Chapter 4 (Final F) 18 March 07

7/30/2019 Chapter 4 (Final F) 18 March 07

1/40

F-1

Welding (chap, 5)

Structural

Connections

Fasteners

Bolts

Non-Structural

(A-307)

Structural

A325

A449

A490

Rivets (obsolete)


2/40

F-2

Bolt

Head

Shank

ThreadedSection


3/40

F-3

The AISC specifications recognize two types of high-strength bolted connection:-

A) Bearing Type Connection:-

Where connected parts are allowed to

bear directly on the bolt shank. Hence both shear

strength and bearing strength need to be

checked.

B) Slip-Critical Type Connection (formerly

known as friction-type connection):-

Where connected parts are not allowed to

slip, and shear forces are transmitted by friction

forces rather than direct shear. This connection

requires high pre-tensioning force of bolts upto

their proof load which equals 70% of tensile

strength.


4/40

F-4

When slip resistance is required, the pre-

tensioning of bolts should as high as theirProof Load, which is equivalent to the

yielding stress of the bolts material as

obtaining by the 0.2% offset method or the

0.5% extension method. The proof load

stress is a minimum of 70% and 80% of the

minimum tensile strength of A-325 bolts, and

the A-490 bolts respectively.

The proof load can be achieved by using the

turn-of-nut method, which is turn of nut from

the snug-fit position. Otherwise a calibrated

wrench must be used to achieve the proof

load for bolts.


5/40

F-5

(70 % Fu)

(70 % Fu)

(80 % Fu)


6/40

F-6

The most typical types of bolted connections are shown below:-

In shear connection, threads could be:-

A) In the shearing plane (N), or

B) Outside the shearing plane (X).

P

P

Thread outside (X)

Lap Joint

(single shear)

Butt Joint

(double shear)

(a) Shear Connections

(b) Eccentric shear

Connection.

W section

Structural tee

(c) Tension Connections

(d) Combined shear and tension connections

P

P

Thread inside (N)


7/40

F-7

There are several limit-

states or failure modes that

may control the strength

of a bolted connection.

As can be seen from the

above figure, three failuremodes (b, d & g) are relevant

to the connected plates and

not relevant to the fasteners.


8/40

F-8

The nominal tensile strength of a bolt (Rn) can be found as:

Where

Fub= Tensile strength of the bolt material.

An = Bolt net area through threaded zone*.This net area 0.75 to 0.79 of the bolt gross

area Ab (area of the shank).Thus:

297430

7850

ndb

..

Where, db = dia. of bolt (shank)

n = number of threads per inch.

Note: Tensile strength values Fub are given in LRFD

table J3.2 page (16.1)-104.

n

b

un AFR

bb

un AFR 750.

Thread cross section area*


9/40

F-9

The nominal-shear strength for one fastener (Rn) will be

Rn = m AbuWhere,

= (1) for single shear

m = number of shear planes

= (2) for double shearAb = Bolt gross area.

u = Ultimate shear strength for the bolt material,

found experimentally to be around (0.62 Fub) of the

ultimate tensile strength (Fub).

Thus:

Rn = m Ab (0.62 Fub)


10/40

F-10

The bearing failure near the edge of the plate is related to the

shear failure (tear-out) as shown in figure 4.6.4 b & d (page F-7).

Assuming angle () 0 as shown in figure 4.6.5 below:

pu

den LtR 22

Where

up = Shear strength of plate material = 0.62 Fu

Fu = Ultimate tensile strength of plate material.

Le = Distance from center of last bolt to the edge

of plate along direction of force.d = Nominal Bolt diameter.

2

1241

d

LdtFR eun .


11/40

F-11

Thus:

Since LRFD recommends spacing of bolts inthe direction of force to be (2.67 d) [see

LRFD J 3.3]. Consider (Le 2.67 d) above:

The lower resistance value guarantees that no tear-outdue to bearing shall occur below this (Rn) value, and no

Deformation at the hole shall occur at service load level.

(AISC J3-10 Page16.1-111)

* Rn = 2.4 Fu dt

or

* Rn = 3.0 Fu dt

(When deformation of the hole at service

load is a design consideration). (Eq J3-6a)

(When deformation of the hole at service

load is not a design consideration). (Eq J3-6b)


12/40

F-12

(1) Threads are included in shear plane (N)

(2) Threads are excluded from shear plane (X)

In general:

n

Rn

iQ

i

Where, Rn = Nominal resistance (strength)

n = Strength reduction factor (resistance factor)

iQi = Factored load on structure (ASCE 7)

for a single bolt :

Rn PuWhere,

u = 0.75 for fracture in tension, shear or bearing.

Rn = Nominal strength of one fastener

Pu = factored load on one fastener

The strength of a fastener is based on:

1- Tension capacity

2- Shear capacity

3- Bearing capacity


13/40

F-13

)75.0(75.075.0 bb

ub

b

un AFAFR

b

b

un AFR ).(. 750750

Reference is made to (LRFD) J 3.6), and page (F-8):

or

Where

= 0.75 as stipulated in AISC J3.6 (page 16.1-108)

b

uF= Tensile strength of bolt material

(120 ksi for A 325 Bolts & 150 ksi for A 490 Bolts).

Ab = Gross area of the bolt (Shank section).


14/40

F-14

A No th reads in shear plane (X):-

Rn = 0.75 (Rn) (AISC J3-6)Rn = 0.62m AbFu

b (page F-9)

For long Connections (upto 50 inch long), a 20% reduction is needed

(strength of long connection < of strength of individual bolts).

Thus:-

Rn = (Reduction due connection length)(0.62 Fub)m Ab

= 0.8 x 0.62 Fub . m Ab = 0.5 Fu

b m Ab

Rn = (0.75)(0.5) Fub m Ab

Where,

Fub

= Ultimate strength of bolt material:120 ksi for A 325 ; 150 ksi for A 490

m = 1 for single shear , 2 for double shear

Ab = Area of the shank portion of bolt

(See LRFD J 3.6 & table J 3.2)


15/40

F-15

B - Threads in Shear Plane (N):-

The area of the threaded section of the bolt is about 75%

of the gross area at the shank section,

Thus:

Rn = [Reduction for length] (0.62 Fub) m (0.75 Ab)

= (0.8)(0.62 Fub) m (0.75 Ab)

= 0.37 Fub m Ab

(0.40 Fub) m Ab

Hence:Rn = 0.75 (0.4 Fu

b) m Ab

(see Table J-3.2)


16/40

F-16

Nominal bearing strength (Rn) was developed earlier (page F-11).

However LRFD-J3.10 further reduces (Rn) in order to prevent boltelongations exceeding 0.25 in.

Hence several categories are presented here below:

1: Deformation Lim it State (Standard Holes):-

Where Le 1.5 d and c/c spacing 3d, and there are two or more bolts:

Rn = (2.4 dt Fu)..(Equ.. J3-6a) = 0.75

d = Nominal diameter of bolt at shank section.

t = Thickness of the least connected plate.

Fu = Ultimate tensile strength of the connected plate.


17/40

F-17

2: Deformation Lim it State (Long -Slotted Holes):

Same as previous conditions.

Rn = (2.0 dt Fu) (Equ. J2 6c)Where = 0.75 ......... (AISC - J3 10 Page 16-1-111)

3: Streng th Limit State (ho le elongat ion 0.25 in.):

Rn = (3.0 dt Fu). (Equ. J3 6b)See page (F-11) of these notes.


18/40

F-18

A) Minimum Spacing Bol ts along l ine of Force:

2.67 d s 3d (AISC J3.3)where,

s = spacing of bolt along force line.

d = diameter of bolts-shank section

B) Minimum Edge Distance along Line of force:

Normally min. edge distance (rolled edge)

Normally min. edge distance (sheared edge)

Where d = Nominal diameter of the bolt.

(Reference table J3-4 page 16.1 107)

d4

11

d4

31


19/40

F-19

C) Maximum spacing for Bol ts:

for Painted members:

s 24t but 12 in.for unpainted members:

s 14t but 7 in.t = thickness of the thinner connected plates.

D) Maximum Edge Distance:

max. edge distance = 12 t but 6 ...

(AISC J 3.5 Page 16.1 106)

t = thickness of the thinner connected plates.

Reference: AISC J 3.5.

(AISC page 16.1 108)


20/40

F-20

Example F-1

Determine the required number of in. dia.A490 bolts for the connection below.

Assuming DL = 14 kips, LL = 126 kips?

Solution:-a) Calculate the factored load:

Tu = 1.2 DL + 1.6 LL= 1.2 14 + 1.6 126 = 218 kips.

b) Design strength Rn per bolt:

Rn = 0.75 Fub (0.75 Ab)

= 0.75 113 0.4418 (AISC Table J 3.2)

= 37.3 kips.

c) Number of bolts required:

(say 6 bolts)

use 6 in A 490 bolts.

5.837.3

218

R

Tn

n

u


21/40

F-21

Example F-2Compute the tensile factored load capacity for the bearing type connection

shown below, Assume (A-572 Grade 50) steel plates, with (7/8) diameter

A-325 bolts in standard holes:

A) Bolt threads are excluded from shear plane (X).

B) Bolt threads are included from shear plane (N).

Solution:-Check Plate Capacity:

.ConnectedFullyin.sq..AA

in.sq...A

in.sq...

ne

n

52

52625026

75362506

8

1

8

7

U

Ag

Tension Member Bearing-Type Connection-LFRD

Tn = 0.90 50 3.75 = 169 kips.or

Tn = 0.75 65 2.50 = 122 kips. (controls)


22/40

F-22

A) Strength of A-325-X bolts:

Shear:

Rn = (0.5 Fub) m Ab

= 0.75 0.5 120 1 0.6013

= 27.1 kips per bolt.

Bearing:Since Le 1.5d and spacing S > 3d, the bearing strength of a bolt is:

Rn = (2.4 Fu dt)

= 0.75 2.4 65 0.875 0.625

= 64 kips per bolt.

Thus shear controls, the capacity of the connection:

Tn = 4 27.1 = 108 kips < 122 kips

So, Connection capacity = 108 kips.


23/40

F-23

B) Strength of A-325-N bolts:

Shear:

Rn = (0.4 Fub) m Ab= 0.75 0.4 120 1 0.6013


No need to check bearing again, Since

Shear will control:

Tn = 4 21.6 = 86.4 kips

Total Connection capacity = 86.4 kips


24/40

F-24

Example F-3

Determine the number and spacing of ( in.) dia. (A-490-N) bolts

required to develop the full strength of the A-572 Grade 65 plates shown

below. Assume double rows of bolts with standard holes?

Solution:-Consider middle plate:

in.sq.1.590.37526A 8143n

Ae

= An

= 1.59 sq. in. (well connected).

max. An = 0.85 Ag = 0.85 6 0.375 = 1.91 sq. in.

Tn = Fy Ag = 0.90 65 6 0.375 = 132 kips.or

Tn = FuAe = 0.75 80 1.59 = 96 kips. (controls)

(check T-16 of these notes)


25/40

F-25

Bolt shear capacity for in. A-490-N bolt:

Rn = (0.4 Fub) m Ab

= 0.75 0.4 150 2 0.4418

= 39.8 kips per bolt. (controls)

Bearing Strength on the 3/8 in. plate:

Rn = (2.4 Fu dt)

= 0.75 2.4 80 0.75 0.375


Number of bolts = =2.4 bolts.

Use 4- in. (A-490-N) bolts. With:End distance must be at least

and spacing of bolts

39.8

96

.23d41 in

.4

1in1


26/40

F-26

Example F-4Design the connection below to develop the full strength of double

angles (back-to-back), using two lines of ( in.) dia. (A-325-N) bolts

in the bearing type connection of 2Ls 6 4 (A-36) material ?8

3

+

+

Pu

(A-36 Steel)

Number of bolts

To be determined.

Gusset Plate

( Thick) (A-36)

2Ls 6 x 4 x 3/8


27/40

F-27

U = 0.80 (assuming 4 bolts or more in one row)

Ae = U An = 0.80 2.95 = 2.36 sq. in.

= 0.9 3.61 36 = 117 kips.

Angle Capacity (Pu)

= 0.75 2.36 58 = 102.7 kips. (controls)

(Pu) for two angles = 205.4 kips.

Solution:-

For a single angle:

Ag = 3.61 sq. in.

in.sq.2.95x23.61A83

81

43

n


28/40

F-28

Shearing capacity each bolt = (cotrols)kips.... 831212040750 243

4

Bearing capacity each bolt = 0.75 2.4 0.75 58 = 39.2 kips(Gusset plate more critical)

Required No. of bolts

Use 8 bolts (Four in line). Use spacing as shown below:

bolts6.4531.8

205.4

2

2

2

1

1 2 2

6

4x

Re-Check:

from L data:

used..

..

..

.

.

80860

11

756222

9330

759

9330

4

1

4

1

4

1

L

xU

inL

inx

Higher value of (U) is always on the safe side.


29/40

F-29

In slip-critical type connection (page F-3), plates are not allowed to

slip against each other, hence there is no practical shearing or

bearing contact between the bolts and the joining plates (or shapes).

This necessities the need for high quality control over the bolt

tension (proof load).

The main performance criteria of the slip-critical type connection is

that no slippage of the connected parts should occur at the service

load level at the structure.

However, to make the design of the slip-critical type connection

compatible with the design of the bearing type connection, AISC

stipulates the design of slip-critical connection at the factored load

level of the connected elements, as follows:-


30/40

F-30

Rn = Du hsc Tb Ns (AISC J 3.8) (Page 16.1 109)where:

= 1.0 for connections which resist slip at the service load level.

Rn = Nominal slip resistance of a (sc) bolt in kips.

= Friction coefficient for class A & B surfaces:

= 0.35 for class (A) surface (clean mill scale).

= 0.50 for class (B) surface (blast clean surface).

Du = 1.13 a statistical factor related to bolt pretension.

Hsc = Hole factor hsc =1.0 for standard holes.

hsc = 0.85 for Oversized holes.

hsc = 0.70 for long-slotted holes.

Ns = Number of slip planes (1 or 2) as (m) before.

Tb = Minimum fasteners tension in kips. (Table J3.1).


31/40

F-31

Example F -5

The connection shown below uses ( inch) A

325 SC bolts. No slippage is permitted. All

connected members are (A-36) steel. Determine

strength capacity of this connection (Tu)

Solution :-

Even the connection is specified as slip-

critical connection, the direct shear and

bearing capacity must be checked (in the

event of accidental slippage).

Tu

1.5

1.5

3

1.51.5

3

(a)

PlateGusset8

3

6 x


32/40

F-32

Tu for four bolts = 15.90 4 = 63.6 kips.

2) Bearing Capacity of (A325N) bolts.

Since the gusset plate is thinner (3/8 < ), It will control bearing :

Rn = (2.4 d t Fu)

58x

8

3x

4

3x2.4x0.75

= 29.36 kips

Bearing for four bolts : Rn = 4x29.36 = 117.45 kits

1) Shear capacity of bolts (A-325 N):

4418.04875.0FR vn bA



33/40

F-33

3) Slip-Critical Connection:-

Rn = Du hsc Tb Ns

where,

= 1.0

= 0.35 assume class (A) surface.

Du = 1.13

hsc = 1.0 standard holes

Tb = 28 kips (Table J 3.1)(Table 16.1 103)

Ns 1 slip plane.

Rn = 1 0.35 1.13 1 28 1 = 11.1 kips per bolt

For four bolts Tu = 4 11.1 = 44.4 kips.


34/40

F-34

4) Tension on the plate :-

in.sq.2.1252.125x1UAA

in.sq.2.125x4.25A

in4.2526Winsq,36xA

ne

21

n

n

21

g

8

1

4

3

Maximum Capacity Tu = 44.4 kips.

0.9 Fy Ag

Rn=

0.75 Fu Ae

(controls)

kips92.42.125x58x0.75R

or

kips97.23x36x0.9R

n

n


35/40

F-35

Example F-6:

A 13-foot-long tension member and its

connection must be designed for a service

dead load of 8 kips and a service live load of

22 kips. No slip of the connection

is permitted. The connection will be to a 3/8-

inch-thick gusset plate, as shown. Use a

single angle for the tension member. Use

A325 SC bolts and A572 Grade 50 steel forboth the tension member and the gusset plate.

D = 8 kips

L = 22 kips

83t

A 325 SC bolts

A572 Grade 50 steel

Gusset plate

Angle

(A 572 Gr 50)


36/40

F-36

The factored load to be resisted is

Pu = 1.2D + 1.6L = 1.2(8) + 1.6(22) = 44.8 kipsBecause the bolt size and layout will affect the net area of the tension member, we will

begin with selection of the bolts. The strategy will be to select a bolt size for trial,

determine the number required, and then try a different size if the number is too large

or too small. Bolt diameter typically range from inch to 1 inches in 1/8-inch

increments.

Try 5/8 inch bolts. The nominal bolt area is

22

in.3068.04

)8/5(

gA

The shear strength is

Rn = FvAb = 0.75(48)(0.3068)

= 11.04 kips/bolt

(assuming that the threads are in the shear plane)

No slip is permitted, so this connection is slip-critical. We will assume class (A)

surfaces, and for a 5/8-inch-diameter A325 SC bolt, the minimum tension is Tb = 19

kips (from AISC Table J3.1).

Rn = Du hsc Tb Ns

= 1 0.35 1.13 1 19 1

= 7.5 kips (controls)

Solution:


37/40

F-37

Say 6 bolts ;This can be achieved in two rows, which requires long leg forthe angle, or select a larger size bolt to have less number of bolts, in order

to fit in one line.

Try ( inch) diameter bolts.

kips15.90x48x0.441875.0R

in.sq.4418.043

4

n

2

bA

(assuming thread in shear plane after slip have occurred).

The Slip-Critical bolt capacity :

Rn = 1 0.35 1.13 1 28 1

= 11.07 kips. (controls)

bolts9.5boltsofNumber5.7

8.44


38/40

F-38

bolts4.011.07

44.8bolts.

4

3ofNo.

in

2y

u

1gin.0.996

0.9(50)

44.8

0.9F

PA

Select four bolts of ( in.) diameter (A-325-SC)

Minimum spacing 3 d = 2 in. say (2 in.)

Minimum Edge distance = 1.5 in (assume sheared edge).

Check Tension on Member :

and the required effective net area is

2

u

ue in.0.9190

0.75(65)

44.8

0.75F

PA

U

AA en

112 22


39/40

F-39

For the layout shown in Figure above, with four bolts in the line of force,

the value of U from the AISC table D3.1 (page 16.1-29) is 0.80. (Once

a member has been selected, U can be computed with AISC.

Therefore

.in1.15

0.80

0.9190A

2

n

t1.15A161

43

2g

Assume t = in thick, (Ag)2 = 1.35 in.2

in.0.52300

13x12

300

Lr

req.min.

(controls)


40/40

F-40

angle23LTry41

21

21

Ag = 1.44 in2 > 1.35 in2 (OK)

rmin = rz = 0.544 in. > 0.52 in. (OK)

Use an with

Long leg connected. Use A325-SC bolts as shown.

4

1

2

1

2

1

23L

434

3

2

x

2

2

2 2141

161

430101

9209190

92057

607011

inAinA

L

xU

gn .....

..

.

Our selected Ag = 1.44 in2 is greater then (Ag)2 Then OK.

Chapter 4 (Final F) 18 March 07

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Transcript of Chapter 4 (Final F) 18 March 07