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Draw shear force and bending moment diagrams for the given simply supported beam.

Draw a free body diagram converting the UDL into a point load.

As UDL is givens to be 5kN and it is acting on a length of 2m .

Point load will act on the center of the UDL at 1 m towards the left from point "B" and shall be calulated as : 5 x 2 = 10 kN.

Now, Lets caluclate the resulting reactions on this beam that is R_{A} and R_{B}.

Condition for equilibrium is Anticlockwise moments = Clockwide moments

Taking moments about "A"(moment = Force x Perpendicular Distance

R_{B} x 5 = ( 10 x 1) + ( 10 x 4)

R_{B} x 5 = 50

**R _{B} = 50/5 = 10 kN.**

Summation of upward force = Summation of downward force

R_{A}+ R_{B} = 10 + 10

R_{A}+10 = 20

R_{A} = 20 -10

**R _{A} = 10 kN**

Shear force calculations:

Shear force is the algebraic sum of unbalanced vertical forces to left or right of the section.

So, if we start calculating from left side then upward force will be positive and downward force will be negative and if we start calculating from the right side of the section downward force will be positive and downwards force will be negative.

In this case we will start calculating from left side.

Shear force @ A = + 10

Shear force @ C = +10-10 = 0

Shear force @ D = 0

Shear force @ B = 0 -10 = -10

**Note: Shear force on UDL shall be represented as a inclined line.**

Bending moment calculations:

Bending moment @ A = Ma=0

Bending moment @ B = Mb= 0

Bending moment @ C = Mc= ( 10 x 1)= 10 kN.m

Bending moment @ D = Md = (10 x 3) – (10 x 2) = 30-20 = 10 kN.m

**Note: Bending moment on UDL shall be represented as a parabolic curve.**

**Hence,** **"Shear force"** and **"Bending moment"** diagram's are drawn for the given simply supported beam.