# Simply supported beam

Simply supported beam on the left side with articulated joint and right articulated sliding. The total length of the beam is 11 meters. Two concentrated forces work on it.

We see a beam that consists of only one shield. It rests on two articulated supports. Let’s see in the figure below what the reactions that work on supports look like.

Let’s start by calculating the horizontal reaction of Ha. In order to know the value of this support force, it is necessary to determine all forces acting horizontally on the horizontal axis, so we must make a sum of force projections on the X axis. In the mathematical notation it looks like this: ΣX (sum of projections of forces on the X axis) – this is a contractual notation.

That the structure does not collapse Σ X (the sum of projections of forces on the X axis) must be zero (must be in balance). Let’s see.

$\Sigma&space;_{x}=0$
$H_{a}=0$

In this case, no horizontal force is active, therefore the value of the horizontal reaction Ha is equal to O. We already have one reaction. There are two vertical left Va and right Rb.

In this case, the sum of the force projections must also equal zero on the Y axis, so now ΣY.

$\Sigma&space;_{y}=0$
$V_{a}+3[kN]-5[kN]+R_{b}=0$, sort the unknowns, leave on the left, move the known values to the right.$V_{a}+R_{b}=-3[kN]+5[kN]$

We have two unknowns, in which case the sum of moments to the selected point a (support) or b (support) should be performed. I choose point b, so ΣMb = O.

$\Sigma&space;M_{b}=0$
$V_{a}*11[m]+3[kN]*9[m]-5[kN]*6[m]=0$
$11*V_{a}+27[kN]-30[kN]=0$
$11V_{a}-3[kN]=0$
$11V_{a}=3/:11$
$V_{a}=0,272kN$

$V_{a}=0,272kN$ – and this way we got rid of the first unknown and know the value of the support Va.

Now we can go back.

$V_{a}+R_{b}=-3[kN]+5[kN]$, but instead of the unknown Va we substitute the reaction value calculated above.
$0,272+R_{b}=-3[kN]+5[kN]$, we order:
$R_{b}=-3[kN]+5[kN]-0,272[kN]$
$R_{b}=1,723kN$

And that’s it! We have just calculated the support reactions of simply supported beam, which in this case are.

$H_{a}=0kN$
$V_{a}=0,272kN$
$R_{b}=1,723kN$

We will now check if our calculations are correct.

The sum of the projections of forces on the X axis must be zero.

$\Sigma&space;_{x}=0$
$H_{a}=0$
$0=0&space;-&space;ok!$

The sum of the projections of forces on the Y axis must be zero.

$V_{a}+3[kN]-5[kN]+R_{b}=0$
$0,272[kN]+3[kN]-5[kN]+1,728[kN]=0$
$0=0&space;-&space;ok!$