Last Updated on July 19, 2025 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. We will consider the elastic behavior of the column first.
G values for pinned columns and Fixed columns.
There is an essential remark if the end of a column is fixed (G=1) or pinned G=10.00, the value of G at that end should not be multiplied by the stiffness reduction factor τb. Therefore, we do not use τb for the hinged or fixed-end column; however, it can be used for all other cases.
Solved problem 4-14-effective length factor.
Let us refer to solved problem 4-14, page 153, from Prof. Segui, where we have a multi-story frame. A rigid, un-braced 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 is provided at each joint by simply connecting bracing in the direction perpendicular to the plane of the frame.
For a global view, X represents the horizontal distance of the multistory frame, Z represents vertical elevation, and there is a 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 is W10x33 at joint A. There are two beams, both with a W12 x 14 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 a long, short, or intermediate column. 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, no need for the stiffness reduction factors τb, estimate τb.
Since τb is stress reduction due to the inelastic behavior of the column.
If the K*L/r value for column <4.71(sqrt(E/Fy), then we proceed in estimating τb for the LRFD and ASD designs.
Estimation of Pult, Pns, and α values.
In such a case for LRFD, we need to evaluate the ultimate load Pult, Pns, and α values for the general equation or through the table.
First, we assume that the column is long, so E 29000 ksi. For the step, calculate Ga and Gb of column AB, and get the K value. In the Y direction, the column is braced so ky=1.0
LRFD part for the effective length factor column AB.
Estimate the GA value for joint A of the column.
For column AB, where A is at the top and the B 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, we have two beams framing joint A with different span lengths, each having a W12x14 section, with a gross area of Ag = 4.16 in², an Ix of 88.60 in⁴, and a 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=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 terms E for column and E for beams will cancel each other; the numerator =Ix/L for 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.
Go to the next post, Solved problem-4-14 part 2.
A very useful external link. A Beginner’s Guide to the Steel Construction Manual, 15th ed.