Team TeachingStructural Design

Civil Engineering Department2013/2014

Steel Structural Design Tension Member

Tension Members

Structural members that are subjected to axial tensile force (truss members, cables in suspension bridges, bracing for buildings,).

Any cross-sectional configuration may be used, since the only determinant of strength is the cross-sectional area.

Circular rods and rolled angle shapes are commonly used.

Examples

truss members, cables in suspension and cable stayed

bridges, bracing for buildings and bridges cable for suspended roof systems

Section Property

The stress in a tension member is uniform throughout the cross-section except:1. near the point of application of load, and2. at the cross-section with holes for bolts or other

discontinuities, etc.

Stress stress in an axially loaded tension member is:

APf

where P = magnitude of loadA = cross sectional area

b b

aa

8 x in. bar

Gusset plate

7/8 in. diameter hole

Section a-a

Section b-bb b

aa

8 x in. bar

Gusset plate

7/8 in. diameter holeb b

aa

8 x in. bar

Gusset plate

7/8 in. diameter hole

Section a-a

Section b-b

Section a-a

Section b-b

Example :

Area of a a = 8 x = 4 in2

Area of b b =(8 2 x 7/8 ) x = 3.12 in2

The reduced area of section b b will be subjected to higher stresses and therefore the higher stresses will be localized around section b b.

Section area with hole :

T T

Fy

NnNn

Elastic stress Yield stress

Reach Nominal Strength

Steel Stress Strain Curve

Strain, y uy u

S

t

r

e

s

s

,

f

Fy

Fu

E

Strain, y uy u

S

t

r

e

s

s

,

f

Fy

Fu

y uy u

S

t

r

e

s

s

,

f

Fy

Fu

E

1. Deformations are caused by the strain ()

2. Small deformation if < y3. Large deformatin if > y

Fy is the yield stress and Fu is the ultimate stress

y is the yield strain and u is the ultimate strain

A tension member can fail by reaching one of two limit states:

(1) Excessive deformation due to the yielding of the gross section (section a-a)

(2) Fracture of the net section can occur if the stress at the net section (section b-b) reaches the ultimate stress Fu.

The objective of design is to prevent these failure before reaching the ultimate loads on the structure.

Limit states of tension member

A tension member can fail by reaching one of the two limiting states: yielding or fracture :

Nominal Strength: Pn

1. Nominal strength in Yielding,

2. Nominal strength in Fracture

ygn FAP

Where : Fy = yield stressAg = gross areaAe = effective net area= U AnAn = net area U = efficiency factor

uen FAP

(Strength Reduction Factor) SRF for yielding,

t = 0.90 Pn = 0.9 Fy Ag

SRF for fracture failure, t = 0.75 Pn = 0.75 Fu Ae

Mechanical Properties of Steel in Indonesia

Mechanical Properties of Steel in America

Steel Type Ultimate Stress Fu (Mpa) Yield Stress Fy (Mpa)Minimum

Elongation (%)

BJ 34 340 210 22

BJ 37 370 240 20

BJ 41 410 250 18

BJ 50 500 290 16

BJ 55 550 410 13

Steel Type Fu (Ksi) Fy (Ksi)

A36 58 36A242 63 - 70 42 - 50A572 65 50A992 65 50

Designed Nominal Capacity Pn Axial factored loads Pu should meet the

requirement :Pu < Pn

Where Pn is designed nominal capacity. The value is the smallest between the two limit states yield strength and fracture strength

Hole diameter

The usual practice is to drill or punch standard holes with a diameter 1/16 in. larger than the fastener diameter.

To account for possible roughness around the edges of the hole, AISC requires the addition of 1/16 in. to the actual hole diameter.

Thus :d hole = d bolt + 1/8 in. or +1/16 d hole = d bolt + 3.2mm

Net section area is determined based on two conditions and the smallest among them :

1. The maximum hole allowed is 15% of total gross area, therefore :

An > 0.85 Ag

2. An is calculated by deducting gross area with hole area.

Net Section Area (An)

2. Net Section Area (An)

uu

s

1

32Nu Nu

thickness = t

Section 1 3 :

Section 1-2-3 :

utsndtAA

ndtAA

gnt

gnt

4

2

Example

Ag = 5 x = 2.5 in2

An = (5-2 x ) x = 1.75 in2

The connecting bolts can be staggered for several reasons:(1) To get more capacity by increasing the effective net area(2) To achieve a smaller connection length(3) To fit the geometry of the tension connection itself

An of Staggered Bolts Holes

utsndtAA gnt 42

An = wn x twn = wg d + s2 / 4g

wn = net width wg = gross width d = sum of the hole diameters s (pitch) = spacing of two adjacent holes

(parallel to direction of the load) g (gage) = transverse spacing of lines of

bolts

3 in.

5 in.

5 in.

3 in.

3 in. 3 in. 3 in. 3 in. 3 in. 3 in.

b

a

c

d

e

i

j

f

h

3 in.

5 in.

5 in.

3 in.

3 in. 3 in. 3 in. 3 in. 3 in.3 in. 3 in. 3 in. 3 in. 3 in. 3 in.

b

a

c

d

e

i

j

f

h

Hole diameter: 1 + 1/8 = 1.125 in.

Line : a-b-d-ewn = 16.0 2 (1.125) = 13.75 in.

Line : a-b-c-d-ewn = 16.0 3 (1.125) + 2 x 32/ (4 x 5)

= 13.52 in.Line : a-b-c-f-h

wn = 16.0 3 (1.125) + 2 x 32/ (4 x 5)= 13.52 in.

a-b-c-d-e is the smallest net width thus:An = t wn = 0.75 (13.52) = 10.14 in2

Possible line fracture :

Staggered bolts in angles If staggered lines of bolts are present in both legs of an

angle, then the net area is found by first unfolding the angle to obtain an equivalent plate

Gage line crossing the heel of the angle should be reduced by an amount equal to the angle thickness

g = 2 + 3 t = 5 =4.5 inch If the thickness of flange and web is different (tf tw)

g = g + g2 - tw or

Determine design strength of this tension member (using A36 and bolt with 7/8in diameter)

Line abcdeg :

Determinethesmallestnetarea(boltdiameterstw =0.550andtf =0.622)

27

Effective Net Area (Ae)

When all elements of the cross section are not connected ( Ex: only one leg of an angle is bolted to a gusset plate), shear lag occurs.

The connected element becomes overloaded and the unconnected part is not fully stressed.

This can be accounted for by using a reduced, or effective, net area.

Examples of Shear Lag

W 8 x 24

in. diameter bolts

3 in. 3 in. 3 in.

Holes in beam flange

W 8 x 24

in. diameter bolts

W 8 x 24

in. diameter bolts

3 in. 3 in. 3 in.

Holes in beam flange

3 in. 3 in. 3 in.3 in. 3 in. 3 in.

Holes in beam flange

29

Effective Net Area (Ae)

For Bolted Connections: Ae = U An

If all elements of the cross section are connected, then, U = 1

If not, use the recommended values of the reduction factor, U (see next slide)

Recommended U value for bolt

connections

Effective Net Area

For welded connections: Ae = U Ag

For any W-, M-, S-, or tee shape connected by transverse weld alone:

Ae = area of connected element

Recommended U value for welded connections

For a rectangular bar or plate Ae = An For plates or bars connected by longitudinal

welds at their ends:U=1 l > 2wU=0.87 2w > l > 1.5wU=0.75 1.5w > l > w

l = length of the pair of welds > w

w = distance between the welds

U = Reduction Factor

L

xU 1

Where :An = net areaU = reduction factor due to efficiencyx = the distance from the centroid of the connected area

to the shear plane of the connectionL = length of the connection in the axial load direction

the length between the edge bolt or the length of welding

U = 1- of bolt connections

(a)

x

x 2

1

Gusset Plat e

CG

Gusset Plat e

Gusset Plat e

CG

x

(b)

(c(

x x

Lx

Example: L = 9 inch

U = 1- of weld connectionsLx

x = 1.68 inchL = 5.5 inch

Ag = 2.86 in2 (dari tabel profil) Net section area = An;

Bolt diameter = 5/8 in. Hole diameter for calculating net area = 5/8 + 1/8 = 3/4 in. Net section area = Ag (3/4) x 3/8 = 2.86 3/4 x 3/8 = 2.579 in2

1. Determine Nominal Strength of L 4 x 4 x 3/8 in made from A36 steel (fy= 36 ksi, Fu=58 ksi) connected to gusset plate with 5/8 in bolts as shown in figure, where bolt spacing is 3 inch (centre to centre)

L 4 x 4 x 3/8 db = 5/8 in.

Gusset plate

a

a

L 4 x 4 x 3/8 db = 5/8 in.

Gusset plate

a

a

L 4 x 4 x 3/8 db = 5/8 in.

Gusset plate

L 4 x 4 x 3/8 db = 5/8 in.

Gusset plate

a

a x

L 4 x 4 x 3/8

x

L 4 x 4 x 3/8

Effective Net Area Aex = jarak titik berat ke pelat sambung,

dari tabel profil x = 1.13 in.L = panjang sambungan = 2 x 3.0 in. = 6.0 in. U = 1- x/L = 1- 1.13/6 = 0.8116 in.

Ae = 0.8116 x 2.579 in2 = 2.093 in2

Yielding design strength : Pn = Ag Fy = 0.9 x 2.86 in2 x 36 ksi = 92.664 kips

Fracture design Strength : Pn = Ae Fu = 0.75 x 2.093 in2 x 58 ksi = 91.045 kips

Design strength = 91.045 kips (net section fracture governs)

An = Ag = 5.00 in2

Ae = U AnU = 1 x/L x= 1.68 in L = 5.5 inU = 1- 1.68/5.5 = 0.79

2. Welded single angle L 6x 6 x tension member made from A36 (fy = 36 ksi, Fu=58 ksi) steel shown below. Calculate the tension design strength.

Gross yielding design strength :Pn= Fy Ag = 0.9 x 36 x 5.00 = 162 kipsNet section fracture strength :Pn= Fu Ae = 0.75 x 58 x 0.79 x 5.00 = 171.825 kipsDesign strength = 162 kips (gross yielding governs)

Determine design strength of tension member made from A36 steel and connected with 7/8 bolts.

Line abcdeg :

Block ShearFor some connection configurations, the tension membercan fail due to tear-out of material at the connected end called block shear

TT

TT

Shear failure

Tension failure

Shear failure

Tension failure

Shear failure

Tension failure

Failure mode

(a) Shear yield and tension fracture: if Fu Ant > 0.6 Fu Anv

]60,0[ ntugvyn AFAFR (b) Shear fracture and tension yield: if Fu Ant < 0.6 Fu Anv

]60,0[ gtynvun AFAFR

75,0

Upper Limit : the block shear strength can not exceed the maximum strength :

Shear failure

Tension failure

Shear failure

Tension failure

Shear failure

Tension failure

]60,0[ ntunvu AFAF

Rn = 0.6 Fu Anv + Ubs Fu Ant < upper limitRn -- Upper limit : 0.6 Fy Agv + Ubs Fu Ant

AISC 2007

Ubs = 1 if tensile stress is uniform (angle, gusset plates and most coped beams)

Ubs = 0.5 if tension stress non uniform (coped beams with two lines of bolts or withnonstandard distance from bolts to end of beam)

..\hasil download purdue univ\block shear failure.mpg

..\hasil download purdue univ\block shear areas.mpg

Block Shear Area

Assuming block shear failure :

x

L 4 x 4 x 3/8

x

L 4 x 4 x 3/8

L 4 x 4 x 3/8 db = 5/8 in.

Gusset plate

a

a

1.5 3.0 3.0

2.0L 4 x 4 x 3/8

db = 5/8 in.

Gusset plate

a

a

L 4 x 4 x 3/8 db = 5/8 in.

Gusset plate

L 4 x 4 x 3/8 db = 5/8 in.

Gusset plate

a

a

1.5 3.0 3.0

1.5 3.0 3.0

2.0

db = 5/8 in.

Gusset plate

a1.5 3.0

3.0

2.0 db = 5/8 in.

Gusset plate

a1.5 3.0

3.0

2.0 db = 5/8 in.

Gusset plate

a

db = 5/8 in.

Gusset plate

db = 5/8 in.

Gusset plate

a1.5 3.0

3.0 1.5 3.0

3.0

2.0

0.6 Fu Anv = 0.6 x 58 x 2.109 = 73.393 kips0.6 Fy Agv = 0.6 x 36 x 2.813 = 60.76 kipsFu Ant = 58 x 0.609 = 35.322 kipsFu Ant < 0.6 Fu Anv : gross fracture (failure mode b) Rn = 0.75 (0.6 Fu Anv + Fy Agt ) Rn = 0.75 (73.393 + 36 x 0.75) = 75.29 kips

Cek upper limit : Rn = 0.75 (73.393 + 35.322) = 81.53 kips Block shear strength : 75.29 kips Yielding strength : 92.664 kips Fracture strength : 91.045 kips Nominal Design Capacity : 75.29 kips

db = 5/8 in.

Gusset plate

a1.5 3.0

3.0

2.0 db = 5/8 in.

Gusset plate

a1.5 3.0

3.0

2.0 db = 5/8 in.

Gusset plate

a

db = 5/8 in.

Gusset plate

db = 5/8 in.

Gusset plate

a1.5 3.0

3.0 1.5 3.0

3.0

2.0

Agt = gross tension area = 2.0 x 3/8 = 0.75 in2

Ant = net tension area = 0.750.5 x (5/8+1/8)x3/8= 0.609 in2

Agv = gross shear area = (3.0 + 3.0 +1.5) x 3/8 = 2.813 in2

Anv = net shear area = 2.8132.5 x (5/8 + 1/8) x 3/8= 2.109 in2

]60,0[ ntunvu AFAF

C6x10.53/8(9.53mm)

152mm

12.6mm

80m

gussetplate 3/8

Determineblockshear strengthofgussetplatewith3/8inch(9.53mm)inthicknessweldedtocanalC6x10.5.SectionpropertiesofC6x10.5:A=1993.5mm2 Bf=50.8mmtf=12.83mmtw=7.94mm

1.5 3 3

3@3=9

centertocenter

T

gussetplate

C15x50

Yielding Strength :

2. Determine the design tension strength for a single channel C15 x 50 (Ag=14.7 in2 tw=0.716in) connected to a 0.5 in. thick gusset plate with 3/4 in. diameter bolts. The plate is made from steel with Fy=50 ksi dan Fu= 65 ksi. (x = 0.798in)

kipsPn 6627.14*50*9.0

219.12716.08747.14 intndAA egn

257.1019.12*6798.011 inA

LxUAA nne

kipsPn 51557.10*65*75.0

Fracture Strength:

Yield strength = 662 kips Fracture strength = 515kips Block Shear = 445kips Design Strength = 445kips

Block Shear

6925.296716.0*8739*65

ntu AF

Rn 0.6Fu Anv Fu Ant 0.75 296.69 296.69 445kips

0.6Fu Anv 0.6Fy Agv

6925.296716.0*87*5.25.7*2*65*6.06.0

nvu AF

0.6Fy Agv 0.6 * 50 * 2* 7.5 * 0.716 322.2kipsFu Ant = 0.6 Fu Anv

6925.296716.0*8739*65

ntu AF

]60,0[ ntugvyn AFAFR kipsRn 16.464]69.2962.322[75.0

Upper Limit control

Slenderness

If the axial load in a slender tension member is removed and small transverse loads are applied, undesirable vibrations or deflections may occur.

Thus AISC recommends:r > L/300 ( not for cables or rods)

Slenderness Ratio L/r

Stiffness limitation to prevent : Deformation due to self weight Vibration due to wind load or machine movement

where : r = minimum radius of gyration of the cross section L = length of the member.

Maximum value of L/r Primary member : 240 secondary member : 300

Threaded Rods and Cables

When slenderness is not a consideration, circular rods and cables are often used (hangers, suspended bridges).

Rods are solid and cables are made from individual strands wound together.

Threaded Rods and Cables

Threaded Rods and Cables

t Pn = 0.75 (0.75 Ab Fu)

Ab = nominal (unthreaded) areaIt is common to use a min diameter of 5/8 in. for rods.

Sag Rods

Sag rods are used to provide lateral support for the purlins (to prevent sag in direction parallel to a sloping roof due to vertical applied loads).