Steel Design.
Tension Members.
Part One.
As this bridge illustrates, a steel structure is a load-carrying system,
consisting of a set of interconnected structural elements.
For design purposes, we have four main structural element types:
tension members,
compression members,
beams,
and beam-columns.
Since the structural elements need to be connected to form the load-carrying system,
connection design is another major consideration when dealing with steel structures.
This lecture focuses on the design of tension members only.
We will discuss the other topics in future lectures.
The term structural design refers to the process through which we ensure
that the member is strong enough to safely, and economically,
carry the maximum forces that it needs to withstand.
In this and other lectures on steel design,
we are going to use the design specifications published by the American Institute of Steel Construction.
More specifically, we use the fifteenth edition of the AISC Steel Manual as our reference guide.
Consider a standard structural steel member subjected to a tensile force.
Standard structural steel comes in several grades.
Suppose our tension member is made of A36 steel.
This particular grade of steel has a yield stress of 36 ksi,
and tensile strength of 58 ksi.
Yield stress and tensile strength are material properties,
determined based on experimental data.
Such data can be visualized graphically using what is called the stress-strain diagram.
Here is the general shape of the stress-strain diagram for structural steel.
This part of the diagram resembles a straight line,
meaning that steel behaves linearly below this point.
We call this the yield point of the material.
The tensile stress at the yield point is called yield stress.
For A36 steel, the yield stress is 36 ksi.
In this region, the material behaves inelastically.
When the stress goes above the yield stress, the rate of member elongation increases rapidly.
Eventually, the material fails to carry the stress, causing the member to fracture.
This peak stress value is called the tensile strength of the material.
This is the maximum stress that the material can experience before it fails.
The tensile strength of A36 steel is 58 ksi.
As you will see shortly, both the yield stress and the tensile strength
play a central role in the design of tension members.
A large axial force affects the tension member in several ways.
Imagine a long tension member having a length of L.
As the axial force in the member increases, so does the axial stress,
causing the member to elongate.
According to the stress-strain diagram for A36 steel,
when the tensile stress reaches 36 ksi,
the axial strain in the member approximately becomes 0.00124.
That is, the member elongates by 0.00124 L, where L is the length of the member.
So, if the member's length is 10 feet, it elongates about 0.15 inches.
This is a reversible elongation.
That is, if we remove the load, the member assumes its undeformed shape.
When the stress in the member moves above the yield point, the elongation rate increases rapidly.
The axial strain here is about three times this value.
So, when the length of the member is 10 feet, it elongates by about 0.45 inches.
When designing tension members, we want to avoid this kind of excessive elongation.
For design purposes, we impose the following condition to limit the elongation of the member.
The axial stress in the main part of the member, away from the connections,
must not exceed the yield stress.
If we refer to the gross area of the cross-section of the member as A_g, we can write:
Here, F_y is the yield stress,
A_g is the cross-sectional area of the member away from the connections,
and T is the applied tensile force in the member.
That is, the yield stress times the gross cross-sectional area of the member
should not be less than the axial tension force in the member.
The left side of this relationship is the tensile strength of the member,
and the right side is the axial force the member needs to carry.
Here, we are simply stating that the axial force being applied to the member
must not exceed the member's strength.
To account for unexpected loads,
misuse of the structure, or the degradation of the system over time,
we need to design the structural members stronger than they need to be.
This is done by applying a factor of safety to the relevant design equations.
We can apply a factor of safety to this expression by decreasing its left side,
and increasing its right side.
That is, we design the member assuming it has less strength than what is given here,
and is subjected to a load larger than T.
More specifically, we employ what is known as the Load and Resistance Factor Design methodology.
The factor for the left side of this equation is given in the AISC manual.
For the right side of the equation, the governing factored load can be determined
using the relevant provisions given in the ASCE 7 design standard.
For the yielding strength in the gross section, the resistance factor is 0.90.
Therefore, the design strength can be written as:
Since we are using A36 steel, F_y equals 36 ksi.
Hence, the design strength becomes 32.4 times A_g.
Suppose we are using a standard angle cross-section for the tension member.
The dimensions of the angle are: 5 inches, by 3 inches, by half an inch.
The designation for this section is: L5x3x1/2.
The gross area of the angle is 3.75 square inches.
Therefore, the design strength of the section equals 121.5 kips.
That is, to ensure that the axial stress in the gross section of the member does not exceed the yield stress,
the factored load acting on the member must not exceed 121.5 kips.
In addition to ensuring that the tension member has adequate strength
against yielding in the gross section,
we need to check the state of tensile stress at the connections.
This is a necessary step, since the cross-sectional area of the member at the connections,
is smaller than (A g).
Let's examine the cross-section of the member along this line.
Here is the plane view of the cross-section.
Note that two holes are present in this section.
Let (d) represent the diameter of each hole, and (t) be the thickness of the angle.
Then, the net crosssectional area of the member can be expressed as:
The strength of the member at the connection can be written as the product
of this cross-sectional area and the permissible axial stress.
For yielding in the gross section, we used the yield stress, F_y,
as the permissible axial stress.
At the connection, however, given that the length of the connection is much smaller
than the overall length of the member,
excessive elongation is not a concern.
Therefore, we can use the tensile strength of steel as the permissible axial stress.
This cross-sectional area is called the effective net area.
It is defined as the product of factor U and the net area.
U is called shear lag factor.
We will discuss the concept of shear lag in the next lecture.
For now, let's assume that for our tension member U equals 0.75.
So, the strength requirement at the connection can be expressed as:
Where phi_t is the resistance factor,
F_u is the tensile strength for A36 steel,
A_e is the effective net area,
and T is the factored load.
We refer to this limit state as the rupture in the net section.
The resistance factor for rupture in the net section is 0.75.
Tensile strength for A36 steel is 58 ksi.
And the effective net area, which equals U times A_n, for this member,
can be written as: 0.75 A_n.
Therefore, the strength of the member against rupture in the net section becomes 32.625 A_n.
To determine A_n, we need to know the bolt diameter.
Let's assume the diameter of each bolt is 3/4 of an inch.
Knowing the bolt diameter, we can determine the nominal hole diameter (d)
using Table J3.3 of AISC manual.
As can be seen in this table, if the bolt has a diameter of less than 1 inch,
the nominal hole diameter should be 1/16 of an inch larger than the bolt diameter.
Therefore, d equals 13/16 of an inch.
To account for possible roughness around the edge of the bolt hole,
AISC requires that we add an additional 1/16 of an inch to the hole diameter.
Therefore, the net area can be written as:
For L5x3x1/2, the thickness of the angle is one-half of an inch.
Also, the gross area for the cross-section is given in the AISC manual.
It equals 3.75 square inches.
Therefore, A_n becomes:
Substituting 2.875 for A_n in this expression, we get:
That is, the strength of the member against rupture in the net section equals 93.8 kips.
Meaning that if the factored load exceeds 93.8 kips,
the rupture failure in the net section becomes imminent.
To summarize, in this lecture we presented two limit states for the design of tension members:
yielding in the gross section, and rupture in the net section.
For our tension member, the yield strength equals 121.5 kips,
and the rupture strength is 93.8 kips.
Therefore, the rupture in the net section governs the design of the member.
That is, if the factored load exceeds 93.8 kips,
the failure of the member at the connection becomes imminent.
We will continue our discussion on the design of tension members in the next lecture.
See if you can solve the following exercise problems.