EASY TO UNDERSTAND OF UDL

Udl is short form of Uniform Distribution Load acting on the beam.What is that mean?.UDL is a load that is distributed or spread across or over the whole length of the beam.The SI unit for UDL is KN/m and for example 50 KN/m of U DL that mean 50 vertical load in KN will acting on the whole 1 m of the beam.Let say 20 KN/m UDL acting on the 10 m length of the beam that mean each of 1 m of the beam will carry the 20 KN of vertical load from starting 0 m until to the end of the beam 10 m.

There are 3 type of UDL

1. that whose magnitude  remains uniform or same value of force throughout the length of the beam.Example if 10 KN/m load is acting on a beam which length is 15m,then 10 KN/m is acting throughout the length of of 15 m can be partly or fully length of the beam
2. Non-uniformly distribution load
1. Triangular shape
1. triangular load or shape is that whose magnitude is zero at one end of span and increasing constantly until to the end the 2nd end of the length.Can be partly or fully of the beam
2. Trapezoidal shape
1. trapezoidal load or shape is that which is acting on the span length in the form of trapezoid shape .It is generally from the combination of UDL and triangular load.Can be partly or fully of the beam.
3. Calculation of the load.
1. For the purpose of calculation all the UDL must convert to point load and conversion of UDL to point load is by simply multiplying the intensity of udl with its loading length.Can be call as Equivalent concentrated load.The point are act at the center of the span.For the triangular shape or triangular UDL load,the center of the load is 2/3 from the starting of the span and for a trapezoidal shape or load the center of load is a resultant distance from the starting of length of the beam.The center of gravity calculation will help or useful to find out the distance of resultant.
2. Example,an UDL load of 10 KN/m acting on the 10 m of simply supported beam AB beam,calculate the shear and bending moment fot the beam.(fully length)
1. step 1-convert the udl to point load
1. 10 k/m x 10 m=100 KN,resultant force distance at mid span
2. Taking moment at  Ma=0,Rb x 10 -100 x(10/2),Rb=500/10=50 KN so Ra=50 KN
3. Max bending moment are at center of the span or 5m,the value is 50 KN x 5m -(10 x 5)(5/2)= 125  KNm.
3. For triangular shape or load,a UDL triangular load starting at A is 0 KN/m until B is 20 KN/m.AB length is 20 m,find the shear force and bending moment and A and B are simply supported beam(fully length)
1. step 1-convert to udl to point load
1. 1/2 x(20 KN/m) x 20 m=200 KN,center of resultant force are 2/3 of the length from A
2. step 2,taking moment Ma=0,Rb x 20 -200 KN x 2/3(20)m=Rb=8000/(3 x 20)=Rb=133.333KN,Ra=200 KN -13.33KN=66.67 KN
3. step 3,max bending moment,are at 13.33 from A,taking moment at center,66.67 KN x 13.33 m -1/2(13.33)(13.33)(1/3*13.33)=493.94 KN/m
4. For trapezoidal shape or load,given beam Ab is simply supported beam,at A is UDL 10 KN/m and at B is 25 KN/m,beam length is 15 m,find out the shear force and bending moment of the beam.(fully length)
1. step 1-convert udl to point load
1. 1/2(10 KN/m +25 KN/m)(15m)=262.5 KN,center of resultant force,calculate using a center of gravity formulae,x= 8.57 m from A
2. step 2,taking moment Ma=0,Rb x 15 m--262.5 KN x 8.57 m,Rb=149.98 KN so Ra=262.5KN - 149.98 KN=112.60 KN
3. step 3,max bending moment,taking moment at resultant force point,383.51 KNm.
5. Note:1.force +ve if upward left of the section and downword if right of the section and - ve if downward left of the section and upward when right of the section
6. Note:1.moment +ve if clockwise on the left of the section and anticlockwise on the right of the section and -ve if anticlockwise on the let of the section and clockwise on the right of the section.
7. Note:1.Shear diagram  + on the top and - on the bottom same with bending moment diagram

The above are the simple exampel for solving the UDL to getting a shear force diagram and bending moment diagram.Thank you