Transcript N.W.F.P. University of Engineering and Technology Peshawar Lecture 06: Tension Members By: Prof Dr.

N.W.F.P. University of Engineering and
Technology Peshawar
Lecture 06: Tension Members
By: Prof Dr. Akhtar Naeem Khan
[email protected]
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Topics to be Addressed
 Types of Steel Structures
 Introductory concepts
 Design Strength
 Net Area at Connection
 Shear Lag Phenomenon
 ASD and LRFD Design of Tension
Members
 Design Examples
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Types of steel structures
 The form of a tension member is
governed to a large extent by

Type of structure of which it is a part

Method of joining it to connecting portions.
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Types of steel structures
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Types of steel structures
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Types of steel structures
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Types of steel structures
CE-409: Lecture 06
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Types of steel structures
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Types of steel structures
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Sections for Tension Members
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Sections for Tension Members
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Design Stresses
for
Base Material
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Introductory Concepts
 Stress: The stress in an axially loaded tension
member is given by Equation
 The stress in a tension member is uniform
throughout the cross-section except:


near the point of application of load, and
at the cross-section with holes for bolts or other
discontinuities, etc.
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Types of steel structures
Gusset plate
b
Section b-b
b
7/8 in. diameter hole
a
a
8 x ½ in. bar
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Section a-a
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Types of steel structures
Gusset plate
b
Section b-b
b
7/8 in. diameter hole
a
a
8 x ½ in. bar
Section a-a
Area of bar at section a – a = 8 x ½ = 4 in2
Area of bar at section b – b = (8 – 2 x 7/8 ) x ½ = 3.12 in2
The unreduced area of the member is called its gross area = Ag
The reduced area of the member is called its net area = An
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Design strength
•
A tension member can fail by reaching one
of two limit states:
1. Excessive deformation
• Yielding at the gross area
2. Fracture
• Fracture at the net area
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Design strength
1. Excessive deformation can occur due to the
yielding of the gross section at section a-a
Gusset plate
b
b
Sectio
7/8 in. diameter hole
a
a
8 x ½ in. bar
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Sectio
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Design strength
2. Fracture of the net section can occur if the stress
at the net section (section b-b) reaches the
ultimate stress Fu
Gusset plate
b
b
Sec
7/8 in. diameter hole
a
a
8 x ½ in. bar
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Se
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Design strength
 Yielding of the gross section will occur when
the stress f reaches Fy
P
f 
 Fy
Ag
Nominal yield strength = Pn = Ag Fy
• Fracture of the net section will occur after the stress
on the net section area reaches the ultimate stress Fu
P
f
 Fu
Ae
Nominal fracture strength = Pn = Ae Fu
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Design strength
• AISC/ASD
Ft = 0.6 Fy
Ft = 0.5 Fu
on Gross Area
on Effective Area
• AISC/LRFD
Design strength for yielding on gross area
øtPn =øt Fy Ag = 0.9 Fy Ag
Design strength for fracture of net section
øtPn = øtFu Ae = 0.75 Fu Ae
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Effective Net Area
• The connection has a significant influence on the
performance of a tension member.
• A connection almost always weakens the member
and a measure of its influence is called joint
efficiency.
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Effective Net Area
•Joint efficiency is a function of:
(a) Material ductility
(b) Fastener spacing
(c) Stress concentration at holes
(d) Fabrication procedure
(e) Shear lag.
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Effective Net Area
Research indicates that shear lag can be accounted for by
using a reduced or effective net area Ae
CG
For Bolted Connections
x1
U 1 
x
L
x2
• For bolted connection, the effective net area is Ae = U An
• For welded connection, the effective net area is Ae = U Ag
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Effective Net Area
• For W, M, and S shapes with width-to-depth ratio of at least
2/3 and for Tee shapes cut from them, if the connection is
through the flanges with at least three fasteners per line in
the direction of applied load ,
U= 0.9
• For all other shapes with at least three fasteners per line ,
U= 0.85
• For all members with only two fasteners per line
U= 0.75
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Net Area Example
Example : A 5 x ½ bar of A572 Gr. 50 steel is used as a tension
member. It is connected to a gusset plate with six 7/8 in. diameter
bolts as shown in below. Assume that the effective net area Ae equals
the actual net area An and compute the tensile design strength of the
member.
Gusset plate
b
b
7/8 in. diameter bolt
a
a
5 x ½ in. bar
A572 Gr. 50
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Net Area Example
Gross section area (Ag):
Ag = 5 x ½ = 2.5 in2
Net section area (An):
Bolt diameter = db = 7/8 in.
Nominal hole diameter = dh = 7/8 + 1/16 in. = 15/16 in.
Hole diameter for calculating net area = 15/16 + 1/16 in. = 1 in.
Net section area = An = (5 – 2 x (1)) x ½ = 1.5 in2
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Net Area Example
Gross yielding design strength:
ft Pn = ft Fy Ag
= 0.9 x 50 ksi x 2.5 in2 = 112.5 kips
Fracture design strength:
ft Pn = ft Fu Ae
= 0.75 x 65 ksi x 1.5 in2 = 73.125 kips
Assume Ae = An (only for this problem)
Therefore, design strength = 73.125 kips (net section fracture
controls).
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Shear Lag in Tension
Members
• Shear lag in tension members arises when all the
elements of a cross section do not participate in the
load transfer at a connection.
•There are two primary phenomena that arise in
these cases:
(i) Non-uniform straining of the web resulting in
biaxial stress states
(ii) Effective area reduction.
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Shear Lag in Tension
Members
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Shear Lag in Tension
Members
Effective area reduction
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Shear Lag in Tension
Members
Design Bottom Line
Shear lag can have a large influence on the
strength of tension members , in essence
reducing the effective area of the section. The
amount of the reduction is related to the length of
the connection and the arrangement of crosssection elements that do not participate directly in
the connection load transfer.
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Block Shear in Tension
Members
Block shear is a combined tensile/shear tearing
out of an entire section of a connection.
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Block Shear in Tension
Members
 A failure in which the member fails in tension
on one section and in shear on the
perpendicular section.
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Block Shear in Tension
Members
(a)
T
(b)
T
Shear failure
(c)
Tension failure
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Block Shear in Tension
Members
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Block Shear in Tension
Members
 For such a failure to occur, there are
two possible mechanisms:
(1) Shear rupture + tensile yielding; and
(2) Shear yielding + tensile rupturing.
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Block Shear in Tension
Members
 AISC/ASD
Ft = 0.6 Fy
on Gross Area
Ft = 0.5 Fu
on Effective Area
Connecting element allowable stresses
where failure may be by shear
Fv = 0.3 Fu
Allowable block shear
F = 0.3 Fu + 0.5 Fu
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Block Shear in Tension
Members
 AISC/LRFD
øtRn = 0.75(0.6 Fy Agv + Fu Ant)
øtRn = 0.75(0.6 Fu Anv + Fy Agt)
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Block Shear in Tension
Members
 Design Bottom Line
As a likely limit state for connections,
block shear must be considered in
design. This can be accomplished by
considering the strength limit states of
the two failure mechanisms outlined
above.
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Design Example 1-ASD
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Design Example 1-ASD
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Design Example 1-ASD
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Design Example 1-ASD
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Design Example 1-ASD
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Design Example 1-ASD
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Design Example 1-ASD
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Design Example 1-ASD
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Design Example 1-LRFD
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Design Example 1-LRFD
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Design Example 1-LRFD
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Design Example 1-LRFD
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Design Example 1-LRFD
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Design Example 1-LRFD
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Design Example 1-LRFD
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Design Example 1-LRFD
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Design Example 2-ASD
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Design Example 2-ASD
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Design Example 2-ASD
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Design Example 2-ASD
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Design Example 2-ASD
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Design Example 2-ASD
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Design Example 2-ASD
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Design Example 2-ASD
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Design Example 2-ASD
Design Alternative 2
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Design Example 2-LRFD
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Design Example 2-LRFD
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Design Example 2-LRFD
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Design Example 2-LRFD
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Design Example 2-LRFD
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Design Example 2-LRFD
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Design Example 2-LRFD
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Design Example 2-LRFD
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