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Section 4.9
Example Problem 4.2
The example problems presented in this section refer to a spreadsheet solution. Click on the link below to get the file:
Chapter 4: Excel Spreadsheet Solutions
Given: A splice plate connection is used to connect two WTs together as shown in Figure 4.9.2.1. The connection is subjected to axial tension. Assume that the load transferred consists of Dead Load (30%) and Seismic Load (70%). The plate steel is ASTM A992. The bolts are fully pretensioned 3/4" diameter ASTM A490-N in standard holes. The faying surfaces are class B.
Figure 4.9.2.1
Splice Connection
Click on image for larger view
Wanted: Determine the capacity of the connection based on bolt strength as specified below. Consider both LRFD and ASD. Express your results in terms of comparable service level loads.
- Determine the slip capacity of the connection.
- Determine the bearing capacity of the connection.
Solution:
There are actually two identical connections in this problem. The each WT transfers the full tension force through the bolts to the splice plates. The splice plates are essentially short tension members with the full force found in the center of the splice plate. As a result, the eleven bolts on either side of the connection carry the full load.
In this problem the bolts are in double shear (i.e. they have two shear planes each).
Since the load combination only includes live and seismic load, ASCE 7 LRFD load combination 5 will be the controlling case. The relationship between factored and unfactored load is:
Pu = 1.2D +1.E = 1.2(0.30 Ps,eq) + 1.6(0.70 Ps,eq) = 1.06 Ps,eq
For ASD, the controlling load combination is ASCE 7 ASD LC5. The relationship between the applied load and the load combination is:
Pa = D + .7E = (0.30 Ps,eq) + .7(.70 Ps,eq) = 0.790 Ps,eq
The solution to this problem is similar to Example 4.1. There are differences in some values due to the change in bolt specification and the fact that the bolts are in double shear.
a) Slip critical condition:
For: m = 0.5, Du = 1.13, hsc = 1.00, Tb = 35 k (from Table J3.1), and Ns = 2 (each bolt has 2 shear planes).
rn = 39.6 k/bolt
Rn = 435 k/connection
for LRFD
Pu = 1.06 Ps,eq < fRn = 1.0(435 k/connection) = 435 k/connection
Ps,eq < 410 k/connection
for ASD:
Ps = 0.79 Ps,eq < Rn/W = (435 k/connection)/1.5 = 290 k/connection
Ps,eq < 367 k/connection
b) Bearing condition:
With Fnv = 60 ksi and Ab = 0.4418 in2:
rn = 26.5 k/shear plane
Rn = 583 k/connection
for LRFD
Pu = 1.06 Ps,eq < fRn = 0.75(583 k/connection) = 437 k/connection
Ps,eq < 413 k/connection
for ASD:
Ps = 0.79 Ps,eq < Rn/W = (583 k/connection)/2.0 = 292 k/connection
Ps,eq < 369 k/connection
Solution Summary
| LRFD | ASD | ||||
| Limit State | fRn | Ps,eq | Rn/W | Ps,eq | |
| (k) | (k) | (k) | |||
| Slip | 435 | 410 | 290 | 367 | |
| Shear Rupture | 437 | 413 | 292 | 369 | |
| Answer | Answer | ||||
Observation:
- Notice that if the extra work is put into creating Class B faying surfaces that the slip capacity approaches the bolt strength (bearing) capacity.