Universitatea Tehnica "Gh. Asachi" IasiFacultatea de Constructii si Instalatii - Sectia C.C.I.A.

Grupa: 3406 Student : Pascal Iulian - 2011 -

Adapost pentru TaurineSa se proiecteze un adapost pentru taurine stiind urmatoarele:

- numarul de ordine : ≔n 36

- numarul de taurine : =≔Nr −170 ⋅2 n 98

- lungimea : ≔L 10.5 m

- traveea : ≔T 4.5 m

- structura acoperisului este din grinzi cu zabrele din lemn cu inaltimea :

≔hg 2.65 m

- peretii adapostului sunt din zidarie de caramida

- inaltimea utila : ≔Hu 3 m

≔Localitate “Bacau”

Etape :

I. Dimensionarea tehnologica a adapostului si determinarea gabaritului adapostului.

II. Dimensionarea structurii acoperisului ( calculul panelor si evidentierea barelor intinse si comprimate )

III. Planse : - sectiune transversala prin adapost

- planul adapostului

I. Dimensionarea tehnologica a adapostului si determinarea gabaritului adapostului.

Adapostul va fi cu stabulatie fixa pentru vaci rasa baltata cu negru.

=L 10.5 m

＝L1 +L 10 cm =≔L1 +170 cm 10 cm 180 cm

≔B 120 cm

=NTravee 17

1 - alee de circulatie2 - rigole3 - stand scurt4 - iesle5 - alee de furajare

II. Dimensionarea structurii acoperisului ( calculul panelor si evidentierea barelor intinse si comprimate )

1. Dimensionarea panelor

=dp 1.96 m

=dp 1.96 m

=dp 1.96 m

=a 1.75 m =a 1.75 m =a 1.75 m =a 1.75 m =a 1.75 m =a 1.75 m

=L 10.5 m

=T 4.5 m

=T 4.5 m

=T 4.5 m

Incarcari Permanentea) Greutatea acoperisului

=qacop 0.2 ――kN

m2

Incarcari Variabileb) Incarcari din zapada

≔s ⋅⋅⋅μi Ce Ct s0.k =μi 0.8

=Ce 1

=Ct 1

=s0.k 2.5 ――kN

m2=zona “III”

=μi 0.8 =Ce 1 =Ct 1 =zona “III” =s0.k 2.5 ――kN

m2

=≔s ⋅⋅⋅μi Ce Ct s0.k 2 ――kN

m2

=≔s' ⋅⋅⋅―μi2

Ce Ct s0.k 1 ――kN

m2

=≔s'' ⋅⋅⋅⎛⎜⎝

+μi ――23.43

30

⎞⎟⎠Ce Ct s0.k 3.953 ――

kN

m2

b) Incarcari din vant

＝W ⋅⎛⎜⎝

⋅⋅―1

2ρ Vb

2⎞⎟⎠

CeZ Cp

WH WI

=≔Vb ⋅⋅Ct Cs Vb.0 35 ―m

s=Localitate “Bacau” =zona “III”

=Vb.0 35 ―m

s=CeZ 1.5

=Ct 1 =Ce 1

≔ρ 1.25 ――kg

m3

=≔W ⋅⎛⎜⎝

⋅⋅―1

2ρ Vb

2⎞⎟⎠

CeZ Cp 1148.438 ――N

m2Cp

≔CpH −0.2438 ≔CpI −0.4

=≔WH ⋅W CpH −0.28 ――kN

m2=≔WI ⋅W CpI −0.459 ――

kN

m2

Ipoteze de incarcare

1 : P + Z --> 1.35 P + 1.5 Zunif

2 : P + V + S' --> nu se ia in consideratie pentru ca avem suctiune3 : P + U --> 1.35 P + 1.5 UConsideram pana ca o grinda simplu rezemata

Ipoteza 1: - incarcari permanente pe pana : =≔qp ⋅qacop ――⋅dp T

T0.392 ――

kN

m

- incarcari din zapada pe pana : =≔Sp ⋅s ――⋅a T

T3.5 ――

kN

m

q1.d.z q1.d.y

=T 4.5 m =T 4.5 m

kN

=≔q1.d +⋅1.35 qp ⋅1.5 Sp 5.779 ――kN

m

=≔q1.d.z ⋅q1.d cos ((α)) 5.303 ――kN

m=≔q1.d.y ⋅q1.d sin ((α)) 2.298 ――

kN

m

=≔My ―――⋅q1.d.z l2

813.423 ⋅kN m =≔Mz ―――

⋅q1.d.y l2

85.817 ⋅kN m

Solicitare - incovoiere oblica dupa doua directii:

≤+――Γm.y.d

fm.y.d

⋅km ――Γm.z.d

fm.z.d

1

≔km 0.7 - pentru sectiuni rectangulare

≤+⋅km ――Γm.y.d

fm.y.d

――Γm.z.d

fm.z.d

1

Predimensionarea panei :

=≔fm.d ⋅fm.k ――Kmod

γi8615.385 ――

kN

m2

＝Γm.y.d ――Mz

Wy

＝Wy ――⋅b h2

6 ＝＝fm.y.d fm.z.d fm.d - rezistenta la incovoiere

＝Γm.z.d ――My

Wz

＝Wz ――⋅b2 h

6

≤+――Γm.y.d

fm.y.d

⋅km ――Γm.z.d

fm.z.d

1 ≤⋅⎛⎜⎝

+――Mz

Wy

⋅km ――My

Wz

⎞⎟⎠――

1

fm.d

1 ≤+――⋅6 Mz

⋅b h2⋅km ――

⋅6 My

⋅h b2fm.d

--> -->

≤+⋅km ――Γm.y.d

fm.y.d

――Γm.z.d

fm.z.d

1 ≤⋅⎛⎜⎝

+⋅km ――Mz

Wy

――My

Wz

⎞⎟⎠――

1

fm.d

1 ≤+⋅km ――⋅6 Mz

⋅b h2――

⋅6 My

⋅h b2fm.d

- impunem : ＝―h

b1.5 ＝h 1.5 b

≤+―――⋅6 Mz

⋅2.25 b3⋅km ―――

⋅6 My

⋅1.5 b3fm.d ≤+⋅2.667 ――

Mz

b3⋅2.8 ――

My

b3fm.d ≤⋅―

1

b3⎛⎝ +2.667 Mz ⋅2.8 My

⎞⎠ fm.d

--> -->

≤+⋅km ―――⋅6 Mz

⋅2.25 b3―――

⋅6 My

⋅1.5 b3fm.d ≤+⋅1.867 ――

Mz

b3⋅4 ――

My

b3fm.d ≤⋅―

1

b3⎛⎝ +1.867 Mz ⋅4 My

⎞⎠ fm.d

=≔b1‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾3

―――――――+⋅2.667 Mz ⋅2.8 My

fm.d

18.334 cm =≔h1 ⋅1.5 b1 27.501 cm

=≔b2‾‾‾‾‾‾‾‾‾‾‾‾‾‾3

――――――+⋅1.867 Mz ⋅4 My

fm.d

19.568 cm =≔h2 ⋅1.5 b2 29.352 cm

=≔b max ⎛⎝ ,b1 b2⎞⎠ 19.568 cm

Adoptam : =b 20 cm =h 30 cm

=≔Wy ――⋅b h2

63000 cm3 =≔Wz ――

⋅b2 h

62000 cm3

=⋅⎛⎜⎝

+――Mz

Wy

⋅km ――My

Wz

⎞⎟⎠――

1

fm.d0.77 < 1 =Obs “Inegalitatea este indeplinita”

=⋅⎛⎜⎝

+⋅km ――Mz

Wy

――My

Wz

⎞⎟⎠――

1

fm.d0.937 < 1 =Obs “Inegalitatea este indeplinita”

Verificam b si h in conditia de rigiditate :

＝<umax uadm ――l

200=――

l

20022.5 mm

＝umax +++u1 u2 u3 u4 ＝＝u3 u4 0

＝u1 ⋅uinst.g⎛⎝ +1 Kdef

⎞⎠

＝u2 ⋅uinst.g⎛⎝ +1 ⋅ψ2i Kdef

⎞⎠ ≔ψ2i 0.4

≔Kdef 0.8

＝umax‾‾‾‾‾‾‾‾‾‾‾‾+umax.y

2 umax.z2 ＝umax.y +ug.y us.y ＝ug.y ⋅uinst.g.y

⎛⎝ +1 Kdef⎞⎠

＝us.y ⋅uinst.s.y⎛⎝ +1 ⋅ψ2i Kdef

⎞⎠

=≔uinst.g.y ⋅――5

384―――

⋅gp.y l4

⋅E Iz0.828 mm =gp.y 0.217 ――

kN

m

=≔uinst.s.y ⋅――5

384―――

⋅Sp.y l4

⋅E Iz

7.582 mm =gp.y 0.217 ――kN

m

=≔ug.y ⋅uinst.g.y⎛⎝ +1 Kdef

⎞⎠ 1.49 mm =≔us.y ⋅uinst.s.y⎛⎝ +1 ⋅ψ2i Kdef

⎞⎠ 10.008 mm

=≔umax.y +ug.y us.y 11.498 mm

g＝umax.z +ug.z us.z =≔uinst.g.z ⋅――

5

384―――

⋅gp.z l4

⋅E Iy

0.848 mm =gp.z 0.5 ――kN

m

g=≔uinst.s.z ⋅――

5

384―――

⋅gs.z l4

⋅E Iy

7.777 mm =gs.z 4.588 ――kN

m

=≔ug.z ⋅uinst.g.z⎛⎝ +1 Kdef

⎞⎠ 1.526 mm =≔us.z ⋅uinst.s.z⎛⎝ +1 ⋅ψ2i Kdef

⎞⎠ 10.265 mm

=≔umax.z +ug.z us.z 11.791 mm

=≔umax‾‾‾‾‾‾‾‾‾‾‾‾+umax.y

2 umax.z2 16.469 mm < =uadm 22.5 mm

=Obs “Inegalitatea este indeplinita”

Ipoteza 3 :

uz uy

gp.z gp.y

=l 450 cm =l 450 cm

≔u 1 kN

=≔uz ⋅u cos ((α)) 0.918 kN =≔uy ⋅u sin ((α)) 0.398 kN

≔gp.z 7.557 ――kN

m≔gp.y 3.275 ――

kN

m

=≔My +―――⋅gp.z l2

8――

⋅uz l

420.161 ⋅kN m =≔Mz +―――

⋅gp.y l2

8――

⋅uy l

48.737 ⋅kN m

Identificarea elementelor intinse si comprimate

=≔Qc.p +⋅qp a gpanta 1.07 kN =≔Qm.p ――Qc.p

20.535 kN

=≔Qc.s ⋅Sp a 6.125 kN =≔Qm.s ――Qc.s

23.063 kN

=≔Qc +⋅1.35 Qc.p 1.5 Qc.s 10.632 kN =≔Qm ―Qc

25.316 kN