONS—SEISMIC AND WIND APPLICATIONS, 2ND EDITION / 31

0.522 = 22.54 in. 2 0.522 h 1 = 20.8 − 0.522 − 2 − = 18.02 in. 2

h 0 = 20.8 + 2 −

3.

t p Req'd = =

Determine the Required Bolt Diameter (ASTM A490) db Req'd =

(3.5)

7.

8.

Use db = 1¼ in. (ASTM A490)

(d

M uc b

− t fb

)

=

8039 = 396 kips (3.11) 20.8 − 0.522

Check Shear Yielding of Extended Portion of End Plate φRn = 0.9(0.6 Fyp)bptp

(3.12)

= 0.9(0.6)(50)(9.0)(1.25)

Bolt Tensile Strength  π (1.25)2  Pt = Ft A b = 113   = 138.7 kips (3.9)   4 M np = 2 Pt (h 0 + h 1 )

= 304 kips Check Inequality 3.12 F fu

(3.7)

2

= 2(138.7)(22.54+18.02) 9.

= 11,251 k-in. φMnp = 0.75(11,251) = 8438 k-in. > Muc = 8039 k-in. OK

1 1 9.0(5.5) bp g = 2 2 = 3.52 in. > p fi = 2 in.

= [9.0 − 2(1.25 + 0.125)] 1.25

 1  1 bp   1 1  h1  +  + h0 −  2   p fi s   p fo  2   2 +  h 1 p fi + s  (Table 3.1)  g

)

9.0  1   1  1  1 + + 22.54  − 18.02    2.0  2  2  2.0 3.52 

2 18.02 (2.0 + 3.52 ) 5.5  = 148.2 in.

Required End Plate Thickness

= 7.81 in2 φRn = 0.75(0.6 Fuf)An

(3.13)

= 0.75(0.6)(65)(7.81)

(Table 3.1)

Yp =

+

396 = 198 kips ≤ φRn = 304 kips OK 2

An = [bp − 2(db + 1/8)]tp

End Plate Yield Line Mechanism Parameter s=

=

Check Shear Rupture of Extended Portion of End Plate

Determine the Required End Plate Thickness

=

0.9 (50 )(148.2 )

Calculate the Factored Beam Flange Force F fu =

Select Trial Bolt Diameter and Calculate the No Prying Bolt Moment

(

1.11 (0.75 )(11, 251 )

Select End Plate Thickness

= 1.22 in.

5.

φ b Fyp Yp

USE tp = 11/4 in. (ASTM A572 Gr. 50 steel)

2 (8039 ) M np = 2 Pt (h 0 + h 1 ) π (0.75 )(113 )(22.54 + 18.02 )

4.

(3.10)

= 1.19 in.

6.

2 M uc πφFt (h 0 + h 1 )

1.11 φ M np

= 228 kips Check Inequality 3.13 F fu 2

= 198 kips ≤ φRn = 228 kips OK

10. End plate is unstiffened, therefore this step is not required. 11. Check Compression Bolts Shear Rupture Strength Vu = 40 kips Vu ≤ φRn = φnbFvAb  π (1.25)2  φRn = 0.75 (4 )(60 )    4

= 221 kips 32 / DESIGN GUIDE 4 / EXTENDED END-PLATE MOMENT CONNECTIONS—SEISMIC AND WIND APPLICATIONS, 2ND EDITION

(3.17)

Vu = 40 kips < φRn = 221 kips OK

Effective length of weld = db /2 − tfb = 20.8/2 − 0.522 = 9.88 in.

12. Check Compression Bolts Bearing/Tearout i)

End Plate Vu = 40 kips ≤ φRn = ni(φRn)i + no(φRn)o ni = 2 no = 2

(3.18)

D=

40 = 1.45 sixteenths 2(1.392)(9.88)

USE 5/16 in. Fillet Welds

Bearing Strength = 2.4 dbtpFu = 2.4(1.25)(1.25)(65)

Column-Side Design

= 244 kips/bolt 14. Check the Column Flange for Flexural Yielding

Tearout Outer Bolts: Lc = (2.0 + 0.522 + 2.0)−(1.25 + 1/16) = 3.21 in. Rn,inner = 1.2 LctpFu = 1.2(3.21)(1.25)(65)

(3.19)

= 313 kips > 244 kips

s=

1 1 14.6(5.5) = 4.48 in. bfc g = 2 2

c = pf o + tfb + pfi = 2.0 + 0.522 + 2.0 = 4.52 in.

By inspection, bearing controls for the inner bolts. φRn = 4(0.75 × 244) = 732 kips > Vu = 40 kips OK

 0.860   Fyp = 50  φR n = 732    1.25   Fyc = 50  = 504 kips > Vu = 40 kips OK

+

Beam Flanges to End-Plate Weld

Use CJP welds and the procedure in Figure 2.10. ii) Beam Web to End-Plate Weld Minimum weld size for 1¼ in. end plate is 5/16 in. The required weld to develop the bending stress in the beam web near the tension bolts using E70 electrodes is D=

0.6 Fyb twb 2(1.392)

2  3c   h 1  s +  + h 0  s +    4 g 

(Table 3.4) c  c2  g + + 4  2  2

=

14.6   1   1  + 22.54  18.02     4.48   2  4.48

+

 3 (4.52 ) 2  4.52  4.52 2  5.5  + 22.54 4.48 + + 18.02  4.48 + +   5.5  4  4  2  2 

= 170.1 in.

13. Design Welds i)

b fc   1   1  + h0  h 1  s  2   s 

Yc =

ii) Column Flange tfc = 0.860 in.

(Table 3.4)

=

0.6(50)(0.375) 2(1.392)

= 4.04 sixteenths USE 5/16 in. Fillet Welds The applied shear is to be resisted by weld between the minimum of the mid-depth of the beam and the compression flange or the inner row of tension bolts plus two bolt diameters and the compression flange. By inspection the former governs for this example.

Required Unstiffened Column Flange Thickness t fc Req'd = =

1.11 φ M np

(3.20)

φ b Fyc Yc 1.11 (0.75 )(11, 251 ) 0.9 (50 )(170.1)

= 1.10 in. > t fc = 0.860 in.

∴ Add Flange Stiffeners Assume ½ in. Stiffener Plates ts = ½ in. pso = psi =

c − t s 4.52 − 0.5 = = 2.01 in. 2 2

For Stiffened Column Flange b fc   1 1  1 1  + h0 + h 1  +   2   s psi   s pso   2 +  h 1 (s + psi )+ h 0 (s + pso ) (Table 3.4) g

Yc =

DESIGN GUIDE 4 / EXTENDED END-PLATE MOMENT CONNECTIONS—SEISMIC AND WIND APPLICATIONS, 2ND EDITION / 33