Dimenzionisanje

Dimenzionisanje temeljne konstrukcijeTemeljna traka (kraća strana)

Vg = 127,65

kNm'

Vp = 25,66

kNm'

V = V g + V P = 153 ,31 kN

m,

γ=18 KN/m3

c=4 MpaΦ=24 〫

pretpostavimo

Df = 1,2 mDOPUŠTENA NOSIVOST TLA :

σ dop =12

⋅γ⋅B⋅N γ ⋅Sγ ⋅iγ + (cm + q⋅tg φm )⋅Nc⋅Sc⋅dc⋅ic + q

cm = cFc

= 42,5

= 1,6 kN

m2

tg φm =tg φϕ

Fϕ

= tg 241,5

= 0 ,2968 → φm = 16 ,53° → N c = 12; N γ = 1,9

q = γ⋅D f = 21,6kN

m2; ic = 1 iγ = 1

Sc = 1 + 0,2 ⋅ BL

= 1 ,0 ; Sγ= 1 − 0,4 ⋅ BL

= 1 ,0

dc = 1,0+0 ,35⋅D f

B=1,0

σ dop = (1,6 + 21 ,6⋅ 0 ,2968 ) ⋅12 ⋅ 1 ⋅1⋅1 + 21 ,6= 117 ,52 kN

m2

Određivanje širine temelja :

1.Iteracija

σ n = σdop − β⋅ D f ⋅γB = 117 ,62 − 0 ,85 ⋅1,2⋅25 = 92 ,23kN

m2

A =Nσ n

=153 ,3192 ,23

= 1,6m

B⋅0 ,65 → B1 = 1 ,07 m

Sc = 1 + 0,2 ⋅ BL== 1,0 ; Sγ= 1 − 0,4 ⋅B

L= 1,0

dc = 1,0+0 ,35⋅

D f

B=1,0+0 ,35⋅ 1,2

1,07=1,35

σ dop =12

⋅ 8 ⋅1 ,07 ⋅1,9 ⋅ 1 ⋅1 + (1,6 + 21 ,6⋅0 ,2968 ) ⋅12 ⋅ 1,0 ⋅ 1 ,35 ⋅ 1 + 21 ,6 = 159 ,5 kN

m2

2. Iteracija

σ n = σdop − β⋅ D f ⋅γB = 159 ,5 − 0 ,85 ⋅1,2⋅25 = 134kN

m2

B =Nσn

=153 ,31134

= 1 ,14 m → B = 1,2m

B = 1,2+1 ,072

=1 ,135m

usvojena širina temelja B=1,2m

σ n =∑ V

A=153,311,2

=127,75kN /m2

Određivanje visine temeljne trake

0,25 1.2

1,2

0,6

σnet=127,75kN/m2

max

M= σnet*0.62/2=23kNm/m’

γ sr =1,35 ⋅V g + 1,5 ⋅V p

V g + V p

= 1,38

M sd = Mmax ɣ∙ SR =16,25 1,38 =31,74∙

kNmm'

h=a/2 = 47,5/2=23,75 usvojeno h=25 cmh=d+Φ/2+c --> d= 21,5cm

μsd =M sd

b ⋅d2⋅ f cd

=23⋅100120⋅21 ,5⋅21 ,5⋅1 ,667

=0 ,024→ ρ=0 ,98

Usvajanje armature :

pot A s1= Msd

ρ⋅d⋅fyd= 23⋅1000 ,98⋅21 ,5⋅43 ,5

=2,5cm2

Asmin = 0,0015･b･d=3,87 usvojeno R424 (As=4,24cm2)

Analiza opterećenja

Opterećenje : 153,31 kN

Vlastita težina temelja : 1,2∙0,25∙25= 7,5 kN

Opterećenja tla: 0,95･0,95･18= 16,24 KN

Podna obloga parket d=2 : 0,14･0.95=0.133KN

Korisno opterećenje : 0,95･2=1,9KN

Podna ploča : 0,95･0,1･24=2,28KN

Termoizolacija d=5 cm : 0,02･0,95=0,019KN

Cementni estrih d=5cm :1,05･0,95=0,99KN

∑V = 184,65 kN

σ =∑V

A= 153 ,875

kN

m2< 159 ,50

kN

m2

odstupanje 3,5 %

Temeljna stopa

Vg = 993,16 kNVp = 317,8kN

V = V g + V P = 1310 ,96 kNUSVAJAMO :Df = 1,2 m Stub 35x35

DOPUŠTENA NOSIVOST TLA :

σ dop =12


cm = cFc

= 42,5

= 1,6 kN

m2

tg φm =tg ϕϕ

Fϕϕ

=tg 241,5

= 0 ,2968 → ϕm = 16 ,53° → N c = 12 ; N γ = 1,9

q = γ⋅D f = 18⋅1,2=21 ,6kN

m2; ic = 1 iγ = 1

Sc = 1 + 0,2 ⋅ BL

= 1 ,2 ; Sγ= 1 − 0,4 ⋅ BL

= 0,6

Određivanje širine temelja1. IteracijaPretpostavimo : B= 3,0 m

dc = 1 + 0 ,35 ⋅Df

B= 1 ,14

σ dop =0,5⋅8⋅3⋅1,9⋅0,6⋅1,0 (1,6 + 21 ,6 x 0 ,2968 ) ⋅12 ⋅1,2 ⋅ 1,0⋅1 ,14 ⋅ 1,0 + 21 ,6= 166 ,78 kN

m2

potA =Nσn

=1310 ,96141 ,28

= 9 ,27 m ;

A = B2 → B1 = √9 ,27 = 3 ,04 m

dc = 1+0 ,35 1,23,1

=1 ,13

σ dop =12

⋅ 8⋅3,1⋅0,6 ⋅1 + (1,6+21 ,6⋅0 ,2968 ) ⋅12 ⋅ 1,2 ⋅ 1,13 ⋅ 1 + 21 ,6 = 166 ,08kN

m2

σ n = σdop − β⋅ D f ⋅γB = 166 ,08− 0 ,85 ⋅1,2⋅25 = 140 ,58kN

m2

potA = Nσ n

= 1310 ,96140 ,58

= 9 ,32 m

A = B2 → B2 = √9 ,32 = 3 ,05 m

usvojeno : B=3,1 m

Određivanje visine temelja:

M I = M II =V ⋅( B − b )

8=1310 ,96⋅( 3,1−0 ,35 )

8= 450 ,64 kNm

T I = T II =V ⋅(B − b )2⋅B

=1310 .96⋅( 3,1−0 ,35 )

2⋅3,1= 581 ,47 kN

γ sr =1,35 ⋅V g + 1,5 ⋅V p

V g + V p

= 1,39

M sd = γ sr⋅M = 1 ,39 ⋅450 ,64 = 626 ,38 kNm

V sd = γ sr⋅T = 1 ,39 ⋅581 ,47 = 808 ,24 kN

Beton C25/30 f cd=

251,5

=16 ,67MPa; Armatura B500

f yd=5001 ,15

=434 ,78MPa

dn = 2,311⋅√α⋅M sd

B⋅ f cd

= 2 ,311⋅√ 2 ,25⋅626 ,38⋅100310⋅1 ,667= 38 ,16 cm

d(V)=ne realni rezultati

h = dn + φ + φ2

+ c = 42 ,9 cm

usvojeno h =43cm

pot A s1=

β⋅M sd

0,9⋅d⋅ f yd

= 1,1⋅626 ,38 ⋅1000,9⋅39⋅43 , 478

= 45 ,53 cm2

usvojeno 12R∅ 14 As =45,62 cm2

Dimenzionisanje temelja na pobijanje:

Vsd = 808,24 KNdkr = 2⋅1,5 ⋅d + b = 2⋅1,5⋅43 + 39 = 152 cm

μkr = dkr⋅π = 477 ,28 cm

vsd =V sd

μkr

= 1 ,69kNcm,

τrd = 0,03

kN

cm2

k = 1,6 - d = 1,6- 0,39= 1,21

ρl =As

b⋅d= 45 ,62310⋅39

= 0 ,0037 ; ρlmin = 0 ,005

V rd1= τ rd⋅k ⋅(1,2 + 40⋅ ρl ) ⋅ d = 0 ,03 ⋅1,2 1⋅(1,2 + 40⋅0 ,005 )⋅39 = 1 ,98 kN

cm,

V rd1> V sd →

nije potrebna dodatna armatura

Kontrola napona

1.2

0.43

3.1


Ukupno opterećenje : 1310,96 kN

Težina temelja: (3,1∙0,34+1,375 ∙0,09+0,35 ∙0,81¿ ∙25= 33,43kN

Težina tla: (2,75∙0,81+1,375 ∙0,09¿ ∙ 18 = 42,32kN

Podna obloga parket d=2cm : 0,14∙1,375∙2=0,38kN

Korisno opterećenje: 2･1,375･2=5,5kN

Podna ploča: 1,375･0,1･24･2=6,6kN

Termoizolacija d=5 cm : 0,02･1,375･2= 0,055kN

Cementni estrih d=5cm: 1,05 ･ 1,375 ･ 2=2,88kN

∑V =¿¿ 1402,12 KN

σ =∑V

A= 145 ,9

kN

m2< 166 ,08

kN

m2 odstupanje 12 %

Za B=2,9 m i H= 0,39m dobijamo

∑V = 1395,97 => odstupanje 0,05

Dimenzionisanje temeljne grede

23 420 420 23

S1 S2 S1

V1=Vg1+Vq1=173,73+14,36=188,09kN

V2=Vg2+Vq2=689,01+113,46=802,47kN

∑V =2V 1+V 2=1178,65 kN

¿4,2m

a=b2+5cm=35

2+5=23cm

L=(a+ξ )∙2=2 ∙ (0,23+4,2 )=8,86m

σ dop = 12

⋅γ⋅B⋅N γ ⋅Sγ ⋅iγ + (cm + q⋅tg ϕm )⋅N c⋅Sc⋅dc⋅ ic + q

cm =cFc

=42,5

= 1,6kN

m2

tg ϕm =tgϕϕ

Fϕ

= tg 241,5

= 0 ,2968 → ϕϕm = 21,83° → N c = 12 ; N γ = 1,9

ic = 1 ; i γ = 1

pretpostavimo

D f =1,2m ; B=1,0m

q = γ⋅D f = 18⋅1,2=21 ,6kN

m2

Sc = 1 + 0,2 ⋅ BL

= 1 + 0,2⋅ 1 ,008 ,86

= 1 ,02 ; Sγ= 1 − 0,4 ⋅BL

= 1 − 0,4⋅ 1,08 ,86

= 0 ,95

dc = 1,0+0 ,35⋅D f

B=1,0+0 ,35⋅1,2

1,0=1 ,35

σ dop =12

⋅ 8 ⋅1,0 ⋅1,9⋅ 0 ,95 ⋅1 + (1,6 + 21 ,6⋅ 0 ,2968 ) ⋅12 ⋅ 1 ,02 ⋅ 1 ,35 ⋅1 + 21 ,6 = 161 , 19 kN

m2

1. Iteracija

σ n = σdop − β⋅ D f ⋅γB = 161 ,19 − 0 ,85 ⋅1⋅25 = 135 ,69kN

m2

B= Nσ n⋅L

= 1178 ,658 ,86⋅135 ,69

= 0 ,98 m → B1= 1 m

Sc = 1 + 0,2 ⋅ BL

= 1 + 0,2⋅ 18 ,86

= 1 ,02 ; Sγ= 1 − 0,4 ⋅BL

= 1 − 0,4⋅ 18 ,86

= 0 ,95

σ dop =12

⋅ 8 ⋅1 ⋅1,9 ⋅ 0 ,95 ⋅1 + (1,6+ 21 ,6 ⋅ 0 ,2968 ) ⋅12⋅ 1 ,02 ⋅1 ,35⋅ 1 + 21,6 = 161 ,19 kNm2

σ n = σdop − β⋅ D f ⋅γB = 161 ,19− 0 ,85 ⋅1⋅25 = 135 ,69kN

m2

2. Iteracija

B= Nσ n⋅L

= 1178 ,658 ,86⋅135 ,69

= 0 ,98 m → B2 = 1 m

usvojeno B = 1 mvisina presjeka usvojena H=1,2m

STATIČKI PRORAČUN Vsd1=1,35∙173,73+1,5 ∙14,36=256,075kN Vsd2=1,35∙689,01+1,5∙113,46=1100,35 kN ∑Vsd=2Vsd 1+Vsd2=1612,49kN

σsd=∑Vsd

B ∙L=1612,498,86 ∙1

=181,99kN /m'

σ '=σsd ∙B=181,99∙1=181,99 kN /m'

23 420 420 23

S1 S2 S1

++

--

41,86

214,21550,175

41,86

214,21

550,175

T

4,81

121,24

710,35

4,81

121,24

-

+

-

M

Proračun potrebne armature

Iznad oslonca:

Msd = 710,35 kNm

μsd =M sd

b ⋅d2⋅ f cd

= 710 ,35 ⋅ 100100 ⋅1142⋅1 ,667

= 0 ,032 → ρ=0 ,975

pot A s1=

M sd

ρ ⋅d⋅ f yd

= 710 ,35⋅1000 ,975⋅114⋅43 ,5

= 14 ,62 cm2

As1,min= 0 ,0015⋅ b ⋅d = 14 ,1 cm2

usvojeno 6R∅ 18 As = 15,27 cm2

U polju :Msd = 121,24kNm

μsd =M sd

b ⋅d2⋅ f cd

= 121 ,24 ⋅ 100100 ⋅1142⋅1 ,667

= 0 ,005 → ρ = 0 ,992

pot A s1=

M sd

ρ ⋅d⋅ f yd

= 121,24⋅1000 ,992⋅114⋅43 ,5

= 2 ,46 cm2

As1,min= 0 ,0015⋅ b ⋅d = 14 ,1 cm2

usvojeno 6R∅ 18 As = 15,27cm2

Određivanje poprečne armature :Vsd = 550,175kN

τrd = 0,03

kN

cm2

k = 1,6 –d = 1,6- 1,14 = 0,46

ρl =15 ,27+15 ,27100⋅120

= 0 ,0025

ρlmin=0 ,005

V rd1= τ rd⋅k ⋅(1,2 + 40⋅ ρl ) bw ¿ d = 0 ,03 ⋅0 ,46⋅ (1,2 + 40⋅0 ,005 )⋅100 ⋅ 114= 220 ,248

kN

m,

V rd1< V sd →

potrebna je dodatna armatura

ν = 0,7 − 25200

= 0 ,575

z = 0,9 d = 0,9 114 = 102,6 cm∙ ∙

V rd 2 = 0,5 ⋅ ν⋅ f cd⋅bw⋅z = 0,5 ⋅ 0 ,575⋅1 ,667⋅100⋅102 ,6 = 5463 ,59kNsmax = 30 cm,ϕ10 → Asw = 0,79 cm

sw =As

w⋅ f yd⋅m⋅z

V sd

= 0 ,79⋅43 ,5 ⋅2⋅102 ,6550 ,175

= 12 ,8 cmusvojenoRϕ10/15 cm

Analiza opterećenjaOpterećenje : 1178,65 kNVlastita težina temelja: 1∙1,2 ∙ 25 = 30 kN ∑V = 1208,65 kN

σ =∑V

A= 136 ,41

kN

m2< 161 ,19

kN

m2

odstupanje 15% -->→→→→→→→→→→

smanjujemo širinu temelja

B=0,88 m

Opterećenje : 1178,65 kNVlastita težina temelja: 0,88 ∙ 1,2 ∙ 25 = 26,4 kN ∑V = 1205,05 kN

σ =∑V

A= 154 ,55

kN

m2< 161 ,19

kN

m2

odstupanje 4,11%

Temeljna traka (duža strana)

Vg = 126,94

kNm'

Vp = 23,13

kNm'

V = V g + V P = 150 ,07 kN

m,

γ=18 KN/m3

c=4 Mpa Φ=24 〫

pretpostavimo

Df = 1,2 mDOPUŠTENA NOSIVOST TLA :

σ dop =12


cm = cFc

= 42,5

= 1,6 kN

m2

tg φm =tg φϕ

Fϕ

= tg 241,5

= 0 ,2968 → φm = 16 ,53° → N c = 12; N γ = 1,9

q = γ⋅D f = 21,6kN

m2; ic = 1 iγ = 1

Sc = 1 + 0,2 ⋅ BL

= 1 ,0 ; Sγ= 1 − 0,4 ⋅ BL

= 1,0

dc = 1,0+0 ,35⋅D f

B=1,0

σ dop = (1,6 + 21 ,6⋅ 0 ,2968 ) ⋅12 ⋅ 1 ⋅1⋅1 + 21 ,6= 117 ,52 kN

m2

Određivanje širine temelja :

1.Iteracija

σ n = σdop − β⋅ D f ⋅γB = 117 ,62 − 0 ,85 ⋅1,2⋅25 = 92 ,23kN

m2

B=Nσn

=150 ,0792 ,23

= 1 ,62m

B⋅0 ,65 → B1 = 1 ,05 m

Sc = 1 + 0,2 ⋅ BL== 1,0 ; Sγ= 1 − 0,4 ⋅B

L= 1,0

dc = 1,0+0 ,35⋅

D f

B=1,0+0 ,35⋅ 1,2

1,05=1 ,35

σ dop =12

⋅ 8 ⋅1 ,05 ⋅1,9 ⋅ 1 ⋅1 + (1,6 + 21 ,6⋅0 ,2968 ) ⋅12 ⋅ 1,0 ⋅ 1 ,35 ⋅ 1 + 21 ,6 = 159 ,35 kN

m2

2. Iteracija

σ n = σdop − β⋅ D f ⋅γB = 159 ,35 − 0 ,85 ⋅1,2⋅25 = 133 ,85kN

m2

B =Nσn

=150 ,07133 ,85

= 1 ,12 m

B = 1 ,12+1,052

=1,095m

usvojena širina temelja B=1,1m

σ n =∑ V

A=150,071,1

=136,42kN /m2

Određivanje visine temeljne trake

0,25 1.2

1,1

0,55

σnet=136,42kN/m2

max

M= σnet*0.552/2=20,63kNm/m’

γ sr =1,35 ⋅V g + 1,5 ⋅V p

V g + V p

= 1,37

M sd = Mmax ɣ∙ SR =20,63 1,37 =28,26∙

kNmm'

h=a/2 = 42,5/2=21,25 usvojeno h=25 cmh=d+Φ/2+c --> d= 21,5cm

μsd =M sd

b ⋅d2⋅ f cd

=20 ,63⋅100110⋅21 ,5⋅21 ,5⋅1 ,667

=0 ,024→ ρ=0 ,98

Usvajanje armature :

pot A s1= Msd

ρ⋅d⋅fyd= 20 ,63⋅1000 ,98⋅21 ,5⋅43 ,5

=2 ,25cm2

Asmin = 0,0015･b･d=3,54cm2 usvojeno R424 (As=4,24cm2)


Opterećenje : 150,07 kN

Vlastita težina temelja : 1,1∙0,25∙25= 6,87 kN

Opterećenja tla: 0,85･0,95･18= 14,53KN

Podna obloga parket d=2 : 0,14･0.85=0.119KN

Korisno opterećenje : 0,85･2=1,7KN

Podna ploča : 0,85･0,1･24=2,04KN

Termoizolacija d=5 cm : 0,02･0,85=0,017KN

Cementni estrih d=5cm :1,05･0,85=0,89KN

∑V = 176,23 kN

σ =∑V

A= 160 ,2

kN

m2> 159 ,350

kN

m2

odstupanje 0,5 %

Documents

Dimenzionisanje