Design of Tension Member Lrfd

7/30/2019 Design of Tension Member Lrfd

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DESIGN OF TENSION MEMBER


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In a design process a structural

engineer must combine the following:

analysis, engineering judgment,experience, construction method,

economic design etc. Design cannot bedone if engineer does not know the basic

concept of structural analysis. There are

two consideration in the design oftension member:


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Strength

the tension member must be

adequate to resist the ultimate axial

tensile load.


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Stiffness

the tension member must notfail due to serviceability

requirements


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Design of Steel Tension Members

Equations for strength of tension members:

a) For yielding in the gross section:

b) For fracture in the net section:

gytnt AFP

eutnt AFP


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P P

Design of Steel Tension Members

Yielding in the gross section:

Max stress FyP P

Max stress Fu


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Variable Definitions

Resistance factor,

= 0.90 for yielding

= 0.75 for fracture

Fy = Yield Strength

Fu = Tensile or Ultimate strength

:t


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Design Requirements

Ag Gross cross-sectional area

Ae Effective net area

If tension load is transmitted directly to each of the

cross-sectional elements by fasteners or welds:

Ae = An

An = Net cross-sectional area

(gross-section minus bolt holes)


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Design Requirements

If tension load transmitted through some but

not all of the cross-sectional elements:

by fasteners or bolts,

Ae = AnU

by welds,Ae = AgU or Ae = AU


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TENSION MEMBERS ARE

CONNECTED BY:

Rivets

Bolts

Welds


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Advantages of Bolted Connections

Do not require highly skilled workers

Holes can be done in the shop

Less equipment needed

Quality of work can be easily

controlled Faster construction


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Advantages of Welded Connections

Minimize construction noise since no impact

wrenches are used as in Bolted connection

Members can be easily installed

Easily be repaired when dimensions are

unmatched

Joints are very rigid

Shop and construction detailing is much

simplier


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Simple Bolted Connections

There are different types of bolted

connections. They can be categorized

based on the type of loading.

Tension member connection and splice. It

subjects the bolts to forces that tend to

shear the shank.

Beam end simple connection. It subjects the

bolts to forces that tend to shear the shank.

Hanger connection. The hanger connection

puts the bolts in tension


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68402/61420 Slide # 14

Simple Bolted Connections

The bolts are subjected to shear or tension loading.

In most bolted connection, the bolts are subjected to shear.

Bolts can fail in shear or in tension.

You can calculate the shear strength or the tensile strength of a bolt

Simple connection: If the line of action of the force acting on the

connection passes through the center of gravity of the connection,

then each bolt can be assumed to resist an equal share of the load.

The strength of the simple connection will be equal to the sum of

the strengths of the individual bolts in the connection.


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The bolts are subjected to shear or tension loading.

In most bolted connection, the bolts are subjected to shear.

Bolts can fail in shear or in tension.

You can calculate the shear strength or the tensile strength of a

bolt

Simple connection: If the line of action of the force acting on

the connection passes through the center of gravity of

the connection, then each bolt can be assumed to resist

an equal share of the load.

The strength of the simple connection will be equal to thesum of the strengths of the individual bolts in the

connection.


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Bolt Types & Materials

A307 - Unfinished(Ordinaryor Common) boltslow carbon steel A36, Fu = 413 MPa,

for light structures under static load

A325 - High strength bolts, heat-treated medium

carbon steel, Fu = 827 MPa,for structural joints

A490 - High strength bolts, Quenched andTempered Alloy steel, Fu = 1033 MPa

for structural jointsA449 - High strength bolts with diameter > 32mm

anchor bolts, lifting hooks, tie-downs


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Common Bolts

ASTM A307 bolts

Common bolts are no longer

common for current structural designbut are still available Common Bolts

nu RP 75.0

boltvn AfR MPafv 165


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HSB Bearing Type Connections

The shear strength of bolts shall be determined as follows

boltvn AfR

75.0nu RP

517413A490

413330A325

Type X ThreadType N ThreadType

If the level of threads is not known, it is conservative to assumethat the threads are type N.

AISC Table J 3.2

The table bellow shows the values of fv (MPa) for different types of bolts


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Bolted Shear Connections

In designing the bolted shear connections the factored designstrength (Rn) should be greater than or equal to the factored

load. Rn Pu

So, examining the various possible failure modes and calculate

the corresponding design strengths is a must.

Possible failure modes are:

Shear failure of the bolts

Failure of member being connected due to fracture or yielding or .

Edge tearing or fracture of the connected plate

Tearing or fracture of the connected plate between two bolt holes

Excessive bearing deformation at the bolt hole


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Failure Modes of Bolted Connections

Bolt Shearing

Tension Fracture

Plate Bearing

Block Shear


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Shear failure of bolts

Average shearing stress in the bolt = fv = P/A = P/(db2/4)

P is the load acting on an individual bolt

A is the area of the bolt and db is its diameter

Strength of the bolt = P = fv x (db2/4) where fv = shear yield

stress = 0.6Fy

Bolts can be in single shear or double shear as shown above.

When the bolt is in double shear, two cross-sections are effective inresisting the load. The bolt in double shearwill have the twice the shear

strength of a bolt in single shear.


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Failure of connected member

Tension Member can fail due to tension fracture or yielding.

Bearing failure of connected/connecting part due to bearing

from bolt holes Hole is slightly larger than the fastener and the fastener is loosely placed in

hole

Contact between the fastener and the connected part over approximatelyhalf the circumference of the fastener

As such the stress will be highest at the radial contact point (A). However,the average stress can be calculated as the applied force divided by the

projected area of contact


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Average bearing stress fp = P/(db t), where P is the force applied to the fastener.

The bearing stress state can be complicated by the presence of nearby bolt oredge. The bolt spacing and edge distance will have an effect on the bearing

strength.

Bearing stress effects are independent of the bolt type because the bearingstress acts on the connected plate not the bolt.

A possible failure mode resulting from excessive bearing close to the edge ofthe connected element is shear tear-out as shown below. This type of shear

tear-out can also occur between two holes in the direction of the bearing load.

Rn = 2 x 0.6 Fu Lc t = 1.2 Fu Lc t


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Spacing and Edge-distance requirements

The AISC code gives guidance for edge distance and spacing to avoid tearout shear

2

hLL ec

SeLdiameterholetheish mmdh bolt 6.1

NOTE: The actual hole diameter is 1.6 mm bigger than the bolt,

we use another 1.6 mm for tolerance when we calculate net area. Here use 1.6 mm only not 3.2

boltdS 322

Bolt spacing is a function of the bolt diameter

Common we assume

The AISC minimum spacing is

eL

boltdS 3

AISC Table J 3.4


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Bolt Spacings & Edge Distances

Bolt Spacings- Painted members or members not subject to corrosion:

2 2/3d Bolt Spacings 24tor 305 mm

(LRFD J3.3) (LRFD J3.5)

- Unpainted members subject to corrosion:

3d Bolt Spacings 14tor 178 mm

Edge Distance

Values in Table J3.4MEdge Distance 12tor 152 mm

(LRFD J3.4) (LRFD J3.5)

d - bolt diameter

t - thickness of thinner plate


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Important Notes

Lc Clear distance


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Design Provisions for Bolted Shear

Connections

In a bolted shear connection, the bolts are subjected to shearand the connecting/connected plates are subjected to bearing

stresses.

Bolt in shear

Bearing stresses in plate


TT

T

T

Bolt in shear



Bolt in shear



TT

T

T


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Connections

The shear strength of all bolts = shear strength of one bolt x

number of bolts

The bearing strength of the connecting / connected plates can

be calculated using equations given by AISC specifications.

The tension strength of the connecting / connected plates can

be calculated as discussed in tension members.


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Connections

In a simple connection, all bolts share the load equally.

TT

T/n T/n

T/n T/n

T/n T/n

TT

T/n T/n

T/n T/n

T/n T/n


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AISC Design Provisions

Chapter J of the AISC Specifications focuses on connections.

Section J3 focuses on bolts and threaded parts

AISC Specification J3.3 indicates that the minimum distance (s)

between the centers of bolt holes is 2.67. A distance of 3db ispreferred.

AISC Specification J3.4 indicates that the minimum edge distance

(Le) from the center of the bolt to the edge of the connected

part is given in Table J3.4. Table J3.4 specifies minimum edgedistances for sheared edges, edges of rolled shapes, and gas cut

edges.


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AISC Specification indicates that the maximum edge distance forbolt holes is 12 times the thickness of the connected part (but

not more than 152 mm). The maximum spacing for bolt holes is

24 times the thickness of the thinner part (but not more than

305 mm).

Specification J3.6 indicates that the design tension or shear

strength of bolts is FnAb

= 0.75

Table J3.2, gives the values of Fn Ab is the unthreaded area of bolt.

In Table J3.2, there are different types of bolts A325 and A490.


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The shear strength of the bolts depends on whether threads are included orexcluded from the shear planes. If threads are included in the shear planes thenthe strength is lower.

We will always assume that threads are included in the shear

plane, therefore less strength to be conservative.

We will look at specifications J3.7 J3.9 later.

AISC Specification J3.10 indicates the bearing strength of plates at bolt holes.

The design bearing strength at bolt holes is Rn Rn = 1.2 Lc t Fu 2.4 db t Fu - deformation at the bolt holes is a design

consideration


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Common Types of Bolted Connections


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TYPES OF BOLT HOLES

Standard

Oversized

Short-slotted

Long-slotted

Connection types

Bearing Slip-critical


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36

Failure Mode of Bolted Shear Connections

Two types of bolted

connector failure are

considered in this

section

Failure of the

connector

Failure of theconnected parts


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Failure Mode of Bolted Shear Connections(cont.)

Connector failure

Single shear connection Single shear plane. P = fvA,

where fv is the average shearing stress and A is the

connectors cross-sectional area.

Double shear connection Double shear plane. P = 2fvA


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Failure of the connected parts , separated into two categories.

1. Failure resulting from excessive tension, shear, or bending

in the parts being connected

For a tension member must consider tension on the net area,

tension on the gross area, and block shear

For beam-beam or beam-column connections, must consider block

shear

Gusset plates and framing angles must be checked for P, M, and V


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2. Failure of the connected part because of bearing

exerted by the fastener (average bearing stress is fp =P/dt)

If the hole is slightly larger than the fastener and the fastener

is assumed to be placed loosely in the hole (rarely the case),

contact between the fastener and the connected part will exist

over approximately 50% of the circumference of the fastener.

The bearing problem is affected by the edge distance and bolt

spacing


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40

LRFD Fasteners (cont)

Design bearing strength1. Usual conditions based on the deformation limit state, according to

LRFD-Formula (J3-1a). This applies for all holes except long-slottedholes perpendicular to the line of force, where end distance is at least

1.5d, the center-to-center spacings is at least 3d, and there are two ormore bolts in the line of force.

where = 0.75

d = nominal diameter of bolt at unthreaded area

t= thickness of part against which bolt bears

= tensile strength of connected part against which bolt bears

= distance along line of force from the edge of the connected part tothe center of a standard hole or the center of a short- and long-slotted hole

perpendicular to the line of force.

uF

eL

)4.2()2.1(uuen

dtFtFLR


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Design shear strengthno threads in shear planes

where = 0.75, the standard value for shear = tensile strength of the bolt material (120 ksi for A325 bolts;

150 ksi for A490 bolts)

= the number of shear planes participating [usually one

(single shear) or two (double shear) ]

= gross cross-sectional area across the unthreaded shank of the bolt

Design shear strengththreads in shear planes

b

b

un mAFR )50.0(75.0

b

uF

m

bA

b

b

un mAFR )40.0(75.0

Sequi Examples 7.1 & 7.2


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LRFDFasteners

(cont)


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Minimum edge distance requirement (AISC J3.4)


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LRFD Slip-critical Connections

A connection with high-strength bolts is classified as either abearing or slip-critical connection.

Bearing connections - the bolt is brought to a snug-tightcondition so that the surfaces of the connected parts are in firmcontact. Slippage is acceptable

Shear and bearing on the connector

Slip-critical connections - no slippage is permitted and thefriction force described earlier must not be exceeded. Slippage is not acceptable (Proper installation and tensioning is key)

Must have sufficient shear an d bearing strength in the event of overloadthat causes slip. AISC J3.8 for details.


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Spacing, Edge and End Distances

Center to Center (pitch) 24 x thickness of

thinner plate 300mm

Edge distance 12 x thickness of part

150mm


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Bolt Shear Strength

FnAb Table J3.2

Fn Table J3.2

Ab Area of bolt

Tabulated on page 7-33, Table 7-10


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Bearing Strength at Bolt Holes

J3.10 Rn = 0.75

Rn given by equations J3-2

Tabulated on page 7-34,35


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Ex. 1.0 - Design Strength

Calculate the design strength of the simple connection shownbelow and check whether it is adequate for carrying the

factored load of 300 kN.

1.25 2.50 1.25

1.25

2.50

1.25 65 k

A36

A36 5 x

3/8 in.

in. bolts

1.25 2.50 1.25

1.25

2.50

1.25 65 k

A36

A36 5 x

3/8 in.

in. bolts

63 k

10 mm

120x15 mm

20 mm A325-N bolts

30 mm

60 mm

30 mm

30 mm 60 mm 30 mm

300 kN


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Solution:

The design shear strength of one bolt in shear= Fn Ab = 0.75 x 330 x x 202/4000 = 77.8 kN

Fn Ab = 77.8 kN per bolt

Shear strength of connection = 4 x 77.8 =311.2 kN



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Minimum edge distance and spacing requirements

See Table J3.4M, minimum edge distance = 26 mm for rollededges of plates

The given edge distances (30 mm) > 26 mm. Therefore,

minimum edge distance requirements are satisfied.

Minimum spacing = 2.67 db = 2.67 x 20 = 53.4 mm.

(AISC Specifications J3.3)

Preferred spacing = 3.0 db = 3.0 x 20 = 60 mm.

The given spacing (60 mm) = 60 mm. Therefore, spacingrequirements are satisfied.



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Bearing strength at bolt holes. Bearing strength at bolt holes in connected part (120x15 mm plate)

At edges, Lc = 30 hole diameter/2 = 30 (20 + 1.6)/2 = 19.2

Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x19.2 x15x400)/1000 = 103.7 kN But, Rn 0.75 (2.4 db t Fu) = 0.75 x (2.4 x 20x15x400)/1000 = 216 kN

Therefore, Rn = 103.7 kN at edge holes.

At other holes, s = 60 mm, Lc = 60 (20 + 1.6) = 38.4 mm.

Rn = 0.75 x (1.2 Lc t Fu) = 0.75x(1.2 x 38.4 x15 x400)/1000 = 207.4 kN

But, Rn 0.75 (2.4 db t Fu) = 216 kN. Therefore Rn = 207.4 kN


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At other holes, s = 60 mm, Lc = 60 (20 +1.6) = 38.4 mm. Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x 38.4 x 10x 400)/1000 = 138.2 kN

But, Rn 0.75 (2.4 db t Fu) = 144 kN

Therefore, Rn = 138.2 kN at other holes

Therefore, bearing strength at holes = 2 x 69.1 + 2 x 138.2 = 414.6 kN Bearing strength of the connection is the smaller of the bearing strengths = 414.6

kN


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Connection Strength

Shear strength = 311.2 kN

Bearing strength (plate) = 622.2 kN

Bearing strength (gusset) = 414.6 kN

Connection strength (Rn) > applied factored loads (gQ).

311.2 > 300 Therefore ok.

Only connections is designed here

Need to design tension member and gusset plate


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Example problem

2.0 Calculate the tensile capacity of two A36

plate spliced using 20mm dia. A307 bolts asshown in the figure. Use LRFD design provisionassuming that the Live load is twice the Dead

load. Determine also the distances a and b


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Example problem

Tension member

P P

a a75mm

250mm

P

P

10mm.

10mm.

b

b


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Bolt Design Example

Design a pair of splice plates to connect

the two parts of the channel tension

member shown. The forces in the

member are 500kN live load and 280kN

dead load. The bolts in the surrounding

joints are 25mm diameter A325N. All

steel is A36. Slip is not critical.


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Bolt Design Example

Design of Tension Member Lrfd

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