The calculation is performed by the following methodology:
1. First you need to replace distributed load with a resultant load which is a concentrated force. For a uniformly distributed load resultant load is equivalent to the product of the intensity q of the load on the lenght of a segment L where it operates. Fq = q*L.
2. Then you should denote supports. It is generally accepted to name them by A and B. Simple beam had one pin support along with one roller support.
3. Let's replace supports action to the beam with reactions of the supports.
Support reactions under such a load will only be vertical.
4. Let's compose equations:
MA = 0; MB = 0,
A moment arm or lever arm is the perpendicular distance between the line of action of the force and the center of moments.
5. Checks the solution. Let's compose equilibrium equation for this:
Y = 0,
If it is fulfilled then we found the right supports reactions. And if not then there is a mistake in the decision.
6. Let's build shear forces epure of Qx. To do this we determine values of shear forces in the specific points. Let us remind that the shear force in the cross section is the sum of the projections of all forces stationed only from the left or only from the right of the section under consideration on an axis which is perpendicular to the axis of the element. The force to the left of the section under consideration and pointed up is considered to be positive (with a "plus"), and the force which is pointed down is considered to be negative (with the sign "minus"). It is contrary to the right part of the beam. In the sections corresponding to the points of application of the concentrated forces including the points of application of support reaction we should determine 2 values of the transverse force: slightly left to the considering poind and slightly right to the considering point. Transverce forces in this sections are denoted as Qleft and Qright. The obtained values of transverse forces in the characteristic points are deposited in a scale from zero line. These values are connected by straight lines according to the following rules: a) if a part of the beam doesnt suffer from a distributed load then the value of this part of transverse forces have to be connected by a straight parallel line which is parallel to the zero line. b) if a part of the beam suffer from a distributed load then the values of transverse forces below that part of the beam have to be connected by a line which is inclined to the zero line. It can cross or not cross the zero line. We will obtain a grapth of the transverse forces along the lenght of the beam when we combine all the values of the transverse forces using rules which is mentioned above. Such graph is called Epures Qx.
7. Let's build a bending moment of Mx graph. To do this we have to define the bending monents in the specific sections. The bending moment in this section is equal to the sum of all the torques (distributed, centered, including supporting reactions and external concentrated moments) located only to the left or only to the right of this section. If any of these force applications tends to turn the left side of the beam in clockwise direction then it is considered positive (with a "plus") and if it is against a clockwise direction then it is considered to be negative (with the sign "minus"). Vice versa for the right side. In the sections whish corresponds to the points of application of concentrated moments it isnecessary to determine 2 values of bending moment: a little to the left of the point under consideration and a little to the right of it. Bending monents at these points denoted as Mleft and Mright. At the points of applications of forces one value of the bending monent is determined. The obtained values are postponed in a scale from a zero line. These values are combined in accordance with the following rules: a) if the section of the beam doesnt suffer from a distributed load then under this section of the beam two neighboring values of the bending moments are connected by a straight line; b) if the section of the beam suffer from distributive load then the value of bending moments under this section of the beam should to be connected be as a parabola.