4(Tension Members)

7/24/2019 4(Tension Members)

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T-1

These are found predominantly as members of plane or space

trusses (2D & 3D), as members in transmission towers and as windbracing (single or double) for single story or high rise steel structures.

Among the common shapes used as tension members:

ound bar !lat bar Angle Double angle "tarred angle

Double

channel

#hannel $atticed

channels%section

(wideflange)"section

(American

"tandard)

'uiltup bo

sections

Cross-section of typical tension members.Cross-section of typical tension members.



T-2

he strength of a tension member is controlled by the lowest

of the following limiting states:

*et

Area (An)

+ross

Area (Ag)

A) Yielding of the Gross Area (Ag):

!n !y Ag

B) Failre (!ltimate strength) on the "et Area (An):

!n !u Ae

%here Ae effecti-e net Area An

eduction #oefficient.

#) Bloc# $hear Failre throgh the end bolt:



T-3

A hole is drilled (or punched) by /0/1 inch greater thenthe normal diameter of the fastener (ri-et or bolt). ole

punching causes some damage to the edges of the

hole to the amount of /032 inch from each side.

hus the normal hole diameter

.81 dia. bolt

32

1

32

1

16

1diameter bolt

inch+=

+++=

http://lrfd.pdf/

http://lrfd.pdf/



T-4

%hat is the net area An for the tension member

as shown in the figure&

Solution:

Ag (4.25) /.4 s6 in.

%idth to be deducted for hole An 7%g 8 (width for hole)9 (thicness of plate)

.in8

7

8

1

4

3

=+=

"tandard ole for a in. diam bolt.4

34

4

1× Plate (inches)

Example (T1):

( ) in.sq.78.025.08

74 =

−=



(a) (b)

T-5

!or a group of staggered holes along the tension direction, one

must determine the line that produces smallest ;*et Area<.

=aths of failure

on net section

>!!># ?! "A++>>D ?$>" ?* *> A>A :

'

A

s

g

A

# p

p p

'@n the abo-e diagram:

p =itch or spacing along bolt line

s "tagger 'etween two adacent bolt lines

(usually s =02)

g gage distance trans-erse to the loading.

@n case (a) abo-e : An (+ross width 8 B hole dia.) . t

@n case (b) abo-e : An (+ross width 8 B hole dia.C B s20g) . t



T-6

Determine the minimum net area of the plate shown in fig. 3..2, assuming

in,diam holes are located as shown:

!igure 3..2 >ample 3../

16

15

Example (T2):



T-7

Solution. According to $!D and A"D'2, the width used in deducing for

holes in the hole diameter plus /0/1 in., and the staggered length correction

@s (s20g).

.... insq502250

16

1

16

15212 =

+−

/) =ath AD (two holes) :

2) =ath ABD (three holes two staggers) :

( )( )

....)(

).(

.

.insq432250

44

1252

524

1252

16

1

16

15312

22

=

++

+−

( )( )

....)(

).(

.

.insq422250

44

8751

524

1252

16

1

16

15312

22

=

++

+−

3) =ath ABC (three holes two staggers) :

(controls)



T-8

Angles:

%hen holes are staggered on two legs of an angle, the gage length (g)for use @n the (s20g) epression is obtained by using length between the

centers of the holes measured along the centerline of the angle

thicness, i.e., the distance A-B in Fig: 3.4.3. hus the gage distance g is

t g g t

g t

g g baba −+=−+−=22

+age dimension for an angle



T-9

>-ery rolled angle has a standard -alue for the location of holes(i.e. gage distance ga and gb), depending on the length of the leg

and the number of lines of holes. able shows usual gages for

angles as listed in the A@"# EanualF.



T-10

Determine the net area (An ) for the angle gi-en in figure below

if holes are used&

Angle with legs shown FflattenedF into one plane4

14

2

1

4

12

2

121 =−+=−+ t g g F

Flegs and thicness in mm.

.,16

15diain

Example (T3):

GH<



T-11

Solutions. !or net area calculation the angle may be -isualiIed as being

flattened into a plate as shown in !igure abo-e.

tg

sDt A A

2

gn ∑∑ +−=

where D is the width to be deducted for the hole.

/) =ath AC:

2) =ath ABC:

..75.35.016

1

16

15275.4 in sq=

+−

..96.35.0)25.4(4

)3(

)5.2(4

)3(5.0

16

1

16

15375.4

22

in sq=

++

+−

"ince the smallest An is 3.J5 s6 in., that -alue go-erns.

An

An

G.5K



T-12

%hen some of the cross section (and not all the section) is

connected, we need to use effecti-e net area concept :

Ae An

where, eduction !actor.

%hen all elements of the section are connected, /.4.



%hen not all elements are connected.

i) rans-erse %eld #onnection:

Ae A

/.4

A Area of connected part only

e.g. A 1 /02 3 in2

ii) $ongitudinal %eld #onnection :

Ae Ag

/.4 for $ 2 w

4.LJ for 2w $ /.5 w

4.J5 for /.5w $ w

1<

+usset

plate

Angle1/02

T-13

w

+usset

plate

Angle

1/02

$

%eld



T-14

@n bolted connections, the reduction factor () is a function

of the eccentricity ( ) in the connection.

B3.2)-(!"# 9.01 ≤−= L

xU

hus:

%here: distance between centroids of elements to

the plane of load transfer

$ $ength of the connection in the direction of load.

("ee #ommentary # 8 ' 3./ & # 8 ' 3.2)

x

x



T-15

xDetermination of for .

LFRD Specification for Structural Steel Buildings Dece!ber "# $%%%

A!erican &nstitute of Steel Construction



T-16

(#ommentary =/1./ 8 /JJ A@"#)

!or bolted or ri-eted connections the following -alues

for () !a' be used:

a) %, E or " "hapes with flange width M 203 depth, and structural tees cut

from these shapes, pro-ided connection to the flanges and has M 3

fasteners per line in the direction of force, 4.G4.b) %,E or " "hapes where flanges width N 203 depth, and all other shapes,

that has no fewer than 3 fasteners per line, 4.L5

c) All members ha-ing only two fasteners in the line of stress 4.J5

!or short tension members such as (usset plates the effecti-e net

area e6uals (An), but must not eceed 4.L5 of the gross area (Ag).



)*a!ple +,-4)*a! ple +,-4

#alculate the Ae -alues of the following section:

J0L bolts % L 2LO flange width (1.5<) P 203 depth (L.4<)

O hree bolts 0 line

4.G4

Ag L.2 m2

An gross area 8 hole area

L.2 8 (2 /.4 hole) web t 4.2L5

J.1L in2

Ae QAn 4.G J.1L 1.G/2 in2

hole dia J0L# G /5

only 2 bolts 0 line, 4.J5 Ag ./ m2

An ./ 8 (2 /50/1) 4.2L5 3.LJ5 in2

Ae 4.J5 3.LJ5 2.G4J in2

(i)

(ii)

T-17

web t



(iii)

$ 3 3 30L

3 3 R dia bolt

4.LLL

$ 1 in (3C3)

/ 0$ / 4.LLL01 4.L52 N 4.G

Ag 2.// in2

An 2.// 8 / (30 C /0L) 30L 2.// 4.32L /.JL2 in2

Ae QAn 4.L52 /.JL2 /.5/L in2

Alternati-e -alue of 4.L5 (3 bolts 0 line)

(i-) w /4 33

J0L dia. bolt

All sides connected

/Q4

Ag G.J/ in2

An G.J/ 8 /.4 4.35 8 2 /.4 4.2G4

G.J/ 8 /.J 4.5L J.3G in2

Ae QAn J.3G in2

oles

in flageflage t

hole

oles

in web web t.

T-18



T-19

T his third mode of failure is limited to

thin plates. his failure is a combination of

tearing (shear rupture) and of tensile yielding. @t

is uncommon, but the code pro-ides on etra

limit state of ($!D S .3). @t is usually checed

after design is completed.

(a) !ailure by

tearing out

+usset

=late

"haded area

may tear out

a

cbE -en as tension members are unliely

to be affected by their stiffness ($0r), it is

recommended to limit the maimum

slenderness ratio ($0r) for all tension members

(ecept rods) to T 344.Ea. slenderness $0r min T 344

his is to pre-ent etra sagging and -ibration

due to wind.



he general philosophy of $!D method:

!or tension members:

where

φ t resistance reduction factor for tensile members

, n *ominal strength of the tensile members

, u !actored load on the tensile members.he design strength φtn is the smaller of:

a) Yielding in the gross section'

φ t , n / φ t F ' Ag / 0.% F ' Ag

b) Fractre of the net section'φ t , n / φ t F u Ae / 0.#1 F u Ae

his is to be followed by chec of rupture strength (bloc shear failure),

and limitation of slenderness ratio T 344. T-20



ample (T-*):-

!ind the maimum tensile capacity of a member consisting of

2$s (1 H) can carry for two cases:

(a) welded connection,

(b) bolted connection /K dia bolts

!y 14 si

!u J5 si.

H<

5H

2H<

2<

/R< /R<

T-21



*et area gross area (all sides connected) G.54 in2

Uielding → !t 4.G !y Ag 4.G 14 G.54 5/3

!racture → !t 4.J5 !u Ae 4.J5 J5 G.5 53

hus tension capacity, φt n 5/3 (yielding controls)

+a 2elded Connection+a 2elded Connection

+b Bolted Connection+b Bolted Connection

#onsider one $

VAnW #alculation: %g gross width 1 C 8 H G.5 in.

(cont.) T-22



"traight section : wn G.5 8 2 J.25 in.( )811+

1.12 in. (#ontrols)

(fracture controls)

An 1.12 H 3.3/ in2 for one $

!or 2$s, An 3.3/ 2 1.12 in2

All sides connected, /.4, Ae .An 1.12 in2

#alculation of φt n :

(i) Uielding: 4.G !y Ag 4.G 14 G.54 5/3

(ii) !racture:4.J5 !u Ae 4.J5 J5 1.12 3J2 .

(thicness)

T-23

XigXag ( )

(/.J5)

2.5

(/.J5)/3G.5w

22

L/

n×

+×

+×−=

(2.5C284.5)2H<

<

/.J5< /.J5<

GH<



+esign is an interacti-e procedure (trial & error), as we do

not ha-e the final connection detail, so the selection is made,

connection is detailed, and the member is checed again.

,roposed +esign ,rocedre:-

i) !ind re6uired (Ag) from factored load .

ii) !ind re6uired (Ae) from factored load .

iii) #on-ert (Ae) to (Ag) by assuming connection detail.

i-) !rom (ii) & (iii) chose largest (Ag) -alue

-) !ind re6uired (r min) to satisfy slenderness

-i) "elect a section to satisfy (i-) and (-) abo-e.

-ii) Detail the connection for the selected member.

-iii)echec the member again.

=

y

ug

4.G!

) A

=

u

u

e 4.J5!

) A

= 344

r

$

min

T-24



>ample (1):

A tension member with a length of 5 feet G inches

must resist a ser-ice dead load of /L ips and a ser-ice li-e

load of 52 ips. "elect a member with a rectangular cross

section. se A31 steel and assume a connection with one

line of J0Linchdiameter bolts.

Eember length 5.J5 ft.

T-25



T-26

=u /.2 D C /.1$ /.2(/L) C /.1(52) /4.L ips

2

u

ue

2

y

ug

in.2.4G4.J5(5L)

/4.L

4.J5!

= Ae6uired

in.3.2354.G4(31)

/4.L

4.G4!

= Ae6uired

===

===

'ecause Ae An for this member, the gross area corresponding to

the re6uired net area is

t2.4GtL

/

L

J2.4G

A A A holeng

+=

++=

+=

ry t / in.

Ag 2.4G C /(/) 3.4G in.2



T-27

'ecause 3.4G P 3.235, the re6uired gross area is 3.4G in.2, and

in.3.4G/

3.4G

t

Aw

g

g ===

ound to the nearest /0L inch and try a / × 3 H cross section.

#hec the slenderness ratio:

se a 3 $ bar.

(?Y) 34423G4.2LLJ

5.J5(/2)

r

$ Eaimum

in.4.2LLJ3.5

4.2G/J

A

@

r

obtainwe, Ar @!rom

in.3.5/(3.5) A

in.4.2G/J/2

3.5(/)@

min

min

2

2

3

min

<==

===

=

==

==



$elect a single angle tension member to carry (4

ips D$) and (24 ips $$), member is (/5)ft long and will be

connected to any one leg by single line of J0L< diameter

bolts. se A31 steel.

Solution:

"tep /) !ind e6uired (u):

u /.2 D$ C /.1 $$ u /. D$

/.2 4 C /.1 24 or /. 4

L C 32 L4 51

∴ u L4 (#ontrols) T-28

ample (T-):-



"tep 2) !ind re6uired Ag & Ae:

/g

2

y

ure6.g )(Ain2.J

314.G

L4

4.G!

)(A ==

×==

2ure6.e in/.L

5L4.J5

L4

4.J5!u

)(A =

×==

"tep 3) #on-ert (Ae) to (Ag):

"ince connection to single leg, then use alternati-e

() -alue 4.L5 (more then 3 bolt in a line).

2in16.285.084.1)( ===

U A A e

n

!or single line J0L< bolts Ag An C (/)t 2./1 C t (Ag)2T-29



"tep ) !ind re6uired r min. "tep 5) "elect angle:

'y selecting (t) we get Ag & r min

t (Ag)/ (Ag)2

/0 2.J 2./

30L 2.J 2.53/02 2.J 2.11

select t 30L<

(Ag)2 2.53 in2

T-30

(#ontrols)



"election

L3

2/3$ ××

Ag 2.1J in2 P 2.53 in2 ?Y

r min 4.J2J in P 4.1 ?Y

"tep 1) Design the bolted connection: (chap. ).

"tep J) echec the section.

T-31



"elect a pair of E# as shown to carry a factored ultimate load

of G4 ips in tension. Assume connection as shown. "teel !y 54 si,

!u 15 si (A5J2, grade 54) length 34 ft.

/. u G4 per channel, u 25

2. e6uired, (Ag)/ 25 0 4.G 54 5. in2

e6uired, (Ae) 25 0 4.J5 15 5.43 in2

e6uired, (An) 5.43 in2U

Ae

3. Assume that flange thicness Z 4.5 in and web t. Z 4.3 in. (eperience [)

An (Ag)2 8 2 /.4 4.5 8 2 /.4 4.3

(Ag)2 8 /.14

(Ag)2 An C /.14 5.43 C /.14 1.13 in.(controls) T-32

/4< 2E#

J0L< bolt /.4 (%ell connected)

ample (T-):-



. e6uired. r min (as a buildup section)

5. ry E# /4 25 Ag J.35 in2 tw 4.3L and tf 4.5J5, r 3.LJ in.

1. #hec capacity

An J.35 8 2 /.4 4.5J5 8 2 /.4 4.3L

J.35 8 /.G/4 5. in2.

Ae 5. in2.

(i) Uielding φn 4.G 54 (2 J.35) 11/.5

(ii) !racture φn 4.J5 15 (2 5.) 534.

φ=n 534. P G4 . ?Y

se 2 E# /4 25

in2.1

300

1230

300

=×

=l

→ r min M /.2

T-33

*

'

*

'



!or builtup members, tie plates are re6uired to mae the

members to beha-e as a single unit.∗ 'etween tie plates, each member beha-es as a single.

herefore, l0r between tieplates corresponds to that for

a single member.

!or single , r min r y r y /.4 in T-34

(*.+.)



*ote:

ie=lates must be used at ends. "ee

Eanual for min. siIes.

$ength of tieplate M 203 (dist. between line of connection) LK

hicness of tieplate M /054 (dist. between line of connection) /02K

T-35

/5$

/5$

herefore one tieplate at middle must be used.

"ee $!D D2. (=. /1./2)

[ ]2=/1./D2,$!D

>n los lados abiertos de barras armadas traccionadas

se pueden utiliIar platabandas perforadas o presillas



T-35

$as presillas tendr\n una longitud

mayor o igual a /01 de la distancia

entre los cordones de soldadura o

las l]neas de remaches o pernos 6ue

las unen a los componentes de la

barra armada. >l espesor de las

presillas ser\ mayor o igual a 20*3

de dicha distancia. $a separaci^n

longitudinal de los cordones desoldadura intermitentes o de los

pasadores (pernos o remaches) de

las presillas no deber\ superar los

2* cm. l espaciamiento entre

presillas ser\ tal 6ue la esbelteI

local de los componentes sea,preferiblemente, menor o igual a

133.

4(Tension Members)

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