Last Updated on September 21, 2026 by Maged kamel
Modification to alignment Chart for the Braced Frame.
We will start a new subject for the braced frame. The first point to consider is the Alignment chart or the Nomograph detailing.
We will review this subject quickly. The second point is adjusting columns with different end conditions and how to get the M Value for girder end conditions in braced frames.
The third point is the solved example of 7-2, from Prof. McCormac’s handbook, which includes both braced and unbraced frames.
The fourth point is the French equation for braced frames, which gives the k Value for braced frames and is approximately Close to the k Value from the Nomograph.
The points considered in developing the Nomograph are that the columns must be in the elastic region, following the Euler equation, where Pcr = π^2 EI/(KL)^2, when (kl/r) > 4.71*sqrt(E/fy), as explained earlier.
Point 2: All members have a constant cross-sectional Area.
Point 3: All joints are rigid, for point 4.
For columns, in an inside-sway-inhibited frame. i.e., braced frames, rotations at opposite ends of the restraint beams or girders are equal in magnitude and opposite in direction.
The moments are equal but opposite, producing single-curvature bending. The bending Moment is equal and opposite in direction.
Point 5- This point is for the uninhibited frame, which we have included in our previous videos. the stiffness parameters L*sqrt(P/EI) of all columns are equal.
For point 6: All columns have the same stiffness parameters L, P, E, and I.
For point 7, joint restraints are distributed to the columns above and below the joint in proportion to EI/L for the two columns. We use EI/L for the two columns because we sum the columns above and below the joint.
Therefore, the proportion is determined by the dimensions and properties.
For point 8:: All columns buckle simultaneously, andBucklingg in each column is independent of the others.
For point 9: No significant axial compression force exists in the beams or girders, which means girders have no axial force.
Side sway-inhibited frame chart.
This is the alignment chart for sideways inhibited bracing, with k-values ranging from 0.5 to 1.
The equation is rather difficult to memorize. This is curvature due to bracing, which is a single curvature, like a wave rotating in that direction.
As for girders, it’s like a wave: unbraced behavior is double curvature, as if we have a joint in the middle that changes the Moment sign from positive to negative. The G Value for a pinned support = 10, and the G Value for a fixed support = 1.
I’ve included the Link to the Alignment charts from the same site. The site contains valuable lectures for compression members and other subjects.
For the alignment chart modification, Case 1 is the ideal condition. The far joint has an equal Moment and opposite sign. If it is considered positive at the near end, then it is also positive at the far end.
Case no. 1 in the alignment chart corresponds to m = 1. G, as estimated at the joint, equals the sum (EI/L) for columns/sum (m*EI/L).
For girders at the same joint, whether on the right or left of the joint, or, in the case of single-girder framing, to the joint, the modification m is based on the end condition.
The Value of m=1 is for an ideal condition, but if we have a hinge at the far end, then the Moment at the far end=0, unlike the first case, when we have a moMomentt the far end and a Moment at the near end, but the moment=0 at the far end.
We will consider case No. 1 for single-curvature bending. We have two bending moments, one at the left joint and the other at the right joint. One Moment is clockwise, and the other is anticlockwise.
For the slope Value at joint A, which is the joint connecting the column with the girder near the end, the aAreaof the Moment, which is a rectangular Area, is ML divided by EI. Take the direction downwards, based on the technique we use when the Moment loads act downward. Take half the Value for the left joint; then we have a slope of ML/2EI.
This is the angle of rotation, and then this is the rotation of the column as shown in the sketch and back to the hinge. The formula for the stiffness coefficient K is M = K*α, where α at A = ML/2EL at the near joint; then k = M/α goes with m. 2EI will be at the top K=2EI/L.m= new situation/ ideal case, m=2/2 =1 since this is the perfect case.
Modification to the alignment chart for a side-sway-inhibited frame with the far end pinned.
We continue discussing the modification to the alignment chart for the braced frame; next is case number #2. For case no. 2, we introduced a pin at the far-end joint. Then, at joint A, the Moment is M; the Area of the Moment diagram is EI = 1/2 ML/EI; the slope at A is 2/3 of the Area; A is near the CG of the Load, at a distance of L/3; slope = ML/3EI.
The slope we use to get the Value for the bending-stiffness k coefficient: k2 = M/α, α = ML/3EI; then at the end = 3EI/L. m = new situation Value/the ideal case = 3/2 = 1.5.
Modification to the alignment chart: The case of the side-sway-inhibited frame with the far end fixed.
The last modification to the alignment chart is the fixation at the far-end joint. The fixation creates a bending Moment M. We have a Moment at the far end of the fixed joint; the slope is 0, and for positive M at the near end, there will be M/2 with a negative sign at the fixed support.
We have a superposition. We have a Triangle with M at the left and M=0 at the far joint. The slope here for joint (2/3)*(ML/2EI)at the other Triangle is negative. We have a negative M/2 Moment. The Area represents an upward force that produces a downward reaction. The CG distance at A is 2/3L, and the Area of the Triangle is (1/4) ML^2/EI.
The slope at B is (2/3)*ML/4EI, the slope at α b ML/6EI, while the slope at A=1/3 ML/4EI, αA=ML/12EI, the final α at A=ML/3EI-(ML/12 EI), making the denominator 12 EI, (4-1)ML=3ML/12. EI=1/4 ML/EI. Divide M by α to get the bending stiffness, k = 4EI/L; m3 Value = 4/2 = 2; and m = 2.
AISC modifications for side-sway-inhibited and uninhibited frames.
Refer to the AISC commentary. Adjust the column for different end conditions for alignment charts.
First, for girders with different end conditions, if rotation at the far end is prevented, multiply by 2; this is the case for our final case: fixation at the far end multiplied by 2.
b—If the far end of the girder is pinned, then multiply the girder’s (EI/L) by 1.5, which is case no.2, where m=1.5, for side-sway un-inhabited the figures as shown at the end of our subject.
You can view or download the PDF file used for this Post from the following document.
The Core Teaching Aids for Structural Steel Design Courses include full data, which you can download from the AISC Link. Check the folder for compression members.
This is the next Post:: solved problem 7-2 for k Value for frame 1/2.
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.