These
are found predominantly as members of plane or space
trusses (2D & 3D), as members in transmission towers and as wind
bracing (single or double) for single story or high rise steel structures.
Among the common shapes used as tension members:
Round bar
Channel
Double
channel
Flat bar
Latticed
channels
Angle
Double angle
W-section
(wide-flange)
S-section
(American
Standard)
Cross-section of typical tension members.
Starred angle
Built-up box
sections
T-1
The strength of a tension member is controlled by the lowest
of the following limiting states:
T
T
Net
Area (An)
Gross
Area (Ag)
A) Yielding of the Gross Area (Ag):
Fn = Fy Ag
B) Failure (Ultimate strength) on the Net Area (An):
Fn = Fu Ae
Where Ae = effective net Area = UAn
U = Reduction Coefficient.
C) Block Shear Failure through the end bolt:
T
T-2
A hole is drilled (or punched) by 1/16 inch greater then
the normal diameter of the fastener (rivet or bolt). Hole
punching causes some damage to the edges of the
hole to the amount of 1/32 inch from each side.
Thus the normal hole diameter
1 1
1
 bolt diameter   
16 32 32
1
 bolt dia.  inch.
8
T-3
Example (T1):
What is the net area An for the tension member
as shown in the figure?
1
Plate  4 (inches)
4
Standard Hole for a 3 -in. diam bolt.
T
T
Solution:
4
Ag = 4(0.25) = 1.0 sq in.
Width to be deducted for hole  34  18  78 in.
An = [Wg – (width for hole)] (thickness of plate)
 7
 4   0.25  0.78 sq. in.
 8
T-4
For a group of staggered holes along the tension direction, one
must determine the line that produces smallest “Net Area”.
EFFECT OF STAGGERED HOLES ON NET AREA :A
Paths of failure
T
on net section
T
B
p
p
(a)
A
g
T
In the above diagram:
p = Pitch or spacing along bolt line
s = Stagger Between two adjacent bolt lines
(usually s = P/2)
g = gage distance transverse to the loading.
p
s
C B
In case (a) above : An = (Gross width – Σ hole dia.) . t
In case (b) above : An = (Gross width – Σ hole dia.+ Σ s2/4g) . t
T
(b)
T-5
Example (T2):
Determine the minimum net area of the plate shown in fig. 3.4.2, assuming
15
in,-diam holes are located as shown:
16
Figure 3.4.2
Example 3.4.1
T-6
Solution. According to LRFD and ASD-B2, the width used in deducing for
holes in the hole diameter plus 1/16 in., and the staggered length correction
Is (s2/4g).
1)
Path AD (two holes) :

 15 1 
12

2
   0.25  2.50sq.in.

 16 16 

2)
Path ABD (three holes; two staggers) :
2

( 2.125)2 
 15 1  2.125

12  3   
 0.25  2.43sq.in.
42.5
4( 4) 
 16 16 

3)
Path ABC (three holes; two staggers) :
2

(1.875)2 
 15 1  2.125

12  3   
 0.25  2.42sq.in.
42.5
4( 4) 
 16 16 

(controls)
T-7
Angles:
When holes are staggered on two legs of an angle, the gage length (g)
for use In the (s2/4g) expression is obtained by using length between the
centers of the holes measured along the centerline of the angle
thickness, i.e., the distance A-B in Fig: 3.4.3. Thus the gage distance g is
t
t
g  ga   gb   ga  gb  t
2
2
Gage dimension for an angle
T-8
Every rolled angle has a standard value 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. Table shows usual gages for
angles as listed in the AISC Manual*.
T-9
Example (T3):
Determine the net area (An ) for the angle given in figure below
if
 15

 in , dia.  holes
 16

are used?
9½”
*legs and thickness in mm.
1
1 1
1
* g  g1  t  2  2   4
2
4 2
4
Angle with legs shown *flattened* into one plane
T-10
Solutions. For net area calculation the angle may be visualized as being
flattened into a plate as shown in Figure above.
s2
A n  A g   Dt   t
4g
where D is the width to be deducted for the hole.
1) Path AC:
 15 1 
An = 4.75  2  0.5  3.75 sq.in.
 16 16 
9.5"
2) Path ABC:
 (3) 2
(3) 2 
 15 1 

An = 4.75  3  0.5  
0.5  3.96 sq.in.
 16 16 
 4(2.5) 4(4.25) 
Since the smallest An is 3.75 sq in., that value governs.
T-11
 When some of the cross section (and not all the section) is
connected, we need to use effective net area concept :-
Ae = U An
where,
U = Reduction Factor.
 When all elements of the section are connected, U = 1.0.
T-12
When not all elements are connected.
Gusset
plate
i) Transverse Weld Connection:Ae = UA
U = 1.0
A = Area of connected part only
e.g. A = 6 x 1/2 = 3
6”
T
Angle
6x4x1/2
in2
ii) Longitudinal Weld Connection :-
Gusset
plate
Ae = Ag U
U = 1.0
U = 0.87
U = 0.75
for
for
for
L 2w
2w  L  1.5 w
1.5w  L  w
Weld
w
T
Angle
6x4x1/2
L
T-13
In bolted connections, the reduction factor (U) is a function
of the eccentricity ( x ) in the connection.
Thus:Where:
x
U  1   0.9
L
(LRFD - B3.2)
x = distance between centroids of elements to
the plane of load transfer
L = Length of the connection in the direction of load.
(See Commentary C – B 3.1 & C – B 3.2)
T-14
Determination of x for U.
LFRD Specification for Structural Steel Buildings, December 27, 1999
American Institute of Steel Construction
T-15
(Commentary P16.1 – 177 AISC)
 For bolted or riveted connections the following values
for (U) may be used:a)
W, M or S Shapes with flange width ≥ 2/3 depth, and structural tees cut
from these shapes, provided connection to the flanges and has ≥ 3
fasteners per line in the direction of force, U = 0.90.
b) W,M or S Shapes where flanges width < 2/3 depth, and all other shapes,
that has no fewer than 3 fasteners per line, U = 0.85
c) All members having only two fasteners in the line of stress
U = 0.75
 For short tension members such as Gusset plates the effective net
area equals (An), but must not exceed 0.85 of the gross area (Ag).
T-16
Example (T-4)
Calculate the Ae values of the following section:(i)
(ii)
→ flange width (6.54”) > 2/3 x depth (8.0”)
→ Three bolts / line
U = 0.90
Ag = 8.24 m2
An = gross area – hole area
= 8.24 – (2 x 1.0 hole) x web tk 0.285
= 7.68 in2
Ae = U·An = 0.9 x 7.68 = 6.912 in2
7/8 bolts W 8 x 28
hole dia = 7/8
C 9 x 15
only 2 bolts / line, U = 0.75
Ag = 4.41 m2
web tk
An = 4.41 – (2 x 15/16) 0.285 = 3.875 in2
Ae = 0.75 x 3.875 = 2.907 in2
T-17
(iii)
L 3 x 3 x 3/8
x
x = 0.888
L = 6 in (3+3)
U = 1 - x /L = 1 -0.888/6 = 0.852 < 0.9
3 3
(iv)
¾ dia bolt
Ag = 2.11 in2
An = 2.11 – 1 x (3/4 + 1/8) x 3/8 = 2.11 -0.328 = 1.782 in2
Ae = U·An = 0.852 x 1.782 = 1.518 in2
Alternative value of U = 0.85 (3 bolts / line)
w 10 x 33
All sides connected
U = 1·0
Holes
Ag = 9.71 in2
hole
in web
web tk.
An = 9.71 – 4 x 1.0 x 0.435 – 2 x 1.0 x 0.290
7/8 dia. bolt
Holes
in flage
flage tk
= 9.71 – 1.74 - 0.58 = 7.39 in2
Ae = U·An = 7.39 in2
T-18
This third mode of failure is limited to
thin plates. This failure is a combination of
tearing (shear rupture) and of tensile yielding. It
is uncommon, but the code provides on extra
limit state of (LRFD J 4.3). It is usually checked
after design is completed.
c
Even as tension members are unlikely
to be affected by their stiffness (L/r), it is
recommended
to
limit
the
maximum
slenderness ratio (L/r) for all tension members
(except rods) to ≤ 300.
Max. slenderness = L/rmin ≤ 300
This is to prevent extra sagging and vibration
due to wind.
Gusset
Plate
a
b
Shaded area
may tear out
T
(a) Failure by
tearing out
T-19
The general philosophy of LRFD method:
For tension members:
tTn  Tu
Rn   iQi
where
t = resistance reduction factor for tensile members
Tn = Nominal strength of the tensile members
Tu = Factored load on the tensile members.
The design strength tTn is the smaller of:
a) Yielding in the gross section;
t Tn = t Fy Ag = 0.9 Fy Ag
b) Fracture of the net section;
t Tn = t Fu Ae = 0.75 Fu Ae
This is to be followed by check of rupture strength (block shear failure),
and limitation of slenderness ratio ≤ 300.
T-20
Example (T-5):Find the maximum tensile capacity of a member consisting of
2Ls (6 x 4 x ½) can carry for two cases:
(a) welded connection,
(b) bolted connection
½”
1" dia bolts
Fy = 60 ksi
2½”
Fu = 75 ksi.
2”
5½
1¾” 1¾”
T-21
(a) welded Connection
Net area = gross area (all sides connected)
= 9.50 in2
Yielding  Ft = 0.9 Fy Ag = 0.9 x 60 x 9.50 = 513 k
Fracture  Ft = 0.75 Fu Ae = 0.75 x 75 x 9.5 = 534 k
Thus tension capacity, t Tn = 513 k (yielding controls)
(b) Bolted Connection
Consider one L
‘An’ Calculation:
Wg = gross width = 6 + 4 – ½ = 9.5 in.
(cont.)
T-22
Straight section : wn = 9.5 – 2 x 1 18  = 7.25 in.
2
2
(1.75)
(1.75)
Zig-Zag = w n  9.5  3  1 81  

4  2.5
4 4
2½”
(2.5+2–0.5)
9½”
= 6.62 in. (Controls)
4”
(thickness)
An = 6.62 x ½ = 3.31 in2
for one L
For 2Ls, An = 3.31 x 2 = 6.62 in2
All sides connected, U = 1.0, Ae = U.An = 6.62 in2
1.75” 1.75”
Calculation of t Tn :(i) Yielding: 0.9 Fy Ag = 0.9 x 60 x 9.50 = 513 k
(ii) Fracture:0.75 Fu Ae = 0.75 x 75 x 6.62 = 372 k.
(fracture controls)
T-23
Design is an interactive procedure (trial & error), as we do
not have the final connection detail, so the selection is made,
connection is detailed, and the member is checked again.
Proposed Design Procedure:i)


 A g  Tu 
Find required (Ag) from factored load 
0.9Fy. 

ii) Find required (Ae) from factored load  A e


Tu .

0.75Fu 
iii) Convert (Ae) to (Ag) by assuming connection detail.
iv) From (ii) & (iii) chose largest (Ag) value
 L




300
v) Find required (rmin) to satisfy slenderness 
 rmin

vi) Select a section to satisfy (iv) and (v) above.
vii) Detail the connection for the selected member.
viii) Re-check the member again.
T-24
Example (T-6):A tension member with a length of 5 feet 9 inches
must resist a service dead load of 18 kips and a service live
load of 52 kips. Select a member with a rectangular cross
section. Use A36 steel and assume a connection with one
line of 7/8-inch-diameter bolts.
Member length = 5.75 ft.
T-25
Pu = 1.2 D + 1.6L = 1.2(18) + 1.6(52) = 104.8 kips
Required A g 
Pu
104.8

 3.235 in.2
0.90Fy 0.90(36)
Required A e 
Pu
104.8

 2.409 in.2
0.75Fu 0.75(58)
Because Ae = An for this member, the gross area corresponding to
the required net area is
A g  A n  A hole
 7 1
 2.409    t  2.409  t
8 8
Try t = 1 in.
Ag = 2.409 + 1(1) = 3.409 in.2
T-26
Because 3.409 > 3.235, the required gross area is 3.409 in.2, and
Ag
3.409
 3.409 in.
t
1
Round to the nearest 1/8 inch and try a 1  3 ½ cross section.
Check the slenderness ratio:
wg 

3.5(1) 3
Imin 
 0.2917 in.4
12
A  1(3.5)  3.5 in.2
From I  Ar 2 , we obtain
Imin
0.2917

 0.2887 in.
A
3.5
L 5.75(12)
Maximum 
 239  300 (OK)
r
0.2887
rmin 
Use a 3 ½  1 bar.
T-27
Example (T-7):Select a single angle tension member to carry (40
kips DL) and (20 kips LL), member is (15)ft long and will be
connected to any one leg by single line of 7/8” diameter
bolts. Use A-36 steel.
Solution:
Step 1) Find Required (Tu):Tu = 1.2 DL + 1.6 LL
= 1.2 x 40 + 1.6 x 20
= 48 + 32 = 80k
 Tu = 80k
(Controls)
Tu = 1.4 DL
or
= 1.4 x 40
= 56k
T-28
Step 2) Find required Ag & Ae:
Tu
80
(A g )req. 

 2.47 in2  (A g )1
0.9Fy 0.9  36
Tu
80
(A e )req. 

 1.84 in2
0.75Fu 0.75  58
Step 3) Convert (Ae) to (Ag):
Since connection to single leg, then use alternative
(U) value = 0.85 (more then 3 bolt in a line).
Ae 1.84
( An ) 

 2.16 in 2
U 0.85
For single line 7/8” bolts ; Ag = An + (1)t = 2.16 + t = (Ag)2
T-29
Step 4)
Find required rmin.
rmin
Step 5)
L
15  12


 0.6 in.
300
300
Select angle:
By selecting (t) we get Ag & rmin
t
(Ag)1
(Ag)2
1/4
2.47
2.41
3/8
2.47
2.53
1/2
2.47
2.66
select t = 3/8”
(Ag)2 = 2.53 in2
(Controls)
T-30
Selection
L4  3 21  3 8
Ag = 2.67 in2 > 2.53 in2
rmin = 0.727 in > 0.6
Step 6)
Design the bolted connection:
(chap. 4).
Step 7)
Re-check the section.
OK
OK
T-31
Example (T-8):Select a pair of MC as shown to carry a factored ultimate load
of 490 kips in tension. Assume connection as shown. Steel Fy = 50 ksi,
Fu = 65 ksi (A572, grade 50) length = 30 ft.
10”
2MC
1. Tu = 490 k; per channel, Tu = 245 k
2. Required, (Ag)1 = 245 / 0.9 x 50 = 5.44 in2
Required, (Ae) = 245 / 0.75 x 65 = 5.03 in2
A
Required, (An) = e = 5.03 in2
U
7/8” bolt
U = 1.0 (Well connected)
3. Assume that flange thickness ~ 0.5 in and web tk. ~ 0.3 in. (experience !)
An = (Ag)2 – 2 x 1.0 x 0.5 – 2 x 1.0 x 0.3
= (Ag)2 – 1.60

(Ag)2 = An + 1.60 = 5.03 + 1.60 = 6.63 in.
(controls)
T-32
4. Required. rmin =
l
30  12

 1.2 in (as a buildup section)
300
300
 rmin ≥ 1.2
5. Try MC 10 x 25 ; Ag = 7.35 in2 ; tw = 0.38 and tf = 0.575, rx = 3.87 in.
y
6. Check capacity
An = 7.35 – 2 x 1.0 x 0.575 – 2 x 1.0 x 0.38
= 7.35 – 1.910 = 5.44 in2.
Ae = 5.44 in2.
(i) Yielding Tn = 0.9 x 50 x (2 x 7.35) = 661.5 k
(ii) Fracture Tn = 0.75 x 65 x (2 x 5.44) = 530.4 k
Pn = 530.4 k > 490 k.
x
x
y
OK
Use 2 MC 10 x 25
T-33
For built-up members, tie plates are required to make the
members to behave as a single unit.
 Between tie plates, each member behaves as a single.
Therefore, l/r between tie-plates corresponds to that for
a single member.
For single , rmin = ry ; ry = 1.0 in
300  1.0
Max. l 
ft  25.0ft  30ft. (N.G.)
12
T-34
Therefore one tie-plate at middle must be used.
Note:
Tie-Plates must be used at ends. See
15'
Manual for min. sizes. LRFD D2, P16.1- 24
Length of tie-plate ≥ 2/3 (dist. between line of connection) = 8"
Thickness of tie-plate ≥ 1/50 (dist. between line of connection) = 1/2"
15'
See LFRD D2. (P. 16.1-24)
T-35