Steel Design.
Tension Members.
Shear Lag.
In the previous lecture, we introduced two limit states
for the design of tension members:
yielding of the gross section,
and rupture of the net section.
See Lecture SD1 for a description of these limit states.
In this expression, A_e E is defined in terms of A_n and U,
where A_n is the net cross-sectional area of the member at the connection,
and U is called the shear lag factor.
In this lecture, we are going to explain what shear lag means,
and discuss how to calculate U (the shear lag factor)
according to the AISC specification.
Consider a tension member having a rectangular cross-section.
Suppose the applied force is distributed uniformly at the ends of the member,
as shown here:
This causes a uniform tensile stress distribution in the member.
So, if we cut the member here,
we end up with an axial stress distribution, like this:
We get the same uniform stress distribution,
if we cut the member anywhere else.
For example, if we cut the member here,
we can see that the axial stress is distributed uniformly
in proportion to the applied force.
So in this case, the entire cross-section of the plate
participates in transferring the applied force
from one end of the member to its other end.
Put it differently, we can say that the axial stress in the member
flows from one end of the member to its other end along straight lines,
like this:
Suppose the axial force is acting only on a part of the member, like this:
This part of the cross-section is loaded,
while this part,
and this part,
are load-free.
Therefore, at the ends of the member where the axial force is applied,
only the loaded part is stressed.
Since no load is present in this part,
no axial stress develops in the region near the corner of the plate.
Similarly, we can argue that the axial stress vanishes in the regions
near the other corners of the plate.
Put it differently, the axial stress flows inward
from the loaded regions and spreads to the side edges of the plate,
like this:
If we cut the member near its midpoint,
where the stress flows through the entire cross-section,
we get a uniform stress distribution.
This is identical to the stress distribution for this case.
In this region of the member, however,
the axial stress is NOT distributed over the entire cross-section.
The stress is concentrated in the middle part of the cross-section.
The ends of the section carry no axial stress,
since they fall in the stress-free zones.
Clearly, the effective cross-sectional area here
is smaller than the gross cross-sectional area of the member.
Next, let us consider a tension member having an L-shaped cross-section.
Assume the axial force is applied to one leg of the angle only.
The stress flow for this member looks like this:
Here, the outer regions of the unloaded leg of the angle are stress-free.
Like the two previous cases,
if we cut the member near its midpoint,
we get a uniform stress distribution.
However, if we cut the member around here,
we get a stress distribution like this,
where a part of the cross-section is stress-free.
So, the effective area here is less than the gross area of the cross-section.
It is important to note that the effective area
is always less than or equal to the gross area.
That is, the effective area equals (the gross area) times (the shear lag factor),
where the shear lag factor is <= 1.
Why do we call this partial stress distribution
over the cross-section of the member shear lag?
The axial stress due to the applied force travels from this leg,
to the other leg of the angle,
in the form of shear stress,
through the rectangular area that the two plates share, as shown here.
As the stress propagates from the central part of the member toward its ends,
the shear stress falls behind in magnitude.
We refer to this “falling behind” phenomenon as shear lag.
A tension member has either bolted connections or welded connections.
For a bolted connection, the effective area equals:
the shear lag factor, times, the net area,
where the net area equals:
the gross area of the cross-section,
minus, the area due to the holes that accommodate the bolts.
For a welded connection, since no holes are present,
the effective area equals, the shear lag factor,
times, the gross area of the cross-section.
Both of these cases share the same shear lag factor,
to be determined using the relevant provisions of the AISC specification.
Let's start our discussion by determining the shear lag factor
for single-angle tension members.
Consider this tension member.
It uses the standard angle section: L6x4x1/2.
Each end of the member is attached to a gusset plate using four bolts,
arranged in two columns, as shown here:
The center-to-center distance between the bolt columns is 3 inches.
Here is the side view of the connection.
In this example, the long leg of the angle is attached to the gusset plate.
The angle's short leg remains unattached.
According to the AISC specification,
when only one leg of the angle is connected to the gusset plate,
the shear lag factor can be calculated using this equation:
Here, L is the center-to-center distance between the connection's first and last columns of bolts.
In this example, L = 3”.
X-bar is the distance between the angle's centroid and the plane of the connection.
Here is the plane of the connection.
And this is the axis that passes through the centroid of the angle.
So, x-bar is this distance.
We can get the x-bar value from Table 1.7 of the AISC manual.
For L6x4x1/2, the Table gives 0.981 inches for x-bar.
Therefore, the shear lag factor can be written as:
Furthermore, Chapter D of the AISC manual
provides another provision for determining the shear lag factor.
In summary, the provision states that the shear lag factor need not be less than,
the ratio of the gross area of the connected leg of the angle,
to the member's gross area.
That is, U, can be greater than or equal to this ratio.
In this case, the gross area of the connected leg of the angle equals:
6 inches times half an inch.
The gross area of the angle is given in Table 1.7.
It equals 4.75 square inches.
So, this ratio evaluates to 0.632.
We can use 0.632 for shear lag factor,
only if the calculated value here is less than 0.632.
In this case, we use 0.673 for U.
Let's consider another case.
Suppose we have attached the short leg of the angle to the gusset plate, like this:
How does this change in the orientation of the angle affect the shear lag factor?
The main parameter that changes in our formulation is x-bar.
When we attach the short leg of the angle to the gusset plate,
when the angle's long leg is protruding outward from the plate,
the geometric center of the angle moves further away from the plane of the connection.
That is, x-bar increases in value.
Table 1.7 defines the location of the centroid of the cross-section,
relative to the edges of the angle, using two values.
For L6x4x1/2, these values are:
1.98 inches, and 0.981 inches.
The smaller value, 0.981, is the distance from the center of the cross-section,
to the plane that defines the edge of the long leg of the angle.
We use this value when the angle’s long leg is attached to the gusset plate.
The larger value, 1.98,
is the distance from the center of the cross-section to the plane
that defines the edge of the short leg of the angle.
We use this value when the short leg is attached to the gusset plate.
Therefore, in this case, since the short leg of the angle
is attached to the gusset plate,
we use 1.98 inches for x-bar.
Then, according to Table D3.1 of the AISC manual,
the shear lag factor becomes 0.34.
The other provision in Chapter D gives the lower limit for the shear lag factor.
This limit is: 0.42.
Since 0.34 is less than 0.42, we can conclude that U = 0.42.
In summary, if the long leg of the angle is attached to the gusset plate,
the shear lag factor equals 0.673.
If the short leg is attached to the plate, we get a U of 0.42.
Assuming both designs use the same hole size,
we can see that the effective net area for this arrangement,
is larger than, the effective net area for this arrangement.
Therefore, we can conclude that given an angle with unequal legs,
connecting the long leg to the gusset plate is a better option.
It results in a stronger tension member.
We will continue this discussion in the next lecture.
See if you can solve the following exercise problems.