In the given diagram ACB we just need to draw a free body. Here, x, y and P and the angle are known. In addition, the weight of the bar is considered as negligible and thus it is neglected.

If we make it a perfect diagram with free body, then it would be just like the following

What are the unknown numbers here? These are 3 numbers as R_{By,} R_{AX} and R_{Ay.}

Now, the number of equations available are –

**Σ** M = 0

**Σ** fy = 0

**Σ** fx = 0

These are also aviable in 3.

It means the number of unknowns = Number of equations available.

These kind of problems are known as statically determinate problems. The exact problem is prvded in the first figure.

To make the solution ins a proper way the following two diagrams are important to understand –

Here the number of known quantities is 4 as P, x, θ and y. The weight is neglected.

Now, the number of unknowns are R_{Ax}, R_{By,} R_{Ay}‘, R_{Bx} = There are 4 unknowns for this

The number of equations = 3

These 3 equations are – **Σ** fx = 0,** Σ** fy = 0 and **Σ** M = 0

Here, if you compare

Number of unknowns > number of equation availability

This kind of problem is known as statically indeterminate.

Now, **Degree of Indeterminacy**

= Number of unknowns – Number of equations available

= 4 – 3

= 1

It indicates that **Degree of Indeterminacy is 1. **

**It can also be said as degree one of the **statically indeterminate problem.

The following diagram shows some other problem as

Statics equations are available = 3

Number of unknowns = 5 = R_{Bx}, R_{By,} R_{Ax}, R_{Ay}‘and M_{B. }

Now the degree of indeterminacy can easily be evaluated as

Degree of indeterminacy = 5-3 = 2

**Solutions related to the problems **

If this figure is the problem, that you need to solve.

Now, go through carefully and you will get that Unknown numbers = 4

Avilable equations = 3

Degree of determinacy = 4-3 =1

By neglecting the bar’s weight the R_{Ay }addition of vertical reactions will be zero.

Now, consider the give case

**Σ** f_{x} = 0 =>

R_{AX} + P – R_{BX} = 0

R_{AX }– R_{BX} = – P

Here it is important to find out another equation.

Elastic deformation for bar ACB

Two supports = Perfectly rigid (Assumed)

The total deformation for the total length (a + b) = zero.

δ_{1} = -R_{Ax} x a / AE

δ_{2 }= – R_{Bx} x b / AE

δ_{1 }+ δ_{2 }= 0

R_{Ax} x a + R_{Bx} x b = 0

R_{Bx }= – R_{Ax }x a /b

Now, after getting through both equations

R_{Ax} x a / b = p

R_{Ax }( a +b / b) = p

R_{Ax} = pb / a+b

R_{Bx} x a /b = – pa / a +b

It means the exact direction of R _{ax }will be completely opposite to the direction of P.

After understanding this you can easily solve the different problems.