Module 4 - Design of Tension Members
Module 4 - Design of Tension Members
Module 4 - Design of Tension Members
1.0 Introduction
The Tension member considered for the design is a linear member which
carries an axial pull. The members undergo extension due to this axial pull. This
is one of the common types of force transmitted in the structural system.
Tension members are very efficient since the entire cross section carries
uniform stress unlike flexural members. The tension members do not buckle
even when stressed beyond the elastic limit. Hence the design is not effected
by the type of section used i.e., Plastic, Compact or Semi-compact. Some of the
common examples of tension members in structures are; Bottom chord of pin
jointed roof trusses, bridges, transmission line and communication towers,
wind bracing system in multi-storey buildings, etc.
The tension members may be made of single structural shapes. The standard
structural shapes of typical tension members are:
Single angle members are economical but the connection produces eccentric
force in the member. These are generally used in towers and in trusses. Double
angle members are more rigid than single angle members. They are used in
roof trusses. Since there exists a gap of about 6 to 10 mm between the two
members (which depends on the thickness of the gusset plate), they are
generally interconnected at regular intervals so that they act as one integral
member. In the members of bridge trusses the tensile forces developed are
very large and hence require more rigid members. In these structures single
channel, single I-section, built-up channels, or built-up I-sections will be
generally used.
Apart from strength requirement, the tension members have to be checked for
minimum stiffness by stipulating the limiting maximum slenderness ratio of the
member. This is required to prevent undesirable lateral movement or
excessive vibration. The slenderness limits specified in IS: 800-2007 for tension
members are given in Table 1.
The shear lag reduces the effectiveness of the component plates of a tension
member that are not connected directly to a gusset plate. The efficiency of a
tension member can be increased by reducing the area of such components
which are not directly connected at the ends. The shear lag effect reduces with
increase in the connection length.
Steel members (plates, angles, etc.) without bolt holes can sustain loads up to
the ultimate load without failure. However, the members will elongate
considerably (10 to 15 % of its original length) at this load, and hence make the
structure unserviceable. Hence the design strength Tdg is limited to the yielding
of gross cross section which is given by
Tdg = fy Ag /m0
where
fy = yield strength of the material in MPa
Ag = gross area of cross section in mm2
m0 = 1.10 = partial safety factor for failure at yielding
This occurs when the tension member is connected to the main or other
members by bolts. The holes made in members for bolts will reduce the cross
section, and hence net area will govern the failure in this case. Holes in
members cause stress concentration at service loads. From the theory of
elasticity, the tensile stress adjacent to a hole will be about two to three times
the average stress on the net area (Fig. 2a). This depends on the ratio of the
diameter of the hole to the width of the plate normal to the direction of the
stress.
(a)
(b)
(c)
Fig. 2 Stress-distribution in a plate adjacent to hole due to tensile force.
When the tension member with a hole is loaded statically, the point adjacent
to the hole reaches the yield stress fy first (Fig. 2b). On further loading, the
stress in other fibers away from the hole progressively reaches the yield stress
fy. Deformations of the member continue with increasing load until final
rupture of the member occurs when the entire net cross section of the
member reaches the ultimate stress fu (Fig. 2c).
where
g
dh
g
b
g
given by
where
= 1.4 – 0.076 (w/t) (fy /fu) (bs /Lc ) (fu m0 / fy m1)
0.7
where
w w
w1
bs = w + w1 - t bs = w
Fig. 4 Angles with single leg connections
Tdn = An fu /m1
where
= 0.6 for one or two bolts, 0.7 for three bolts and 0.8 for four or more
bolts along the length in the end connection or equivalent weld
length
The tearing strength, Tdn, of the double angles, channels, I sections and other
rolled steel sections, connected by one or more elements to an end gusset is
also governed by shear lag effects. The design tensile strength of such sections
as governed by tearing of net section may also be calculated using equation in
6.2.3, where is calculated based on the shear lag distance, bs taken from the
farthest edge of the outstanding leg to the nearest bolt/weld line in the
connected leg of the cross section.
AD
BC
Fig. 6 Block shear failure in angle with Fig. 7 Block shear failure of gusset plate
bolted connection in welded connections
The block shear failure is also seen in welded connections. A typical failure of a
gusset in the welded connection is shown in Fig. 7. The planes of failure are
chosen around the weld. Here plane B-C is under tension and planes A-B and C-
D are in shear.
The block shear strength, Tdb, of connection shall be taken as the smaller of
Atn /m1 ) or
Avg, Avn = minimum gross and net area in shear along a line of transmitted
force, respectively (1-2 and 3–4 as shown in Fig. 8 and 1-2 as
shown in Fig. 9)
Atg, Atn = minimum gross and net area in tension from the bolt hole to the
toe of the angle, end bolt line, perpendicular to the line of force
(2-3 as shown in Figs. 8 and 9)
3
1 2
3
1 2
Fig. 8 Block shear failure in plate Fig. 9 Block shear failure in angle
The block shear strength, Tdb, shall be checked for welded connections by
taking an appropriate section in the member around the end weld, which can
shear off as a block.
Lug angles are short angles used to connect the gusset and the outstanding leg
of the main member as shown in Fig. 10. The lug angles help to increase the
efficiency of the outstanding leg of angles or channels. They are normally
provided when the tension member carries a very large load. Higher load
results in a larger end connection which can be reduced by providing lug
angles. It is ideal to place the lug angle at the beginning of the connection than
at any other position.
Lug
Angle
Problem 1
Determine the design tensile strength of the plate 120 mm x 8 mm connected to a 12
mm thick gusset plate with bolt holes as shown in Fig. 11. The yield strength and
ultimate strength of the steel used are 250 MPa and 400 MPa. The diameter of the
bolts used is 16 mm.
Gusset 12 mm thick
Plat
e
3
0
6
0
3
0
3060 6030
Solution
The design tensile strength Td of the plate is calculated based on the following
criteria.
(i) Gross section yielding
The design strength Tdg of plate limited to the yielding of gross cross section Ag
is given by
Tdg = fy Ag /m0
0
3
0
6
0
3
30 60 60 30
Fig. 12 Failure of plate in block shear
4
0
75 x8
6
100 x
0
50 50
30 50 50 50
5
7
2
Fig. 13 Details of end connection 1
Solution
The design tensile strength Td of the angle is calculated based on the following
criteria.
= 1.4 – 0.076 (w/t) (fy /fu) (bs /Lc ) (fu m0 / fy m1)
Here fu = 400 MPa, fy = 250 MPa, m1 = 1.25, and m0 = 1.10
w = 75 mm, t = 8 mm, bs = (75 + 60 – 8) = 127 mm, Lc = 250 mm
Further, diameter of bolt hole = 20 + 2 = 22 mm.
Anc = (100 – 8/2 – 22) 8 = 592 mm2, Ago= (75 – 8/2) 8 = 568 mm2
Hence, = 1.17. Since 0.7 1.41 , = 1.17
4
0
3050 5050
5050
75 x 75 x 6
5
7 6
5
7
Solution
The design tensile strength Td of the angles is calculated based on the following
criteria.
Tdn = An fu /m1
Design of bolts
Bolts are in double shear.
Hence, strength of single 20 mm dia bolt = 2 x 45.3 = 90.6 kN
For the strength of connection to be larger than the strength of member,
Number of bolts required = 384.15 / 90.6 = 4.24
Hence provide 5 numbers of 20 mm bolts. Hence the connection is safe.
Assume edge and end distances = 35 mm and pitch = 50 mm
3
5
3550 5050
0
5
Avg = 235 x 6 = 1410 mm2, Avn = (235 – 4.5 x 22) x 6 = 816 mm2,
Problem 4
Design a suitable angle section to carry a factored tensile force of 210 kN assuming a
single row of M20 bolts. The yield strength and ultimate strength of the material is
250 MPa and 410 MPa, respectively. The length of the member is 3 m.
Solution
Step 1:
Obtain the net area, An, required to carry the design load Tu from the equation
using the ultimate stress.
Tu = fu An / m1
Here, Tu = 210 kN, fu = 410 MPa, and m1 = 1.25
Therefore, An = 619.8 mm2
Increase the net area, An, by 25 percent to obtain the gross area.
Hence, Ag= 774.8 mm2
Step 2:
Obtain the gross area, Ag, required to carry the design load Tu from the
equation using the yield stress.
Tu = fy Ag /m0
Step 3:
From steps 1 and 2,
Required gross area Ag,req. = 924.0 mm2 (max. value)
Select an angle 65 x 65 x 8 with Ag = 976 mm2 ( > 924.0 mm2)
Step 4:
The strength of 20 mm diameter bolts in single shear = 45.3 kN
Hence required number of bolts = 210/45.3 = 4.64
Provide 5 bolts at a pitch of 60 mm
Step 5:
The design strength Tdg of plate limited to the yielding of gross cross section Ag
is given by
Tdg = fy Ag /m0
Here fy = 250 MPa, Ag = 976 mm2 and m0 = 1.10
Hence Tdg = 221.80 kN
Step 6:
The design strength Tdn of angle governed by rupture of net cross sectional
area, An, is given by
Tdn =0.9 fu Anc / m1 + Ago fy /m0
= 1.4 – 0.076 (w/t) (fy /fu) (bs /Lc ) (fu m0 / fy m1)
Here fu = 410 MPa, fy = 250 MPa, m1 = 1.25, and m0 = 1.10
w = 65 mm, t = 8 mm, bs = (65 + 35 – 8) = 92 mm,
Lc = 4 x 60 = 240 mm
Further, diameter of bolt hole = 20 + 2 = 22 mm
Anc = (65 – 8/2 – 22) 8 = 312 mm2, Ago= (65 – 8/2) 8 = 488 mm2
Hence, = 1.26. Since 0.7 1.44 , = 1.26
Hence, Tdn = 231.85 kN
Step 7:
Step 8:
The tensile capacity of member ISA 65 x 65 x 8 with 5 bolts of 20 mm diameter
is the least of Tdg, Tdn and Tdb.
Step 9:
Check for stiffness.
L = 3000 mm, rmin = 12.5 mm
L/ rxx = 240 < 250
Hence the section is safe.
Problem 5
A single angle member carries a factored axial force of 400 kN. Design the member
and the connection with a gusset plate and a lug angle. The yield strength and
ultimate strength of the material is 250 MPa and 410 MPa, respectively.
Solution
Sizing of Single Angle
Factored load = 400 kN
For preliminary sizing of single angle use the relation (Cl. 6.3.3 of IS 800 : 2007)
where Tdn = 500 kN, α = 0.8 ( 4 bolts), fu = 410 MPa and m1 = 1.25
Hence, Required net area is An = 1524.4 mm2
The gross area is arrived by increasing the net area by 15% (say)
Therefore, Required gross area is Ag = 1753.1 mm2
Therefore provide ISA 125 x 75 x 10
Hence actual gross area Ag = 1902 mm2
Here, the 125 mm side is connected to the gusset and 75 mm side is the outstanding
leg.
The design strength Tdg of angle limited to the yielding of gross cross section Ag is
given by
Tdg = fy Ag /m0
For preliminary sizing of lug angle use the relation (Cl. 6.3.3 of IS 800 : 2007)
where Tdn = 176.64 kN, α = 0.8 ( 4 bolts), fu = 410 MPa and m1 = 1.25
Hence, Required net area is An = 673.17 mm2
The gross area is arrived by increasing the net area by 15% (say)
Therefore, Required gross area is Ag = 774.15 mm2
Therefore provide ISA 75 x 75 x 8
Hence actual gross area Ag = 1140 mm2
Design of connections
Assume one row of 20 mm diameter bolt. Use a pitch of 2.5 x 20 = 50 mm, and an
edge distance of 30 mm (Cl. 10.2.2 and Cl. 10.2.4.2 of IS 800 : 2007)
Strength of 20 mm bolt in single shear = 45.30 kN
Load carried by the connecting leg of the main member is proportional to its area in
comparison with the total area.
= 176.74 kN
= 206.08 kN
Gusset
Main angle
ISA
125x75x10
Lug angle
ISA 125x75x10
Fig. 20 Details of connection of main angle with lug angle and gusset
9.0 References