233 Projekt wykonawczy.

3.1 Płyta

3.1.1 Dane wyjściowe

- beton f_cd=21,4 Mpa

- stal f_yd=420Mpa B500SP

- otulina XC3 c_min=25mm

- otulina REI90 a=25mm

3.1.2 Zestawienie obciążeń:

Wyszczególnienie	Obc. Char.	γf	Obc. obliczeniowe
		>1	=1
Gładz cementowa2,5cm 0,025*21	0,52	1,35	1
Płytki PVC	0,07	1,35	1
Styropian 0,03*0,45	0,014	1,35	1
Płyta żelbetowa 0,14	3,5	1,35	1
Tynk cem-wap 19*0,02	0,38	1,35	1
SUMA	gk=4,484
Obc zmienne	qk=8	1,5	1

g_k=4,484kN/m

g_d1=6,036kN/m

g_d2=4,484kN/m

q_k=8kN/m

q_d=12kN/m

Δg= g_d1- g_d2=6,036-4,484=1,552kN/m

$\overline{q}$=q_d +Δg=13,552kN/m

3.1.3 Schemat statyczny

1. M_{Ed, A − B}^max = 3, 998kNm

2. M_{Ed, B − C}^max = 3, 044kNm

3. M_{Ed, C − C}^max = 3, 215kNm

4. M_Ed, B^min = −6, 603kNm

5. M_Ed, C^min = −6, 259kNm

6. M_Ed, Bkr^min = −4, 827kNm

7. M_Ed, Ckr^min = −4, 517kNm

$$8.\ {\overline{M}}_{B - C} = \frac{1}{3}\left( M_{Ed,B - C}^{\min} + max\left\{ M_{Ed,B}^{\min} + M_{Ed,C}^{\min} \right\} \right) = \frac{1}{3}\left( 3,044 + 6,603 \right) = 3,216kNm$$

$$9.\ {\overline{M}}_{C - C} = \frac{1}{3}\left( M_{Ed,C - C}^{\min} + M_{Ed,C}^{\min} \right) = \frac{1}{3}\left( 3,215 + 6,259 \right) = 3,158kNm$$

3.1.4 Wymiarowanie ze względu na zginanie:

$$A_{S,min} = 0,26*\left( \frac{f_{\text{ctm}}}{f_{\text{yk}}} \right)*b*d \geq 0,0013*b*d$$

$$0,26*\left( \frac{2,9}{500} \right)*100*10,1 \geq 0,0013*100*10,1$$

1, 53cm² ≥ 1, 31cm² ok.

$$\sum_{}^{}{M_{As1,A - B} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right)} - M_{\text{Ed}} = 0$$

21400kPa*1,0m*x_eff*(0,101m-0,5*x_eff)-3,998kNm=0

X_eff=0,0015m

$$\xi_{\text{eff}} = \frac{0,0015m}{0,101} = 0,015$$

$$\xi_{eff,lim} = 0,8*\frac{\varepsilon_{\text{ck}}}{\varepsilon_{\text{ck}} + \varepsilon_{\text{yd}}} = 0,8*\frac{3,5}{3,5 + 2,15} = 0,496$$

ξ_eff, lim > ξ_eff ok.

f_cd * b * x_eff = f_yd * A_S1

$A_{S1} = \frac{f_{\text{cd}}}{f_{\text{yd}}}*b*x_{\text{eff}} = \frac{21,4}{420}*1,0*0,0015 =$0,95cm²

A_S, min = 1, 53cm²

Przyjęto 6Φ6 =1,69cm²

$$\sum_{}^{}{M_{As1,B - C} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right)} - M_{\text{Ed}} = 0$$

21400kPa*1,0m*x_eff*(0,101m-0,5*x_eff)-3,044kNm=0

X_eff=0,0014m

$$\xi_{\text{eff}} = \frac{0,0014m}{0,101} = 0,014$$

$$\xi_{eff,lim} = 0,8*\frac{\varepsilon_{\text{ck}}}{\varepsilon_{\text{ck}} + \varepsilon_{\text{yd}}} = 0,8*\frac{3,5}{3,5 + 2,15} = 0,496$$

ξ_eff, lim > ξ_eff ok.

f_cd * b * x_eff = f_yd * A_S1

$A_{S1} = \frac{f_{\text{cd}}}{f_{\text{yd}}}*b*x_{\text{eff}} = \frac{21,4}{420}*1,0*0,0014 =$0,72cm²

A_S, min = 1, 53cm²

Przyjęto 6Φ6=1,69cm²

$$\sum_{}^{}{M_{As1,C - C} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right)} - M_{\text{Ed}} = 0$$

21400kPa*1,0m*x_eff*(0,101m-0,5*x_eff)-3,215kNm=0

X_eff=0,0015m

$$\xi_{\text{eff}} = \frac{0,0015m}{0,101} = 0,015$$

$$\xi_{eff,lim} = 0,8*\frac{\varepsilon_{\text{ck}}}{\varepsilon_{\text{ck}} + \varepsilon_{\text{yd}}} = 0,8*\frac{3,5}{3,5 + 2,15} = 0,496$$

ξ_eff, lim > ξ_eff ok.

f_cd * b * x_eff = f_yd * A_S1

$A_{S1} = \frac{f_{\text{cd}}}{f_{\text{yd}}}*b*x_{\text{eff}} = \frac{21,4}{420}*1,0*0,0015 =$0,76cm²

A_S, min = 1, 53cm²

Przyjęto 6Φ6=1,69cm²

$$\sum_{}^{}{M_{As1,B} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right)} - M_{\text{Ed}} = 0$$

21400kPa*1,0m*x_eff*(0,101m-0,5*x_eff)-6,603kNm=0

X_eff=0,0031m

$$\xi_{\text{eff}} = \frac{0,0031m}{0,101} = 0,0307$$

$$\xi_{eff,lim} = 0,8*\frac{\varepsilon_{\text{ck}}}{\varepsilon_{\text{ck}} + \varepsilon_{\text{yd}}} = 0,8*\frac{3,5}{3,5 + 2,15} = 0,496$$

ξ_eff, lim > ξ_eff ok.

f_cd * b * x_eff = f_yd * A_S1

$A_{S1} = \frac{f_{\text{cd}}}{f_{\text{yd}}}*b*x_{\text{eff}} = \frac{21,4}{420}*1,0*0,00307 = 1$,58cm²

A_S, min = 1, 53cm²

Przyjęto 6Φ6 =1,69cm².

$$\sum_{}^{}{M_{As1,C} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right)} - M_{\text{Ed}} = 0$$

21400kPa*1,0m*x_eff*(0,101m-0,5*x_eff)-6,259kNm=0

X_eff=0,0029m

$$\xi_{\text{eff}} = \frac{0,0014m}{0,101} = 0,029$$

$$\xi_{eff,lim} = 0,8*\frac{\varepsilon_{\text{ck}}}{\varepsilon_{\text{ck}} + \varepsilon_{\text{yd}}} = 0,8*\frac{3,5}{3,5 + 2,15} = 0,496$$

ξ_eff, lim > ξ_eff ok.

f_cd * b * x_eff = f_yd * A_S1

$A_{S1} = \frac{f_{\text{cd}}}{f_{\text{yd}}}*b*x_{\text{eff}} = \frac{21,4}{420}*1,0*0,0029 = 1$,50cm²

A_S, min = 1, 53cm²

Przyjęto 6Φ6=1,69cm²

$$\sum_{}^{}{M_{As1,Bkr} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right)} - M_{\text{Ed}} = 0$$

21400kPa*1,0m*x_eff*(0,101m-0,5*x_eff)-4,827kNm=0

X_eff=0,0022m

$$\xi_{\text{eff}} = \frac{0,0022m}{0,101} = 0,022$$

$$\xi_{eff,lim} = 0,8*\frac{\varepsilon_{\text{ck}}}{\varepsilon_{\text{ck}} + \varepsilon_{\text{yd}}} = 0,8*\frac{3,5}{3,5 + 2,15} = 0,496$$

ξ_eff, lim > ξ_eff ok.

f_cd * b * x_eff = f_yd * A_S1

$$A_{S1} = \frac{f_{\text{cd}}}{f_{\text{yd}}}*b*x_{\text{eff}} = \frac{21,4}{420}*1,0*0,0022 = 1,15\text{cm}^{2}$$

A_S, min = 1, 53cm²

Przyjęto 6Φ6=1,69cm²

$$\sum_{}^{}{M_{As1,Ckr} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right)} - M_{\text{Ed}} = 0$$

21400kPa*1,0m*x_eff*(0,101m-0,5*x_eff)-4,517kNm=0

X_eff=0,0021m

$$\xi_{\text{eff}} = \frac{0,0021m}{0,101} = 0,021$$

$$\xi_{eff,lim} = 0,8*\frac{\varepsilon_{\text{ck}}}{\varepsilon_{\text{ck}} + \varepsilon_{\text{yd}}} = 0,8*\frac{3,5}{3,5 + 2,15} = 0,496$$

ξ_eff, lim > ξ_eff ok.

f_cd * b * x_eff = f_yd * A_S1

$$A_{S1} = \frac{f_{\text{cd}}}{f_{\text{yd}}}*b*x_{\text{eff}} = \frac{21,4}{420}*1,0*0,0021 = 1,08\text{cm}^{2}$$

A_S, min = 1, 53cm²

Przyjęto 6Φ6=1,69cm²

Pręty główne:

S_max,slabs={2h;250mm}=250mm

Pręty rozdzielcze:

S_max,slabs={3h;400mm}=400mm

3.1.5 Sprawdzenie ze względu na ścinanie:

V_Ed, max = 19, 554kN

$$k = 1 + \sqrt{\frac{200}{d}} = 1 + \sqrt{\frac{200}{101}} = 2,41 > 2 \rightarrow k = 2.$$

$$\rho_{L} = \frac{A_{\text{SL}}}{d*b} = \frac{1,69\text{cm}^{2}}{10,1*100} = 0,00167$$

$$C_{Rd,c} = \frac{0,18}{\gamma_{c}} = \frac{0,18}{1,4} = 0,129$$

$$V_{Rd,c} = C_{Rd,c}*k\left( 100*\rho_{L}*f_{\text{ck}} \right)^{\frac{1}{3}} = 0,129*{2*\left( 100*0,00167*30 \right)}^{\frac{1}{3}}*b*d = 44,62kN > V_{Ed,max}$$

Przekrój nie wymaga dodatkowego zbrojenia na ścinanie.

3.1.6 Sprawdzenie płyty ze względu na ugięcie:

Dla kat E (pow. magazynowe) wsp. Ψ₂=0,8

M_Ek, max = 2, 412kNm

ξ = 0, 9

$$\sigma_{s} = \frac{M_{Ek,max}}{\xi*d*A_{S1}} = \frac{2,412}{0,9*0,101*1,69*10^{- 4}}157MPa$$

K=1,5

$$\left( \frac{l_{\text{eff}}}{d} \right) \leq \left( \frac{l_{\text{eff}}}{d} \right)_{\max}*\frac{310}{\sigma_{s}}*K$$

$$\left( \frac{2050}{101} \right)20,29 \leq \left( \frac{2050}{101} \right)_{\max}*\frac{310}{157}*1,5 = 60,12\ \ \ ok.$$

4. Żebro ż2

4.1. dane wyjściowe

- beton C30/37; f_cd=21,4 MPa; f_ck=30 MPa; f_ctm=2,9 Mpa; f_ctk=2,0 Mpa; E_cm=32GPa

- stal f_yd=420Mpa; f_yk=500MPa

- E_s=200GPa

- otulina XC3 c_min=25mm

- otulina REI90 a=40mm; a_sd=50mm

4.2 Zestawienie obciążeń

Wyszczególnienie	Obc. Char.	γf	Obc. obliczeniowe
		>1	=1
gk,pl*l2,pł	9,225
gk,pl*l2,pł
Żebro 0,25x0,2*25	1,25	1,35	1
Tynk 0,25*0,02*2*19	0,19	1,35	1
Suma	gk=10,66
Obc. zmienne	qk=16,4	1,5	1

4.3 Schemat statyczny

g_k=10,66kN/m

g_d1=14,33kN/m

g_d2=10,66kN/m

q_k=16,4kN/m

q_d=24,6kN/m

Δg= g_d1- g_d2=14,33-10,66=3,67kN/m

$\overline{q}$=q_d +Δg=24,6+3,67=28,27kN/m

Obwiednia momentów:

1. M_{Ed, A − B}^max = 99, 66kNm

2. M_{Ed, B − C}^max = 95, 54kNm

3. M_{Ed, C − C}^max = 100, 50kNm

4. M_Ed, B^min = −140, 33kNm

5. M_Ed, C^min = −144, 85kNm

6. M_Ed, Bkr^min = −125, 55kNm

7. M_Ed, Ckr^min = −130, 79kNm

4.4 Wymiarowanie na moment:

Na podporze:

$$b_{1} = b_{2} = \frac{b_{}{- b}_{w}}{2} = \frac{2050 - 250}{2} = 900mm$$

l₀ = 0, 85 * l₁ = 0, 85 * 4925 = 4186mm

b_eff, i = 0, 2b_i + 0, 1l₀    ≤ {0, 2l₀;b}

b_eff, 1 = 0, 2 * 0, 9 + 0, 1 * 4, 186 = 0, 599  < {0,84;0,9}

b_eff, 1 = b_eff, 2 = 0, 599

b_eff = 2 * b_eff, 1 + b_w = 1, 448m

W przęśle środkowym:

$$b_{1} = b_{2} = \frac{b_{}{- b}_{w}}{2} = \frac{2050 - 250}{2} = 900mm$$

l₀ = 0, 7 * l₁ = 0, 7 * 5800 = 4060mm

b_eff, i = 0, 2b_i + 0, 1l₀    ≤ {0, 2l₀;b}

b_eff, 1 = 0, 2 * 0, 9 + 0, 1 * 4, 06 = 0, 586  < {0,812;0,9}

b_eff, 1 = b_eff, 2 = 0, 586

b_eff = 2 * b_eff, 1 + b_w = 1, 422m

Wysokośc użyteczna żebra:

a=40mm;c_nom=25mm;⌀_strz = 8mm; ⌀ = 20mm

a₁=c_nom+Φ_strz+Φ/2=25+8+10=43=45mm

d=h_ż-a₁=390-45=345mm

d_wew=345+0,3333*150=395mm

Sprawdzenie rzeczywistego kształtu rozkładu momentów w żebrze:

M_F = b_eff * h_f * f_cd * (d−0,5h_f)

Dla przęsła skrajnego:

M_F = b_eff * h_f * f_cd * (d−0,5h_f) = 1, 488m * 0, 14m * 21400kPa * (0,345−0,5*0,14) = 1226kNm > 145kNm

Dla przęseł srodkowych:

M_F = b_eff * h_f * f_cd * (d−0,5h_f) = 1, 422m * 0, 14m * 21400kPa * (0,345−0,5*0,14) = 1172kNm > 145kNm

Przekrój jest pozornie teowy.

$$\mu_{\text{sc}} = \frac{M_{\text{Ed}}}{f_{\text{cd}}*b_{\text{eff}}*d^{2}} = \frac{99,66}{21400*1,488*{0,345}^{2}} = 0,027 = \zeta = 0,986;$$

$$A_{s1} = \frac{M_{\text{Ed}}}{f_{\text{yd}}*d*\zeta} = \frac{99,66}{420000*0,345*0,986} = 0,000698m^{2} = 6,98cm^{2}$$

$$\mu_{\text{sc}} = \frac{M_{\text{Ed}}}{f_{\text{cd}}*b_{\text{eff}}*d^{2}} = \frac{95,54}{21400*1,422*{0,345}^{2}} = 0,026 = \zeta = 0,986;$$

$$A_{s1} = \frac{M_{\text{Ed}}}{f_{\text{yd}}*d*\zeta} = \frac{95,54}{420000*0,345*0,986} = 0,000669m^{2} = 6,69cm^{2}$$

$$\mu_{\text{sc}} = \frac{M_{\text{Ed}}}{f_{\text{cd}}*b_{\text{eff}}*d^{2}} = \frac{100,5}{21400*1,422*{0,345}^{2}} = 0,028 = \zeta = 0,985;$$

$$A_{s1} = \frac{M_{\text{Ed}}}{f_{\text{yd}}*d*\zeta} = \frac{100,5}{420000*0,345*0,985} = 0,000704m^{2} = 7,04cm^{2}$$

$$\mu_{\text{sc}} = \frac{M_{\text{Ed}}}{f_{\text{cd}}*b_{\text{eff}}*d^{2}} = \frac{140,33}{21400*0,2*{0,395}^{2}} = 0,21 = \zeta = 0,88;$$

$$A_{s1} = \frac{M_{\text{Ed}}}{f_{\text{yd}}*d*\zeta} = \frac{140,33}{420000*0,395*0,88} = 0,000961m^{2} = 9,61cm^{2}$$

$$\mu_{\text{sc}} = \frac{M_{\text{Ed}}}{f_{\text{cd}}*b_{\text{eff}}*d^{2}} = \frac{144,85}{21400*0,2*{0,395}^{2}} = 0,217 = \zeta = 0,876;$$

$$A_{s1} = \frac{M_{\text{Ed}}}{f_{\text{yd}}*d*\zeta} = \frac{144,85}{420000*0,395*0,876} = 0,000997m^{2} = 9,97cm^{2}$$

$$\mu_{\text{sc}} = \frac{M_{\text{Ed}}}{f_{\text{cd}}*b_{\text{eff}}*d^{2}} = \frac{125,55}{21400*0,2*{0,345}^{2}} = 0,246 = \zeta = 0,856;$$

$$A_{s1} = \frac{M_{\text{Ed}}}{f_{\text{yd}}*d*\zeta} = \frac{125,55}{420000*0,345*0,856} = 0,00101m^{2} = 10,01cm^{2}$$

$$\mu_{\text{sc}} = \frac{M_{\text{Ed}}}{f_{\text{cd}}*b_{\text{eff}}*d^{2}} = \frac{130,79}{21400*0,2*{0,345}^{2}} = 0,258 = \zeta = 0,848;$$

$$A_{s1} = \frac{M_{\text{Ed}}}{f_{\text{yd}}*d*\zeta} = \frac{130,79}{420000*0,345*0,848} = 0,00106m^{2} = 10,6cm^{2}$$

Minimalna powierzchnia zbrojenia na zginanie:

f_ctm=2,9Mpa

$$A_{s,min} = 0,26*\frac{f_{\text{ctm}}}{f_{yk}}*b*d \geq 0,0013b*d;$$

$$A_{s,min} = 0,26*\frac{2,9}{500}*0,2*0,345 \geq 0,0013*0,2*0,345$$

A_s, min = 1, 04cm² ≥ 0, 897cm²;

Minimalna odległośc między prętami:

Max(Φ,20mm,d_g+5mm)=max(20,20,21mm)=21mm

Zbrojenie górne4Φ20=12,57cm²;

Zbrojenie dolne3Φ20=9,42cm²;

Rozmieszczenie prętów zbrojeniowych:

Nośnośc momentowa:

4 pręty:

$$x_{\text{eff}} = \frac{A_{s1}*f_{\text{yd}}}{f_{\text{cd}}*b_{w}} = \frac{0,001257*420}{21,4*0,2} = 0,1234m$$

S=3,14*4,5=14,14cm³

A=12,57cm²

$$x_{s} = \frac{S}{A} = \frac{14,14}{12,57} = 1,13cm$$

d=0,345-0,011=0,334m

M_Rd=A_s1*f_yd*(d-0,5x_eff)=0,001257*420000*(0,334-0,5*0,1234)=143,75kNm>130,79kNm

3 pręty

$$x_{\text{eff}} = \frac{A_{s1}*f_{\text{yd}}}{f_{\text{cd}}*b_{w}} = \frac{0,000943*420}{21,4*0,2} = 0,0925m$$

M_Rd=A_s1*f_yd*(d-0,5x_eff)=0,000943*420000*(0,345-0,5*0,0925)=118,26kNm

2 pręty:

$$x_{\text{eff}} = \frac{A_{s1}*f_{\text{yd}}}{f_{\text{cd}}*b_{w}} = \frac{0,000628*420}{21,4*0,2} = 0,0617m$$

M_Rd=A_s1*f_yd*(d-0,5x_eff)=0,000628*420000*(0,345-0,5*0,0617)=82,91kNm

Zbrojenie dołem:

Podpora A:

3 pręty:

$$x_{\text{eff}} = \frac{A_{s1}*f_{\text{yd}}}{f_{\text{cd}}*b_{w}} = \frac{0,000943*420}{21,4*1,488} = 0,1244m$$

M_Rd=A_s1*f_yd*(d-0,5x_eff)=0,000943*420000*(0,345-0,5*0,1244)=134,18kNm

2 pręty

$$x_{\text{eff}} = \frac{A_{s1}*f_{\text{yd}}}{f_{\text{cd}}*b_{w}} = \frac{0,000628*420}{21,4*1,488} = 0,0828m$$

M_Rd=A_s1*f_yd*(d-0,5x_eff)=0,000628*420000*(0,345-0,5*0,0828)=89,9kNm

Podpora B:

3 pręty:

$$x_{\text{eff}} = \frac{A_{s1}*f_{\text{yd}}}{f_{\text{cd}}*b_{w}} = \frac{0,000943*420}{21,4*1,422} = 0,1301m$$

M_Rd=A_s1*f_yd*(d-0,5x_eff)=0,000943*420000*(0,345-0,5*0,1301)=134,06kNm

2 pręty

$$x_{\text{eff}} = \frac{A_{s1}*f_{\text{yd}}}{f_{\text{cd}}*b_{w}} = \frac{0,000628*420}{21,4*1,422} = 0,0867m$$

M_Rd=A_s1*f_yd*(d-0,5x_eff)=0,000628*420000*(0,345-0,5*0,0828)=89,85kNm

Wyznaczenie a_L

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 1, 0

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*1*1*1,4=3,15Mpa

⌀ = 20mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,prov}} = 420*\frac{1,67}{6,28} = 111,63MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{20}{4}*\frac{111,63}{3,15} = 177,19mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 200mm

α₁ − α₅ = 1, 0

l_bd = l_b, min = 200mm

Długośc zakotwienia prętów na przęśle:

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 0, 7

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*0,7*1*1,4=2,21Mpa

⌀ = 20mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,req}} = 420*\frac{6,98}{9,42} = 279,99MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{20}{4}*\frac{279,99}{2,21} = 634mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 200mm

α₁ − α₅ = 1, 0

l_bd = l_b, rqd = 640mm

Długośc zakotwienia prętów górnych nad podpora A

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 0, 7

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*0,7*1*1,4=2,21Mpa

⌀ = 20mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,req}} = 420*\frac{1,67}{9,42} = 74,45MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{20}{4}*\frac{74,45}{2,21} = 168mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 200mm

α₁ − α₅ = 1, 0

l_bd = l_b, min = 200mm

Długośc zakotwienia prętów górnych nad podpora B

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 0, 7

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*0,7*1*1,4=2,21Mpa

⌀ = 20mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,req}} = 420*\frac{6,28}{9,42} = 279,99MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{20}{4}*\frac{279,99}{2,21} = 634mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 200mm

α₁ − α₅ = 1, 0

l_bd = l_b, min = 640mm

Wymiarowanie na ścinanie:

Obwiednia sił tnących:

V_Ed^A, kr = 83, 22kN

V_Ed^A′ = 69, 8kN

V_Ed^B, L, kr = 122, 93kN

V_Ed^B′ = 109, 5kN

V_Ed^B, P, kr = 126, 2kN

V_Ed^B′, P = 112, 8kN

V_Ed^C, L, kr = 124, 7kN

V_Ed^C′ = 111, 3kN

V_Ed^C, P, kr = 126, 98kN

V_Ed^C′, P = 113, 5kN

Podpora skrajna:

$$V_{Rd,C}^{} = \left\lbrack C_{Rd,c}*k*\left( 100*\varrho_{L}*f_{\text{ck}} \right)^{\frac{1}{3}} \right\rbrack*b_{c}*d$$

$$C_{Rd,c} = \frac{0,18}{\gamma_{C}} = \frac{0,18}{1,4} = 0,129$$

$$k = 1 + \sqrt{\frac{200}{d}} = 1 + \sqrt{\frac{200}{345}} = 1,76 < 2,0$$

$$\varrho_{L} = \frac{A_{\text{SL}}}{b*d} = \frac{9,42}{20*34,5} = 0,01365$$

$$V_{Rd,C}^{} = \left\lbrack C_{Rd,c}*h*\left( 100*\varrho_{L}*f_{\text{ck}} \right)^{\frac{1}{3}} \right\rbrack*b_{c}*d =$$

$\left\lbrack 0,129*1,76*\left( 100*0,01365*30 \right)^{\frac{1}{3}} \right\rbrack*200*345 = 54kN > \nu_{\min}*b_{w}*d$=30,981kN ok

$$\nu_{\min} = 0,035*k^{\frac{3}{2}}*f_{\text{ck}}^{\frac{1}{2}} = 0,035*2,34*5,48 = 0,449$$

V_Rd, C^A < V_Ed^A′

Podpora B,C (ze względu na podobnieństwo wartości sił przekrojowych oraz te same przekroje zbrojenia-licze raz dla max sil przekrojowych):

$$V_{Rd,C}^{} = \left\lbrack C_{Rd,c}*k*\left( 100*\varrho_{L}*f_{\text{ck}} \right)^{\frac{1}{3}} \right\rbrack*b_{c}*d$$

$$C_{Rd,c} = \frac{0,18}{\gamma_{C}} = \frac{0,18}{1,4} = 0,129$$

$$k = 1 + \sqrt{\frac{200}{d}} = 1 + \sqrt{\frac{200}{345}} = 1,76 < 2,0$$

$$\varrho_{L} = \frac{A_{\text{SL}}}{b*d} = \frac{12,57}{20*34,5} = 0,0182$$

$$V_{Rd,C}^{} = \left\lbrack C_{Rd,c}*h*\left( 100*\varrho_{L}*f_{\text{ck}} \right)^{\frac{1}{3}} \right\rbrack*b_{c}*d =$$

$\left\lbrack 0,129*1,76*\left( 100*0,0182*30 \right)^{\frac{1}{3}} \right\rbrack*200*345 = 59,43kN > \nu_{\min}*b_{w}*d$=30,981kN ok

$$\nu_{\min} = 0,035*k^{\frac{3}{2}}*f_{\text{ck}}^{\frac{1}{2}} = 0,035*2,34*5,48 = 0,449$$

V_Rd, C^B < V_Ed^B′, C′

Na wszystkich podporach należy obliczyc dodatkowe zbrojenie na ścinanie.

Strzemiona 90 ̊, Φ8.

Podpora skrajna:

$$l_{t} = \frac{V_{\text{Ed}}^{\text{kr}} - V_{Rd,c}^{}}{g + q} = \frac{83,22 - 54}{10,66 + 16,4} = 1,07m$$

1,8d=0,621m

V_0,621=63,91kN

$$s \leq \frac{A_{\text{sw}}}{V_{\text{Ed}}}*z*f_{\text{ywd}}*ctg\theta = \frac{0,0001005}{63,91}*0,311*420000*2 = 0,411m$$

S_l,max=0,75*d*(1+ctgα)=0,75*0,345*1=0,2587m

$$\upsilon = 0,6*\left( 1 - \frac{f_{\text{ck}}}{350} \right) = 0,6*\left( 1 - \frac{30}{350} \right) = 0,55$$

$${V_{\text{Rd}}^{\max}}^{'} = \frac{b_{w}*z*\upsilon*f_{\text{cd}}}{ctg\theta + tg\theta} = \frac{0,2*0,311*0,55*21400}{2 + 0,5} = 291,84kN$$

Strzemiona Φ8 co 250mm.

Podpora B/C:

$$l_{t} = \frac{V_{\text{Ed}}^{\text{kr}} - V_{Rd,c}^{}}{g + q} = \frac{126,98 - 54}{10,66 + 16,4} = 2,69m$$

1,8d=0,621m

V_0,621=107,68kN

$$s \leq \frac{A_{\text{sw}}}{V_{\text{Ed}}}*z*f_{\text{ywd}}*ctg\theta = \frac{0,0001005}{107,68}*0,311*420000*2 = 0,244m$$

S_l,max=0,75*d*(1+ctgα)=0,75*0,345*1=0,2587m

Strzemiona Φ8 co 240mm.

V_2*0,621=83,5kN

$$s \leq \frac{A_{\text{sw}}}{V_{\text{Ed}}}*z*f_{\text{ywd}}*ctg\theta = \frac{0,0001005}{83,5}*0,311*420000*2 = 0,314m$$

S_l,max=0,75*d*(1+ctgα)=0,75*0,345*1=0,2587m

Strzemiona Φ8 co 240mm. Jeden rozstaw.

$$\upsilon = 0,6*\left( 1 - \frac{f_{\text{ck}}}{350} \right) = 0,6*\left( 1 - \frac{30}{350} \right) = 0,55$$

$${V_{\text{Rd}}^{\max}}^{'} = \frac{b_{w}*z*\upsilon*f_{\text{cd}}}{ctg\theta + tg\theta} = \frac{0,2*0,311*0,55*21400}{2 + 0,5} = 291,84kN$$

SGU:

b_eff=1,488m;b_w=0,2m;h_f=0,14m;h=0,39m

$$g_{k}^{z} = \frac{10,66kN}{m}$$

$$q_{k}^{z} = \frac{16,4kN}{m}$$

$$q_{k}^{z}*\psi_{2} = 16,4*0,8 = \frac{13,12kN}{m}$$

Obwiednia momentów:

M_max=57,089kNm; przęsło A

A_s1=9,42cm²;

u=b_eff,1+(h-h_f)+b_w+(h-h_f)+ b_eff,2=0,599+0,25+0,25+0,25+0,599=1,948m

A_c=b_eff*h_f­+(h-h_f)*b_w=1,448*0,14+0,25*0,2=0,253m²

h₀=2A_c/u=2*0,253/1,948=0,259m

φ(∞,t₀) = 2, 2

$$E_{c,eff}^{} = \frac{E_{\text{cm}}^{}}{1 + \varphi\left( \infty,t_{0} \right)} = \frac{32}{1 + 2,2} = 10GPa$$

$$\propto_{e}^{} = \frac{E_{s}}{E_{c,eff}} = \frac{200}{10} = 20$$

Faza I – przekrój niezarysowany

A_cs = A_c + α_e • A_s = 0, 253 + 20 • 0, 000942 = 0, 272 m²;

S_cs = (b_eff − b_w)*h_f * 0, 5 * h_f + b_w * h * 0, 5 * h + α_e * A_s1 * d = (0,599−0,2) * 0, 14 * 0, 5 *  0, 14 + 0, 2 * 0, 39 * 0, 5 * 0, 39 + 20 * 0, 000942 * 0, 345 = 0, 0256m³

Środek ciężkości przekroju:

$$x_{I} = \frac{S_{\text{cs}}}{A_{\text{cs}}} = \frac{0,0256}{0,272} = 0,0942m$$

Moment bezwładności przekroju niezarysowanego:

$$I_{I} = \lbrack\frac{\left( b_{\text{eff}} - b_{w} \right) \bullet h_{f}^{3}}{12} + \left( b_{\text{eff}} - b_{w} \right) \bullet h_{f} \bullet \left( x_{I} - 0,5h_{f})^{2} \right\rbrack + \lbrack\frac{b_{w} \bullet x_{I}^{3}}{12} + b_{w} \bullet x_{I} \bullet \left( 0,5x_{I})^{2} \right\rbrack + \lbrack\frac{b_{w} \bullet (h - x_{I})^{3}}{12} + b_{w} \bullet \left( h - x_{I} \right) \bullet \left( \frac{h - x_{I}}{2})^{2} \right\rbrack + \lbrack\alpha_{e} \bullet A_{s1} \bullet \left( d - x_{I})^{2} \right\rbrack = = \lbrack\frac{\left( 1,448 - 0,2 \right) \bullet {0,14}^{3}}{12} + (1,448 - 0,2) \bullet 0,14 \bullet \left( 0,0942 - 0,5 \bullet 0,14)^{2} \right\rbrack + \lbrack\frac{0,2 \bullet {0,0942}^{3}}{12} + 0,2 \bullet 0,0942 \bullet \left( 0,5 \bullet {0,0942)}^{2} \right\rbrack + \left\lbrack \frac{0,2 \bullet (0,39 - {0,0942)}^{3}}{12} + 0,2 \bullet \left( 0,39 - 0,0942 \right) \bullet \left( \frac{0,39 - 0,0942}{2} \right)^{2} \right\rbrack + \lbrack 20 \bullet 0,000942 \bullet \left( 0,345 - 0,0942 \right)^{2} = 0,003354\ m^{4};$$

Wskaznik na zginanie:

$$W_{\text{cs}} = \frac{I_{I}}{h - x_{I}} = \frac{0,003354}{0,39 - 0,0942} = 0,0113m^{3}$$

Moment rysujacy:

M_cr = W_cs * f_ctm = 0, 0113 * 2900 = 32, 77kNm < 57, 09kNm

Belka jest zarysowana.

Ugięcie belki w fazie I

$$\propto_{I} = \propto *\frac{M_{eq,p}*l_{\text{eff}}^{2}}{E_{c,eff}^{}*I_{I}^{}} = \frac{1}{12}*\frac{57,09*5,175}{10000000*0,000942} = 0,0314 = 3,14mm$$

Faza II

1, 448 * x_II * 0, 5 * x_II − 0, 000942 * 20 * (0,345−x_II) = 0

0, 724 * x_II² + 0, 01884x_II² − 0, 0065 = 0

x_II = 0, 082m < h_f P.T.

$$I_{\text{II}} = \left\lbrack \frac{b_{\text{eff}}*x_{\text{II}}^{3}}{12} + b_{\text{eff}}*x_{\text{II}}^{}*\left( 0,5*x_{\text{II}}^{} \right)^{2} \right\rbrack + A_{S1}* \propto_{e}*\left( d - x_{\text{II}}^{} \right)^{2} = \left\lbrack \frac{1,448*0,082}{12} + 1,448*0,082*\left( 0,5*0,082 \right)^{2} \right\rbrack + 0,000942*20*\left( 0,345 - 0,082 \right)^{2} = 0,00157m^{4}$$

Ugięcie belki w fazie II

$$a_{\text{II}} = a_{1}*\frac{0,003354}{0,00157} = 0,0314*2,136 = 6,71mm$$

Ugięcie żebra:

$$\zeta = 1 - 0,5*\left( \frac{M_{\text{cr}}}{M_{Eq,p}} \right)^{2} = 1 - 0,5*\left( \frac{32,77}{57,09} \right)^{2} = 0,835$$

a = a_II * ζ + a_I * (1−ζ) = 6, 71 * 0, 835 + 3, 14 * 0, 165 = 6, 12

$a_{\lim} = \frac{l_{\text{eff}}}{250} = \frac{5175}{250} = 20,7$ ok.

Szerokosc rysy:

$$A_{c,\text{eff}} = * \bullet \min\left\{ \begin{matrix} 2,5*\left( h - d \right) \\ \frac{h - x_{I}}{3} \\ \end{matrix} \right.\ = 25*\min\left\{ \begin{matrix} 2,5*\left( 39 - 34,5 \right) = 11,25 \\ \frac{39 - 9,42}{3} = 9,86 \\ \end{matrix} \right.\ = 25*9,86 = 246,5cm^{2}$$

$$\rho_{\text{eff}} = \frac{A_{S1}}{A_{c,eff}} = \frac{9,42}{246,5} = 0,038$$

$$S_{r,max} = k_{3}*c + k_{1}*k_{2}*k_{4}*\frac{\phi}{\rho_{\text{eff}}} = 3,4*30 + 0,8*0,5*0,425*\frac{20}{0,038} = 191,47mm$$

$$\sigma_{s} = \frac{\alpha_{e}*M_{eq,p}}{I_{\text{II}}}*\left( d - x_{\text{II}} \right) = \frac{20*57,09}{0,00157}*\left( 0,345 - 0,082 \right) = 191,3MPa$$

$$\varepsilon_{\text{sm}} - \varepsilon_{\text{cm}} = \frac{\sigma_{s} - k_{t}*\frac{f_{ct,eff}}{\rho_{\text{eff}}}*\left( 1 + \alpha_{e}*\rho_{\text{eff}} \right)}{E_{s}} = \frac{191,3 - 0,4*\frac{2,9}{0,038}*\left( 1 + 20*0,038 \right)}{200000} = 0,000688$$

w_k = S_r, max * (ε_sm−ε_cm) = 191, 3MPa * 0, 000688 = 0, 1316mm < 0, 3mm ok.

2.4 Podciąg

2.4.1 Dane wyjściowe:

C30/37

- beton C30/37; f_cd=21,4 MPa; f_ck=30 MPa; f_ctm=2,9 Mpa; f_ctk=2,0 Mpa; E_cm=32GPa

- stal f_yd=420Mpa; f_yk=500MPa

- E_s=200GPa

- otulina XC3 c_min=25mm

- otulina REI90 a=40mm; a_sd=50mm

Wyszczególnienie	Obc. Char.	γf	Obc. obliczeniowe
		>1	=1
gk,ż*l2,ż=10,66*6,05	64,5
gd1,ż*l2,ż
Podciąg(0,64-0,14)*0,25x6,05*25	18,9	1,35	1
Tynk 2*(0,64-0,39)*0,02*19*6,05	2,3	1,35	1
Suma	Gk=85,7
Obc. zmienne	Qk=99,22	1,5	1

2.4.3 Schemat statyczny

Zawsze G_d2;

Ewentualnie:

G = G_d1 − G_d2 = 115, 3 − 85, 7 = 29, 6kN

$$\overset{\overline{}}{Q} = Q_{d} + G = 148,83 + 29,6 = 178,43kN$$

2.4.4 obliczenia statyczne:

1. M_{Ed, A − B}^max = 428, 95kNm

2. M_{Ed, B − C}^max = 318, 94kNm

3. M_Ed, B^min = −504, 45kNm

5. M_Ed, C^min = −437, 43kNm

6. M_Ed, Bkr^min = −457, 08kNm

7. M_Ed, Ckr^min = −393, 72kNm

2.4.5 Wymiarowanie na moment:

Przeslo skrajne:

$$b_{1} = b_{2} = \frac{b_{}{- b}_{w}}{2} = \frac{5,85 - 0,25}{2} = 2,8mm$$

l₀ = 0, 85 * l₁ = 0, 85 * 5, 85 = 4973m

b_eff, i = 0, 2b_i + 0, 1l₀    ≤ {0, 2l₀;b}

b_eff, 1 = 0, 2 * 2, 8 + 0, 1 * 4, 973 = 1, 057 > {0,995;2,775}

b_eff, 1 = b_eff, 2 = 0, 995

b_eff = 2 * b_eff, 1 + b_w = 2, 24m

W przęśle środkowym:

$$b_{1} = b_{2} = \frac{b_{}{- b}_{w}}{2} = \frac{6,15 - 0,25}{2} = 2,95m$$

l₀ = 0, 7 * l₁ = 0, 7 * 6150 = 4305mm

b_eff, i = 0, 2b_i + 0, 1l₀    ≤ {0, 2l₀;b}

b_eff, 1 = 0, 2 * 2, 95 + 0, 1 * 4, 305 = 1, 02 > {0,861;2,775}

b_eff, 1 = b_eff, 2 = 0, 861

b_eff = 2 * b_eff, 1 + b_w = 1, 972m

Wysokośc użyteczna podciągu:

a=40mm;c_nom=25mm;⌀_strz = 8mm; ⌀ = 28mm

a₁=c_nom+Φ_strz+Φ/2=25+8+14=47=50mm

d=h_p- c_nom-Φ_strz-Φ-0,5* Φ =640-(25+8+25+12,5)=640-71=569mm

d_kr=h-c_nom-Φ_pl-Φ_ż-Φ_p-0,5*25=640-25-8-20-28-14=545mm

d_oś=545+300/6=595mm

Sprawdzenie rzeczywistego kształtu rozkładu momentów w podciagu:

M_F = b_eff * h_f * f_cd * (d−0,5h_f)

Dla przęsła skrajnego:

M_F = b_eff * h_f * f_cd * (d−0,5h_f) = 2, 29m * 0, 14m * 21400kPa * (0,569−0,5*0,14) = 3568kNm > 428, 95kNm

Dla przęseł srodkowych:

M_F = b_eff * h_f * f_cd * (d−0,5h_f) = 2, 022m * 0, 14m * 21400kPa * (0,569−0,5*0,14) = 3150kNm > 318, 94kNm

Przekrój jest pozornie teowy.

1. M_{Ed, A − B}^max = 428, 95kNm

x_eff=0,016m

As1=18,2cm²

2. M_{Ed, B − C}^max = 318, 94kNm

x_eff=0,01344m

As1=13,5cm²

3. M_Ed, B^min = −504, 45kNm

x_eff=0,188m

As1=23,98cm²

4. M_Ed, C^min = −437, 43kNm

x_eff=0,159m

As1=20,19cm²

5. M_Ed, Bkr^min = −457, 08kNm

x_eff=0,19m

As1=24,18cm²

6. M_Ed, Ckr^min = −393, 72kNm

x_eff=0,158m

As1=20,11cm²

Minimalna powierzchnia zbrojenia na zginanie:

f_ctm=2,9Mpa

$$A_{s,min} = 0,26*\frac{f_{\text{ctm}}}{f_{\text{yk}}}*b*d \geq 0,0013b*d;$$

$$A_{s,min} = 0,26*\frac{2,9}{500}*0,25*0,545 \geq 0,0013*0,25*0,545$$

A_s, min = 2, 05cm² ≥ 1, 8cm²;

Minimalna odległośc między prętami:

Max(Φ,20mm,d_g+5mm)=max(28,20,21mm)=28mm

Zbrojenie górne6Φ25=29,45cm²;

Zbrojenie dolne5Φ25=24,54cm²;

Określenie nośności prętów oraz długości zakotwienia:

A_s1=29,45cm^2

$$x_{\text{eff}} = \frac{f_{\text{yd}}*A_{s1}}{f_{\text{cd}}*b} = \frac{420}{21,4}*\frac{29,45}{25} = 23,12cm$$

M_Rd = f_yd * A_s1 * (d−0,5x_eff) = 420 * 29, 45 * (59,5−0,5*231,12) = 592, 96

Dla 6Φ25 592,96kNm

Dla 5Φ25 513,94kNm

Dla 4Φ25 427,12kNm

Dla 3Φ25 332,25kNm

Dla 2Φ25 229,45kNm

Długośc zakotwienia prętów dolnych na podporze skrajnej

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 1, 0

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*1,0*1,0*1,4=3,15Mpa

⌀ = 25mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,prov}} = 420*\frac{1}{9,82} = 42,8MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{25}{4}*\frac{42,8}{3,15} = 85mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 250mm

α₁ − α₅ = 1, 0

l_bd = l_b, min = 250mm

Długośc zakotwienia prętów dolnych w przęsłach

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 0, 7

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*1,0*1,0*1,4=2,21Mpa

⌀ = 25mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,prov}} = 420*\frac{18,2}{24,54} = 311,5MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{25}{4}*\frac{311,5}{2,21} = 880mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 264mm

α₁ − α₅ = 1, 0

l_bd = l_b, min = 880mm

Długośc zakotwienia prętów górnych

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 0, 7

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*1,0*1,0*1,4=2,21Mpa

⌀ = 25mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,prov}} = 420*\frac{24,18}{29,45} = 344,8MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{25}{4}*\frac{344,8}{2,21} = 975mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 292mm

α₁ − α₅ = 1, 0

l_bd = l_b, min = 980mm

Długośc zakotwienia prętów górnych na podporze skrajnej:

10⌀ = 250mm

Długośc zakotwienia pretow na podporami posrednimi dolem:

2*7*⌀ = 2 * 7 * 25 = 350mm

Ścinanie:

Podpora skrajna:

$$V_{Rd,C}^{} = \left\lbrack C_{Rd,c}*k*\left( 100*\varrho_{L}*f_{\text{ck}} \right)^{\frac{1}{3}} \right\rbrack*b_{c}*d$$

$$C_{Rd,c} = \frac{0,18}{\gamma_{C}} = \frac{0,18}{1,4} = 0,129$$

$$k = 1 + \sqrt{\frac{200}{d}} = 1 + \sqrt{\frac{200}{545}} = 1,61 < 2,0$$

$$\varrho_{L} = \frac{A_{\text{SL}}}{b*d} = \frac{24,54}{25*54,5} = 0,018$$

$$V_{Rd,C}^{} = \left\lbrack C_{Rd,c}*k*\left( 100*\varrho_{L}*f_{\text{ck}} \right)^{\frac{1}{3}} \right\rbrack*b_{c}*d = 0,129*1,61*\left( 100*0,018*30 \right)^{\frac{1}{3}}*250*545 = 107kN$$

υ_min = 0, 035 * k^1, 5 * f_ck^0, 5 = 0, 035 * 1, 61^1, 5 * 30^0, 5 = 0, 392

υ_min * b_w * d = 0, 392 * 250 * 545 = 53, 36kN

Ok.

V_Ed = 228, 73kN

S=545+50mm=595mm

Z=0,9d=490mm

f_ywd=500Mpa

α = 45

$$cot\theta = \frac{595}{490} = 1,214$$

cotα = 1

sinα = 0, 707

$$V_{Rd,s2} = \frac{A_{\text{sw}}}{s}*z*f_{\text{ywd}}*\left( cot\theta + cot\alpha \right)*sin\alpha = \frac{4,91}{0,595}*0,49*\frac{500MPa}{1,15}*\left( 1,214 + 1 \right)*0,707 = 275kN$$

V_Ed − V_Rd, s2 = 228, 73 − 275 < 0 → V_Rd, s1 = 0, 5 * V_Ed = 115kN

$$s \leq \frac{A_{\text{sw}}}{V_{Rds1}}*z*f_{\text{ywd}}*ctg\theta = \frac{{2*0,004}^{2}*\pi}{115kN}*0,49m*420000kPa*2 = 0,35m$$

s_l, max = 0, 75 * d * (1+ctgα) = 0, 41m

S=0,35m

V_Ed = 228, 73kN

S_bmax=0,6d*(1+ctgα)=0,6*545*(1+1)=654650mm

Z=0,9d=490mm

f_ywd=500Mpa

α = 45

$$cot\theta = \frac{650}{490} = 1,327$$

cotα = 1

sinα = 0, 707

$$V_{Rd,s2} = \frac{A_{\text{sw}}}{s}*z*f_{\text{ywd}}*\left( cot\theta + cot\alpha \right)*sin\alpha = \frac{4,91}{0,65}*0,49*\frac{500MPa}{1,15}*\left( 1,327 + 1 \right)*0,707 = 256kN$$

V_Ed − V_Rd, s2 = 228, 73 − 256 < 0 → V_Rd, s1 = 0, 5 * V_Ed = 115kN

$$s \leq \frac{A_{\text{sw}}}{V_{Rds1}}*z*f_{\text{ywd}}*ctg\theta = \frac{{2*0,004}^{2}*\pi}{115kN}*0,49m*420000kPa*2 = 0,35m$$

s_l, max = 0, 75 * d * (1+ctgα) = 0, 41m

S=0,35m

V_Ed = 228, 73kN

S_bmax=0,6d*(1+ctgα)=0,6*545*(1+1)=654650mm

Z=0,9d=490mm

f_ywd=500Mpa

α = 45

$$cot\theta = \frac{650}{490} = 1,327$$

cotα = 1

sinα = 0, 707

$$V_{Rd,s2} = \frac{A_{\text{sw}}}{s}*z*f_{\text{ywd}}*\left( cot\theta + cot\alpha \right)*sin\alpha = \frac{4,91}{0,65}*0,49*\frac{500MPa}{1,15}*\left( 1,327 + 1 \right)*0,707 = 256kN$$

V_Ed − V_Rd, s2 = 228, 73 − 256 < 0 → V_Rd, s1 = 0, 5 * V_Ed = 115kN

$$s \leq \frac{A_{\text{sw}}}{V_{Rds1}}*z*f_{\text{ywd}}*ctg\theta = \frac{{2*0,004}^{2}*\pi}{115kN}*0,49m*420000kPa*2 = 0,35m$$

s_l, max = 0, 75 * d * (1+ctgα) = 0, 41m

S=0,35m

Podpora środkowa:

V_Ed = 333, 84kN

S=545+50mm=595mm

Z=0,9d=490mm

f_ywd=500Mpa

α = 45

$$cot\theta = \frac{595}{490} = 1,214$$

cotα = 1

sinα = 0, 707

$$V_{Rd,s2} = \frac{A_{\text{sw}}}{s}*z*f_{\text{ywd}}*\left( cot\theta + cot\alpha \right)*sin\alpha = \frac{4,91}{0,595}*0,49*\frac{500MPa}{1,15}*\left( 1,214 + 1 \right)*0,707 = 275kN$$

V_Ed − V_Rd, s2 = 333, 84 − 275 = 58, 7 < 0, 5V_Ed → V_Rd, s1 = 0, 5 * V_Ed = 167kN

$$s \leq \frac{A_{\text{sw}}}{V_{Rds1}}*z*f_{\text{ywd}}*ctg\theta = \frac{{2*0,004}^{2}*\pi}{167kN}*0,49m*420000kPa*2 = 0,25m$$

s_l, max = 0, 75 * d * (1+ctgα) = 0, 41m

S=0,25m

V_Ed = 333, 84kN

S_bmax=0,6d*(1+ctgα)=0,6*545*(1+1)=654650mm

Z=0,9d=490mm

f_ywd=500Mpa

α = 45

$$cot\theta = \frac{650}{490} = 1,327$$

cotα = 1

sinα = 0, 707

$$V_{Rd,s2} = \frac{A_{\text{sw}}}{s}*z*f_{\text{ywd}}*\left( cot\theta + cot\alpha \right)*sin\alpha = \frac{4,91}{0,65}*0,49*\frac{500MPa}{1,15}*\left( 1,327 + 1 \right)*0,707 = 256kN$$

V_Ed − V_Rd, s2 = 333, 84 − 256 = 77, 84 < 0, 5 * V_Ed → V_Rd, s1 = 0, 5 * V_Ed = 167kN

$$s \leq \frac{A_{\text{sw}}}{V_{Rds1}}*z*f_{\text{ywd}}*ctg\theta = \frac{{2*0,004}^{2}*\pi}{167kN}*0,49m*420000kPa*2 = 0,35m$$

s_l, max = 0, 75 * d * (1+ctgα) = 0, 41m

S=0,35m

V_Ed = 333, 84kN

S_bmax=0,6d*(1+ctgα)=0,6*545*(1+1)=654650mm

Z=0,9d=490mm

f_ywd=500Mpa

α = 45

$$cot\theta = \frac{650}{490} = 1,327$$

cotα = 1

sinα = 0, 707

$$V_{Rd,s2} = \frac{A_{\text{sw}}}{s}*z*f_{\text{ywd}}*\left( cot\theta + cot\alpha \right)*sin\alpha = \frac{4,91}{0,65}*0,49*\frac{500MPa}{1,15}*\left( 1,327 + 1 \right)*0,707 = 256kN$$

V_Ed − V_Rd, s2 = 333, 84 − 256 = 77, 84 < 0, 5 * V_Ed → V_Rd, s1 = 0, 5 * V_Ed = 167kN

$$s \leq \frac{A_{\text{sw}}}{V_{Rds1}}*z*f_{\text{ywd}}*ctg\theta = \frac{{2*0,004}^{2}*\pi}{167kN}*0,49m*420000kPa*2 = 0,35m$$

s_l, max = 0, 75 * d * (1+ctgα) = 0, 41m

S=0,35m

Koniec ścinania.

Rysunek.

2.4 Słup

2.4.1 Dane wyjściowe:

- beton C30/37; f_cd=21,4 MPa; f_ck=30 MPa; f_ctm=2,9 Mpa; f_ctk=2,0 Mpa; E_cm=32GPa

- stal f_yd=420Mpa; f_yk=500MPa

- E_s=200GPa

- otulina XC3 c_min=25mm

- otulina REI90 a=40mm; a_sd=50mm

2.4.2 Zestawienie obciążeń:

- stałe ze stropodachu N₁=158kN

- śnieg N₂=24,1kN

- obciążenie ze stropów N₃=1931kN

- ciężar żeber N₄=88,1kN

- ciężar tynku na żebrach N₅=13,38kN

- ciężar podciągu N₆ = [(h_p−h_f)*b^p*l₂^p*25] * 1, 35 * 1 * 3 = [(0,64−0,14)*0,3*6,15*25] * 1, 35 * 1 * 3 = 93, 4kN

- ciężar tynku na podciągu N₇ = [2*(h_p−h_f)*0,02*l₂^p*19] * 1, 35 * 1 * 3 = [(0,64−0,14)*0,02*6,15*25] * 1, 35 * 1 * 3 = 1, 42kN

- ciężar słupa N₈ = [(h−h_p)*b*h*25] * 1, 35 * 3 = [(4−0,64)*0,3*0,3*25] * 1, 35 * 1 * 3 = 30, 62kN

N_Ed=2340kN

M_Ed=0

2.4.3 Schemat statyczny:

$$I = \frac{bh^{3}}{12} = \frac{30^{4}}{12} = 67500cm^{4}$$

$$i = \sqrt{\frac{I}{A}} = \sqrt{\frac{67500}{900}} = 8,67cm$$

l₀ = 0, 7 * 422 = 295, 4cm

$$\lambda = \frac{l_{0}}{i} = \frac{295,4}{8,67} = 34,07$$

$$\lambda_{\lim} = \frac{20*A*B*C}{\sqrt{n}} =$$

$$A = \frac{1}{1 + 0,2\varphi_{\text{ef}}} = 0,7$$

$$B = \sqrt{1 + 2\omega}$$

$$\omega = \frac{A_{S}*f_{\text{yd}}}{Ac*f_{\text{cd}}}$$

$$\varrho = \frac{A_{s}}{b*h} = > A_{s} = 0,02*b*h = 0,02*30*30 = 18cm^{2}$$

$$\omega = \frac{A_{S}*f_{\text{yd}}}{Ac*f_{\text{cd}}} = \frac{18*420}{900*21,4} = 0,393$$

$$B = \sqrt{1 + 2\omega} = \sqrt{1 + 2*0,393} = 1,34$$

C = 1, 7 − r_m

$$r_{m} = \frac{M_{01}}{M_{02}} = 0$$

C = 0, 7

$$n = \frac{N_{\text{Ed}}}{A_{c}*f_{d}} = \frac{2340kN}{0,09m\hat{}2*21400kPa} = 1,215$$

$$\lambda_{\lim} = \frac{20*A*B*C}{\sqrt{n}} = \frac{20*0,7*1,34*0,7}{\sqrt{1,215}} = 11,91$$

λ > λ_lim

Należy uwzględnic wpływ efektów 2 rzędu.

2.4.5 Mimośród I rzędu:

$$e_{0} = \left( \frac{M_{\text{Ed}}^{1}}{N_{\text{Ed}}^{1}} + e_{1} \right) \geq max\left\{ 20mm;\frac{h}{30} \right\}$$

$$e_{0} = \left( \frac{M_{\text{Ed}}^{1}}{N_{\text{Ed}}^{1}} + e_{1} \right) = \left( 0 + \frac{l_{0}}{400} \right) = \left( 0 + \frac{2954}{400} \right) = 7,385 \geq max\left\{ 20mm;\frac{300}{30} = 10mm \right\}$$

e₀ = 20mm

M_0, Ed = N_Ed * e₀ = 2340 * 0, 02 = 46, 8kNm

2.4.6 Wyznaczenie siły krytycznej

$$N_{B} = \frac{9,6*E_{\text{cd}},eff*I_{I}}{l_{0}^{2}}$$

u=2*b+2*h=4*0,3=1,2m

A_c=b²=0,09m²

h₀=2A_c/u=2*0,09/1,2=0,18m

φ(∞,t₀) = 2, 6

$$E_{c,eff}^{} = \frac{E_{\text{cm}}^{}}{1 + \varphi\left( \infty,t_{0} \right)} = \frac{32}{1 + 2,6} = 8,89GPa$$

$$\propto_{e}^{} = \frac{E_{s}}{E_{c,eff}} = \frac{200}{8,89} = 22,5$$

$$E_{cd,eff}^{} = \frac{E_{c,eff}^{}}{1,2} = \frac{8,89GPa}{1,2} = 7408kPa$$

$$A_{s} = 0,02*0,3*0,3 = 0,0018m\hat{}2$$

$$I_{I} = \frac{bh\hat{}3}{12} + \alpha_{e}*A_{s}*\left( \frac{h}{2} - a \right)^{2} = \frac{{0,3}^{4}}{12} + 22,5*0,0018*\left( \frac{0,3}{2} - 0,045 \right)^{2} = 0,001122m^{4}$$

$$N_{B} = \frac{9,6*E_{\text{cd}},eff*I_{I}}{l_{0}^{2}} = \frac{9,6*7408kPa*0,001122m^{4}}{(2,954{m)}^{2}} = 9,14kN$$

Moment II rzędu:

$$M_{\text{Ed}} = M_{0Ed}*\left( 1 + \frac{\beta}{\frac{N_{B}}{N_{\text{Ed}}} - 1} \right) = 46,8*\left( 1 + \frac{\frac{\pi}{c_{0}}}{\frac{9,14}{2340} - 1} \right) = 31,43kNm$$

$$e_{\text{II}} = \frac{M_{\text{Ed}}^{\text{II}}}{N_{\text{Ed}}} = \frac{31,43}{2340} = 0,0134m$$

2.4.8 Wymiarowanie zbrojenia z wykorzystaniem nomogramów:

$$\frac{M_{\text{Ed}}^{\text{II}}}{b*h^{2}*f_{\text{ck}}} = 0,0388$$

$$\frac{N_{\text{Ed}}^{}}{b*h^{}*f_{\text{ck}}} = 0,867$$

Zbrojenie:

$$A_{s} = \zeta*b*h*\frac{f_{\text{ck}}}{f_{\text{yk}}} = 0,7*0,3*0,3*\frac{30}{500} = 0,00378cm^{2}$$

A_s1 = A_s2 = 18, 9cm²

$$\varrho_{\text{prov}} = \frac{A_{s,prov}}{b*d} = \frac{19,63}{30*30} = 0,0218$$

$${\Delta\varrho}_{} = \frac{\left| \varrho_{}{- \varrho}_{\text{prov}} \right|}{\varrho_{}} = \frac{\left| 0,02 - 0,0218 \right|}{0,02} = 0,9\ \ ok.$$

Długośc zakotwienia prętów:

l_bd = α₁ * α₂ * α₃ * α₄ * l_b, rqd ≥ l_b, min

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}};$$

f_bd = 2, 25 * η₁ * η₁ * f_ctd

η₁ = 0, 7

η₂ = 1, 0

f_ctd = 1, 4MPa

f_bd = 2, 25 * η₁ * η₁ * f_ctd=2,25*1,0*1,0*1,4=2,21Mpa

⌀ = 25mm

$$\sigma_{\text{sd}} = f_{\text{yd}}*\frac{A_{s,req}}{A_{s,prov}} = 420*\frac{18,9}{19,63} = 404,4MPa$$

$$l_{b,rqd} = \frac{\varnothing}{4}*\frac{\sigma_{\text{sd}}}{f_{\text{bd}}} = \frac{25}{4}*\frac{344,8}{2,21} = 1142mm$$

l_b, min = max{0, 3l_b, rqd;10ϕ;100mm} = 342mm

α₁ − α₅ = 1, 0

l_bd = l_b, min = 1142mm

Rozmieszczenie prętów:

e=20mm<0,15b=0,15*300=45mm

Metoda A.

l_0, fi = 0, 7 * 4, 22 = 2, 954m ≤ 3m

REI90a=45mm

2.5 Stopa

2.5.1 Dane wyjściowe:

- beton C30/37; f_cd=21,4 MPa; f_ck=30 MPa; f_ctm=2,9 Mpa; f_ctk=2,0 Mpa; E_cm=32GPa

- stal f_yd=420Mpa; f_yk=500MPa

- E_s=200GPa

- otulina XC3 c_min=25mm

- otulina REI90 a=40mm; a_sd=50mm

$$\sigma = \frac{N_{\text{Ed}}}{B*L} = \frac{2340}{{2,9}^{2}} = 278,2kPa \leq \sigma_{\lim} = 300kPa$$

Wymiarowanie ze względu na zginanie:

a=1,3m

b=0,3m

$$x_{0} = \frac{\left( a*b*0,5*a \right) + \left( a*a*\frac{2}{3}*a \right)}{\left( a*b \right) + \left( a*a \right)} = \frac{\left( 1,3*0,3*0,5*1,3 \right) + \left( 1,3*1,3*\frac{2}{3}*1,3 \right)}{\left( 1,3*0,3 \right) + \left( 1,3*1,3 \right)} = 0,826m$$

$$g_{r} = \frac{N_{\text{Ed}}}{B*L} = \frac{2340}{2,9*2,9} = 278,24kPa$$

$$N_{r} = g_{r}*A = \frac{278kN}{m}*\frac{\left( 2,9m + 0,3m \right)*0,826m}{2} = 278*1,32 = 367,4kN$$

M_∝ = N_r * x₀ = 367kN * 0, 826m = 303, 5kNm

$$\sum_{}^{}{M_{As1} = f_{\text{cd}}*b*x_{\text{eff}}*\left( d - 0,5x_{\text{eff}} \right) - M_{\text{Ed}} = 0}$$

x_eff = 1, 43cm

$$\sum_{}^{}{X = f_{\text{cd}}*b*x_{\text{eff}} - f_{\text{yd}}*A_{s1} = 0}$$

A_s1 = 21, 08cm²

Minimalna powierzchnia zbrojenia na zginanie:

f_ctm=2,9Mpa

$$A_{s,min} = 0,26*\frac{f_{\text{ctm}}}{f_{\text{yk}}}*b*d \geq 0,0013b*d;$$

$$A_{s,min} = 0,26*\frac{2,9}{500}*2,9*0,35 \geq 0,0013*2,9*0,35$$

A_s, min = 1, 53cm² ≥ 1, 32cm²;

Minimalna odległośc między prętami:

Max(Φ,20mm,d_g+5mm)=max(12,20,21mm)=21mm

Przyjęto 19Φ1221,48cm²

Sprawdzenie stopy na przebicie:

Jest konieczne projektowanie zbrojenia na przebicie.