In the previous lectures, we talked about two governing limit states for the design
of tension members: yielding of the gross section and rupturing of the net section. Note that the
rupture strength is a function of the net area of the member's cross-section. For tension members
that use bolted connections, the determination of the net area could become rather involved.
That is particularly true when the bolt holes are staggered as shown here. This lecture focuses on
determining the net area due to staggered holes in the rupture strength equation.
When the bolt holes are not staggered, when they are placed along straight lines,
for rupture of the net section, failure occurs along the straight
line that passes through the holes closest to the center of the member.
The net area to be used for calculating rupture strength, therefore, equals: In this equation,
Ag is the gross area of the cross-section, N is the number of holes in the path of failure,
d is the whole diameter, and t is the thickness of the connected member. This net area is subjected to
the axial stress the tension force (t) produces. We can also consider this path as the path of failure.
However, if we examine the area of the cross-section along this line,
we find it to be subjected to the tensile force of T/2. Why T/2?
The axial force in the member transfers to the gusset plate via the bolts.
Each bolt is assumed to be responsible for carrying an equal share of the force.
Since the connection is formed using 6 bolts, each bolt carries a force of T/6.
If we cut the tension member along this line, we end up with this free body diagram.
Here we can see that the cross-section of the member is subjected to an axial
force of T/2, since half of the load has already been transferred to the gusset plate
by these bolts. From this free body diagram we can conclude that the axial stress along this path
is significantly less than the axial stress along this path. Hence the path further away
from the member's end is considered the critical path for rupture of the net section.
However, determining the critical path of failure may not be straightforward when the holes are staggered.
Suppose a tension member is connected to a gusset plate at each end using two bolts as shown here.
This hole configuration offers two possible failure paths.
The rupture of the net section could occur along this path, or along this path.
The path that produces the smaller net area governs the design.
We know how to determine the net area along Path 1. Here is a three-dimensional view of the cut
plate along Path 1. If we refer to the width of the plate as w, the thickness of the plate
as t, and the effective diameter of the hole as d, the net area of the cross-section can be written as:
Alternatively, we can express An as: where Ag is the gross area of the cross-section.
Here is the three-dimensional view of the cut plate along Path 2. To determine the
net area along this path, we need to know the center-to-center distances between the holes.
Let's label this distance s,
and use g to denote this distance.
The AISC specification references a simple procedure for calculating the net area
along such a diagonal path. Victor Cochrane presented a derivation of this procedure
in the November issue of Engineering News-Record in 1922. For your reading pleasure, we have captured
the relevant part of the original Cocnrane's article in pdf format, the link is provided in the video
description field. The procedure involves two steps: First, we subtract the area due to the holes along
the path from the gross area of the cross-section. In this case, since there are 2 holes along the
path, we subtract 2 hole diameters times the thickness of the plate from the gross area.
Second, to compensate for the additional area along the inclined line between the two holes,
we add the plate's thickness times s^2/4g to that area. We compare the values
obtained from these two equations. The path with a smaller net area is the critical path of failure.
Suppose the plate has a width of 6 inches and a thickness of 1/2 inch, and the holes are made
for 3/4-inch bolt diameter. Therefore, the effective hole diameter equals:
3/4 + 1/16 + 1/16 = 7/8"
See the previous lectures to learn why we need to add these
additional fractions to the bolt diameter to arrive at the effective hole diameter.
For this tension member, the horizontal spacing between the two bolts is 3 inches,
g equals 3 inches as well.
given these dimensions, the net area for Path 1 becomes 2.56 in^2
And the net area for Path 2 is 2.5 in.
Since Path 2 has a smaller net area it controls the design. This means we need to use 2.5 for
calculating the rupture strength of the tension member. The rupture strength equals 108.75 kips
Suppose a tension member's connection to a gusset plate consists of 10 bolts
staggered with the spacings shown here. The standard holes are made for 7/8" diameter
bolts, and the thickness of each plate is 5/8 of an inch. Here we can identify three possible
failure paths for the net section rupture. The tension member could fail along this path,
this path,
or this path.
The path with the smallest net area governs the design.
Let's calculate the net area for each path. The effective hole diameter is:
For Path 1, the net area can be expressed as the gross area minus the number of holes,
(there are three holes along the path) times the hole diameter
times the thickness of the plate. The net area equals 7.5 in^2.
For path two, the net area can be expressed as: Since the path goes through 4 bolts, we subtract
4 times d times t from the gross area. Then we add 2 times t times square of s divided by
4 g to the expression. Here we use 2 because there are 2 inclined lines along the path.
This equation yields 7.3 in^2. For the net area, however, the axial stress along this path
is not caused by a force of t, rather a force of 9T/10 causes it.
We assume that each bolt carries an equal share of the axial force, T/10, therefore, this
bolt has transferred a force of T/10 to the gusset plate before the stress reaches Path 2.
That is, the member along Path 2 is subjected to a force of 9T/10, not a force of T. Since in our
design equation we compare the member's strength against T, for comparing this net area with the
other net areas, we can multiply the calculated net area 7.3 by 10/9. So the net area along Path 2,
assuming the member along the path is subjected to the full force of T, equals 8.1 in^2.
For path 3, the net area can be written as:
Here, there are 5 holes and 4inclined lines along the path.
In summary, the 3 net areas are:
Since path 3 produces the smallest net area, it controls the design. Therefore, the rupture
strength for the tension member can be written as: For rectangular plates the shear lag factor
(u) equals 1. Hence the rupture strength for the tension member is 308 kips.
See if you can solve the following exercise problems.