A curved beam AB embedded in a wall is acted upon by

Calculation of the embedding moment on a curved beam

Given: curved beam AB, embedded in a wall, with distributed loads of intensity q1 = 5 N/m and q2 = 3 N/m. Lengths BC = 3 m, AD = 5 m.

Find: closing moment.

Answer:

Let's calculate the ground reaction force at point B:

The sum of the moments of forces acting on the beam relative to point A is equal to zero:

Answer: the closing moment is 26.25 N*m.

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This product is a curved beam AB, which is embedded in the wall. It is subject to distributed loads of intensity q1 = 5 N/m and q2 = 3 N/m. The lengths BC and AD are 3 m and 5 m, respectively.

It is necessary to determine the moment of embedding the beam. To solve the problem, the following formulas and laws are used:

1. Equilibrium law: the sum of all moments of forces acting on the beam must be equal to zero.

2. The formula for the moment of force: M = F * L, where M is the moment of force, F is the force, L is the leverage of the force.

3. Distributed load formula: q = F / L, where q is the intensity of the distributed load, F is the force, L is the length of the section of the beam on which the load acts.

4. Formula for the moment of inertia: I = (b * h^3) / 12, where I is the moment of inertia, b is the width of the beam section, h is the height of the beam section.

After carrying out the necessary calculations, we get the answer to the problem: the moment of embedding the beam is 137.5 N * m.

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