Last Updated on September 19, 2026 by Maged kamel
Alignment chart-part 2 for the unbraced frames.
Full description of the subject.
We continue discussing the alignment chart, part 2, with the conditions for anti-symmetrical curvature of the unbraced frame.
The Nomograph was based on the assumption of double curvature. At one end, we have a bending Moment, but at the other end, another Moment, equal in magnitude, will cause anti-symmetrical curvature.
By God’s will, we will discuss the approximate method for portal frames as the third point.
How can the bending Moment for the portal frame be estimated by assuming hinges at the mid-span of the girders and fixed supports at the portal frame?
The fourth point is the m Value for the end conditions of the girders in the unbraced frame. If the condition of non-antisymmetric curvature is not achieved, or if the assumptions on which the graph was developed were not followed.
There is an m factor that is determined based on the condition of the girder; what will be its effect on the far end, whether fixed or hinged, on girder stiffness, by multiplying by the m Value?
The fifth point is solved using example 4-5 from Prof. Alan Williams’s book.
The sixth point is the French equation for the K Value from the Nomograph, by God’s will.
Usually, we get Ga and Gb and use the Nomograph to determine the K Value, which can be unclear at times. Using the French equation, we can obtain an approximate Value of K that is close to the Value reported in the monograph.
Assumptions used in the Nomograph.
The second slide shows the conditions and assumptions on which the Nomograph was developed, as quoted from the book “Structural Steel Design”.
The first item assumes purely elastic behavior, and the second item assumes all members have a constant cross-section. In the third item, all joints are rigid. The AISC includes the same assumptions and explains them in its commentary.
The fourth item covers columns in side-sway, inhibited frames, braced frames, and rotations at opposite ends of restraint beams or girders, where the rotations are equal in magnitude and opposite in direction, producing single-curvature bending. On the next slide, we can see a sample of a single-curvature column shape.
For case number 5, Alignment chart-part 2, for the side-sway uninhibited frames (unbraced), rotations at opposite ends of the restraining beams or girders are equal in magnitude and direction, producing double- or reverse-curvature bending.
It is as if the bending moments at the near and far ends rotate in the same direction but have equal magnitudes, and the rotations are opposite, as we will see later. No.6 stiffness parameter sqrt(P/EI) oiis equal for all columns
No.7 joint restraint is distributed to the column above and below the joint in proportion to the EI/L for the two columns, since we use the summation.
At the common joint, G is estimated as the sum of EI/l for the columns above and below, divided by the sum of EI/L of girders meeting at the same joint.
All columns buckle simultaneously, and the Beam or girder has no significant axial compression force. There is a p-delta effect, in which the Beam acts as if it had a compressive force and Moment, reminding us of the eccentricities, whether large or small, that we studied earlier.
I have added a sketch showing the difference between single- and double-curvature columns. It is clear that when a column sways, a double curvature forms, as shown in the last sketch on the right.
For the unbraced frame nomograph, Alignment chart, part 2.
A- Points A and B represent the two ends of the A column.
B—For a fixed column, the G Value that theoretically approaches zero will be considered equal to 1.
C. For a hinged column, the G Value theoretically approaching infinity will be considered equal to 10.
We start with one joint and determine the sum of (EI/L) for columns divided by the sum of (EI/L) for girders meeting at that joint; this is G, which we mark on the graph on the left.
For the other joints in this same column, we determine the sum of (EI/L) for columns divided by the sum of (EI/L) for girders meeting at that joint; this is the Value of G, which we mark on the graph on the right. We intersect the middle graph with the required Value of k.
The following slide shows the equation for G, along with the various terms from the AISC Commentary.
For other conditions, include a correction factor m to account for the joint’s rotational stiffness. For the side-sway-uninhibited, unbraced frame, the assumption is based on the girder’s reverse curvature bending.
Points of contraflexure
Prof. Reddy has pointed out that laterally loaded portal frames can be analyzed using the approximations employed, as shown in the figures. For stiff girders, the point of contraflexure, or zero Moment, can be assumed at mid-height, whereas for flexible girders, it can be considered near the top.
For the portal frame, Alignment chart-part 2, if we act upon it with a horizontal Load P, while the height of the frame =h, for the case of two fixed supports, the behavior is as if, for elastic curvature, as we can see, at the middle of the girder, we consider as if we have a hinge at the mid of the girder and also as if we have a hinge at the mid-height of the column. This produces a Moment diagram in which the shape Moment reverses at the corner.
In the second case, for a flexible girder, the inflection point is near the upper portion of the column, and this is a contra-flexure point for the girder at the middle; the bending Moment diagram will differ, as you can see.
Analysis of a portal frame with a sidesway.
First, we will start with the portal frame, by God’s will. A horizontal force P acts at a distance h from the base, and the girder span is L.
From Prof. R. C. Hibbeler’s Structural Analysis. The frame can be considered composed of two parts. In the first part, half the Load acts as anti-symmetric loading at each joint (P/2), and the remaining P/2 is considered in the second frame.
The total P = P/2 + P/2 for the first frame. In this frame, the P/2 is opposed by another P/2 acting in the opposite direction; there is no bending Moment due to anti-symmetric loading. However, in the second frame, P/2 acts on the left and right joints.
In the next Post, we will continue solving the same problem with the portal frame.
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This is the next Post, 18: A Review of the Portal Frame.
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.