## Modification to Alignment Chart for Braced Frame.

### Modification to Alignment Chart for Braced Frame video.

This subject we are going to review quickly, the second point to consider is the adjustment of columns of different end conditions .and how to get the- m value for end conditions for girders of braced frames.

The third point is the solved example of 7-2, from Prof. Mccormac’s handbook, the example includes both braced and unbraced frames.

The fourth point is the French equation for the braced frame, which gives the k value for braced frames, which is approximately Close to the k value by the Nomograph. The video has a subtitle and closed caption in English.

## Full description of the content.

We will start a new subject for the braced frame, the first point to consider is the Alignment chart or the Nomograph detailing.

This subject we are going to review quickly, the second point to consider is the adjustment of columns of different end conditions .and how to get the- m value for end conditions for girders of braced frames.

The third point is the solved example of 7-2, from Prof. Mccormac’s handbook, the example includes both braced and unbraced frames.

The fourth point is the French equation for the braced frame, which gives the k value for braced frames, which is approximately Close to the k value by the Nomograph.

The points which were taken into consideration while developing the Nomograph, the columns must be in the elastic region, which follows the Euler equation, where Pcr=π^2 EI/(KL)^2, when the(kl/r) >4.71*sqrt(E/fy) as was explained earlier.

point 2-all All members have a constant cross-section area.

point 3-all All joints are rigid, for point 4.

For columns, inside-sway inhibited frames (i.e braced frames), rotations, and rotations at opposite ends of the restraint beams or girders are equal in magnitude and opposite in direction.

The moments are equal but in opposite directions, producing single curvature bending, the bending moment is supposed to be 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 stiffness parameters L, P,E, I of all columns are equal.

For point 7- joint restraints are distributed to the columns above and below the joint in proportion to EI/L for the two columns. EI/L for the two columns, because we use summation for the columns above and below the joints.

Therefore, the proportion is according to dimensions and properties.

For point 8- all columns buckle simultaneously buckling for each column is separate from the other column.

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, starting k-value, starts from 0.5 to k=1.

The equation is rather difficult to memorize this is curvature due to bracing, which is single curvature same as a wave rotating in that direction.

As for girders, the same as a wave is also, different from the unbraced behavior, which is a double curvature as if we have a joint in the middle that covers the moment sign from positive to negative. The G value for pinned Support =10 and the G value for fixed Support =1.

From the same site, I have written the link as shown for the Alignment charts. the site contains valuable lectures for compression members and other subjects as well.

For the modification to the alignment chart, the first case is the ideal condition, the far joint has an equal moment and opposite sign. if considered as positive at the near end, then it is also positive at the far end.

Case no.1 for the alignment chart, is for the case of m value =1, G, as estimated at the joint, will be equal to the sum(EI/L) for columns/sum (m*EI/L).

For girders at the same joint, whether at 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 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 at the far end the moment= moment at the near end, there is a moment at the near end, but 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 is at the right joint, one moment is in the clockwise direction the other moment is in the anticlockwise direction.

For the slope value at joint A, which is the joint connecting the column with the girder, near the end, the area of the moment, which is a rectangular area area=ML divided by EI. Take direction as downwards, based on the technique which we use the moment loads acting down. half of the value for the left joint then, we have a slope=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 coefficient, for stiffness K, M=K*α, 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 ideal case.

### Modification to alignment chart, case of a side-sway Inhibited frame with the far end pinned.

We continue to discuss the modification to the alignment chart for the braced frame, next is case number#2. For case no.2, we have introduced a pin at the far end joint, then the moment alone, then at joint A the moment=M, the area of the moment diagram EI=1/*2M*L/EI, the slope at A=2/3 of the area, A is near the CG of the load with a distance=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 alignment chart, case of side-sway Inhibited frame with the far end is fixed.

The last case of modification to the alignment chart is the case of fixation at the far end joint in the alignment charts. We have a bending moment M due to fixation we have a moment at the far joint at the fixed joint, the slope is =0, the for positive M at the near end, there will be M/2 of negative value at the fixed support.

We have a superposition, we have a triangle with M at left and at far joint M=0, the slope here for joint (2/3)*(ML/2EI)at the other triangle, we have a negative M/2 moment, the area is represented by an upward force which gives a downward reaction, the CG distance at A=2/3L the area of the triangle=(1/4)*ML*M/EI.

The slope at B (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 denominator as 12 EI, (4-1)ML=3ML/12. EI=1/4 ML/EI. Divide M/ α to get the bending stiffness, k=4 EI/L, m3 value=4/2=2, and the multiplication m= 2.

### Commentary of AISC regarding for side-sway inhibited and uninhibited frames.

Let us refer to the commentary of the AISC. adjustment of the column with different end conditions for alignment charts.

First, the adjustment for girders with different end conditions a- if rotation at the far end of the girder is prevented, multiply by 2, this is the case of our final case, fixation at the far end multiply by 2.

b- If the far end of the girder is pinned, then multiply (EI/L) of the girder by 1.5, which is case no.2 as shown where m=1.5, for side-sway un-inhabited the figures as shown this the end of our subject.

This is the pdf file used in the illustration of this post.

A very useful external link is **Chapter 7** – Concentrically Loaded Compression,** Members**.

The Core Teaching Aids for Structural Steel Design Courses very use full data that can be downloaded 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.