Ćwiczenie projektowe z Konstrukcji Betonowych

Sem.V

Temat nr 214 - Projekt hali żelbetowej

Projekt wstępny – przyjęcie wymiarów

L=m

B=m

n=

Sprawdzenie płyty

$$\frac{B}{L} = \frac{}{} = > 2 \rightarrow plyta\ jednokierunkowa$$

Grubość płyty

$$h_{p} = \left\langle \frac{1}{40} \div \frac{1}{30} \right\rangle L = \left\langle cm;\text{cm} \right\rangle \rightarrow przyjeto\text{cm}$$

nastąpiła zmiana grubości płyty ze względu na nie spełnienie warunku :

(w rysunkach jest stara wartość 10 cm, z projektu wstępnego)

Żebro

1.3.1. Wysokość żebra

$$h_{z} = \left\langle \frac{1}{15} \div \frac{1}{12} \right\rangle B = \left\langle cm;\text{cm} \right\rangle \rightarrow przyjeto\ \text{cm}$$

1.3.2. Szerokość żebra

$$b_{z} = \left\langle \frac{3}{10} \div \frac{5}{10} \right\rangle h_{z} = \left\langle cm;\text{cm} \right\rangle \rightarrow przyjeto\text{cm}$$

Rygiel

1.4.1.Wysokość rygla

$$h_{r} = \left\langle \frac{1}{12} \div \frac{1}{10} \right\rangle nL = \left\langle cm;\text{cm} \right\rangle \rightarrow przyjeto\ \text{cm}$$

1.4.2. Szerokość rygla

$$b_{r} = \left\langle \frac{3}{10} \div \frac{5}{10} \right\rangle h_{r} = \left\langle cm;\text{cm} \right\rangle \rightarrow przyjeto\ \text{cm}$$

SÅ‚up

1.5.1. Wysokość przekroju

$$h_{sl} = \left\langle \frac{6}{10} \div \frac{7}{10} \right\rangle h_{r} = \left\langle cm;\text{cm} \right\rangle \rightarrow przyjeto\ \text{cm}$$

1.5.2. Szerokość przekroju

b_sl = b_r = cm

Fundament

2.1. Zebranie obciążeń ze stropu na 1mb

Dla obciazenia stalego

Lp.	Obciazenie	$$g_{\text{k\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$	γf	$$g_{\text{o\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$
1.	Terakota m •  kN/m^3 • 1m		1, 35
2.	Gladz cementowa  m •  kN/m^3 • 1m		1, 35
3.	Styropian  m • kN/m^3 • 1m		1, 35
4.	Folia PE	−	−	−
5.	Plyta stropowa  m •  kN/m^3 • 1m		1, 35
6.	Tynk cementowo − wapienny  m • kN/m^3 • 1m		1, 35
	$$\sum_{}^{}:$$		−

Dla obciazenia zmiennego

Lp.	Nazwa obciazenia	$$p_{\text{k\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$	γf	$$p_{\text{o\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$
1.	Obciazenie uzytkowe kN/m^2 • 1m		1, 5

p_o = kN/m

2.2. Zebranie obciążeń ze stropodachu na 1 mb

Dla obciazenia stalego

Lp.	Obciazenie	$$q_{\text{k\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$	γf	$$q_{\text{o\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$
1.	2 x papa termoizolacyjna 2 • 0, 11 kN/m^2 • 1m	0, 200	1, 35	0, 270
2.	Gladz cementowa 0, 03 m • 21 kN/m^3 • 1m	0, 630	1, 35	0, 851
3.	Styropian 0, 15 m • 0, 45 kN/m^3 • 1m	0, 067	1, 35	0, 091
4.	Folia PE	−	−	−
5.	Plyta stropowa 0, 10 m • 25 kN/m^3 • 1m	2, 5	1, 35	3, 375
6.	Tynk cementowo − wapienny 0, 015 m • 19 kN/m^3 • 1m	0, 285	1, 35	0, 385
	$$\sum_{}^{}:$$	gk ≈ 3, 682Â	−	go ≈ 4, 972Â

Dla obciazenia zmiennego

Lp.	Nazwa obciazenia	$$p_{\text{k\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$	γf	$$p_{\text{o\ }}\lbrack\frac{\text{kN}}{m}\rbrack$$
1.	Obciazenie sniegiem − strefa 3 1, 2 kN/m^2 • 0, 8 • 1m	0, 96	1, 5	1, 44
	$$\sum_{}^{}:$$	pk ≈ 0, 96	−	po ≈ 1, 44

$$\Sigma{(p + g)}^{\text{obl}} = 4,972 + 1,440 = 6,412\frac{\text{kN}}{m}$$

2.3. Zebranie obciążeń na stopę fundamentową

1. $Strop:\ \ 21,997\frac{\text{kN}}{\text{mmb}} \bullet 7,7m \bullet 0,5 \bullet \left( 11,4m - 2 \bullet 0,65m \right) = 855,353kN$

2. $Stropodach:\ \ 6,412\frac{\text{kN}}{m}mb \bullet 7,7m \bullet 0,5 \bullet 11,4m = 281,423\ kN$

3. $Slup:\ 8,8m \bullet 25\frac{\text{kN}}{m^{3} \bullet 0,65m \bullet 0,4m = 57,2\ kN}$

4. $Fundament:2,8m \bullet 3,8m \bullet 25\frac{\text{kN}}{m^{3} \bullet 1m = 266,00\ kN}$

Σ : 1459, 976 kN

$$0,81 \bullet 200kN/m^{2} \geq \frac{1,4 \bullet 1459,976\ kN}{2,8m \bullet 3,8m}$$

$$162\frac{\text{kN}}{m^{2}} \geq 192,102\frac{\text{kN}}{m^{2}} \rightarrow \mathbf{Nie\ spelniono\ warunku,\ zwiekszenie\ stopy}$$

1. $Fundament:3,2\ m \bullet 4,2m \bullet 25\frac{\text{kN}}{m^{3} \bullet 1m = 336\ kN}$

Σ : 1529, 976 kN

$$162\frac{\text{kN}}{m^{2}} \geq \frac{1,4 \bullet 1593,976\ kN}{3,2m \bullet 4,2m} = 159,373\frac{\text{kN}}{m^{2}} \rightarrow \mathbf{Warunek\ spelniono}$$

Przyjęto stopę fundamentową o wymiarach 3,2 m x 4,2 m

1. Obliczenia statyczne

   1. Momenty

3.1.1. Momenty przęsłowe

3.1.1.1.maxM_AB= maxM_CD

kN/m

kN/m	^+	kN/m

M [kNm]

$$\mathbf{\text{maxM}}_{\mathbf{\text{AB}}} = 0,080 \bullet g_{o} \bullet L^{2} + 0,101 \bullet p_{o} \bullet L^{2} = 0,080 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m} + 0,101 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m}$$

maxM_AB= maxM_CD=kNm

3.1.1.2. minM_AB= minM_CD

kN/m

	^+ kN/m

M [kNm]

$$\mathbf{\text{minM}}_{\mathbf{\text{AB}}} = 0,080 \bullet g_{o} \bullet L^{2} - 0,025 \bullet p_{o} \bullet L^{2} = 0,080 \bullet \frac{\text{kN}}{m} \bullet \left( m \right)^{2} - 0,025 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m}$$

minM_AB = minM_CD= kNm

  3.1.1.3. maxM_BC

kN/m

	^+ kN/m

		M [kNm]

$$\mathbf{\text{maxM}}_{\mathbf{\text{BC}}} = 0,025 \bullet g_{o} \bullet L^{2} + 0,075 \bullet p_{o} \bullet L^{2} = 0,025 \bullet \frac{\text{kN}}{m} \bullet \left( m \right)^{2} + 0,075 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m}$$

maxM_BC= kNm

3.1.1.4. minM_BC

kN/m

kN/m	^+	kN/m

M [kNm]

$$\mathbf{\text{minM}}_{\mathbf{\text{BC}}} = 0,025 \bullet g_{o} \bullet L^{2} - 0,050 \bullet p_{o} \bullet L^{2} = 0,025 \bullet \frac{\text{kN}}{m} \bullet \left( m \right)^{2} - 0,050 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m}$$

minM_BC= kNm

3.1.2. Momenty podporowe

3.1.2.1. maxM_B=maxM_C

kN/m

	^+	kN/m

M [kNm]

$$\mathbf{\text{maxM}}_{\mathbf{B}}\mathbf{=} - 0,1 \bullet g_{o} \bullet L^{2} + 0,017 \bullet p_{o} \bullet L^{2} = - 0,1 \bullet \frac{\text{kN}}{m} \bullet \left( m \right)^{2} + 0,017 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m}$$

maxM_B=maxM_C=kNm

3.1.2.2. minM_B=minM_C

kN/m

kN/m	^+

M [kNm]

$$\mathbf{\text{minM}}_{\mathbf{B}} = - 0,1 \bullet g_{o} \bullet L^{2} - 0,117 \bullet p_{o} \bullet L^{2} = - 0,1 \bullet \frac{\text{kN}}{m} \bullet \left( m \right)^{2} - 0,117 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m}$$

minM_B = minM_C=kNm

4.1.2.3.  M_A=M_D=0

maxM_A = minM_A = 0

maxM_D = minM_D = 0

Obwiednia momentów

Momenty krawędziowe

[M_B] = max|M_B| − min{|odpT_B^L|, |odpT_B^P| • 0, 5 • b_z + (g_o + p_o)•0, 5 • b_z • 0, 25 • b_z

$$T_{B}^{L} = - 0,6 \bullet g_{o} \bullet L - 0,617 \bullet p_{o} \bullet L = - 0,6 \bullet \frac{\text{kN}}{m} \bullet - 0,617 \bullet \frac{\text{kN}}{m} \bullet$$

T_B^L=kN

$$T_{B}^{P} = 0,5 \bullet g_{o} \bullet L + 0,583 \bullet p_{o} \bullet L = 0,5 \bullet \frac{\text{kN}}{m} \bullet m + 0,583 \bullet \frac{\text{kN}}{m} \bullet m$$

T_B^P=kN

$$\left\lbrack \mathbf{M}_{\mathbf{B}} \right\rbrack = \left| \text{kNm} \right| - kN \bullet 0,5 \bullet cm \bullet 0,01 + \left( + \right)\frac{\text{kN}}{m} \bullet \frac{1}{8} \bullet cm \bullet 0,01$$

[M_B]=[M_C]=kNm

Momenty uśrednione

3.1.5.1.  usrM_BC

$$usrM_{\text{BC}} = 0,5 \bullet \left( \min M_{\text{BC}} + \frac{\text{odp}M_{B} + odpM_{C}}{2} \right)$$

$$\text{odp}M_{B} = - 0,1 \bullet g_{o} \bullet L^{2} - 0,05 \bullet p_{o} \bullet L^{2} = - 0,1\frac{\text{kN}}{m} \bullet \left( m \right)^{2} - 0,05 \bullet \left( m \right)^{2} \bullet \frac{\text{kN}}{m}$$

odpM_B = odpM_C=kNm

$$\mathbf{usr}\mathbf{M}_{\mathbf{\text{BC}}} = 0,5 \bullet \left( kNm + \frac{\text{kNm}\text{kNm}}{2} \right)$$

usrM_BC=kNm

3.1.5.2.  usrM_AB=usrM_CD

$$usrM_{\text{AB}} = 0,5 \bullet \left( \min M_{\text{AB}} + \frac{\text{odp}M_{A} + odpM_{B}}{2} \right)$$

odpM_B=kNm

odpM_A = 0 kNm

$$\mathbf{usr}\mathbf{M}_{\mathbf{\text{AB}}} = 0,5 \bullet \left( kNm + \frac{0\ \ kNm\text{kNm}}{2} \right)$$

usrM_AB = usrM_CD=kNm

Reakcje podporowe

3.2.1.  maxR_A=maxR_D

$$\mathbf{\text{maxR}}_{\mathbf{A}} = 0,4 \bullet g_{o} \bullet L + 0,45 \bullet p_{o} \bullet L = 0,4 \bullet \frac{\text{kN}}{m} \bullet m + 0,45 \bullet \frac{\text{kN}}{m} \bullet m$$

max R_A =max R_D =kN

3.2.2.  minR_A=minR_D

$$\text{minR}_{A} = 0,4 \bullet g_{o} \bullet L - 0,05 \bullet p_{o} \bullet L = 0,4 \bullet \frac{\text{kN}}{m} \bullet m - 0,05 \bullet \frac{\text{kN}}{m} \bullet m$$

min R_A =min R_D =kN

 3.2.3.  maxR_B=maxR_C

$$\text{maxR}_{B} = 1,1 \bullet g_{o} \bullet L + 1,2 \bullet p_{o} \bullet L = 1,1 \bullet \frac{\text{kN}}{m} \bullet m + 1,2 \bullet \frac{\text{kN}}{m} \bullet m$$

maxR_B=maxR_C=kN

3.2.4.   minR_B=minR_C

$$\text{minR}_{B} = 1,1 \bullet g_{o} \bullet L - 0,1 \bullet p_{o} \bullet L = 1,1 \bullet \frac{\text{kN}}{m} \bullet m - 0,1 \bullet \frac{\text{kN}}{m} \bullet m$$

minR_B =minR_C =kN

Siły tnące

3.3.1.  T_B^L/T_C^P

minT_B^L=kN (pkt.3.1.4)

$$\text{maxT}_{B}^{L} = - 0,6 \bullet g_{o} \bullet L + 0,017 \bullet p_{o} \bullet L = - 0,6 \bullet \frac{\text{kN}}{m} \bullet m + 0,017 \bullet \frac{\text{kN}}{m} \bullet m$$

maxT_B^L=kN

maxT_C^P=−minT_B^L=kN

minT_C^P=−maxT_B^L=kN

3.3.2.  T_B^P/T_C^L

$$\text{minT}_{B}^{P} = 0,500 \bullet g_{o} \bullet L - 0,083 \bullet p_{o} \bullet L = 0,500 \bullet \frac{\text{kN}}{m} \bullet m - 0,083 \bullet \frac{\text{kN}}{m} \bullet m$$

minT_B^P=kN

maxT_B^P = kN (pkt.3.1.4)

maxT_C^L=−minT_B^P=kN

minT_C^L=−maxT_B^P=kN

1. Wymiarowanie

   1. Obliczenia wstępne

Beton C/

f_ck = MPa

$$f_{\text{cd}} = \frac{f_{\text{ck}}}{\gamma} = \frac{\text{MPa}}{1,5} = \text{\ MPa}$$

f_ctm = MPa

Stal A III N

f_yk = MPa

$$f_{\text{yd}} = \frac{f_{\text{yk}}}{\gamma_{s}} = \frac{\text{MPa}}{1,15} = \text{MPa}$$

E_s = GPa = •10³MPa

$\varepsilon_{S1} = \frac{f_{\text{yd}}}{E_{s}} = \frac{}{\bullet 10^{3}} =$

Otulina

wedlug normy PN − EN 1992 − 1 − 1 2008 przyjeto klase ekspozycji XC1 -> Klasa S3 -> projektowy okres użytkowania to 100 lat.

c = c_nom = c_min + c_dev

c_min = 15 mm

c_dev = 5 mm

c = c_nom = 15 mm + 5 mm = 20 mm=cm

Zbrojenie przęseł

5.2.1.  M_AB=M_CD

Zastotowanie zbrojenia :     ⌀=mm

$$\mu_{\text{eff}} = \frac{|M_{\text{AB}}|}{b \bullet d^{2} \bullet f_{\text{cd}}}$$

a₁ = c + 0, 5 • ⌀ = cm + 0, 5 •  cm =  cm

d = h_p − a₁ =  cm −  cm = cm

$$\mu_{\text{eff}} = \frac{\mathbf{}kNm \bullet 100}{100\ cm \bullet \left( \text{cm} \right)^{2} \bullet \text{\ MPa} \bullet 0,1} =$$

$\xi_{\text{eff}} = 1 - \sqrt{1 - 2 \bullet \mu_{\text{eff}}} = 1 - \sqrt{1 - 2 \bullet} =$

$$\xi_{eff,lim} = 0,8 \bullet \frac{\varepsilon_{\text{cu}}}{\varepsilon_{\text{cu}} + \frac{f_{\text{yd}}}{E_{s}}}$$

ε_cu = 3, 5 %0 = 3, 5 • 10^−3

$$\xi_{eff,lim} = 0,8 \bullet \frac{3,5 \bullet 10^{- 3}}{3,5 \bullet 10^{- 3} + \frac{}{\bullet 10^{3}}} =$$

ξ_eff < ξ_{eff, lim}

$$A_{S1} = \xi_{\text{eff}} \bullet b \bullet d \bullet \frac{f_{\text{cd}}}{f_{\text{yd}}} = \bullet 100\ cm \bullet \ cm \bullet \frac{\text{\ MPa}}{\text{MPa}} = \text{cm}^{2}$$

$$A_{S1,min} = \max\left\{ \begin{matrix} 0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d \\ 0,0013 \bullet b \bullet d \\ \end{matrix} \right.\ $$

$$0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d = 0,26 \bullet \frac{\text{MPa}}{\text{MPa}} \bullet 100\ cm \bullet \text{\ cm}$$

0, 0013 • b • d = 0, 0013 • 100 cm •  cm

A_{S1, min} = cm²

=A_S1 > A_{S1, min}=

$$n = \frac{A_{S1}}{A_{\varnothing}} = \frac{\text{cm}^{2}}{\text{cm}^{2}} = \approx$$

$$A_{\varnothing} = \frac{\pi \bullet^{2}}{4} = \text{cm}^{2}$$

$$r = \frac{100cm}{} \approx \text{\ cm}$$

$$A_{S1,prov} = \frac{100cm}{r} \bullet A_{\varnothing} = \frac{100\ cm}{\text{\ cm}} \bullet \text{cm}^{2} = \text{cm}^{2}$$

$$\rho_{L} = \frac{A_{S1,prov}}{b \bullet d} = \frac{\text{cm}^{2}}{100\ cm \bullet \text{\ cm}} = \ \%$$

4.2.2.  M_BC

Zastotowanie zbrojenia :     ⌀=mm

$$\mu_{\text{eff}} = \frac{|M_{\text{AB}}|}{b \bullet d^{2} \bullet f_{\text{cd}}}$$

a₁ = c + 0, 5 • ⌀ = cm + 0, 5 •  cm =  cm

d = h_p − a₁ =  cm −  cm = cm

$$\mu_{\text{eff}} = \frac{kNm \bullet 100}{100\ cm \bullet \left( \text{cm} \right)^{2} \bullet \text{\ MPa} \bullet 0,1} =$$

$\xi_{\text{eff}} = 1 - \sqrt{1 - 2 \bullet \mu_{\text{eff}}} = 1 - \sqrt{1 - 2 \bullet} =$

ξ_{eff, lim}=

ξ_eff < ξ_{eff, lim}

$$A_{S1} = \xi_{\text{eff}} \bullet b \bullet d \bullet \frac{f_{\text{cd}}}{f_{\text{yd}}} = \bullet 100\ cm \bullet \ cm \bullet \frac{\text{\ MPa}}{\text{MPa}} = \text{cm}^{2}$$

$$A_{S1,min} = \max\left\{ \begin{matrix} 0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d \\ 0,0013 \bullet b \bullet d \\ \end{matrix} \right.\ $$

$$0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d = 0,26 \bullet \frac{\text{MPa}}{\text{MPa}} \bullet 100\ cm \bullet \text{\ cm}$$

0, 0013 • b • d = 0, 0013 • 100 cm •  cm

A_{S1, min} = cm²

=A_S1 > A_{S1, min}=

$$n = \frac{A_{S1}}{A_{\varnothing}} = \frac{\text{cm}^{2}}{\text{cm}^{2}} = \approx$$

$$A_{\varnothing} = \frac{\pi \bullet^{2}}{4} = \text{cm}^{2}$$

$$r = \frac{100cm}{} \approx \text{\ cm}$$

$$A_{S1,prov} = \frac{100cm}{r} \bullet A_{\varnothing} = \frac{100\ cm}{\text{\ cm}} \bullet \text{cm}^{2} = \text{cm}^{2}$$

$$\rho_{L} = \frac{A_{S1,prov}}{b \bullet d} = \frac{\text{cm}^{2}}{100\ cm \bullet \text{\ cm}} = \ \%$$

Zbrojenie podpory górą

4.3.1. M_B=M_C

Zastotowanie zbrojenia :     ⌀=mm

$$\mu_{\text{eff}} = \frac{|M_{B}|}{b \bullet d^{'2} \bullet f_{\text{cd}}}$$

$$d^{'} = d + \frac{b_{z}}{6} = \ cm + \frac{\text{cm}}{6} \approx \text{cm}$$

a₁ = c + 0, 5 • ⌀ = cm + 0, 5 • cm = cm

d = h_p − a₁ = cm −  cm = cm

$$\mu_{\text{eff}} = \frac{|M_{B}|}{b \bullet d^{'2} \bullet f_{\text{cd}}} = \frac{||\ kNm \bullet 100}{100\ cm \bullet {(cm)}^{2} \bullet \text{\ MPa} \bullet 0,1} =$$

$$\xi_{\text{eff}} = 1 - \sqrt{1 - 2 \bullet \mu_{\text{eff}}} = 1 - \sqrt{1 - 2 \bullet} =$$

ξ_{eff, lim}=

ξ_eff < ξ_{eff, lim}

$${A_{S1}}^{a} = \xi_{\text{eff}} \bullet b \bullet d' \bullet \frac{f_{\text{cd}}}{f_{\text{yd}}} = \bullet 100\ cm \bullet \ cm \bullet \frac{\text{\ MPa}}{\text{MPa}} = \text{cm}^{2}$$

$$A_{S1,min} = \max\left\{ \begin{matrix} 0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d' \\ 0,0013 \bullet b \bullet d' \\ \end{matrix} \right.\ $$

$$0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d' = 0,26 \bullet \frac{\text{MPa}}{\text{MPa}} \bullet 100\ cm \bullet \text{\ cm}$$

0, 0013 • b • d^′ = 0, 0013 • 100 cm • cm

A_{S1, min}=cm²

=A_S1 > A_{S1, min}=

$$n = \frac{A_{S1}}{A_{\varnothing}} = \frac{\text{cm}^{2}}{\text{cm}^{2}} = \approx$$

$$A_{\varnothing} = \frac{\pi \bullet^{2}}{4} = \text{cm}^{2}$$

$$r = \frac{100cm}{} \approx \text{\ cm}$$

$$A_{S1,prov} = \frac{100cm}{r} \bullet A_{\varnothing} = \frac{100\ cm}{\text{cm}} \bullet \text{cm}^{2} = \text{cm}^{2}$$

$$\rho_{L} = \frac{A_{S1,prov}}{b \bullet d'} = \frac{\text{cm}^{2}}{100\ cm \bullet \text{\ cm}} = \%$$

4.3.2.  M_A=M_D

Zastotowanie zbrojenia :     ⌀=mm

$$\mu_{\text{eff}} = \frac{|M_{A}|}{b \bullet d^{'2} \bullet f_{\text{cd}}}$$

$$d^{'} = d + \frac{b_{z}}{6} = \ cm + \frac{\text{cm}}{6} \approx \text{cm}$$

a₁ = c + 0, 5 • ⌀ = cm + 0, 5 • cm = cm

d = h_p − a₁ = cm −  cm = cm

$$\mu_{\text{eff}} = \frac{|M_{A}|}{b \bullet d^{'2} \bullet f_{\text{cd}}} = \frac{\text{\ kNcm}}{100\ cm \bullet {(cm)}^{2} \bullet \text{\ MPa} \bullet 0,1} =$$

$$\xi_{\text{eff}} = 1 - \sqrt{1 - 2 \bullet \mu_{\text{eff}}} = 1 - \sqrt{1 - 2 \bullet} =$$

ξ_{eff, lim}=

ξ_eff < ξ_{eff, lim}

$${A_{S1}}^{a} = \xi_{\text{eff}} \bullet b \bullet d' \bullet \frac{f_{\text{cd}}}{f_{\text{yd}}} = \bullet 100\ cm \bullet \ cm \bullet \frac{\text{\ MPa}}{\text{MPa}} = \text{cm}^{2}$$

$$A_{S1,min} = \max\left\{ \begin{matrix} 0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d' \\ 0,0013 \bullet b \bullet d' \\ \end{matrix} \right.\ $$

$$0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d' = 0,26 \bullet \frac{\text{MPa}}{\text{MPa}} \bullet 100\ cm \bullet \text{\ cm}$$

0, 0013 • b • d^′ = 0, 0013 • 100 cm • cm

A_{S1, min} = cm²

=A_S1 < A_{S1, min}=

$$n = \frac{A_{S1,min}}{A_{\varnothing}} = \frac{\text{cm}^{2}}{\text{cm}^{2}} = \approx$$

$$A_{\varnothing} = \frac{\pi \bullet^{2}}{4} = \text{cm}^{2}$$

$$r = \frac{100cm}{} \approx \text{\ cm}$$

$$A_{S1,prov} = \frac{100cm}{r} \bullet A_{\varnothing} = \frac{100\ cm}{\text{cm}} \bullet \text{cm}^{2} = \text{cm}^{2}$$

$$\rho_{L} = \frac{A_{S1,prov}}{b \bullet d'} = \frac{\text{cm}^{2}}{100\ cm \bullet \text{\ cm}} = \%$$

Zbrojenie przęseł górą

$$M_{\text{cr\ }} = W_{c} \bullet f_{\text{ctm}} = \frac{b \bullet h_{p}^{2}}{6} \bullet f_{\text{ctm}} = \frac{100cm \bullet \left( \text{\ cm} \right)^{2}}{6}\ \bullet \text{\ MPa}\ \bullet 0,1 = \text{\ kNcm\ }$$

|usrM_AB| = |usrM_CD| =  kNcm

=|usrM_AB| ≤ M_cr =

|usrM_BC|= kNcm

=|usrM_BC| ≤ M_cr =

Zastotowanie zbrojenia :     ⌀=mm

$$\mu_{\text{eff}} = \frac{|{usrM}_{\text{BC}}|}{b \bullet d^{2} \bullet f_{\text{cd}}}$$

a₁ = c + 0, 5 • ⌀ = cm + 0, 5 •  cm =  cm

d = h_p − a₁ =  cm −  cm = cm

$$\mu_{\text{eff}} = \frac{\text{kNcm}}{100\ cm \bullet \left( \text{cm} \right)^{2} \bullet \text{\ MPa} \bullet 0,1} =$$

$\xi_{\text{eff}} = 1 - \sqrt{1 - 2 \bullet \mu_{\text{eff}}} = 1 - \sqrt{1 - 2 \bullet} =$

ξ_{eff, lim}=

ξ_eff < ξ_{eff, lim}

$$A_{S1} = \xi_{\text{eff}} \bullet b \bullet d \bullet \frac{f_{\text{cd}}}{f_{\text{yd}}} = \bullet 100\ cm \bullet \ cm \bullet \frac{\text{\ MPa}}{\text{MPa}} = \text{cm}^{2}$$

$$A_{S1,min} = \max\left\{ \begin{matrix} 0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d \\ 0,0013 \bullet b \bullet d \\ \end{matrix} \right.\ $$

$$0,26 \bullet \frac{f_{\text{ctm}}}{f_{\text{yk}}} \bullet b \bullet d = 0,26 \bullet \frac{\text{MPa}}{\text{MPa}} \bullet 100\ cm \bullet \text{\ cm}$$

0, 0013 • b • d = 0, 0013 • 100 cm •  cm

A_{S1, min} = cm²

=A_S1 > A_{S1, min}=

$$n = \frac{A_{S1}}{A_{\varnothing}} = \frac{\text{cm}^{2}}{\text{cm}^{2}} = \approx$$

$$A_{\varnothing} = \frac{\pi \bullet^{2}}{4} = \text{cm}^{2}$$

$$r = \frac{100cm}{} \approx \text{\ cm}$$

$$A_{S1,prov} = \frac{100cm}{r} \bullet A_{\varnothing} = \frac{100\ cm}{\text{\ cm}} \bullet \text{cm}^{2} = \text{cm}^{2}$$

$$\rho_{L} = \frac{A_{S1,prov}}{b \bullet d} = \frac{\text{cm}^{2}}{100\ cm \bullet \text{\ cm}} = \ \%$$

1. Stan graniczny użytkowalności

   1. Dane materiałowe

Beton C/

f_ck = MPa

$$f_{\text{cd}} = \frac{f_{\text{ck}}}{\gamma} = \frac{\text{MPa}}{1,5} = \text{\ MPa}$$

f_ctm = MPa

E_cm = GPa

Stal A III N

f_yk = MPa

$$f_{\text{yd}} = \frac{f_{\text{yk}}}{\gamma_{s}} = \frac{\text{MPa}}{1,15} = \text{MPa}$$

E_s = GPa = •10³ MPa

$\varepsilon_{S1} = \frac{f_{\text{yd}}}{E_{s}} = \frac{}{\bullet 10^{3}} =$

ξ_{eff, lim}=

Dla przęsła M_AB=M_CD

maxM_AB = (k₁•g_k+0,8•k₂•p_k) • L²â€„= (0,080•+0,8•0,101•)•²

maxM_AB=kNm

minM_B=kNm

maxM_B=kNm

⌀_{∞, to} = 3 

$$E_{c,eff} = \frac{E_{\text{cm}}}{1 + \varnothing_{\infty,to}} = \frac{}{1 + 3} = \text{GPa}$$

$$\alpha_{e,t} = \frac{E_{s}}{E_{c,eff}} = \frac{}{} =$$

$$A_{s1} = \frac{\pi \bullet \phi^{2}}{4} \bullet n = \frac{\pi \bullet^{2}}{4} \bullet = \text{cm}^{2}$$

$$S_{y} = \alpha_{e,t} \bullet A_{s1} \bullet d + b \bullet h_{p} \bullet \frac{h_{p}}{2} = \bullet \bullet + 100 \bullet \bullet \frac{}{2} = \mathbf{}\text{cm}^{3}$$

$${A_{\text{cs}} = \alpha_{e,t} \bullet A_{s1} + b \bullet h_{p} = \bullet + 100 \bullet = \mathbf{}\text{cm}^{2}\backslash n}{x_{I} = \frac{S_{y}}{A_{\text{cs}}} = \frac{\mathbf{}}{\mathbf{}} = \mathbf{}\text{cm}}$$

$$I_{I} = \frac{b \bullet x_{I}^{3}}{3} + \frac{b \bullet \left( h_{p} - x_{I} \right)^{3}}{3} + \alpha_{e,t} \bullet A_{s1} \bullet \left( d - x_{I} \right)^{2}$$

$$I_{I} = \frac{100 \bullet \mathbf{}^{3}}{3} + \frac{100 \bullet {( - \mathbf{})}^{3}}{3} + \bullet \bullet {( - \mathbf{})}^{2}$$

I_I = cm⁴

$$\rho = \frac{A_{s1}}{b \bullet d} = \frac{}{100 \bullet} =$$

$$x_{\text{II}} = d \bullet \left\lbrack \sqrt{\rho \bullet \alpha_{e,t} \bullet (2 + \rho \bullet \alpha_{e,t})} - \rho \bullet \alpha_{e,t} \right\rbrack$$

$$x_{\text{II}} = \bullet \left\lbrack \sqrt{\bullet \bullet (2 + \bullet )} - \bullet \right\rbrack = \text{cm}$$

$$I_{\text{II}} = \frac{b \bullet x_{\text{II}}^{3}}{3} + \alpha_{e,t} \bullet A_{s1} \bullet {(d - x_{\text{II}})}^{2} = \frac{100 \bullet^{3}}{3} + \ \bullet \bullet {( - )}^{2}$$

I_II =  cm⁴

$$\alpha_{k} = \frac{5}{48} \bullet \left( 1 - \frac{\left| M_{A} \right| + \left| M_{B} \right|}{10 \bullet \left| M_{\text{AB}} \right|} \right) = \frac{5}{48} \bullet \left( 1 - \frac{0 + \left| \right|}{10 \bullet \left| \mathbf{} \right|} \right) =$$

$$a_{I} = \alpha_{k} \bullet \frac{M_{Ed,qp} \bullet L_{\text{eff}}^{2}}{E_{c,\text{eff}} \bullet I_{I}} = \bullet \frac{\mathbf{}\mathbf{\bullet}^{2}}{\bullet} \bullet 10^{2} = m$$

$$a_{\text{II}} = \alpha_{k} \bullet \frac{M_{Ed,qp} \bullet L_{\text{eff}}^{2}}{E_{c,eff} \bullet I_{\text{II}}} = \bullet \frac{\mathbf{}\mathbf{\bullet}^{2}}{\bullet} \bullet 10^{2} = m$$

a = ζ • a_II + (1−ζ) • a_I

$$\zeta = 1 - \beta \bullet \left( \frac{M_{\text{cr}}}{M_{Ed,qp}} \right)^{2} = 1 - 0,5 \bullet \left( \frac{}{\mathbf{}\mathbf{\bullet}10^{2}} \right)^{2} =$$

β = 0, 5 dla obciazen dlugotrwalych

a = ζ • a_II + (1−ζ) • a_I = • + (1−) • =m

a = cm

$$\frac{L_{\text{eff}}}{250} = \frac{\bullet 10^{2}}{250} = \text{\ cm}$$

L_eff ≤ a

Dla podpory M_B

maxM_BB = (k₁•g_k+0, 8 • k₂•p_k) • L²â€„= (0,025•+0,8•0,075•)•²

maxM_BB=kNm

$$A_{s1} = \frac{\pi \bullet \phi^{2}}{4} \bullet n = \frac{\pi \bullet \mathbf{}^{2}}{4} \bullet = \mathbf{}\text{cm}^{2}$$

$$S_{y} = \alpha_{e,t} \bullet A_{s1} \bullet d + b \bullet h_{p} \bullet \frac{h_{p}}{2} = \bullet \bullet + 100 \bullet \bullet \frac{}{2} = \mathbf{}\text{cm}^{3}$$

A_cs = α_e, t • A_s1 + b • h_p = • + 100 • =cm²

$$x_{I} = \frac{S_{y}}{A_{\text{cs}}} = \frac{\mathbf{}}{} = \mathbf{}\text{cm}$$

$$I_{I} = \frac{b \bullet x_{I}^{3}}{3} + \frac{b \bullet \left( h_{p} - x_{I} \right)^{3}}{3} + \alpha_{e,t} \bullet A_{s1} \bullet \left( d - x_{I} \right)^{2}$$

$$I_{I} = \frac{100 \bullet \mathbf{}^{3}}{3} + \frac{100 \bullet {( - \mathbf{})}^{3}}{3} + \bullet \bullet {( - \mathbf{})}^{2}$$

I_I = cm⁴

$$\rho = \frac{A_{s1}}{b \bullet d} = \frac{}{100 \bullet} = \mathbf{}$$

$$x_{\text{II}} = d \bullet \left\lbrack \sqrt{\rho \bullet \alpha_{e,t} \bullet (2 + \rho \bullet \alpha_{e,t})} - \rho \bullet \alpha_{e,t} \right\rbrack$$

$x_{\text{II}} = \bullet \left\lbrack \sqrt{\mathbf{} \bullet \bullet \left( 2 + \mathbf{} \bullet \right)} - \mathbf{} \bullet \right\rbrack =$cm

$$I_{\text{II}} = \frac{b \bullet x_{\text{II}}^{3}}{3} + \alpha_{e,t} \bullet A_{s1} \bullet {(d - x_{\text{II}})}^{2} = \frac{100 \bullet \mathbf{}^{3}}{3} + \bullet \bullet {( - \mathbf{})}^{2}$$

$$\zeta = 1 - \beta \bullet \left( \frac{M_{\text{cr}}}{M_{Ed,qp}} \right)^{2} = 1 - 0,5 \bullet \left( \frac{}{\mathbf{\bullet}10^{2}} \right)^{2} = \mathbf{}$$

$$\alpha_{k} = \frac{5}{48} \bullet \left( 1 - \frac{\left| M_{B} \right| + \left| M_{B} \right|}{10 \bullet \left| M_{\text{BB}} \right|} \right) = \frac{5}{48} \bullet \left( 1 - \frac{\left| \right| + \left| \right|}{10 \bullet \left| \right|} \right) = \mathbf{}\backslash n$$

$$a_{I} = \alpha_{k} \bullet \frac{M_{Ed,qp} \bullet L_{\text{eff}}^{2}}{E_{c,\text{eff}} \bullet I_{I}} = \mathbf{} \bullet \frac{\mathbf{\bullet}^{2}}{\bullet \mathbf{}} \bullet 10^{2} = \mathbf{}m$$

$$a_{\text{II}} = \alpha_{k} \bullet \frac{M_{Ed,qp} \bullet L_{\text{eff}}^{2}}{E_{c,eff} \bullet I_{\text{II}}} = \mathbf{} \bullet \frac{\mathbf{\bullet}^{2}}{\bullet} \bullet 10^{2} = \mathbf{}m$$

a = ζ • a_II + (1−ζ) • a_I =  •  + (1−) •  = m = cm

$$\frac{L_{\text{eff}}}{250} = \frac{\bullet 10^{2}}{250} = \text{\ cm}$$

L_eff ≤ a

Sprawdzenie szerokości rysy

W_k = S_{r, maxÂ} â€…• (ε_sm − ε_cm)

$$S_{r,max\ } = 1,7 \bullet (2 \bullet c + 0,1 \bullet \frac{\varnothing}{\rho_{p,eff}})$$

$$\rho_{p,eff}\mathbf{=}\frac{A_{s}}{A_{c,eff}}$$

A_{c, eff} = b • min{2,5•(h−d);(h−x_II)/3}

$$\varepsilon_{\text{sm}} - \varepsilon_{\text{cm}} = \frac{\sigma_{s} - k_{t} \bullet \frac{f_{ct,eff}}{\rho_{p,eff}} \bullet (1 + \alpha_{e,t} \bullet \rho_{p,eff)}}{E_{s}} \geq 0,6 \bullet \frac{\sigma_{s}}{E_{s}}$$

$$\sigma_{s} = \alpha_{e,t} \bullet \frac{M_{Ed,qp}}{I_{\text{II}}} \bullet (d - x_{\text{II}})$$

k_t = 0, 4 dla obciazen dlugotrwalych 

Dla przęsła M_AB

x_II = cm

I_II =  cm⁴

$$\sigma_{s} = \alpha_{e,t} \bullet \frac{M_{Ed,qp}}{I_{\text{II}}} \bullet \left( d - x_{\text{II}} \right) = \bullet \frac{\mathbf{}\mathbf{\bullet}10^{2}}{} \bullet \left( - \right) = \mathbf{}\frac{\text{kN}}{\text{cm}^{2}}$$

A_{c, eff} = b • min{2,5•(h_p−d);(h_p−x_II)/3}

A_{c, eff} = 100 • min{2,5•(−);(−)/3} = cm²

$$\rho_{p,eff}\mathbf{=}\frac{A_{s1}}{A_{c,eff}}\mathbf{=}\frac{}{}\mathbf{=}$$

$$\varepsilon_{\text{sm}} - \varepsilon_{\text{cm}} = \frac{\sigma_{s} - k_{t} \bullet \frac{f_{ct,eff}}{\rho_{p,eff}} \bullet \left( 1 + \alpha_{e,t} \bullet \rho_{p,eff} \right)}{E_{s}} \geq 0,6 \bullet \frac{\sigma_{s}}{E_{s}}$$

$$\varepsilon_{\text{sm}} - \varepsilon_{\text{cm}} = \frac{\mathbf{} - 0,4 \bullet \frac{\mathbf{}}{} \bullet 0,1 \bullet (1 + \bullet )}{\bullet 10^{2}} = \mathbf{}$$

$$0,6 \bullet \frac{\sigma_{s}}{E_{s}} = 0,6 \bullet \frac{\mathbf{}}{\bullet 10^{2}} = \mathbf{}$$

 ≥ 

$$S_{r,max\ } = 1,7 \bullet \left( 2 \bullet c + 0,1 \bullet \frac{\varnothing}{\rho_{p,eff}} \right) = 1,7 \bullet \left( 2 \bullet \bullet 10 + 0,1 \bullet \frac{}{} \right) = \mathbf{}\text{\ mm\ }$$

W_k = S_{r, maxÂ} â€…• (ε_sm − ε_cm)= •  =  mm

W_max = mm

W_k≤W_max

Dla podpory M_B

minM_B=kNm

$$\sigma_{s} = \alpha_{e,t} \bullet \frac{M_{Ed,qp}}{I_{\text{II}}} \bullet \left( d - x_{\text{II}} \right) = \bullet \frac{\left| \right|\mathbf{\bullet}10^{2}}{} \bullet \left( - \right) = \frac{\text{kN}}{\text{cm}^{2}}$$

A_{c, eff} = b • min{2,5•(h_p−d);(h_p−x_II)/3}

A_{c, eff} = 100 • min{2,5•(−);(−)/3} = cm²

$$\rho_{p,eff}\mathbf{=}\frac{A_{s1}}{A_{c,eff}}\mathbf{=}\frac{\mathbf{}}{}\mathbf{=}\mathbf{}$$

$$\varepsilon_{\text{sm}} - \varepsilon_{\text{cm}} = \frac{\sigma_{s} - k_{t} \bullet \frac{f_{ct,eff}}{\rho_{p,eff}} \bullet \left( 1 + \alpha_{e,t} \bullet \rho_{p,eff} \right)}{E_{s}} \geq 0,6 \bullet \frac{\sigma_{s}}{E_{s}}$$

$$\varepsilon_{\text{sm}} - \varepsilon_{\text{cm}} = \frac{- 0,4 \bullet \frac{\mathbf{}}{\mathbf{}} \bullet 0,1 \bullet (1 + \bullet \mathbf{})}{\bullet 10^{2}} = \text{mm}$$

$$0,6 \bullet \frac{\sigma_{s}}{E_{s}} = 0,6 \bullet \frac{}{\bullet 10^{2}} = \text{mm}\backslash n$$

≥

$$S_{r,max\ } = 1,7 \bullet \left( 2 \bullet c + 0,1 \bullet \frac{\varnothing}{\rho_{p,eff}} \right) = 1,7 \bullet \left( 2 \bullet \bullet 10 + 0,1 \bullet \frac{}{\mathbf{}} \right) = \text{\ mm\ }$$

W_k = S_{r, maxÂ} â€…• (ε_sm − ε_cm)= • = mm

W_max = mm

W_k≤W_max