# Transfer of moments in beams with internal hinges

We have a beam with 1 internal hinge just like the picture. We know the internal hinge can not transfer moments between AB and BD. We know the 30 kN force makes some moments on BD & AB. If B (internal hinge) can not transfer moments between the two elements, then why does the 30 kN (which is applied on BD) cause moments in AB?

I don't know if i'm clear enough but the 30 kN force makes some moments on AB while we know that no moments can transfer between AB & BD.

Imagine if AB didn't exist, so all we have is BD with the pinned support at D and the force at C. In this case, BD is hypostatic and becomes a mechanism, rotating around D.

Obviously, we know that doesn't happen when AB is there. This tells us that the hinge is supporting BD, apparently generating an upwards vertical force at B.

However, Newton's Third Law tells us that every action causes an equal and opposite reaction. So if BD "feels" an upwards vertical force at B, then AB will "feel" an equal downwards vertical force at B as well.

This "downwards force" at B (as seen by AB) will then cause a bending moment in AB.

For a more visual demonstration, notice that this structure can be replaced by two individual beams AB and BD. BD is a simply-supported beam, and AB is a cantilever with a concentrated load equal to the reaction found in BD's support at B.

Diagram obtained with Ftool, a free 2D frame analysis program.

• i understand what you're saying. but what does it mean that the internal hinge can not transfer the moment between the two beams? doesn't it mean that BD can not apply any moments on AB? – user14854 Feb 8 '18 at 12:55
• @Farhad - BD doesn't apply any moments on AB, but it does apply a shear force, and what this shear force does to AB is completely unrelated to the fact that the hinge can't transfer moment. All the hinge does is mean that the moment at B is 0, not that there is no moment anywhere. – AndyT Feb 8 '18 at 14:32
• @Farhad, a hinge allows a beam to have different rotations to each side of the hinge. Since the rotation isn't resisted around the hinge, no bending moment appears around it. That's why the bending moment at B is zero. That's all that's meant by "cannot transfer moment". – Wasabi Feb 8 '18 at 14:49

The moment at the hinge is sought:

$$M=-R_1\times X_1+W_2\times X_2-W_3\times X_3+W_4\times X4$$

where the sign of the moment is $$+CCW$$:

\begin{align} P_1&=\text{reaction left}\\ P_2&=\text{weight left}\\ P_3&=\text{weight right}\\ P_4&=\text{reaction left} \end{align}

$$X_1$$ through $$X_4$$ are the absolute values of the distances from the hinge.

Moment at the hinge is the sum of the moments about it.