Calculul elementelor de sarpantaAmplasament : BUCURESTI Clasa de importanta a constructiei : IV a=2.0 ; b=2.2 ; c=1.2 ; d=1.4 ; e=0.25 ; f=0.45 ;

I. Stabilirea incarcarilor :Incarcari permanente : greutate proprie invelitoare : aleg : Tabla tip LINDAB 300 daN/m2 (inclusive astereala si capriorii ) gpn = 300 N/m2 = 0.5 KN/m2 ; gpc = 300 * 1.2 = 360 N/m2 = 0.36 KN/m2 ; Incarcari din zapada : pzc = Ce * Cz * gz * F Ce=0.8 ( conditii normale de exploatare) ; CZ=1.25 ( coeficient de aglomerare ) ; gz=1.5 ( zona C ) ; = a- ( 0.4 * gp ) / ( Ce * gz ) 0.3 * a 2.0 - 0.4 * 300 / 0.8 * 1500 = 1.90 > 0.60 F = 1.90 > 0.60F

pzc=1500*1.25*0.8*1.90 =2850 N/m2 Incarcari din vint : pnc = * Cni * Ch (z) * gv * F =1.6 gv=420 N/m2 = 0.42 KN/m2 F=1.90 Ch(z)=0.65

1

20

0.5 -0.40 +0.02 +0.30

h1 / l = 0.72 0.67 -0.122

1.0 -0.7 -0.40 -0.2

32 40

h1=6.85m l=9.45m h1/l=0.72

Cni = - 0.122 Observatie ! Deoarece are un efect de suctiune incarcarea din vant nu se ia in calcul . = 32 : sin = 0.53 cos = 0.848 Incarcarea utila : Pn = 1000 N = 1 KN ; Pc = Pn * n = 1.2 KN ;

2

Calculul sipcilor :Conform alegerii facute tabla tip LINDAB , distanta aferenta dintre o sipca , pe o suprafata inclinata este de c = 400 mm ;

3

Schema de calcul a sipcilor : = 32 : sin = 0.53 cos = 0.848 Incarcarea permanenta : qsp= gp * c , de unde se scade incarcarea capriorilor 80 N/m2 gp = 300 80 = 220 N/m2 gpc = 220 * 1.2 = 264 N/m2 qps x = gpc * c *sin = 264 * 0.4 * 0.53 = 56 N/m2 qps y = gpc * c * cos = 264 * 0.4 * 0.848 = 89.5 N/m2

Incarcari din zapada : pzc=1500 * 1.25 * 0.8 * 1.90 =2850 N/m2 ; c = 400 mm = 0.4 m ; gzs = pzc * c * cos = 2850 * 0.4 * 0.848 = 967 N/m2 ; qzs x = gsz * sin = 966.72 * 0.53 = 513 N/m qzsy = gsz * cos = 966.72 * 0.848 = 820 N/m Incarcare din vant : nu se ia in considerare deoarece a rezultat suctiune ; Incarcare utila : nu se ia in considerare la calculul sipcilor ; Ipoteze de incarcare : Observatie ! Deoarece la calculul sipcilor se ia in considerare numai incarcarea permanenta sic ea din zapada , ipoteza de calcul este numai una : IPOTEZA 1 qs1 x = qsp x + qsz x = 56 +513 =569 N/m qs1 y = qsp y + qsz y = 89.5 +820 = 910. N/m Calculul momentelor : d1 = 1.00 m4

M = Ms1y =

1 s x

569 *1.00 2 = 71 .20 Nm 8 8 q 1, y * d12 910 *1.00 2 s = = 114 Nm ; 8 8 =

q1, x * d12 s

;

Verificarea capacitatii portante :s M ef , x

M rs, x

s M ef , y

M rs, y

1.00

-

Aleg sectiunea sipcilor 58 x 38 : Msr,x = Ric * Wcalc,x * mTi Msr,y = Ric * Wcalc,y * mTi Unde : mTi = 0.9 ; Wcalc,x = Wcalc,y =b h 2 58 38 2 = = 13958 .67 mm3 6 6 b h 2 38 58 2 = = 21305 .33 mm ; 6 6

;

gp = 220 N/m2 ; gz * ce = 1500 * 0.8 = 1200 N/m2=0.55 220 + 0.65 1200 = 0.63 220 + 1200

Observatie! Conform tabelului de rezistente de calcul ale lemnului la incovoiere statica , se face interpolari ale valorilor coeficientului : Ric 0.55 . 10.8 0.63 ..? 0.7013.7 0.15 2.9 0.08 .x x = 1.55 c Ri = 10.8 + 1.55 = 12.35 N/mm

5

Msr,x = 12.35 * 13958.67 * 0.9 = 155151 N/mm Msr,y = 12.35 * 21305.33 * 0.9 = 236809 N/mms M ef , x s M ef , y

M

s r,x

M

s r,y

1.00 =

71 .2 114 1.00 155151 236809

0.46 0.49 = 0.95 < 1.00

Verificarea rigiditatii la incovoiere : Incarcari permanente : gp = 220 N/m2 c = 0.4 m qp = gp * c sn q p x = qs np * sin = 220 * 0.4 * 0.53 = 46.7 N/m qs np y = qs n * cos = 220 * 0.4 * 0.848 = 74.6 N/m Incarcari din zapada : qzs n = gs * ce * cz * gz * c * o

o

= c 0.2 c c

gp

e

gz

0.3 c

= 1.2 ;300 0.36

o

= 1.2 0.2 0.8 1500

qz

sn

o = 1.15 0.36 = 1500 * 0.8 * 1.25 * 1.15 * 0.40 * 0.848 = 585.12 N/m = 586 N/m suprafata inclinata

qzs n x = qzs n * sin = 585.12 * 0.53 = 310.1 N/m qzs n y = qzs n * cos = 585.12 * 0.848 = 496.18 N/m

6

Deformatiile datorate incarcarilor permanente : Iy = Iy =b3 h 58 38 3 = = 265214 .67 mm4 12 12 b h 3 58 3 38 = = 617854 .67 mm4 12 12

E= 11300 N/mm2 d1= 1000 mm fp inst x = 384 5 q s ,,n d14 p x E Iys ,n 4

=

5 46 .7 10 3 1000 4 384 11300 265214 .67

=0.203 mm = 0.139 mm

q p , y d1 5 5 74 .6 10 3 1000 4 = fp inst y = 384 E Ix 384 11300 617854 .67

fp x= fp inst x * ( 1 + kzdef ) = 0.203 * ( 1+ 0.5 ) =0.305 mm fp y= fp inst y * ( 1 + kzdef ) = 0.139 * ( 1+ 0.5 ) =0.209 mm Deformatiile datorate incarcarilor din zapada :q s ,,n d14 5 5 310 .1 10 3 1000 4 p x = 384 E I = 384 11300 265214 .67 ys ,n 4

fz inst x fz inst y

= 1.35 mm = 0.93 mm

q p , y d1 5 5 496 .18 10 3 1000 4 = = 384 E Ix 384 11300 617854 .67

fp x= fp inst x * ( 1 + kzdef ) = 1.35 * ( 1+ 0.5 ) =2.03 mm fp y= fp inst y * ( 1 + kzdef ) = 0.139 * ( 1+ 0.5 ) =1.40 mm Ipoteza 1 ! fl x = fp x + fz x = 0.305 + 2.03 = 2.34 mm fl y = fp y + fz y = 0.209 + 1.40 = 1.61 mm fmax final =2 f l ,2x + f ly =

2.34 2 +1.61 2

= 2.84 mm

7

fadm =

lc 1000 = 150 150

= 6.67 mm ; lc = 1000 mm

fmax final < fadm 2.84 < 6.67 mm - SIPCA va avea sectiunea de : 58 x 38II.

Calculul capriorilor :

sipca : 28 x 58 c = 40 cm ; caprior : 120 x 120 d1 = 100 cm ; Incarcarea permanenta : = 32 : sin = 0.53 cos = 0.848 c q p = gp * d1 * n unde : n = 1.2 d1 = 100 cm = 1.00 m gp = 300 N/m2 gcp = 300 * 1.00 *1.2 = 360 N/m qcp n = gcp * cos = 360 * 0.848 = 305.25 306 N/m Incarcare din zapada :8

qcz = pcz * d1 d1 = 100 cm = 1.00 m Pzc= 2850 N/m2 pcz = 2850 * 1.00 = 2850 N/m qcz n = qcz * cos2 = 2850 * 0.8482 = 2049.45 N/m Observatie ! Incarcare din vant nu se ia in considerare deoarece a rezultat efect de suctiune ! Incarcarea utila : p c n = pn * n ; pn = 1000 N ; n =1.2 ; c p n = 1000 * 1.2 = 1200 N Pc c n = p c n * cos Pc c n = 1200 * 0.848 = 1017.6 N Ipoteze de incarcare : Observatie ! Intrucat nu se ia in considerare incarcarea din vant avem 2 ipoteze de calcul : Ipoteza 1 ! qc1 = qcp n + qcz n = 306 + 2049.45 = 2355.45 N/m Ipoteza 2 ! 1. Incarcare uniform distribuita qc2 = qcp n = 305.25 306 N/m 2. Incarcare concentrate P2 = Pc c n = 1017.6 N Calculul momentelor : - deschiderea de calcul a capriorilor pe directia inclinata este de 3.00 m ; Ipoteza 1 ! l2 = 3.00 m ;9

2 q1c l2 2355 .45 3.00 2 M1= = = 2650 Nm 8 8

c

Mc1 = 2650 Nm Ipoteza 2 ! M c2 = M c2 =c 2 q2 l2 Pc l + 2 2 = 4 8

306 3.00 2 8

+

1017 .6 3.00 4

= 1107.45 Nm

Mcmax = max (Mc1 ; Mc2 ) Mcmax = max (2650 ; 1107.45 ) Mc1 = 2650Nm Verificarea capacitatii portante : Mcmax Mr ; Ric = 12.35 N/mm mTi = 0.9 ( lemn tratat pe suprafata ) Mr = 12.35 * 0.9 * Wcalc Mr = 11.10 * Wcalc , de unde rezulta : Wnec c M max 2650 10 3 = = 238738.8 mm 11 .1 11 .1

. se alege caprior cu sectiune de : 100 x 120 mm Wef =100 120 2 6

= 240000 mm3 > Wnec = 238739 mm3

Verificarea rigiditatii capriorilor la incovoiere : fmax final fadm lc = 3000 mm10

fadm =

lc 200

=

3000 200

= 15 mm

Incarcarea permanenta : qc np n = gp * d1 * cos unde : cos = 0.848 d1 = 100 cm = 1.00 m gp = 300 N/m2 qc np n = 300 * 1.00 * 0.848= 254.4 N/m Incarcari din zapada : qzs n = gz * ce * cz * d1 * o * cos 2 o = 1.15 sn qz = 1500 * 0.8 * 1.25 * 1.15 * 1.00 * 0.8482 = 1335.84 N/m = 1336 N/m Incarcarea utila : Pc nn = p c * cos pc = 1000 N ; cos = 0.848 cn P n = 1000 * 0.848 = 848 N Deformatiile datorate incarcarii permanente : E= 11300 N/mm2 l2 = 3.00 m ; kpdef = 0.5 fpc = fpc inst * (1+ kpdef ) fc p inst4 q c ,,n l2 5 p n = 384 E I

I=

b h3 100 120 3 = 12 12

= 14400000 mm4 = 1.6 mm

fpc inst =

fpc = 1.65 * (1+ 0.5) = 2.4 mm11

5 254 .4 10 3 3000 4 384 11300 14400000

Deformatiile datorate incarcarii din zapada : E= 11300 N/mm2 d1= 3000 mm kpdef = 0.5 fzc = fzc inst * (1+ kzdef ) fzc inst = I=q c,n l 4 5 z ,n 2 384 E I

b h3 100 120 3 = 12 12

= 14400000 mm4 =8.6 mm

fzc inst =

5 1335 .84 10 3 3000 4 384 11300 14400000

fzc = 8.6 * (1+ 0.5) = 12.9 mm Deformatiile datorate incarcarii utile : E= 11300 N/mm2 l2= 3000 mm kpdef = 0.00 fuc = fuc inst * (1+ kzdef ) fc u instc ,n 3 1 P,n l2 = 48 E I

I=

b h3 100 120 3 = 12 12

=14400000 mm4 =2.93 mm

fuc inst =

1 848 3000 3 48 11300 14400000

fuc = 2.93 mm Ipoteza 1 ! f1c = fpc + fzc12

f1c = 2.48 + 13 = 15.48 mm fmax final > fadm se mareste sectiunea la : 120 x 120 =>b h3 120 120 3 = I= = 17280000 mm4 12 12 120 120 2 Wef = = 288000 mm3 > Wnec 6

= 238739 mm3

Deformatiile datorate incarcarii permanente : fpc inst = fpc inst =4 q c ,,n l2 5 p n 384 E I

5 254 .4 10 3 3000 4 384 11300 17280000

= 1.37mm

fpc = 1.37 * (1+ 0.5) = 2.10 mm Deformatiile datorate incarcarii din zapada : E= 11300 N/mm2 d1= 3000 mm kpdef = 0.5 fzc = fzc inst * (1+ kzdef ) fzc inst =q c,n l 4 5 z ,n 2 384 E I

fzc inst =

5 1335 .84 10 3 3000 4 384 11300 17280000

=7.22 mm

fzc = 7.22 * (1+ 0.5) = 10.83 mm Deformatiile datorate incarcarii utile : E= 11300 N/mm2 l2= 3000 mm kpdef = 0.00 fuc = fuc inst * (1+ kzdef )13

f

c u inst

c ,n 3 1 P,n l2 = 48 E I

I=

b h3 100 120 3 = 12 12

=14400000 mm4 =2.44 mm

fuc inst =

1 848 3000 3 48 11300 17280000

fuc = 2.44 mm Ipoteza 1 ! f1c = fpc + fzc f1c =2.10 + 10.83 = 12.93 mm Ipoteza 2 ! f1c = fpc + fuc f1c = 2.10 + 2.44 = 4.54 mm fadm = 15 mm > fmax = 12.93 mm - CAPRIORUL va avea o sectiune de 120 x 120 .

III.

Calculul panei centrale :

14

Incarcarea permanenta :

15

lc = 2.10 0.75 = 1.35 m = 1350 mm gp = 300 N/m2 d2= 2.25 m = 2250 mm ( deschideri inegale d2 d2 ) cos = 0.848 n = 1.2 n1 = 1.1 - aleg lemn de rasinoase ( brad ) cu sectiunea de : 120 x 120 mm : lemn = 480 daN/m3 = 4800 N/m3 bp = 120 mm hp = 120 mm qpp = gp * n * d2 * cos + bp * hp * n1 * 1lemn

=

=300 * 1.2 * 2.25 * 1.18 + 0.12 * 0.12 * 1.1 * 4800 = 1031.83 N/m p qp = 1032 N/m

Incarcarea din zapada : qzp = pzc * d2 = 2850 *2.25 = 6412.5 N/m = 6413 N/m

Incarcarea utila : P=1000N N=1.2 Pp = P * n = 1000 * 1.2 = 1200 N Ipoteza 1 ! q1 = qpp + qzp = 1032 + 6413 = 7445 N/m Ipoteza 2- nu se ia in considerare deoarece efectul de incovoiere produs de incarcarea de 1200 N este mult mai mica decat incarcarea uniform distribuita .

16

Calculul momentelor :q1p l c2 7445 1.35 2 = = = 1696.06 Nm 8 8

M

p 1

Verificarea capacitatii portante la incovoiere a panei centrale : Mpmax Mr ; Mr = Rci * Wcalc * mTi gpn = 300 N/m2 + bp * hp * 4800 * d gpn = 300 + 0.12 * 0.12 * 4800 * gz * ce = 1500 * 0.8 = 1200 N/m2=0.55 331 + 0.65 1200 = 0.63 331 +1200 12

= = 330.72 N/m2

1 2.25

Ric = 12.35 N/mm mTi = 0.9 ( lemn tratat pe suprafata ) Wef =120 120 2 6

= 288000 mm3

Mr = Rci * Wcalc * mTi Mr = 12.35* 288000 * 0.9 = 3201120 Nmm Mr = 3201.12 Nm 3202 Nm Mcmax Mr ;Mcmax = 1697 Nm Mr = 3202 Nm

Verificarea rigiditatii la incovoiere a panei centrale : fmax final fadm lc = 1350 mm

17

fadm=

lc 200

=

1350 200

= 6.75 mm

Incarcarea permanenta : gppny = gp * d2 * cos + bp * hp * gp n p y

1

lemn

=

=300 * 2.25 * 1.18 + 0.12 * 0.12 * 4800 = 865.62 = 866 N/m

Incarcarea din zapada : gczny = cz * gz * ce * 0 * d2 Ce=0.8 0 = 1.15 CZ=1.25 d2 = 2.25 gz=1500 gczny = 1500 * 1.25 * 0.8 * 1.15 * 2.25 = 3881.25 N/m gczny = 3882 N/m Deformatiatii datorate incarcarilor permanente : E= 11300 N/mm2 lc= 1350 mm kpdef = 0.5 fpp = fpp inst * (1+ kpdef ) f I=p p instp n q p ,,y lc4 5 = 384 E Ix

b h3 120 120 3 = = 17280000 mm4 12 12 5 866 10 3 1350 4 fpp inst = = 0.19 mm 384 11300 17280000

fpp = 0.19 * (1+ 0.5) = 0.285 mm

Deformatiile datorate incarcarii din zapada : E= 11300 N/mm2 lc= 1350 mm kpdef = 0.5 fzp = fzc inst * (1+ kzdef )18

fz I=

p

inst

q zp,,yn lc4 5 = 384 E Ix

b h3 120 120 3 = = 17280000 mm4 12 12 5 3882 10 3 1350 4 fzp inst = = 0.86 mm 384 11300 17280000

fzp = 0.86 * (1+ 0.5) = 1.29 mm f1 = fpp + fzp = 0.285 +1.29 f1 = 1.58 mm < fadm= 6.74 mmIV.

sectiunea PANEI CENTRALE va fi de 120 x 120 mm .

Calculul panei intermediare :

19

Incarcarea permanenta : gp = 300 N/m2 d2 = 2.55 m = 2550 mm ( deschideri inegale d2 d2 ) cos = 0.848 n = 1.2 n1 = 1.1 - aleg lemn de rasinoase ( brad ) cu sectiunea de : 150 x 170 mm : lemn=480 daN/m3 = 4800 N/m3 bp = 150 mm hp = 170 mm qpp = gp * n * d2 * cos + bp * hp * n1 * 1lemn

=

=300 * 1.2 * 2.55 * 1.18 + 0.15 * 0.17 * 1.1 * 4800 = 1217.88 N/m p qp = 1218N/m

Incarcarea din zapada : qzp = pzc * d2 = 2850 *2.55 = 7267.5 N/m = 7268 N/m

20

Incarcarea utila : P=1000N N=1.2 Pp = P * n = 1000 * 1.2 = 1200 N Ipoteza 1 ! q1 = qpp + qzp = 1217.88+ 7267.5 = 8485.38N/m Ipoteza 2- nu se ia in considerare deoarece efectul de incovoiere produs de incarcarea de 1200 N este mult mai mica decat incarcarea uniform distribuita ! Calculul momentelor :q1p l c2 8486 2.55 2 = = 6897 Nm 8 8

M1p =

Verificarea capacitatii portante la incovoiere a panei intermediare : Mpmax Mr ; Mr = Rci * Wcalc * mTi gpn = 300 N/m2 + bp * hp * 4800 * d gpn = 300 + 0.15 * 0.17 * 4800 * gz * ce = 1500 * 0.8 = 1200 N/m2=0.55 348 + 0.65 1200 = 0.63 348 +1200 12

= = 348 N/m2

1 2.55

Ric = 12.35 N/mm

21

mTi = 0.9 ( lemn tratat pe suprafata ) Wef =150 170 2 6

= 722500 mm3

Mr = Rci * Wcalc * mTi Mr = 12.35 * 722500 * 0.9 = 8030587.5 Nmm Mr = 8030.6 Nm 8030 Nm Mcmax Mr ; Mcmax = 6897 Nm Mr = 8030 Nm Verificarea rigiditatii la incovoiere a panei intermediare : fmax final fadm lc = 3000 mm fadm=lc 200

=

2550 200

= 12.75 mm

Incarcarea permanenta : gppny = gp * d2 * cos + bp * hp * gp n p y

1

lemn

=

=300 * 2.55 * 1.18 + 0.15* 0.17 * 4800 = 1025.1 N/m = 1026 N/m

Incarcarea din zapada : gczny = cz * gz * ce * Ce=0.8 CZ=1.25 gz=1500

0

* d2

0 = 1.15 d2 = 2.25

gczny = 1500 * 1.25 * 0.8 * 1.15 * 2.55 = 4398.75 N/m gczny = 4399 N/m22

Deformatiatii datorate incarcarilor permanente : E= 11300 N/mm2 lc= 2550 mm kpdef = 0.5 fpp = fpp inst * (1+ kpdef ) fpp inst = I=q p ,n l 4 5 p, y c 384 E Ix

b h3 150 170 3 = 12 12

=61412500 mm4 = 0.81 mm

fpp inst =

5 1026 10 3 2550 4 384 11300 61412500

fpp = 0.81 * (1+ 0.5) = 1.22 mm Deformatiile datorate incarcarii din zapada : E= 11300 N/mm2 lc= 2550 mm kpdef = 0.5 fzp = fzc inst * (1+ kzdef ) fz I=p inst

q p ,n l 4 5 z, y c = 384 E Ix

b h3 150 170 3 = = 61412500 mm4 12 12 5 4399 10 3 2550 4 fzp inst = =3.49 mm 384 11300 61412500

fzp = 3.49 * (1+ 0.5) = 5.24 mm f1 = fpp + fzp = 1.22 + 5.24 f1 = 6.46 mm < fadm= 12.75 mm - sectiunea PANEI INTERMEDIARE va fi de 150 x 170 mm .

Calculul popului central :

23

Incarcarea permanenta : N =P P

g p n cos

* d2 * t + bp * hp *

lemn

*n1*t +

2 d pop

4

* n1 *

lemn

* hpop

gp=300N/m2 lemn=4800N/m2 n=1.2 n1=1.1 Se alege un pop cu diametrul de 12cm hpop=3.20 - ( 0.15 + 0.12 ) = 2.93 m cos =0.848 d2=2.25 m hp=15cm bp=12cm t=2.10 m NPP=300 1.2 * 2.25 0.848 3.14 0.12 2 + 4

* 2.10 + 0.12 * 0.15 * 4800 *1.1*2.10 * 1.1 * 4800 * 2.93 =

NPP= 2005.9+ 200 + 174.88 = 2380.78N24

Incarcarea din zapada : Nzp = pzc * d2 * t = 2850 * 2.25 * 2.10 = 13466.25N Observatie 1 ! Incarcarea din vant nu se ia in considerare deoarece din calcul a rezultat suctiune . Observatie 2 ! Incarcarea utila este nesemnificativa in comparative cu incarcarea din zapada pentru calculul popului . Ipoteze de incarcare : N1P = NPP + Nzp = 2380.78+ 13466.25 = 15847.03 N Verificarea popului central : N1P = Nmax CR CR = A calcul * RCC * mtc * A calcul = d 24

=

3.14 120 2 4lemn

= 11304 mm21 t

gperma = gp + bp * hp * + d 24

*

+lemn

* hpop *

*

gperma = 300 + 0.12 * 0.15 * 4800 * +3.14 0.12 2 4

1 t 1 2.10

* +

1 d2

* 2.93 * 4800 *

1 2.25

*

1 2.10

=

gperma = 300 + 41.14 + 33.65 gperma = 374.79N gz * ce = 1500 * 0.8 = 1200 N/m2=0.80 374 .79 + 0.85 1200 = 0.83 374 .79 +1200

Rccll=8.3 N/mm2

25

=

lf i

i = 0.25 * dpop = 0.25 * 0.12 = 0.030m lf = 2.93 - 0.8 = 2.13 m =lf i

=

2.13 0.030

= 71 < 752

= 1 0.8 *

d 100

= 1- 0.8 *

71 100

2

= 0.60

CR= 11304 * 8.3 * 0.9 * 0.60 = 50665 N Nmax = 15848 N