**draw shear force and bending moment diagram of given figure.**

**SIGN CONVENTION **

**STEPS FOR FINDING REACTIONS**

Here, In The Above Figure We Have Two Unknown Reaction Force **RA **And **RC.**

To Solve Two Unknowns We Need Two Equations.

**1.** *Apply First Equilibrium Equation*

**∑FY**( Summation Of Vertical Forces)= **0**

Here, TWO Vertical Reaction Forces Are Acting Upward. They Are **+RA** And **+RC** and One Inclined load is acting downward and the angle inclination is 30 degree.

Divide the inclined load into two components.

Horizontal force= **5cos30** and vertical force= **5sin30**.

Take The Summation Of All Vertical Forces Acting On The Beam (Both Upward And Downward)

In the inclined load case, we have to find Horizontal force (FX) also

Take The Summation Of Horizontal Force Acting On The Beam (Both left And right).

Force acting towards right is taken as positive (+) and force acting towards left is taken as negative (-).

**2.** *Apply Second Equilibrium Equation*

**∑M**(Summation Of Moments)= **0**

Formula Of Moment= Force*Distance.

e.g. ∑MB= +(RA*1)= RA

Here We Have Taken ∑M At B.

Take The Summation Of All Moments Acting On The Beam (Both Clock Wise And Anti clock Wise)

(Better See The Sign Convention For Positive And Negative Moment Or Clockwise And Anticlockwise Moment).

By Solving The Two Equations We Can Calculate Both The Support Reaction.

Once We Got The Values Of Reaction We Can Calculate Shear Force And Bending Moment.

**FINDING THE REACTION**

*Vertical Reaction*

∑FY=0

RA-5SIN30+RC=0

RA+RC= 2.5KN – (1)

∑MA=0

(RA*0)+(5SIN30*1)-(RC*3)=0

RC=5SIN30/3= 0.83KN

FROM EQN (1)

RA+RC=2.5KN

RA=2.5-.83=1.667KN

*Horizontal Reaction*

∑FX=0

RAX-5COS3=0

RAX=4.330KN

*To Calculate The End Shear Force*

Take summation of forces acting on node or member.

In the below calculation L indicates left and R indicates right.

**CALCULATION OF END SHEAR FORCE**

FA=RA=1.667KN

FB=1.667-5SIN30=-0.883KN

RC=1.667-5SIN30+0.883=0

*To Calculate The Mid Shear Force*

Take Summation of forces acting at the mid span of the member.

**CALCULATION OF MID SHEAR FORCE**

F(A-B)= 1.667KN

F(B-C)= 1.667-5SIN30= -0.833KN

*To Calculate The End Bending Moment*

Calculate moments acting at the mid span of the member and add the remaining distance up to which node/member you want to calculate the moment.

**CALCULATION OF END BENDING MOMENTS**

MA=(1.667*0)=0

MB=(1.667*1)= 1.667KNm

MC=(1.667*3)-(5SIN30*2)=0

*To Calculate The Mid Bending Moment*

Calculate moments acting at the mid span of the member and add the remaining distance up to which node/member you want to calculate the moment.

**CALCULATION OF MID BENDING MOMENTS**

M(A-B)= (1.667*0.5)= 0.833KNm

M(B-C)= (1.667*2)-(5SIN30*1)= 0.833KNm

*To Calculate The Axial Force*

Take summation of forces acting on node or member.

**CALCULATION OF Axial Force**

FA= RAX= 4.33KN

FB=4.33-5COS3=0

##### shear force and bending moment diagram manually

##### software output

*shear force*

*bending moment*

*axial force*