# Justification of static equilibrium for a cut structure/frame

Axial deformations are ignored. This structure is under the specified force: • Is the internal shear force in column BC equal to that of column DE?
• If we cut the structure at middle of BC and DE columns:
• Internal moment at the middle of DE is zero, right?
• But the moment at the middle of BC is not zero, right?
• So, how it has static equilibrium? I don't get it :( For your first question regarding whether the shear in BC equals the shear in DE: yes, they are the same.

An easy way of seeing this is that some fraction of $$P$$ is going to be absorbed by BC as shear, which will then be transferred to CD as an axial load, only to then become shear in DE again.

As for your main question: you haven't included all the loads acting on that slice of the frame. You've drawn the moments, but forgot about all the other forces acting on that part of the structure. Specifically, the axial and shear forces at the cut points (mid-BC and mid-DE).

If you were to solve this entire structure, you'd find that BC and DE are under axial loads due to the support reactions (ABC under compression, DE under tension). These opposing loads are a force couple which generates a moment equal and opposite to the moment at mid-BC:   In the example above, we have that mid-BC has a bending moment of $$\dfrac{133.2}{2} = 66.65\text{ kNm}$$. However, BC and DE are under 13.33 kN compression/tension, a force couple which generates a moment of $$13.33 \cdot 5 = -66.65\text{ kNm}$$ (negative because it's clockwise), perfectly cancelling out the bending moment at mid-BC.

Diagrams obtained with Ftool, a free 2D frame analysis program

• Thanks =) Just curious, which software application did you use for your simulations? May 14, 2022 at 6:06
• @user3405291: I used Ftool, I added a link to it at the end of my answer. Highly recommend it (disclaimer: I worked developing it a few years ago).
– Wasabi
May 15, 2022 at 3:26
• Looks awesome, thanks :) May 15, 2022 at 3:30