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Section 2.4.4
Example Problem 2.4
Last Revised: 11/04/2014
Given: A transfer girder supports a column at mid-span as shown in Figure 2.4.4.1. The beam has a nominal moment capacity, Mn, equal to 100 ft kips. The column load is (1) part dead load, (2) parts live load, and (1.5) parts snow load. Use an LRFD resistance factor, f, of 0.90 and an ASD factor of safety, W, of 1.67.
Figure 2.4.4.1
Example Problem 2.4
Wanted: Determine the maximum comparable equivalent service load for P considering only the flexure limit state. Consider only LRFD LC3c and ASD LC4b.
Solution:
The basic limit state for this situation can be expressed as:
| LRFD | ASD |
| Mu < f Mn | Ma < Mn / W |
Where
- Mu = Pu L / 4
- Ma = Pa L / 4
Solving for Pu and Pa:
| LRFD | ASD |
| Pu < f Mn 4 / L | Pa < 4 Mn / (L W) |
| Pu < 18.0 k | Pa < 11.98 k |
Before we can compare the two values to see which is more restrictive, we need to express the results in terms of Ps,equiv. To do this we need to determine the composite load factors for the two given load combinations.
|
D = (1/4.5) Ps,equiv |
|
L = (2/4.5) Ps,equiv |
|
S = (1.5/4.5) Ps,equiv |
Substituting these values into the given load combination equations
| LRFD | ASD |
| Pu = 1.2D + 1.6S +.5L | Pa = D + .75L + .75S |
| Pu = 1.022 Ps,equiv | Pa = .8056 Ps,equiv |
With the composite load factors (the coefficients on Ps,equiv) we can now compute the load in comparable terms:
| LRFD | ASD |
| Pu = 1.022 Ps,equiv = 18.0 k | Pa = D + .75L + .75S |
| Ps,equiv = 17.6 k | Ps,equiv = 14.9 k |
The final loads are now comparable. Since LRFD allows a higher load than ASD, it is less restrictive. This means that under ASD the member can support less than under LRFD, in this case.