Last Updated on September 19, 2026 by Maged kamel
A Review of portal frames- pin supports at the base.
We stopped at the end of the previous Post while discussing the approximate method for the portal frame. We discussed how to perform a structural analysis, and we provided an example: a frame with two hinged supports acted upon by a force on the left side.
We analyze the portal frame in two frames.
We can analyze the frame by dividing it into two parts: the first has antisymmetric loading, and the second has symmetric loading, resulting in elastic curvature.
We assume roller support at the girder midspan. In the next slide, after adding roller support at the girder midspan, we will examine the support reactions.
If we consider the portal frame height to be h, the lower joint is A, the upper-left corner is b, the right corner is c, and the last support is D.
Estimation of the portal frame reactions.
The roller support distance is at L/2, and this is the continuation of the frame. Looking at the left-side portion, we have P/2 at the left and another Load of P/2 at the right joint, acting to the right; from the sum of forces in the x-direction ∑, Fx=0, ∑Fy=0, ∑and M=0.
We can determine the magnitude of the reactions for the lower-hinged support and also for the roller support at B’, since the hinge at A can take the horizontal force. Since the roller support at B’ can not take that force, Ax=P/2, so ∑Fx=0 is utilized.
The reaction Ax is to the left. Taking the bending Moment at A =0, the Vertical reaction at B’ can be estimated as B’y*L/2=P/2*h, B’y=(P/2)*h/(L/2)=B’y=P*h/L=P*h/L, due to the summation of moments = 0.
From ∑Fy=0, since there are no external forces in the y direction, Ay = P*h/L, acting downwards; at B, force B’y acts upwards. At the right joint, a force of P*h/L acts downwards.
The reaction at d is P*h/L acting upwards, since a force of P/2 acts to the right at the joint. For ∑Fx=0, the joint force Xd = P/2 acts to the left.
We can find the reactions for the first part: the horizontal reaction at A is P/2, and at D it is also P/2. The vertical reaction at A is Ph/L downwards, and the vertical reaction at D is Ph/L upwards. For the other frame, the joint loads are P/2 and P/2.
After summing the two frames, we can obtain the final situation: the normal force diagram, the Shear force diagram, and the Moment diagrams.
How do we draw internal force diagrams for a portal frame with pin supports at the base?
We have a tension force of p*h/L for column AB at joint B. The compression force for member bc will be (P-p/2)=p/2 acting at joint C. Then member DC has a compression force of P*h/L acting at joint D. That was a review of portal frames with two-hinged supports.
For the Shear force at joint A, we have a positive force of P*h/L acting up to the joint. The anticlockwiseP*h/L will create negative Shear in member BC. From D to C, the negative Shear force is P/2 until joint C.
For the bending Moment diagram, we have a Moment at joint B = P/2 h, and at the corner, there is a Moment. At the Midpoint, then continue to joint C; the Value is Ph/2.
The Value m = 0 at joint D reminds us of the double curvature, where we have two equal moments in the same direction. The same result is shown in these diagrams, but I explain in more detail that the bending Moment diagram is drawn on the positive side.
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This is the next Post: Analysis of portal frame-fixed supports at base.
For a good Beginner’s Guide to the Steel Construction Manual, 14th ed. Chapter 7 – Concentrically Loaded Compression Members.
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