2.0 Poprzecznica.

2.1. Zestawienie obciążeń.

2.1.1. Obciążenia stałe.

Na szerokości jezdni:

g_max=9.31 kN/m²

g_min=6.53 kN/m²

Na szerokości chodnika:

Δg_chmax=4.80 kN/m²

Δg_chmin=2.89 kN/m²

2.1.2. Obciążenie użytkowe.

P_o=181.5 kN

q_o=3.00 kN/m²

q_to=3.25 kN/m²

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

Dla obciążenia równomiernego od Δg_ch

γ_1gmax=4⋅Δg_chmax/1⋅π=4⋅4.80/1⋅π=6.11 kN/m²

γ_1gmin=4⋅Δg_chmin/1⋅π=4⋅2.89/1⋅π=3.67 kN/m²

γ_3gmax=4⋅Δg_chmax/3⋅π=4⋅4.80/3⋅π=2.03 kN/m²

γ_3gmin=4⋅Δg_chmin/3⋅π=4⋅2.89/3⋅π=1.23kN/m²

γ_5gmax=4⋅Δg_chmax/5⋅π=4⋅4.80/5⋅π=1.22 kN/m²

γ_5gmin=4⋅Δg_chmin/5⋅π=4⋅2.89/5⋅π=0.73 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⋅181.5/32.2⋅cos(π⋅0.5/32.2)⋅cos(2⋅0.5⋅π/32.8) ⋅sin(π⋅16.1/32.2)==

=22.54⋅cos0.048⋅cos0.097=22.4 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⋅181.5/32.2⋅cos(3⋅π⋅0.5/32.2)⋅cos(6⋅0.5⋅π/32.2)⋅sin(3π⋅16.1/32.2)==

=22.54⋅cos0.144⋅cos0.289=-21.30 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⋅181.5/32.5⋅cos(5⋅π⋅0.5/32.5)⋅cos(10⋅0.5⋅π/32.2)⋅sin(5π⋅16.1/32.2)==

=22.54⋅cos0.241⋅cos0.484=19.37 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

2.3 Obwiednia momentów .

b=6.46m

m_yn=b⋅sin(n⋅π⋅x_o/L)⋅Σγ_n⋅μ_n

dla przekroju y=0 n=1

m_y1max=6.46⋅sin(π⋅16.1/32.2)⋅[(-0.784)3.67+7.36⋅0.802+22.40(0.249+0.156+0.064)]=

=6.46⋅sin0.5π⋅[(-2.87)+5.90+10.438]=67.00 kN

m_y1min=6.46⋅sin(π⋅16.1/32.2)⋅[(-0.784)6.11-7.36⋅ 0.0056+(-0.304-0.304)⋅4.13]=

=6.46⋅sin0.5π⋅[(-4.79)-2.51-0.0412]=-47.42 kN

dla przekroju y=0 n=3

m_y3max=6.46⋅sin(3π⋅16.1/32.2)⋅[(-0.772)1.23+2.54⋅0.784-21.3(0.247+0.154+0.062)]=

=6.46⋅sin1.5π.⋅[(-0.949)+1.99-9.86.]=36.97 kN

m_y3min=6.46⋅sin(3π⋅16.1/32.2)⋅[(-0.772)2.03-2.54⋅0.0058+(-0.299-0.299)⋅1.37]=

=6.46⋅sin1.5π⋅[(-1.56)-0.819-0.014]=15.46 kN

dla przekroju y=0 n=5

m_y5max=6.46⋅sin(5π⋅16.1/32.2)⋅[(-0.758)0.73+1.52⋅0.773+19.37(0.242+0.15+0.06)]=

=6.46⋅sin2.5π⋅[(-0.553)+1.17+8.75)]= 40.51kN

m_y5min=6.46⋅sin(5π⋅16.1/32.2)⋅[(-0.758)1.22+(-0.289-0.289)⋅0.82-1.52⋅0.0058]=

=6.46⋅sin2.5π⋅[(-0.924)-0.473-0.00881]=-9.08 kN

dla y=0.5b n=1

m_y1max=6.46⋅sin(π⋅16.1/32.2)⋅[(-0.292-0.176)3.67+0.0239⋅6.11+7.63⋅0.44

+22.40(0.143+0.113+0.083+0.053)]=

=6.46⋅sin0.5π⋅[(-1.71)+0.146+3.23+8.78]=47.48 kN

m_y1min=6.46⋅sin(π⋅16.1/32.2)⋅[(-0.292-0.176)6.63-0.239⋅3.67+(-0.239-0.132)⋅4.13

-0.0067⋅7.63]=

=6.46⋅sin0.5π⋅[(-3.10)-0.877-1.53-0.051]=-35.9 kN

dla y=0.5b n=3

m_y3max=6.46⋅sin(3π⋅16.1/32.2)⋅[(-0.288-0.17)1.23+0.026⋅2.03

+2.54⋅0.40-21.30(0.142+0.112+0.081+0.051)]=

=6.46⋅sin1.5π⋅[(-0.51)+0.052+1.016-8.22]=29.50 kN

m_y3min=6.46⋅sin(3π⋅16.1/32.2)⋅[(-0.288-0.17)2.03+0.026⋅1.23-2.54⋅0.0065+

(-0.236-0.127)1.37]=

=6.46⋅sin1.5π⋅[(-0.929)+0.032-0.016-0.62]=9.91 kN

dla przekroju y=0.5b n=5

m_y5max=6.46⋅sin(5π⋅16.1/32.2)⋅[(-0.284-0.163)0.73+0.025⋅1.22+1.52⋅0.42+

19.37(0.14+0.109+0.049+0.079)]

=6.46⋅sin2.5π⋅[(-0.326)+0.03+0.0638+7.30]= 29.37kN

m_y5min=6.46⋅sin(5π⋅16.1/32.2)⋅[(-0.284-0.163)1.22+0.025⋅0.73-1.52⋅0.064+(-0.233-0.122)0.82+]=

=6.46⋅sin2.5π⋅[(-0.545)+0.018-0.097-0.291]=-5.91 kN

Dla przekroju y=0

m_ymax= m_y1max+ m_y3max+ m_y5max=144.48 kN

M_ymax= m_ymax⋅x_o=2326.12 kNm

m_ymin= m_y1min+ m_y3min+ m_y5min=-41.04kN

M_ymin= m_ymin⋅x_o=-660.74 kNm

Dla przekroju y=0.5b

m_ymax= m_y1max+ m_y3max+ m_y5max=106.35 kN

M_ymax= m_ymax⋅x_o=1712.235 kNm

m_ymin= m_y1min+ m_y3min+ m_y5min=-31.9 kN

M_ymin= m_ymin⋅x_o=-513.59 kNm

2.4 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=513.59kNm

h₁=1.45m

b=0.40m

t=0.21m

t/h=0.20

w_z=n*M/R_a*b*h₁²=0.010

m₁=0.159

m_z=0.954

m_x=0.138

x=m_x*h₁=0.22>t

σ_b=m₁*R_a/n=8.129Mpa<R_b=23.1MPa

A_s=M/(R_­a*m_z*h₁)=17.73cm²

6 φ 20 A_a=18.85cm²

μ=18.85/0.40*1.45=0.003 <μ_min=0.004

A_a=0.004*0.4*1.45=24.8cm²

10 φ 20 A_a=31.42cm²

Moment przenoszony przez zbrojenie

M=R_a*m_z*h₁*A_a=1370.59kNm

σ_a=n*σ_b*(h₁-x/x)=25.79Mpa<295MPa

Maksymalny moment ujemny.

M=2326.12 kNm

w_z=0.047

m₁=0.378

m_z=0.908

m_x=0.275

x=m_x*h₁=0.426>t

σ_b=m₁*R_a/n=19.32Mpa<R_b=23.1MPa

A_s=M/(R_­a*m_z*h₁)=56.02cm²

14 φ 24 A_a=63.33cm²

μ=56.54/0.40*1.45=0.009 >μ_min=0.004

Moment przenoszony przez zbrojenie

M=R_a*m_z*h₁*A_a=2629.35kNm

σ_a=n*σ_b*(h₁-x/x)=117.83Mpa<295MPa

---