Solution of problem 2.3.6 from the collection of Kepe O.E.

Solution to problem 2.3.6 from the collection of Kepe O..

This digital product is a solution to problem 2.3.6 from the collection of Kepe O.. on statics. The solution was completed by a qualified specialist and presented in the form of a beautifully designed html document.

The task is to determine the reaction of support A to a homogeneous beam AB, 6 meters long, with a distributed load of intensity q = 0.5 kN/m and weight G = 20 kN. The solution is presented as a sequence of logically related steps, each of which is accompanied by detailed explanations and calculations.

By purchasing this product, you receive a ready-made solution to the problem, which can be used as a sample for performing other statics problems. The beautiful design of the html document makes it easy to use and allows you to easily find the necessary information.

Have a nice work!

This digital product is a solution to problem 2.3.6 from the collection of Kepe O.?. according to statics. The task is to determine the reaction of support A to a homogeneous beam AB, 6 meters long, with a distributed load of intensity q = 0.5 kN/m and weight G = 20 kN.

The solution was completed by a qualified specialist and presented in the form of a beautifully designed HTML document, which includes a sequence of logically related steps, each of which is accompanied by a detailed explanation and calculations.

By purchasing this product, you receive a ready-made solution to the problem, which can be used as a sample for performing other statics problems. The beautiful design of the html document makes it easy to use and allows you to easily find the necessary information.

So, the answer to the problem is 10.4 kN. All necessary calculations and explanations are provided in the purchased digital product. Have a nice work!

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Solution to problem 2.3.6 from the collection of Kepe O.?. is associated with determining the reaction of support A of a homogeneous beam AB, which is subject to a distributed load of intensity q = 0.5 kN/m. The lengths AB and AC are 6 meters, and it is also known that the weight of the beam G = 20 kN.

To solve the problem, it is necessary to use the principle of resultant forces, according to which the sum of all external forces acting on a body is equal to zero. In this case, this means that the reaction of support A must compensate for the weight of the beam and the distributed load.

First you need to determine the force with which the distributed load acts on the beam. To do this, you need to multiply the load intensity by the length of the beam: q * AB = 0.5 kN/m * 6 m = 3 kN.

Then, using the principle of resultant forces, we can write the equation for the balance of forces along the vertical axis: RA + PB = G + F, where RA and PB are the reactions of supports A and B, respectively, G is the weight of the beam, and F is the force with which the distributed load.

Substituting the known values, we get: RA + PB = 20 kN + 3 kN = 23 kN.

Since the beam is symmetrical, the reactions of supports A and B are equal to each other: RA = PB = 23 kN / 2 = 11.5 kN.

Finally, to find the reaction of support A, it is necessary to subtract the weight of the beam from it: PA = 11.5 kN - 20 kN = 10.4 kN.

Thus, the reaction of support A is 10.4 kN.

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