In this entry:

## Beams with hinge - Examples of problem solutions

Below in the drawing you will find the beam that we will solve together. The beam has one **internal hinge** .

The static determinacy check for such beam is as follows. You can find more about checking static determinacy __here__ .

Since we have an internal hinge at point B, we substitute 1 for J in determining the degree of static determinacy. The result we get is zero, so the beam is statically determinate. In the next step, we will calculate the reaction forces from the equilibrium equations.

The beam diagram, all calculations and internal force diagrams are generated in |

We calculate the reaction in the same way as for simply supported beam. Due to the internal hinge, we write an additional equilibrium equation.

The sum of the moments about point B on the right side is zero. The hinge allows us to write the bending moment equation for only one side of the beam, the right or left. The choice is ours. In this example, it is better to choose the right side.

We have calculated the reactions, let's move on to determining __the internal forces__ in spans.

We proceed exactly the same way as in a simply supported beam. In this example we have three spans. The determination of shear forces and bending moments can be found in the drawing below.

Span - 1

Span - 2

Span - 3

Note that the bending moment at the joint at point B is zero. This is due to the fact that a hinge, by definition, does not transfer a bending moment, so in tasks involving a beam with a hinge there should always be zero bending moment at the hinge. |

Once we have determined the values of the shear forces and bending moments at characteristic points, we can proceed to drawing the diagrams.

And this is where we finish solving the example of a beam with one internal hinge.

## An example solution for a multi-span beam

As another example, we will solve a beam with two internal hinges. The procedure is the same as for a beam with one hinge. The difference will be when you write the equilibrium equations. In this case we will have one more equation.

The static determinacy check for such a beam is as follows.

Since we have an internal hinge at points B and D, so J=2 in determining the degree of static determination. The result we get is zero, so the beam is statically determinate. In the next step, we will calculate the reactions from the equilibrium equations.

Due to the joints, we write two additional equilibrium equations.

Similarly to a beam with one joint, we can choose the side with respect to which we determine the bending moment at the joint. In this example it is better to choose the left side. For both joint B and the joint at point D.

We have calculated the reactions, let's move on to determining __the internal forces__.

In an multispan beam, we proceed exactly the same way as in a simply supported beam. In the example discussed, we have six spans. The determination of shear forces and bending moments can be found in the drawing below.

Note that the bending moment at the joint at point B and at joint D is zero. This is due to the fact that a joint, by definition, does not transfer a bending moment, so in tasks involving a beam with a joint there should always be zero bending moment at the joint. |

Finally, we determine the shear forces and bending moments diagrams.

Thank you for making it to the end 😊

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