Last Updated on September 21, 2026 by Maged kamel
Modification to the alignment chart for an unbraced 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 now trace the modification to the unbraced frame for the three end-support cases. The image on the next slide shows the three cases. The bending stiffness is EI/L, and we will derive the Expression for it.
For m, we obtain its Value 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 is (3EI/L)/(6EI/L) = 1/2; multiply this by the girder bending stiffness.
Let us review each case individually. In the first case, with 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 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 upward.
For the slope at the left joint, the reaction Moment acts downward; to find the slope at the near end, take the Moment at the far joint; the distance between forces is LL/3 (see the sketch), and 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 (downward), and the slope is downward at the near end and upward at the far end.
This is the first case of modifying an unbraced frame by connecting the end support. As shown in the elastic curve, the bending stiffness k is the stiffness 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, which is the normal condition. For reversed curvature, let’s examine the second case. Let’s say K = 3EI/. On d is pinned. Moment M acts on the near end, so we have M at support A and zero Moment at B. Then the slope αA = the reaction at A, which is the Moment of 1/2EI multiplied by the arm distance/span.
The reaction at Support A is (1/2M/EI)*(2/3)L;; α-A = ML/3EI. This frame has a hinged end support. The slope at the far support is α-BB = MLL/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 the far end is fixed. 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 bearing 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, 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, examine problem 4-16 from Prof. Alan Williams’s book, Structural Engineering Manual.
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