Continuous beam

Beam Continuous
It consists of three parts. The total length of the beam is 17 meters.

In this case, there will be a lot more calculations than in previous beams, probably you have already figured it out by drawing. First, I will tell you about this beam and I will graphically present characteristic points, arrowheads for the support response arrows and numbering of individual parts, which we have three in this beam.
Let’s get an idea of dividing the structure into parts, this is done in the case of composite beams. The beginnings and ends of individual parts are characterized by the following elements:

  • Beginning of beam
  • Joint
  • End of beam

Part No. 1 is between point E (beginning of the beam), and point D (joint).
Part No. 2 is between point D (joint), and point B (joint).
Part No. 3 is between point B (joint), and point A (end of the beam).
The drawing will be easier to see. In the picture below I will present characteristic points, arrowheads for the support reaction arrows and the numbering of individual parts.

The figure shows us the designation of individual characteristic elements. There was nothing else to do but calculate the reaction.

The part 1 after disconnecting from the whole looks as follows. Returns of the arrowheads in the support and joint are chosen so that it would be easier for us. With time, you’ll know how to set them up to make counting easier.

Very important thing at beginning!
The concentrated force of 8 kN is ideally located above the joint between part 1 and part 2. In this case, we have to decide which part to assign it to. I will take it into account in the calculation of part 1.

We can never take it twice, I mean in part 1 and part 2!
Let’s start with horizontal strength.

\Sigma _{X}=0
-H_{d}=-15,00kN /*(-1,00)

With this fast way, we have a horizontal reaction that works on the joint in point D.
We continue, now vertical reactions.

\Sigma _{Y}=0
V_{d} + R_{e} = 8kN – we have two unknowns.

We use the bending moment equation to point D.

ΣM_{d} = 0
R_{e} = 0kN

In this case, we are lucky, because the force 8kN and reaction on the joint Vd intersects the beam just in point. D.
We return to the calculations with two unknowns and instead of Re we assign it a value of zero.

V_{d} = 8 kN

We already know all the reactions in part No. 1.

As we can see, the part 2 is between the two joints, but the articulation in point D does not hide any secrets from us, because we calculated his reactions earlier in part 1.
We calculate the reactions in the support and articulation in point B.

Another important thing!
I wrote earlier that the forces 8 kN are not taken into account, and here in the figure there is some force 8kN.
Well, this power is no longer in part number 2, these are just reactions from the joint but turned by 180 ° and this is the important thing that you can’t  forget! Whenever you go to the next part, the reactions in the part must be taken exactly the opposite of what you did in the previous part, which means.
In part No. 1 the vertical reaction Vd was directed upwards, in part No. 2 it is directed downwards. In the part No. 1, the horizontal reaction is directed to the left, while in the part No. 2 is directed to the right.


Because the forces in the joint are always in balance, if you do not know what’s going on you have to go back to the previous lessons!

Let’s start counting.
First horizontal reaction.

\Sigma _{X} = 0
- H_{b} + 15,00kN = 0
-H_{b} = - 15,00kN / * (- 1)
H_{b} = 15,00kN – the value of the horizontal reaction Hb.

Now vertical reactions.

\Sigma _{Y} = 0
-V_{b} + R_{b} - 8kN = 0
-V_{b} + R_{b} = 8,00kN – again two unknowns, we already know what to do.

\Sigma M_{b} = 0
-R_{c} * 4,00m + 8,00kN * 8,00m - 6,00kNm = 0
-4R_{c} + 64,00kNm - 6,00kNm = 0
-4R_{c} = -58,00kNm /* (- 4,00)
R_{c} = 14,50kN

We return to the equation with two unknowns, and instead of an unknown Rc, we give 14,50kN.

-V_{b} + 14,50kN = 8,00kN
-Vb = 8,00kN - 14,50kN
-V_{b} = -6,50 kN /* (-1,00)
V_{b} = 6,50 kN

Let’s go to the last stage. part No. 3.

Earlier, I wrote about taking turns of arrowheads to make it easier for us, in this part we will see, for example, in calculating the horizontal response what I meant.

\Sigma _{X} = 0
H_{a} + 15,00kN = 0

H_{a} = -15,00kN -> See?
The reaction came out negative. Here, it does not make a significant difference to us, because this is the end of calculations, but if such a situation happened to us in the No. 1 part, we would have to take into account the value of 15 kN with this minus to the rest of the calculations. It is slightly tiring and under stress, for example during the exam, it is very easy to lose somewhere this minus during calculations.

\Sigma _{Y} = 0
-V_{a} + 6,50kN = 0
-V_{a} = - 6,50kN / * (- 1,00)
V_{a} = 6,50 kN

Finally, the bending moment.

Here is another important thing!
When calculating the bending moment in a multi-jointed beam, we take into account the reactions and forces along the entire length of the beam, and not only the forces that work in the disc No. 3.
Let’s see…

\Sigma M_{a} = 0
M_{a}-6,00kN-14,50kN*6,00m+8,00kN*10,00m = 0
M_{a}-6,00kNm-87,00kNm+80,00kNm = 0
Ma-13,00kNm = 0

I hope I have explained how reactions in different types of beams are calculated.
Remember that you can always go back to this guide and remember!