Last Updated on March 5, 2026 by Maged kamel
Modification to the alignment chart for an un-braced frame.
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The Nomograph equation for the unbraced frame is based on that assumption. This is a portal Unbraced frame. If the far end is considered the near end, with the left joint as the near end and the right joint as the far end, from the first sketch, the girder has a different slope than in the second case.
The slope in the second sketch is due to the hinge at the right joint.
We will start tracing the modification to the un-braced frame for the three end-support cases. The three cases are shown in the image on the next slide. The bending stiffeness=EI/L for which we are going to derive the expression.
For m, its value is obtained by dividing the second case bending stiffness by the original bending stiffness, as in case 1, which is = 6EI/L.
The m value for far-end hinge bending stiffness (3 EI/L) / (6 EI/L )=1/2; this value is to be multiplied by the bending stiffness for the girder.
Let us review each case individually. In the first case, where we have double curvature, the m values are equal at both ends. From the conjugate beam analysis, m is drawn at the tail of the moment. We can obtain the slope at each joint and consider two triangles drawn at each joint.
From M to zero, when the moment acts at the near joint, and from zero to M, when the moment acts at the far end.
The reaction at the moment is the slope value; we react = the area of the triangle, area = 1/2M*L, where L is the span distance; the direction is upwards.
For the slope at the left joint, the reaction of the moment, coming downwards, to get the slope at the near end, take the moment at the far joint, the distance between forces=L/3. Here is the sketch. We have 1/2*ML/EI acting upward and 1/2*ML/EI acting downward. The slope at the near end, A, is M*L/6EI, as shown.
The slope at the far end is B = M*L/6EI, coming downwards = M*L/6EI. The slope is downwards at the near end, and the slope is upward for the far end.
This is the first case of modifying an unbraced frame with the end support connected. This is the first case, as shown in the elastic curve, the bending stiffness k is the stiffness when multiplied by the slope.
Modification to the alignment chart when the far end is pinned for the unbraced frames.
Moment value K =M/α, α=M*L/6EI, M goes with M, K=6EI/L L. This is the normal condition. For the reversed curvature, let’s examine the second case. where K=3 EI/L, we have a far end that is pinned, the near end is acted upon by Moment M, so we have M at support A, zero moments at B. Then the slope α A=the reaction at A, which is the moment of 1/2 /EI multiplied by the arm distance/ span.
The reaction at Support A is (1/2M/EI)*(2/3)L, α-A = ML/3EI; this is the value of the slope near joint A. This is the second case of modifying an unbraced frame with hinged end support. The slope at the far support is α-B=ML/6EI. This is the sway of the portal frame.
The slope here is different than the far end, hence no double curvature as before in case 1, then K the bending stiffness for case 2, if we let M= M2= M 2/αA=M /( ML/3EI) =M /( ML/3EI), k=3EI/L, k=3EI/L, this is the value of k for the second case, where far end is pinned, then m=1/2.
Modification to the alignment chart when the far end is fixed for the unbraced frames.
The third case occurs when we have fixed support at the far end. as if we have a superposition of two instances, The moment at joint A=M, and again moment at support b=M αA= (the area of the moment)*arm distance/span, Area=1/2*MLl/EI /EI *the arm distance =2/3*L, αA=(M L/3EI), acting up, for the second slope at B, αb=(M L/6EI).
For the other part, the area of the triangle is 1/2 ML/EI, and the arm length is L/3.
This is the third case of modification to an unbraced frame with hinged fixed-end support. The remaining distance to A is L/3. The moment value at B is ML/4EI * L/3 = ML^2/12, acting clockwise. The slope is ML/12EI going down. Adding the two cases together, the total slope is ML/12EI.
αA=(ML/3EI)-(ML/12 EI), α-A=(ML/4EI). The k value is M/αA; thus, K = 4EI/L. This is case number 3, m = 4/6 = 2/3. Next time, we will examine problem 4-16 from Prof. Alan Williams‘s book, Structural Engineering Reference Manual.
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