MEMORY OF CALCULATION METALLIC STRUCTURE

10.- CALCULATION OF THE BRACED BEAMS

The mission of the brace is to prevent longitudinal displacements of

the structure due to the wind. The bracing of the roof will be made up of
St. Andrew's crosses, which along with the straps and the lintels, will form a beam.
of the lattice whose calculation is essential for the subsequent sizing of the
diagonals. The verticals will be the straps, the lintels serve the function of
The chords and the diagonals are the additional bars that we will have to calculate.
We will calculate the truss beam with the software program AM2 that will give us
the values of the stresses in the different bars; next we will have to
carry out an additional check on the cover belts and the lintels as
they have some additional efforts to those calculated in their specific sections.

10.1 - Assessment of the loads

The wind load will be transmitted by the head of the small columns to the truss.
the wind load on the facade will be multiplied, due to the area of influence of
each small pillar, to obtain the load per unit length.

According to NTE ECV, the total wind load, as well as the pressure at windward and the
suction downwind on the facade will be:
q= 63.83 kg m2
2
p= ⋅ q= 42.55 kg m2
3
1
s= ⋅ q = 22.28 kg m2
3
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The linear load for each small column is detailed below:

LOAD SUP. WIDTH OF
small pillar LONGITUDE LINEAR LOAD
Barlovento INFLUENCE
1 6.3 m 42.55 kg/m2 2.5 m 106.37 kg/m
2 7.3 m 42.55 kg/m2 5m 212.75 kg/m
3 8.3 m 42.55 kg/m2 5m 212.75 kg/m
4 7.3 m 42.55 kg/m2 5m 212.75 kg/m
5 6.3 m 42.55 kg/m2 2.5 m 106.37 kg/m

The pillar will be calculated as a beam embedded at the base and supported on the
head, so the value of the reaction at the support will be:

3 Pillar R REACTIONv
Rv= ⋅ q ⋅ L
8
1 251.30 kg
2 582.40 kg
3 662.18 kg
4 582.40 kg
5 251.30 kg

We will perform the calculation of the truss disregarding those diagonals that
They are going to work on compression, since we will have solid rounds that are not
compression resistant. The calculations will be carried out using the program
computer AM2 that will give us the stress values in the various bars. The
The system of beams and loads to be calculated is represented by the following graph:

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Results obtained:
EFFORTS IN BARS

i j Axil i j Axil i j Axil

1 2 -1.165 1 3 -0.801 1 4 1.215

2 4 0.000 3 4 -0.913 3 5 -1.092
3 6 0.440 4 6 0.801 5 6 -0.662
5 7 -1.092 6 7 0.440 6 8 0.801
7 8 -0.913 7 9 -0.801 8 9 1.215
8 10 0.000 9 10 -1.165

Because we have resolved

the structure in projection
horizontal, only the studs
they have the real tensions,
while the tensions of the α

diagonals and the laces will have

to be projected.

10.2.- Calculation of the diagonal profiles

We will adopt profiles for the diagonals that make up the wind bracing beam.
solid round bars. We will calculate which section resists the stresses already
calculated. For this we choose the value of the maximum axil.

DIAGONAL AXIL ω PROJECTION

d1-4= d8-9 1,215 t 7.51O 1,225 t

d3-6= d6-7 0.440 t 7.51O 0.444 t

Maximum axil N1-4= N8-9 1.225 kg

We will choose a circle with a diameter of 20 mm, with an area of 3.142 cm.2.

1.5⋅1.225
σ *= = 585.19 kg cm2< 2.600 kg cm2
3,140
COMPLETES

10.3.- Checking the belts

Due to the additional effort exerted by the compression caused by the wind
frontal on the belts, we will have to verify the effect of buckling produced on

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These. The efforts on the belts should not be projected due to their
the direction is parallel to the one we have used for the calculation.

AMOUNT AXIL
m1-2= m9-10 -1,165 t
m3-4= m7-8 -0.913 t
m5-6 -0.662 t

We will study the belt that supports the greater axial load. The most unfavorable ones are m.1-2 y
m9-10.
N = 1.165 kg
IPE-120: 13.2 cm2 Wx= 53 cm3 Wy= 8.65 cm3

Resistance check:
* *
N * Mx M
*
σ = + + y ≤σu
A Wx Wy
1.165⋅1.5 32.256 1.650
σ *= + + = 931.74 kg cm2< 2.600 kg cm2
13.2 53 8.65
FULFILL
Buckling check:
* *
N* M x My
*
σ = ⋅ω + + ≤ σu
A Wx Wy
According to NBE EA-95 in section 3.2.4.2, in the buckling of uprights and
diagonals in the plane of the structure,β = 0.8. Therefore, the length of buckling will be:
lk= β ⋅l
Due to the large distance between pillars, the slenderness will also be great and with
the security strap will extend in the plane of the cover. For that reason, we will have
round straps that divide each strap into two equal parts, achieving the
decrease in buckling length.
5,7
= 2.28 m
l k= 0.8⋅
2
The buckling will occur around the axis of least inertia, the Y-Y. Being the plane
of bending the X-X plane and its slenderness:

l k 228
λ y= = = 157,24
I y 1.45

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Forλ = 157.24 according to table 3.2.7 of the NBE EA-95, the coefficientω it will be
ω 4.31
* *
N* M My
σ *= ⋅ω + x + ≤σu
A Wx Wy
1.165⋅1.5 32.256 1.650
σ *= ⋅ 4.31+ + = 1,369.94 kg cm2< 2.600 kg cm2
13.2 53 8.65
BIRTHDAY

10.4 - Beam check

Due to the fact that the chords of the cross-bracing beam coincide with the lintels of
the portals, we will have to make the relevant checks for the stresses of
additional compressions that occur due to the action of the wind. The actions in
the most requested lintels are:
Mz*= 16,0111 t·m
N*5.9146 t
To determine the most requested cord, we will have to project the
actions that we have obtained through the software program, according to the angle that
form the skirt with the horizontal.
CORD AXIL α PROJECTION

c1-3= c7-9 0.801 t 11.31O 0.817 t

c3-5= c5-7 1,092 t 11.31O 1,114 t
c4-6= c6-8 0.801 t 11.31O 0.817 t

The most requested cord is the C3-5whose compression effort will be:
N3-51.092 kg
Considering a variable load such as that of the wind, we obtain:
N*3-5 1.638 kg
Therefore, the total axial effort that the lintel will bear will be the sum of
these:
N* 5,914.6 + 1,638 = 7,552.6 kg
The characteristics of the lintel profile are:
HEB-220: Wx736 cm3 91 cm2

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Resistance testing:

*
N * Mx
σ *= + ≤σu
A Wx
7,552.6
σ *= + = 2.25934 kg cm2< 2.600 kg cm2
91 736
fulfill

Buckling check:
According to NBE EA-95 in section 3.2.4.2, for the buckling of welds
compressed when it occurs in the plane of the structure,β 1. The length of
pandeo will therefore be:
5
l k= β ⋅l= 1⋅ = 5.1m
cos11.31

The buckling will occur around the y-y axis, with the mechanical slenderness being:

l k 510
λ y= = = 91.23
i y5.59
Forλ = 60.55, according to table 3.2.7 of the NBE EA-95, the coefficientω it will be
ω = 1.77
*
N * Mx
σ *= + ≤σu
A Wx
7,552.6 1,601,110
σ *= ⋅1.77+ = 2,322.32 kg cm2< 2.600 kg cm2
91 736
fulfills

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