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Section 3.10
Example Problem 3.3
Last Revised:
08/07/2017
For this problem, we want to determine the tensile capacity of the WT
sections in the connection shown in
Figure 3.10.2.1 for example problem 3.2. We want to be able
to compare the results from both LRFD and ASD, so the capacity will need to be
in comparable equivalent service load terms for both methods.
Given: The tension splice connection shown in Figure
3.10.2.1 connects two WT sections with two splice plates attached to the web.
Assume that the system supports a tension load that is 40% dead load and 60%
live load.
Wanted: The capacity of the splice plates in terms of
equivalent service load, Ps,eq, using both LRFD and ASD.
Solution: The chapter spreadsheet has a solution to the problem. You will need to
review that solution as this discussion progresses.
In this case, each of the WTs carries the full tension force and they are
mirrored images of one another. Consequently we only need to find the
capacity of one to find their limit on the connection. The provided
spreadsheet solution gives all the relevant computations. Table 3.10.3.1
summarizes the results.
Observations:
- WTs are W shapes that have been cut longitudinally down the middle of the
web, so when looking for
material properties in SCM Table 2-3 you need to look under the W shape
column.
- The section properties for the WT are obtained from SCM pg 1-60.
Since the bolts are only on the web, the only section properties that are
needed are Ag and tw.
- Since we don't know how long the members are, we used the slenderness
limit state to compute a maximum length for the members.
- Figures 3.10.3.1 and 3.10.3.2 show the only valid unique failure paths
for tensile rupture and block shear, respectively. You should be able
to explain why these are the only valid failure paths. Review the
failure path tutorial
if you have difficulties.
- When computing Ae, U is no longer 1.0 since the flange is not
transferring force directly to connectors. In this case, you have two
options for determining U. Our situation falls under either case 2 or
case 7. We chose to use case 7 to simplify the problem. If you use
case 2 you will need to take the time to compute the centroidal distance
x. We can avoid this
since we have at least four bolts in a line in the direction of force.
- The capacity of the WTs is significantly less than the capacity of the
plates that connect them. From a designer's stand point, you would
consider downsizing the splice plates if the WTs have sufficient capacity for
the application.
- Note the controlling limit state is block shear and it is significantly
less than the other limit states. Some consideration should be made to
increasing the
spacing the bolts (i.e. increasing the tension and shear areas) in order to use more of the capacity of the members.
Table 3.10.3.1 Example Problem 3.3 Results
Table of Capacities |
|
|
|
|
|
LRFD |
|
ASD |
Limit State |
fPn |
Pn/Weqiv |
Pn/W |
|
|
(k) |
(k) |
(k) |
Slenderness |
N/A |
N/A |
N/A |
Tensile Yielding |
410.0 |
284.7 |
272.8 |
Tensile Rupture |
293.3 |
203.7 |
195.5 |
Block Shear |
162.5 |
112.9 |
108.4 |
Bolt Bearing |
372.5 |
258.7 |
248.3 |
Controlling Capacity |
162.5 |
112.9 |
108.4 |
Dead Load |
|
45.2 |
43.3 |
Live Load |
|
|
67.7 |
65.0 |
|
|
|
Answer |
Answer |
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