Last Updated on September 20, 2026 by Maged kamel
Solved problem 4-14-effective length factor-part 1.
An introduction to solved problem 4-14-effective length factor-1. We have a given frame, and we need to find the effective length factor for column AB that has a W10x33; the columns are braced in the Y direction. We will focus our estimate on the X- direction. First, we will consider the column’s elastic behavior.
G values for pinned columns and Fixed columns.
Important remark: if the end of a column is fixed (G=1) or pinned (G=10.00), do not multiply the G value at that end by the stiffness reduction factor τb. Therefore, we do not use τb for hinged- or fixed-end columns; however, we can use it for all other cases.
Solved Problem 4- 14: effective length factor.
Let us refer to solved problem 4-14, page 153, from Prof. Segui, which has a multi-story frame. A rigid, unbraced frame is shown in Figure 4.17; the plane of the Web is in the plane of the paper.
The Web of all beams intersects with the Web of columns.
Both beams and columns have W sections. Lateral support at each joint is provided by connecting bracing perpendicular to the plane of the frame.
For a global view, X represents the horizontal distance of the multi-story frame, Z represents vertical elevation, and there is bracing in the y-direction.
Determine the effective length factor in the x-direction, and determine whether the given column behaves elastically or inelastically.
Assuming a Load is acting at the top of column AB, the given loads are as follows: DL = 35.50 kips, LL = 142.0 kips.
For member AB, the column section at joint A is W10x33. There are two beams, both with a W12x14 section. One Beam has a length of 20′, while the other has a length of 18 ft. For columns, the length of each column is 12′.
So far, we do not know whether column AB is long, short, or intermediate. This depends on whether its KL/r is > or <4.71(sqrt(E/Fy)). Recall that if the k*L/r Value for the column is > 4.71(sqrt(E/Fy)), the column is long; in that case, there is no need for the stiffness reduction factor τb; estimate τb.
τb is the stress reduction due to the column’s inelastic behavior.
If the K*L/r Value for the column is < 4.71 (sqrt(E/Fy), then we proceed with estimating τb for the LRFD and ASD designs.
Estimation of Pult, Pns, and α values.
In this case, for LRFD, we need to evaluate the ultimate Load Pult, Pns, and α values using the general equation or a Table.
First, we assume that the column is long, so E 29000 ksi. For this step, calculate Ga and Gb for column AB and obtain the K Value. In the Y direction, the column is braced, so ky=1.0
LRFD part for the effective length factor for column AB.
Estimate the GA Value for joint A of the column.
For column AB, A is at the top, and the joint is at the bottom.
The section of column AB is a W10x33.
From the Excel sheet for W-section data, we obtain the gross Area, Ag = 9.71 in², the Moment of Inertia about the central axis, Ix = 171.0 in⁴, and the radius of Gyration about the x-axis, rx = 4.19 in².
At joint A, two beams frame the joint with different span lengths; each has a W12x14 section, a gross Area of Ag = 4.16 in², an Ix of 88.60 in⁴, and an rx of 4.62 in².
The length of the first Beam is L = 18′ and the second Beam is L = 20′.
The height of column AB is 12 feet, considering E is elastic for the column, and beams are always considered elastic.
GA Value equals ∑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 (171/12) = 14.25.
The denominator equals ∑EI/L for beams, which will be 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 88.60 inch 4.
The GA Value equals E*171/12/(3366.8/360)/9.3522=1.5237. Please refer to the next slider image for more details.
We will continue with the solved problem as part 2 in the next Post.
The PDF file used to illustrate this Post is available for viewing or download below.
Go to the next Post, Solved problem-4-14 part 2.
For a good A Beginner’s Guide to the Steel Construction Manual, 14th ed. Chapter 7 – Concentrically Loaded Compression Members.
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For a good A Beginner’s Guide to the Steel Construction Manual, 16th ed. Chapter 7 – Concentrically Loaded Compression Members.