brief illustration for post 17-comp

17-Alignment chart-part 2 for the unbraced frames.

Spread the love

 Alignment chart-part 2 for the unbraced frames.

Full description of the subject.

We continue discussing the alignment chart-part 2 with conditions,  of anti-symmetrical curvature for the unbraced frame.

The Nomograph was based on the double curvature assumptions. We have a bending moment at one end, but at the other end, there is another moment, equal in magnitude, that will cause anti-symmetrical curvature.

We are going to discuss, by God’s will, the portal frame approximate method as the third point.
 How to estimate the bending moment for the portal frame by assuming places of hinges? at the mid-span of girders and the condition of the portal frame, when having fixed supports.

The fourth point, is the m value for end conditions for girders of the un-braced frame. if the condition of non-antisymmetric curvature is not achieved, or the assumptions, on which the graph was developed were not followed.

There is m factor which is determined based on the condition of the girder, What will be it’s the effect of the far end whether, fixed or hinged on girder stiffness, by multiplying by the m value?

The fifth point is the solved example, 4-5 from Prof Alan Williams’s book.
The 6th point, by God’s will, is the French equation for the K value from the nomograph.

Content of the lecture.

Normally, we get Ga and  Gb and use the nomograph to get the k value, which is sometimes not clear. By using the french equation, we can get the approximate value of K near to k value obtained from the nomograph.

From the second slide, there are conditions of assumptions from which the nomograph was developed in the first item, the behavior is purely elastic, and in the second term, all members have a constant cross-section. The third item, all joints are rigid.

The fourth item, for columns in side-sway, inhibited frames, braced frames, and rotations at opposite ends of the restraint beams or girders are equal in magnitude and opposite in direction, producing single curvature bending.

Alignment chart-part 2-general discussion.

For case, number 5 Alignment chart-part 2, for the side-sway uninhibited frames, rotations at opposite ends of the restraining beams or girders are equal in magnitude and direction, producing double or reverse curvature bending.

As if the bending moments at the near and far end are rotating in the same direction but equal in magnitude rotations are opposite as we are going to see later. No.6 stiffness parameter  L*sqrt(P/EI) of all columns are equal.

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 for the columns.

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 no significant axial compression force exists in the beam or girder, there is a p delta effect, the beam as if having compression force and moment and this reminds us of the eccentricities whether big or small which we have studied earlier.

Assumptions for the Alignment chart part 2.

For the Nomograph for the unbraced frame, Alignment chart-part 2, Ga where a is the upper point Gb, for joint b the lower point for the column, for instance, column AB if we consider the upper point as A, and the lower point as B after the determining of the summations, we mark g value on the graph at the left.

On the other side at the right as we can see in the small sketch the summation is made for the columns passing by the joint at the top and bottom for joint A, estimate sum(EI/L) for columns/ the sum of (EI/L) for girders passing by the joint at the right and the left.

For joint B, the same procedure will be done for column AB, if have an extension as  BC for the alignment chart-part 2. For example, we repeat the same, sum(EI/L) for columns / the sum of (EI/L) for girders for fixed support G value. The code considers G=1 for fixed support, while for hinged support, G= 10.

The graph for the alignment chart for unbraced columns.

For the portal frame, Alignment chart-part 2, if we act upon by a horizontal load P, while the height of the frame =H, for the case of two fixed support, the behavior as if for elastic curvature as we can see, at the middle, at the middle of the girder we consider as if we have a hinge at the mid of girder and also as if we have a hinge at the mid-height of the column. This will produce a moment diagram in which this shape moment will be reversed at the corner.

We have moment due to the rigid joint the upper moment and lower moment are of the same values, and reversed shape will occur, same for the girder as if there is a hinge so we have reversed shape at the corner the moment direction is outside the right joint and due to a hinge again.

.At the mid-height, the moment is reversed. 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 shape of the bending moment will differ as you can see.

Analysis of portal frames.

If the end condition of the unbraced frame or the braced frame, let us review the G value which is the sum of EI/L for columns/sum EI/L for girders. if the end condition is different we will modify the equation by using the m factor for the girder as we can see in the equation, in the normal situation, the m value =1.

For other conditions include correction factor m to account for the actual rotational stiffness at the joint. for the side sway uninhibited, unbraced frame, the assumption is based on the reverse curvature bending of the girder.
First, we will start with the subject of the portal frame, by God’s will. For which it is acted upon by a horizontal force P and at h distance from the base, and the span of girder=L.

From prof. Devdas Menon structural analysis. The frame can be considered as composed of two parts, for the first part, half of the load will act with anti-symmetrical loading at each joint with P/2 and the remaining P/2 will be considered in the second frame.

The total P=P/2+P/2  for the first frame the P/2 will be opposed by other P/2 acting in the reverse direction.

For the first frame due to antisymmetrical loading no bending moment. But for the second frame, P/2 acts on the left joint and at the right joint, then the joint will be considered at the mid of the girder the elastic curvature coming to the right and outside the frame due to side-sway, consider the hinge is roller support since a horizontal movement will occur, as he assumed.

m factor for correction for G a value at the joints.

The example can be resolved as roller support we will continue next time.

Analysis of portal frame how to determine the reaction values?

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

This is the next post, A review of the portal frame.

This is a link to A very useful external link Chapter 7 – Concentrically Loaded Compression Members.

Scroll to Top