AISC-28

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: Tension member design. Bolted connection. SHEET 101 of 131Connection for a 38.3 foot long tension member that has to resist 78 kips of already factored load.No slip is permitted, so this connection is slip critical.

Ag = gross cross-sectional area

Ae = U*An

U = reduction coefficient

Fy = specified (ASTM) minimum yield stress

Fu = specified (ASTM) minimum tensile strength

d = diameter of the bolt

An = net area

Rn = strength

Rstr = slip critical strength

Lc = distance from edge of hole to edge of connected part

t = thickness of connected part

Ab = nominal bolt area

Pu = factored load to be resisted

Fv = ultimate shearing strength

Tb = minimum fastener tension

Agv = gross area acted upon by shear

Agt = gross area acted upon by tension

Anv = net area acted upon by shear

Ant = net area acted upon by tension

Because the bolt size and layout will affect the net area of the tension member, we will begin with the selection of the bolts. Use A325 bolts and A572 Grade 36 steel for both the tension member and the gusset plate (standard holes).

Pu = 76 (kips)t = 3/4 in 0.75

Fv = 48 (ksi)0.33

Fy = 36 (ksi)Fu = 58.0 (ksi)

L = 38.3 (ft)

TRY : 7/8 (in) bolts Tb = 39 (kips) from table J3.1 LRFD

Ab = 0.60 Ns = 2Shear strength:

(assuming that the threads are in the shear plane)21.65 Kips/bolt29.09 Kips/bolt

Shear strength controls 21.65 Kips/bolt

f t = resistance factor relating to tensile strength

m = mean slip coefficient (coefficient of static friction)

f=

m =

(in2)

fRn= fFv Ab

fRn= f Rstr = f (1.13 m Tb b Ns) f Rstr =

f Rn =

A

Section at A

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PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: Bolted connection SHEET 102 of 131

Number of bolts required = 3.51 bolts Try = 4 bolts

Minimum spacing : the distance between centers of standard, oversized, or slotted holes, shallnot be less than 3d (from AISC J3.3).s =3 d= 2.63 insay, s = 3 in

Minimum edge distance : (from AISC table J3.4)Le = 1.5 in

2.34568U = 0.85 (from commentary to the AISC specification)

1.74713

2.05544

minimum radius of gyration:r min = 1.532 in

TRY 2L5x3x1/2LLBBbolts placed in the long leg at the usual gage distance (check LRFD page 10-10)

usual gage = 3 in

Ag = 7.51 > 2.35 OK!r = 1.58 in > 1.53 in OK!t = 0.5 in

An = 7.01 > 2.05544 OK!x = 0.736 inU = 0.91822 < 0.9 use U = 0.9

Ae = 6.309 > 1.74713 OK!Check Bearing strength:

for bearing strength computation use a hole diameter of h = 15/16 inif Lc < 2 dif Lc > 2 d

For the hole nearest the edge of the member :

Lc = 1.03 in2 d = 1.75 in 26.92 kips/bolt

For the other holes:Lc = 2.06 in2 d = 1.75 in 45.68 kips/bolt

Total bearing strength of the connection :

163.94 kips > 76 kips OK!

Ag req. ³ in2

Ae req. ³ in2

An req. ³ in2

in2 in2

in2 in2

in2 in2

f Rn = f (1.2 Lc t Fu)f Rn = f (2.4 d t Fu)

f Rn =

f Rn =

f Rn =

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PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: Bolted connection SHEET 103 of 131

Check Block shear:

Shear areas:

Agv = 5.25

Anv = 3.50

Tension areas:

Agt = 0.50

Ant = 0.25

LRFD J4-191.35 kips

LRFD J4-210.875 kips

The larger fracture term controls: LRFD J4-1 controls

Block shear strength:

104.85 kips < 102.225102.225 kips > Pu/2 = 38 kips OK!

Le = 1.5 inLe = 1.5 in s = 3 in.

g = 3 in

Use a 2L5x3x1/2LLBB with the long leg connected. Use 4 7/8 in. A325 bolts.

in2

in2

in2

in2

Shear rupture design strength fVn: f(.6*Fu)*Anv =

Tension rupture design strength fTn: fFu*Ant =

f Rn = f Rn =

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