 # 6-Definition of the effective area for tension members.

## What are the effective area and shear lag?

A new subject is how to estimate an effective area?  the effective area will be considered if there is a structural steel element that is subjected to tensile force, for that part for example if you have an angle, acted upon by tension force, for which a bolted connection only exists in one leg, while the other leg has no bolts.

For some tension members, such as rolled shapes, that do not have all elements of cross-section connected to the supporting members. The failure load is less than would be predicted by the product An*Fult.

The shape as shown in the sketch, is an angle under force, and has bolts at the lower leg  P is a tension force.

The shaded area stressed very little, which means that most of the stresses occur at the bolts located at the lower leg. and connected to the plate.

As the load applied, transfer of that load to the unbolted portion, will not occur immediately but, after a lag, that is why this phenomenon is called shear lag, the phenomena to which this situation is generally attributed is called shear lag and is illustrated in figure 2-9.
This is a part of the video, which has a subtitle and closed caption in English.

You can click on any picture to enlarge then press the small arrow to review all the other images as a slide show.

### This is an explanation in detail of the content of the subject.

If you have an angle, acted upon tension force, for which a bolted connection exists in one leg while the other leg has no bolts. for some tension members, such as rolled shapes, that do not have all elements of cross-section connected to the supporting members.

The failure load is less than would be predicted by the product of An*Fult.
The shape as shown in the sketch, is an angle under force, has bolts at the lower leg. P is a tension force The shaded area is stressed very little, which means that most of the stresses occur at the bolts located at the lower leg. and connected to the plate.

As the load is applied, the transfer of that load to the unbolted portion, will not occur immediately but, after a lag, that is why this phenomenon is called shear lag.

The phenomena to which this situation is generally attributed to what is called shear lag and is  Illustrated in figure 2-9.Note that the angle is connected alone with only one leg and only one leg is connected to the plate.

This leads to a concentration of stress along that leg, and leaves a part of the unconnected leg part of the un-connected leg unstressed or stressed very little.

This means that stress for the connected portion will be increased, due to no sharing of the unconnected part and may exceed the Fy The farther we move out from the connection, the more uniform the stress becomes.
In the transition region, the shear transfer has lagged and the phenomenon is referred to as shear lag.

Investigators have found that one major of the effectiveness of a member such as an angle connected by one leg is the distance x̅ after investigations, there are two factors the first one is x̅, comparing two shapes as shown, shape a.

For an equal angle and shape b with an unequal angle, where the fasteners exist on the long leg side, the observation is such that when the distance x̅ is smaller, then the effectiveness of the connection is getting better.

The smaller value of x̅, the larger is the effective areas of the member, and thus the larger the member’s strength.

The x̅ is closer since the bigger area has closer  CG, as known from statics. So the angle in shape b is more effective than in shape a.

Shear lag affects both bolted and welded connections, therefore the effective net area concept is applied to both types of connections, then Ae=U *An, and also for welded connections.

Shear lag affects both bolted and welded connections, therefore the effective net area concept is applied to both types of connections, then Ae=U *An, and also for welded connections.

Where x̅ is the distance from the centroid of the connected area to the plane of the connection, and L is the length of the connection.

If the member has two symmetrical planes of connections, x̅ is measured from the centroid of the nearest one-half of the area.

In the case of wide flange connected, with a plate to the upper and lower flanges.

The web area has no connection, because of symmetry, the wide flange section can be converted to two T sections, and hence estimate the CG distance or x̅, for the half of the area.

.

Additional approaches for calculating x̅ for different connections. Types are shown in the AISC manual on page, 16-1-178.

The different values of U value are based on the shapes and the dimensions.

.

The different values of U value are based on the shapes and the dimensions.

.

The different values of U value are based on the shapes and the dimensions.

.

### Solved example 4-1, how to estimate shear lag factor?

It is required to determine the shear lag factor U, the net area An, and the effective area Ae. we have an angle of 5x5x3/8 inch, with 4 bolts, connected to one plate from one side.

We want to estimate the value of U, bolts are given as 4, 3/4 bolts A325 N bolts, so hole dia=3/4+ 1/8, Diameter of the hole=3/8 inch. Our expression U=1- x̅/ L,  x̅/ L is the distance perpendicular to the load, refer to the table then we the get the area for angle 5x5x 3/8 inch, the area=3.65 inch2.

Check the CG distance, the C.G is in the y-direction, but it is referenced as  x̅   and is=1.37 inch, what about the L distance, it is= 3*3 inches is the length of the connection, so U=1-and is=1.37 inch/l=1-(1.37/9)=0.848.

This is the calculation for the U value, but if referring to table U=0.80.

.

For four or more fasteners as the third case of the table as shown then select the bigger value of the two values, which is =0.848, then we estimate the net area as per usual method Anet=Ag- sum(d*t), dia is=7/8 inch.

We can get the y bar for the angle L5xL5x 3/8 from table 1-7. y bar, Anet=Ag -Ag- sum(d*t), Ag =3.65 inch2, dia 7/8 and t=3/8 of the angle,  sum(d*t)=7/8*3/8, only one bolt hole to be deducted, to be deducted from 3.65 inch2.

.

The value is 3.32 inch 2, then A -effective=0.848*3.32=2.82 inch2, which was the first example of the effective area.

.

This is the pdf file used in the expansion of this post.
The next post is A Solved problem for Design strength.
Chapter 3 – Tension Members– A Beginner’s Guide to Structural Engineering is a great external resource.

Scroll to Top