Last Updated on September 20, 2026 by Maged kamel
SolveProblemem 4-1,4Partt 2, for the effective length factor.
Estimate the GB Value for joint B of the column.
At joint B, consider two columns. For each column section, W10x33, we have two beams with section W14x. For each, we estimated the Beam section Area as Ag = 6.49 in², Ix = 199 in⁴, and rx = 5.54 in². The Inertia of the column is < the Inertia of the two beams.
Gb Value equals 2∑EI/L for column AB /∑EI/L for the two framing beams. The E terms for the column and beams cancel; the numerator, Ix/L for the column, equals 2(171/12).
The denominator equals ∑EI/L for the beams, which is the sum of two items. The first item equals (I/L) for beam1, which has a length of 18 ft, while the second Beam has a length of 20 ft.
The two beams have an Inertia of 199.0 inch4. The sum is (199/20 + 199/18). The Gb Value is 1.3567; please refer to the next slide image.
Use the alignment chart to get the K Value for column AB.
Since we have unbraced frames, we can use the unbraced-frame graph to obtain the K factor from the estimated GA and GB. We find that the effective length k is 1.45.
Usingthe French equation, we substitute the values and obtain k = SQRT(.60GAGb + 4(G + GB) + 7.50)/SQRT(GA + GB + 7.50), which yields k = 1.466.
Having estimated the k Value, we check whether K*L/r is smaller or bigger than 4.71*Sqrt(E/Fy).
We proceed, having k=1.45, then KL/r at x=1.45(1212)/rx of the column which=4.19 inch2, KL/r at x=49.83.
For 4.71*(29000/50)=113.34. Since Kl/rx (le/rx) is less than 4.71*(E/Fy) and the column is not long and inelastic, we need to estimate the stiffness reduction factor τb.
LRFD Design procedure for τb estimation.
For the LRFD Design, we start by estimating P_ultimate as 1.20D + 1.60L, where the dead Load D is 35.50 kips, and the Live Load L is 142.0 kips.
Estimate 1.2D + 1.60L = 1.2 (35.50) + 1.6 (142.0) = 269.80 kips and compare with 1.4D; the governing Pult is 269.80 kips.
This Value is considered Pr in the reduction-factor equation.
First, check that αPr/Pns < 0.50. Then τb = 1. If not, τb will be < 1.0.
For the LRFD case, Pr = Pult = 269.80 kips. the Yielding Load Py=Ag*Fy=9.7150=485.50 kips, αPr/Pns will be=(1269.80/485.50) =0.557 >0.50, then τb<1, but if αPr/Pns<0.50, then τb=1.
But in our case τb<1, then use this equation τb= 4(αPr/Pns)*(1-(αPr/Pns)).
We substitute the values as τb= 4*(0.5557)(1-(10.5557), τb=0.9876.
We estimate Pult/Area = 269.8/9.71 = 27.79 ksi; we will use this Value to obtain τb from Table 4-13.
Use Table 4-13 to get the τb Value for the LRFD Design.
We have Pu/A = 27.79. We use this Value in the horizontal direction and intersect the column at Fy = 50 ksi. The stiffness factor is equal to 0.9877.
We need to adjust the k Value, which is done by re-adjusting GA and GB by multiplying them by τb. The adjusted Value of Ga=1.52*0.9877=1.5013, while the adjusted Value of Gb=1.36*0.9877=1.343 for the solved problem 4-14 part 2 for the effective length factor.
The final Kx Value is 1.43 for inelastic column AB, while Ky = 1.0.0.
ASD Design procedure for τb.
For the ASD Design, we will estimate kA and KB for column AB. We will repeat the same procedure used earlier and check whether the column is elastic or inelastic for Kl/rx >= 4.71*Sqrt(E/Fy).
Based on the estimate, the column is inelastic, so we need to determine τb.
In the ASD calculation, we have D = 35.50 kips and L = 142.0 kips. Pt = 35.50 + 142.0 = 35.50+142.0=177.50 kips.
We check if αPr/Pns < 0.50; then τb = 1, α = 1.60, Pr = Ptotal = 177.50 kips, and Pns is the yield Load, which equals A*Fy = 9.71*(50) = 485.50 kips.
Then αPr/Pns=0.585>0.50, τb<1, then use the equation τb= 4(αPr/Pns)*(1-(αPr/Pns)). We calculate a substitute using the known values, and we obtain τb = 0.9703.
From Prof. Segui’s calculation, τb = 0.97033, which Segui extracted from Table 4-13, as shown in the next slide image. How to use Table 4-13 is shown in full detail for the ASD Design.
Use Table 4-13 to obtain the τb-ASD Design.
We have Pt/A = 177.50/9.70 = 18.28. We move along this V Value horizontally and intersect the Fy = 50 ksi column. The stiffness factor is equal to 0.9703.
We need to adjust the k Value by re-adjusting GA and GB, multiplying them by τb.
The adjusted Value of Ga=1.52*0.9703=1.47, while the adjusted Value of GB=1.36*0.9703=1.3432.28 for the solved problem 4-14 part 2 for the effective length factor.
The final Kx Value is 1.43, while Ky = 1.0. I hope this Post contributes to the subject of compressed steel members.
For the previous Post, 27, please refer to Solved problem 4-14-effective length factor-Part 1.
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For a good A Beginner’s Guide to Steel Construction, 14th ed. Chapter 7 – Concentrically Loaded Compression Members.
For a good A Beginner’s Guide to the Steel Construction Manual, 15th ed. Chapter 7 – Concentrically Loaded Compression Members.
For a good A Beginner’s Guide to the Steel Construction Manual, 16th ed. Chapter 7 – Concentrically Loaded Compression Members.