Last Updated on September 20, 2026 by Maged kamel
Stiffness reduction factor for inelastic columns-part 2.
In the previous Post, we talked about the difference between Fcr from the Euler graph and the Fcr for inelastic columns. Please refer to this Link for review.
How do we get the Expression for the Stiffness reduction factor for inelastic columns?
Prof. Segui has introduced several equations to obtain an Expression for τb using the Fcr inelastic term.
Using λ^2 = Fycrelastic based on Galambos’ second equation, then *(Fcr inel/Fcr inelastic), use τb = Fcr in/Fcr elastic, then we get λ^λ^2 = τb*(Fy/Fcr inelastic).
We obtain the final Expression for τb: the stiffness reduction factor τb = 4*(Fcr/Fy)*(1-Fcr/Fy).
Using the stress Expression in terms of loads, the stiffness reduction factor is τb = 4*(Pn/Py)*(1-Pn/Py), where Pn = Fcr*Ag and Py = Fy*Ag.
The Load Expression was further modified by the Load factor, which can be expressed in either the LRFD or ASD format; an additional α factor was introduced.
There is a new factor, termed α; In the case of the LRFD Design, α = 1, and for the ASD Design,
α = 1.6. This is the code provision. First, check that α*Pr/Pns <=0.50. If we use Pns, this is considered Py in all cases, with an α factor of 1 for LRFD, and Pr = Pult = 1.2 D.L + 1.60L.
In the case of ASD then Pt=Pd+PL, refer to τb equation, for the pr expression (α*Ppr)=(1*Pult)=(1*Pult), the Pns =Py, in case of ASD,(α*Ppr)=(1.60*PT).
The new slide shows various types of curves, as quoted from the UMass Link, which includes a handy illustration for compression steel members. The blue curve is the Euler curve, and the dotted yellow curve is the inelastic column curve, where Et < E.
The dark blue curve is the AISC code provision for inelastic columns, and the Fcr equation can be divided into two parts. In the first part, the elastic columns, where Kl/r>4.71(sqrt(E/Fy)), are differentiated from the inelastic columns.
The Value of Fcr for elastic columns equals 0.877 Fe, while Fcr for inelastic columns equals 0.658 raised to λ^2, both multiplied by Fy.
Table 4-13 for stiffness reduction factor.
Table 4-13 is used to determine the stiffness-reduction factor for inelastic columns, τb.
The Table is divided into the upper Row for the different values of the yield stress Fy.
Check two values of Pu/Ag for Fy = 50 ksi (LRFD Design).
I prepared a check to verify the inelastic stiffness reduction factor τb for Fy = 50 ksi. First, Pu/Ag = 45 ksi; I estimated both the elastic and inelastic critical stresses. The final τb is 0.36, the same Value shown in Table 4-13.
The second point is Pu/Ag = 30 ksi; the final τb equals 0.96, which matches the Value in Table 4-13.
Check two values of Pu/Ag for the yield stress of Fy = 36 ksi (LRFD Design).
I prepared a check to verify the inelastic stiffness reduction factor τb for Fy = 36 ksi. First, Pu/Ag = 35 ksi; I estimated both the elastic and inelastic critical stresses. The final τb equals 0.1029, the same Value included in Table 4-13.
The second point is Pu/Ag = 20 ksi; the final τb equals 0.988, which matches the Value in Table 4-13.
Check one Value of PT/Ag for the yield stress of Fy = 50 ksi (ASD Design).
For a given stress of 48 ksi, we divide by alpha, which is 1.60, and we call it Fcr’, which is equal to 30 ksi. Next, we estimate lambda^2 as equal to(4-alpha*fcr’/Fy)=0.16. Next, we estimate fcr elastic=Fy/(alpha*lambda^2)=50/(1.6*0.16)=195.312 ksi.
Finally, τb equals (30/195.312)=0.154. This Value matches that given in Table 4-13.
On the left side of the Table, Pt/Ag, as in the case of ASD, is Pt = Pd + Pl, and in the case of LRFD, we use Pult/Ag.
This slide shows Table 4-13 for Fy = 50 ksi and ASD Design, where fcr/alpha = 30 ksi, and the tau Value is verified to be 0.154. When tau = ttau0/Fcr = 15 ksi, the elastic stress is also 15 ksi.
In the following Post, we will introduce problem 4-13, which estimates the stiffness reduction factor.
The PDF file used to illustrate this Post and the previous Post is available for viewing or download below.
This links to the first part of the stiffness reduction factor.
The following Post covers solved problem 4-13 for the stiffness reduction factor.
For a good A Beginner’s Guide to the Steel Construction Manual, 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.