Last Updated on September 20, 2026 by Maged kamel
Solved problem 7-2 for k Value for columns in Frame 1/3.
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We will discuss how to determine the k factors for each of the Frame’s columns, as shown in Fig. 7-6, for the solved problem 7-2, which provides the K Value for the Frame. Here are the W sections tentatively selected for each frame member, along with their I/L values, as shown in the figure.
The girder data shown are for the upper girder, W18x50.
The second girder is W24x76, but between brackets, the Inertia Ix Value is calculated and then divided by the length for each member, whether columns or girders, to save Time instead of looking up the tables to estimate Inertia for each W section.
Please calculate the Inertia for each member, then divide by the length of the two braced floors.
We have two braced floors; movement was restrained at joints B, C, F, G, I, and J. The unbraced part is only on the upper floor. We want to check.
Estimate the joint rotations and translations in the Frame.
Let us determine the number of translations and rotations for all joints in solved problem 7-2 for the K Value for frame 1/2. What matters is the translation between the two joints, using the Table for the k Value, which represents the sway between them. Let us evaluate the number of joints: we have 10.
For this frame member, each joint has two translations and one rotation: horizontal and vertical translations, and one rotation. How much support do we have?
We have two fixed supports and two hinged supports.
A fixed support restrains both horizontal and vertical translations and one rotation.
So we deduct (2*3) = 6. For the hinges, vertical and horizontal movements are prevented, and one rotation is allowed, so we deduct (2*2) = 4.
For each member, one translation is taken, so we have 5 girders, and we deducted five horizontal translations.
We have six columns, so we deduct six vertical translations. In total, we have 10 joints × 3 = 30 translations and rotations. Let us consider each joint individually.
At E, we have three in total: two translations and one rotation, taken by the fixed support at E.
We have three total at A: two translations and one rotation, with two provided by the hinged support.
We are left with one rotation. Let us inspect point B. We have three total: two translations and one rotation. The two members at B provide two translations, leaving one rotation.
Let us check joint F; we have three total: two translations and one rotation. After removing the two members at F, we are left with one rotation.
We have the fixed joint I, along with all translations and rotations. We have a joint C, and we have three total: two translations and one rotation.
The two members take two translations, so we have one rotation. The same is true at joint G, where we have three total: two translations and one rotation. The two members take two translations, so we have one rotation.
At joint J, we have three total: two translations and one rotation. The hinge requires two translations and one rotation.
Let us proceed to joint D. We have three in total: two translations and one rotation. The two translations come from D and H to A, leaving I
Proceed to joint H. We have three total: two translations and one rotation. Member HG takes one translation. Member GH is already used by joint D, so joint H has sway and rotation.
Let us count the number of rotations for joints A, B, C, F, G, D, J, and H; we have 9 in total, with 8 rotations and 1 translation at H.
We have an unbraced frame for two columns. The remaining columns (B, C, F, G, and J) have no translations. The calculations are shown in the previous slide image.
In the next Post (Part 2), we will continue our discussion of solved problem 7-2, which addresses how to obtain the K-value for the columns of the given Frame.
If a frame has no Shear walls or bracing, it is generally unbraced. But in that example, no data is given.
The PDF file used to illustrate this Post and the following Post is available for viewing or download below.
This is the next Post. Solved problem 7-2 for Frames Part 2.
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.