3.8 Poprzecznica.

3.8.1 Zestawienie obciążeń.

3.8.1.1 Obciążenia stałe.

Na szerokości jezdni:

g_max=9.61 kN/m²

g_min=6.72 kN/m²

Na szerokości chodnika:

Δg_chmax=5.21 kN/m²

Δg_chmin=3.13 kN/m²

3.8.1.2 Obciążenie użytkowe.

P_o=369 kN

q_o=6.00 kN/m²

q_to=3.25 kN/m²

3.8.2 Wyznaczenie współczynników γ dla n=1,3,5.

Dla obciążenia równomiernego od Δg_ch

γ_1gmax=4⋅Δg_chmax/1⋅π=4⋅5.21/1⋅π=6.63 kN/m²

γ_1gmin=4⋅Δg_chmin/1⋅π=4⋅3.13/1⋅π=3.98 kN/m²

γ_3gmax=4⋅Δg_chmax/3⋅π=4⋅5.21/3⋅π=2.21 kN/m²

γ_3gmin=4⋅Δg_chmin/3⋅π=4⋅3.13/3⋅π=1.32 kN/m²

γ_5gmax=4⋅Δg_chmax/5⋅π=4⋅5.21/5⋅π=1.32 kN/m²

γ_5gmin=4⋅Δg_chmin/5⋅π=4⋅3.13/5⋅π=0.79 kN/m²

Dla obciążenia równomiernego taboru samochodowego

γ_1q=4⋅Δq_o/1⋅π=4⋅6/1⋅π=7.63 kN/m²

γ_3q=4⋅Δq_o/3⋅π=4⋅6/3⋅π=2.54 kN/m²

γ_5q=4⋅Δq_o/5⋅π=4⋅6/5⋅π=1.52 kN/m²

Dla obciążenia ciągnikiem K

γ_1p=(8⋅P_o/2)/L⋅cos(1⋅π⋅a/L)⋅cos(2⋅1⋅π⋅a/L)⋅sin(1⋅π⋅x_o/L)=

=4⋅369/28⋅cos(π⋅0.6/28)⋅cos(2⋅0.6⋅π/28)⋅sin(π⋅20/28)=

=52.71⋅cos0.067⋅cos0.134⋅sin2.24=40.87 kN/m

γ_3p=(8⋅P_o/2)/L⋅cos(3⋅π⋅a/L)⋅cos(2⋅3⋅π⋅a/L)⋅sin(3⋅π⋅x_o/L)=

=4⋅369/28⋅cos(3⋅π⋅0.6/28)⋅cos(6⋅0.6⋅π/28)⋅sin(3⋅π⋅20/28)=

=52.71⋅cos0.201⋅cos0.403⋅sin6.73=20.52 kN/m

γ_5p=(8⋅P_o/2)/L⋅cos(5⋅π⋅a/L)⋅cos(2⋅5⋅π⋅a/L)⋅sin(5⋅π⋅x_o/L)=

=4⋅369/28⋅cos(5⋅π⋅0.6/28)⋅cos(10⋅0.6⋅π/28)⋅sin(5⋅π⋅20/28)=

=52.71⋅cos0.336⋅cos0.673⋅sin11.22=-38.101 kN/m

Dla obciążenia od tłumu.

γ_1t=4⋅3.25/π=4.13 kN/m²

γ_3t=4⋅3.25/3π=1.37 kN/m²

γ_5t=4⋅3.25/5π=0.82 kN/m²

3.8.3 Obwiednia momentów .

b=7.11m

m_yn=b⋅sin(n⋅π⋅x_o/L)⋅Σγ_n⋅μ_n

dla przekroju y=0 n=1

m_y1max=7.11⋅sin(π⋅20/28)⋅[(-0.792)3.98+7.36⋅0.883+40.87(0.1448+0.1166+0.0607)]=

=7.11⋅sin2.24⋅[(-3.15)+6.17+20.189]=69.42 kN

m_y1min=7.11⋅sin(π⋅20/28)⋅[(-0.79)6.63+(-0.388-0.25)⋅4.13+

(-0.066)⋅7.63]=

=7.11⋅sin2.24⋅[(-5.23)-2.63-0.503]=-46.6 kN

dla przekroju y=0 n=3

m_y3max=7.11⋅sin(3π⋅20/28)⋅[(-0.698)1.32+2.54⋅0.788+20.52(0.2292+0.1479+0.0969)]=

=7.11⋅sin6.73⋅[(-0.92)+2.00+9.726]=16.81 kN

m_y3min=7.11⋅sin(3π⋅20/28)⋅[(-0.698)2.21-2.54⋅0.61+(-0.343-0.311)⋅1.37+

(-0.061)⋅2.21]=

=7.11⋅sin6.73⋅[(-1.54)-0.895-0.15-0.134]=-8.35 kN

dla przekroju y=0 n=5

m_y5max=7.11⋅sin(5π⋅20/28)⋅[(-0.415)0.79+1.52⋅0.505-38.101(0.1684+0.0972+0.0368)]=

=7.11⋅sin11.22⋅[(-0.32)+0.767-11.52)]= 7.635kN

m_y5min=7.11⋅sin(5π⋅20/28)⋅[(-0.415)1.32+(-0.203-0.125)⋅0.82-1.52⋅0.048]=

=7.11⋅sin11.22⋅[(-5.47)-0.268-0.072]=40.27 kN

dla y=0.5b n=1

m_y1max=7.11⋅sin(π⋅20/28)⋅[(-0.313-0.155)3.98+0.048⋅6.63+0.004⋅4.13

+7.36⋅0.49+40.87(0.1448+0.1166+0.0607)]=

=7.11⋅sin2.24⋅[(-1.86)+0.318+0.016+3.6+13.16]=61.69 kN

m_y1min=7.11⋅sin(π⋅20/28)⋅[(-0.313-0.155)6.63+0.048⋅3.98+(-0.292-0.106)⋅4.13

-0.058⋅7.63]=

=7.11⋅sin2.24⋅[(-3.10)-0.19-1.64-0.443]=-29.9 kN

dla y=0.5b n=3

m_y3max=7.11⋅sin(3π⋅20/28)⋅[(-0.290-0.127)1.32+0.047⋅2.21

+2.54⋅0.43+20.52(0.1389+0.1072+0.0776+0.0500)]=

=7.11⋅sin6.73⋅[(-0.55)+0.103+1.092+7.66]=11.51 kN

m_y3min=7.11⋅sin(3π⋅20/28)⋅[(-0.290-0.127)2.21+0.047⋅1.32-2.54⋅0.048+

(-0.270-0.087)1.37+(-0.0061)⋅2.21]=

=7.11⋅sin6.73⋅[(-0.92)+0.062-0.12-0.489-0.134]=-4.91 kN

dla przekroju y=0.5b n=5

m_y5max=7.11⋅sin(5π⋅20/28)⋅[(-0.211-0.058)0.79+0.042⋅1.32+1.52⋅0.43—

38.101(0.1389+0.1072+0.0776+0.05)]

=7.11⋅sin11.22⋅[(-0.021)+0.0055+0.065-14.23]= 9.829kN

m_y5min=7.11⋅sin(5π⋅20/28)⋅[(-0.211-0.058)1.32+0.042⋅0.79-1.52⋅0.048+(-0.27-0.087)0.82+]=

=7.11⋅sin11.22⋅[(-0.35)+0.038-0.072-0.29]=4.67 kN

Dla przekroju y=0

m_ymax= m_y1max+ m_y3max+ m_y5max=93.86 kN

M_ymax= m_ymax⋅x_o=1877.3 kNm

m_ymin= m_y1min+ m_y3min+ m_y5min=-14.68kN

M_ymin= m_ymin⋅x_o=-293.6 kNm

Dla przekroju y=0.5b

m_ymax= m_y1max+ m_y3max+ m_y5max=83.02 kN

M_ymax= m_ymax⋅x_o=1660.4 kNm

m_ymin= m_y1min+ m_y3min+ m_y5min=-30.14 kN

M_ymin= m_ymin⋅x_o=-602.80 kNm

3.9 Wymiarowanie .

Beton B40 R_b=23.1Mpa E_b=36.4Gpa

Stal AII 18G2 R_a=295Mpa E_a=210Mpa

n=E_a/E_b=5.77

Maksymalny moment ujemny.

M=602.80kNm

h₁=1.65m

b=0.45m

t=0.21m

t/h=0.127

w_z=n*M/R_a*b*h₁²=0.0096

m₁=0.152

m_z=0.956

m_x=0.132

x=m_x*h₁=0.217>t

σ_b=m₁*R_a/n=6.74Mpa<R_b=23.1MPa

A_s=M/(R_­a*m_z*h₁)=12.95cm²

5 φ 20 A_a=15.71cm²

μ=15.71/0.45*1.65=0.0021 <μ_min=0.004

A_a=0.004*0.45*1.65=29.7cm²

10 φ 20 A_a=31.42cm²

Moment przenoszony przez zbrojenie

M=R_a*m_z*h₁*A_a=1462.07kNm

σ_a=n*σ_b*(h₁-x/x)=256.8Mpa<295MPa

Maksymalny moment ujemny.

M=1887.3 kNm

w_z=0.030

m₁=0.287

m_z=0.926

m_x=0.223

x=m_x*h₁=0.367>t

σ_b=m₁*R_a/n=14.17Mpa<R_b=23.1MPa

A_s=M/(R_­a*m_z*h₁)=41.87cm²

10 φ 24 A_a=45.24cm² 14 φ 20 A_a=43.98cm²

μ=45.24/0.45*1.65=0.006 >μ_min=0.004

Moment przenoszony przez zbrojenie

M=R_a*m_z*h₁*A_a=2039.10kNm

σ_a=n*σ_b*(h₁-x/x)=285.82Mpa<295MPa

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