Termocasa System® - Modular, earthqake safety, heat insulated residential units. - 4. Calculation Report.

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Modular, earthqake safety, heat insulated residential units.

TECHNICAL REPORT -
4. Calculation Report

VERSIONE ITALIANA

INTRODUCTION

DESCRIPTION

TECHNICAL REPORT

THE FACTORY

COMMERCIAL GUIDELINES

ECONOMIC AND FINANCIAL ISSUES

INFORMATION

|| Table of Contents || Previous Page || Next Page ||

Index:
|| Intro || Needed materials || Analysis of the covering weights || Slab calculation ||
|| Section test || Outer overhang calculation || Transversal saddle frame ||
|| Analysis of the covering vertical loads || Analysis of the loads due to horizontal force ||
|| Max stresses of the structure || Check of the central line section ||
|| Check of the section at the joint || Check of the head section of the pilaster ||
|| Wind action measured at 120 kg/mt.² || Longitudinal frame || Geometric characteristics ||
|| Coefficients of the division of the moments in the crosses ||
|| Coefficients of the division of the moments in the pilasters ||
|| Analysis of the loads for vertical forces || Analysis of the loads for the horizontal forces ||
|| Calculation of the concrete bed foundation || Horizontal thrust check ||
|| Tangential thrust check ||

Introduction

The building we are going to speak of has a reinforced concrete structure and is described in the enclosed tables. it is made of one floor covered with a 20 degree sloping roof. It is a house to live in.

Considering the geometry of the structure, we have considered a longitudinal frame and a transversal one.

In the calculation of the structure we have taken the seismic laws for the first class 25.11.1952 n. 1684 and the D.M. 3.3.1975 as well as the "Norme tecniche" D.M. 26.3.1980 into due consideration.

For each frame we have separately calculated the moments due to the horizontal stresses and those due to the vertical ones.
The final moments are the results of both of them in the worst conditions.
The actions due to the eccentricity between the barycentre of the rigidity and that of the weights have been taken into consideration.
Moreover we have also checked the pressure-bending of the most loaded pilasters.

Needed Materials

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Cement
type 325 - Concrete R'bk 250 kg/mt² - Steel (Feb 38k) improved adhesion.

The concrete was made of:

sand 0.4 m.³

gravel 0.4 m.³

small gravel 0.4 m.³

cement 300 kg./m.³ for pilasters and girder

200 kg./m.³ for basaments

water 150 litres

The following tensions are allowed:

for pilasters girders subject to bending or pressure bending C =60+ ^{R'bk - 150} = 85 kg./cm.²
^⁴

for pilasters with simple pressure C = 0.7 x 85 = 60 kg./cm².
f = 1600 kg./cm.²

tangential tension
bo = 4 + ^{R'bk - 150} = 5.33 kg./cm.²
⁷⁵

ANALYSIS OF THE COVERING WEIGHTS
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Slab weight
0.05 x 2500 = 125 kg./mt.²

Thickness of the inner plaster work: 1.5 cm. 30 = 30 kg./mt.²

Polystyrene weight 20 x 0.4 = 8 kg./mt.²

Corrugated tiles 20 = 20 kg./mt.²

Snow load 60 x 0.925 = 55.5 kg./mt.²

Wind
p= 60 · 0.40 · 1 = ^{24 kg./mt²}
^{264 kg./mt.²}

SLAB CALCULATION
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We have tested a piece of slab - 5 cm. thick and 1 mt. large - whose distance
between the ledges is 2.70 mt.

In = 2.80 b = 100

We assume the conditions of partial joint:

Joint moment
M = ¹ p · 1² = 172 kgmt.
          ^¹²

Central line moment
M = ¹ p · 1² = 172 kgmt.
          ^¹²

r = ^4.2 = 0.3202
       ¹⁷²⁰⁰⁰
¹⁰⁰

for = 1600 m. = 10 we get = 71    t = 0.002179 A = t Mb = 2.8577 cm.²

Use 10 Ø 4 iron bars mer mt. equal to cm.² 3.14.

Section test
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A  = 3.14
b=100
a=5
h = 4.2

X = ^{10 · 3.14} -1 + 1+ ^{2 · 10 · 4.2} = 1.34
          ^{^{12

10 · 3.14}}

i = ^{10 · 1.34³} + 10 · 3.14 (4.2 - 1.34)² · 2 = 337
          ^³

c = ^{17200 · 1.34}   = 68.39      85 kg./cm.²
          ^³³⁷

f = 10 · ¹⁷²⁰⁰ · ( 4.2 - 1.34) = 1459     1600 kg./cm.²
          ^³³⁷

OUTER OVERHANG CALCULATION
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q =264 kg./mt.²

p = 262 x 1 x 0.67 = 176.9 kg./mt.

Let's increase the load of a 40 % considering the overhanging structure

176.9 x 1.40 = 267.60

The outer overhang has the same kind of iron reinforcement used for the slab. So calculation is not needed.

TRANSVERSAL SADDLE FRAME

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1 = 6281 cm. (see enclosed tables)

2 = 53864 cm.

K = ------- - · ^h= 9.2
          ^{^s}

Analysis of the covering vertical loads
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Weight of the covering and accidental loads
264 x 3 = 792 kg.

Weight of the girder
0.20 x 0.20 x 2500 ^{100 kg.}
          ^{^{p¹ 892 kg.}}

p
= 892 · ^3.12= 968 kg./mt.
          ^{^2.875}

Va = Ve = 2783 kg.

H = 438 kg.

Ma = Me = 570 kgmt.

Mb = Md = - 897 kgmt.

Mc = 2578 kgmt.

Analysis of the loads due to horizontal force
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Evaluation of Gi

Weight of the slab 0.05 x 6.66 x 3.00 x 2500   =   2498 kg.

Weight of the girder 0.20 x 0.20 x 6.66 x 2500   =   666 kg.

Weight to the pilasters 2 x 3.25 x 15.29 + 291   =   390 kg.

Weight of tiles
20 x 6.66 x 3.00   =   ³⁹⁹
                               ^{^{Gi 3953 Kg.}}

Evaluation of Qi

Snow load 55.5 x 6.66 x 3.00  = 1109

Wind
24 x 6.66 x 3.00 =   ^{= 480}
                               ^{^{Qi 1598 Kg.}}

Wi = Gi + 0.33 Qi = 4477 kg.

Fh
= ¹⁰· 4477 = 448 kg.
       ^¹⁰⁰

Considering the shape, this force will be distributed along a height of 1.20 mt.

qu
=   ⁴⁴⁸= 373 kg./mt.
         1.20

from the analysis of the structure we have:

Va = - Ve = 302.3 kg.

H    = 216 kg.

Ma    = - 765 kgmt.

Me    = 738 kgmt.

Md = 14.4 ksmt.

Mb = 10.86 kgmt.

Mc = - 93.8 kgmt.

Max stresses of the structure
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Ma
= 1335 kgmt.

Mb = - 912 kgmt.

MC = 2672 kgmt.

Md = 912 kgmt.

Me = 1335 kgmt.

Va = 3085 kgmt.

Ve = 3085 kgmt.

H = 654 kgmt.

Check of the central line section
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Mb = 2672
b = 138
Bo = 20
h = 23.5
h = 1.5

Let's check the section with reinforcement:

Af = 3 0 16 = 6.03 CM.2    h = 0.06

Af =20 16+14o4=5.78 CM.2

Af = 1600 kg./cm.2 Mt. = 10

Ff * = 6.03 + 5.78 = 11.81

h * = 12.8

x = 3.9 cm. inside the slab. So we can consider the section rectangular:

= 26226

= 39.7 kg./cm.²            70 kg./mt.²

=  1580                       1600 kg./mt.²

Check of the section at the joint
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M = 912 kgmt.
b = 20
h = 23.5
h = 1.5

Af = 3 Ø 16 + 2 Ø 4 = 6.28 cm.²

A'f = 2 Ø 16 = 4.02 cm.²

X = 8.26 cm.

= 20179

= 38 kg./cm.²        85 kg./cm.²

= 688                   1600 kg./cm.²

Check of the head section of the pilaster
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Mb
= - 897 kgmt.

The moment in the more stressed section is

M = 802 kgmt. N = 3085 kg.

e
=   ^M= 26 cm. ¹⁰= ³³⁵· ¹= 10       12
          N                               D              9         15.9

max
=   ³⁰⁸⁵= ⁸⁰²⁰⁰ = 1247    kg./cm²      1600
             19.5                 73.6

WIND ACTION MEASURED AT 120 KG./MT.²
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The actions are supposed to be due to pressures:
p = cq
c = exposure coefficient
q = shape coefficient.

in the worst exposure case we have: q¹ kg. /mt.² = 120

Being the building lower than 10 mt. 10 q = 0.75 · 1.20

The shape coefficient, being a sloping roof e = 23°, with 20° < e < 60° we have:

Ce = + 0.03 - 1 (e in degrees) so that p = 27.9 kg. /mt.²

LONGITUDINAL FRAME
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We are going to check two longitudinal frames, one for the 90 mt.² prefab of 4 spans and one for the 79 Mt.² prefab of 3 spans. For both of them we shall sum up the stresses due to the orizontal and vertical loads.

1) 90 mt.²    prefab: 1 = 6281 cm.² = 22284 cm.²

Geometric characteristics
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SEGMENT
1

/1

AA' 3.35 6281 19

BB' 3.35 6281 19

CC' 3.35 6281 19

DD' 3.35 6281 19

EE' 3.35 6281 19

A'B' 3.00 22284 74

B'C' 3.00 22284 74

C'D' 3.00 22284 74

D'E' 3.00 22284 74

Coefficients of the division of the moments in the crosses

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CROSS
SEGMENT
W
2 W
^W
2 W

A' AA' 19 186 0.10

A'B' 74 186 0.40

B'             A'B' 74 334 0.22

B'C' 74 334 0.22

B'B 19 334 0.06

C' B'C' 74 334

C'C 19 334

C'D' 74 334

D' C'D' 74 334

D'D 19 334

D'E' 74 334

E' D'E' 74 186 0.40

E'E   19 186 0.10

Coefficients of the division of the moments in the pilasters

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PILASTER

AA' - 0.3

BB' - 0.3

CC' - 0.3

DD' - 0.3

EE' - 0.3

Analysis of the loads for vertical forces
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Weight of the girder
0.20 x 0.25 x 2500      125

Overload 100 x 0.25 = ²⁵
^{150 kg./mt.}

Perfect joint moment M = ^QP= 113 kgmt.
¹²

The final results of the calculation of the frame according to Kani method give us the following stresses:

Max M at the girder joint = 26 kgmt.

Max M at the head of the pilaster = 26 kgmt.

Analysis of the loads for the horizontal forces
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Weight of the slab
0.05 x 4 x 2500 x 12 = 6000 kg.

Weight fo the girder 0.20 x 0.25 x 12 x 2500 = 1500 kg.

Weight of the pilasters 5 x 390 = 1950 kg.

Inner plaster work
15 x 4 x 12 = ^{= 720 kg.}
^{Gi = 10170 kg.}

Qi = 55.5 x 12 x 4 = 2664

Wi = 10170+0.33 · 2664 = 11049 kg.

Fh
= ¹⁰= 11049 = 1105 kg. = Qr
¹⁰⁰

Mr
= ^{Qr · hr}= 1234 kgmt.
³

The geometrical and elastic characteristics are similar to the calculated ones. The max stresses, calculated according to Kani method, for horizontal forces are:

Max M at the girder joint =  332 kgmt.

Max M at the head of the pilaster =  380 kgmt.

Max M at the bottom =  392 kgmt.

Previously the pilaster has been checked for higher stresses.
The girder is scantly stressed and it is reinforced with 2 upper Ø 12 iron bars and 2 lower Ø 12 one. They are quite enough.

2) 78 mt.² prefab.

The longitudinal frame is similar to the former one which is not so much stressed. Moreover, being a three span frame, the quantity of the horizontal strength will be less than the former one, while the size of the pilasters is the same. So it is not necessary to check it.

Calculation of the concrete bed foundation
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Analysis of the loads

Weight of the slabs 0.06 x 6.66 x 18 x 2500 = 17982 kg.

Weight of the girders 7 x 0.14 x 0.50 x 6.66 x 2500 = 8159 kg.

Weight of the pilasters 14 x 0.18 x 0.50 x 3.00 x 2500 = 7875 kg.

Weight of the ceiling plaster work 30 x 18 x 6.66 = 3596 kg.

Weight of the wall plaster work 4 x 30 x 18 x 3.00 = 6480 kg.

Weight of the overload
150 x 18 x 6.66 = ^{17982 kg.}
^{80056 kg.}

The unitary pressure on the ground will be
= ⁸⁰⁰⁵⁶= 0.050 kg./cm.² = 1.5
^{800 x 2000}

Being the stricture rather light, it is not necessary to deal with.

The 10 cm. concrete bed must be reinforced by means of an electrosoldered grate with Ø 15 meshes every 10 cm.

Horizontal thrust check
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The horizontal thrust, due to the vertical load which has effect on the sloping pitches, is given by H = 863 kg.

The structure has a transversal girder joined to the longitudinal one, its interaxis is every 6 mt.

The traction stress on the up stated girder will be

St = 863 + ¹· 863 = 1294.5 kg
²

This traction stress will be completely absorbed by the reinforcements which are

4 Ø 16 = 8.04 cm.² = ^1294.5= 161 kg/cm.² 1600 kg./cm.²
^8.04

Considering the symmetry of the structure, the torsion, due to the eccentricity between the barycentre of the rigidides and that of the masses, is negligible.

Tangential thrust check
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Tm = T/b x 0.9 x h

On this matter the standard provides that the acceptable tangential thrusts, without any reinforcement at the breaking point are

4 + ^{R'bk - 150}= 5.33 kg. /cm.²;
⁷⁵
for the external girder we have: Tmax = 3.2 kg. /cm.² 5.33.

Anyhow there are also some stirrups whose pitch is 25 cm. and whose section is 2 Ø 6.

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