Channel. Double channel. Latticed channels. W-section (wide-flange). S-section (American Standard). Built-up box sections. Tension Members. Chap. (3).
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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 Flat bar Angle Double angle Starred angle
Cross-section of typical tension members.
The strength of a tension member is controlled by the lowest
of the following limiting states:
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
Standard Hole for a -in. diam bolt.
Evaluation of Net Area
What is the net area An for the tension member
as shown in the figure?
Ag = 4(0.25) = 1.0 sq in.
Width to be deducted for hole
An = [Wg – (width for hole)] (thickness of plate)
Effect of staggered holes on Net Area :
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 :-
Paths of failure
on net section
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.
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
Determine the minimum net area of the plate shown in fig. 3.4.2, assuming
in,-diam holes are located as shown:
Figure 3.4.2 Example 3.4.1
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
1) Path AD (two holes) :
2) Path ABD (three holes; two staggers) :
3) Path ABC (three holes; two staggers) :
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
Gage dimension for an angle
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*.
Example on Net Area:
Determine the net area (An ) for the angle given in figure below
if holes are used?
Angle with legs shown *flattened* into one plane
Solutions. For net area calculation the angle may be visualized as being
flattened into a plate as shown in Figure above.
where D is the width to be deducted for the hole.
1) Path AC:
2) Path ABC:
Since the smallest An is 3.75 sq in., that value governs.
connected, we need to use effective net area concept :-
Ae = U An
where, U = Reduction Factor.
When not all elements are connected.
i) Transverse Weld Connection:-
Ae = UA
U = 1.0
A = Area of connected part only
e.g. A = 6 x 1/2 = 3 in2
ii) Longitudinal Weld Connection :-
Ae = Ag U
U = 1.0 for L 2 w
U = 0.87 for 2w L 1.5 w
U = 0.75 for 1.5w L w
For Bolted Connections:-
In bolted connections, the reduction factor (U) is a function
of the eccentricity ( ) in the connection.
= 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)
Determination of for U.
LFRD Specification for Structural Steel Buildings, December 27, 1999
American Institute of Steel Construction
(Commentary P16.1 – 177 AISC)
for (U) may be used:-
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
area equals (An), but must not exceed 0.85 of the gross area (Ag).
Example on Effective Net Areas:
Calculate the Ae values of the following section:-
→ 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
hole dia = 7/8
C 9 x 15
only 2 bolts / line, U = 0.75
Ag = 4.41 m2
An = 4.41 – (2 x 15/16) 0.285 = 3.875 in2
Ae = 0.75 x 3.875 = 2.907 in2
L = 6 in (3+3)
¾ dia bolt
U = 1 - /L = 1 -0.888/6 = 0.852 < 0.9
All sides connected
U = 1·0
Ag = 9.71 in2
An = 9.71 – 4 x 1.0 x 0.435 – 2 x 1.0 x 0.290
= 9.71 – 1.74 - 0.58 = 7.39 in2
Ae = U·An = 7.39 in2
7/8 dia. bolt
L 3 x 3 x 3/8
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
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.
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.
may tear out
The general philosophy of LRFD method:
For tension members:
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.
Example of strength calculation (capacity)
(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.
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
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.
= 6.62 in. (Controls)
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:-
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.
Pu = 1.2 D + 1.6L = 1.2(18) + 1.6(52) = 104.8 kips
Because Ae = An for this member, the gross area corresponding to
the required net area is
Try t = 1 in.
Ag = 2.409 + 1(1) = 3.409 in.2
Because 3.409 > 3.235, the required gross area is 3.409 in.2, and
Round to the nearest 1/8 inch and try a 1 3 ½ cross section.
Check the slenderness ratio:
Use a 3 ½ 1 bar.
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.
Step 1) Find Required (Tu):-
Tu = 1.2 DL + 1.6 LL Tu = 1.4 DL
= 1.2 x 40 + 1.6 x 20 or = 1.4 x 40
= 48 + 32 = 80k = 56k
Tu = 80k (Controls)
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).
For single line 7/8” bolts ; Ag = An + (1)t = 2.16 + t = (Ag)2
Step 5) 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
Ag = 2.67 in2 > 2.53 in2 OK
rmin = 0.727 in > 0.6 OK
Step 6) Design the bolted connection:
Step 7) Re-check the section.
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.
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.
Required. rmin = (as a buildup section)
rmin ≥ 1.2
For built-up members, tie plates are required to make the
members to behave as a single unit.
For single , rmin = ry ; ry = 1.0 in
Therefore one tie-plate at middle must be used.
Tie-Plates must be used at ends. See
Manual for min. sizes.
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"
See LFRD D2. (P. 16.1-24)