Aktiver Erddruck analytisch

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1. Übungsbeispiel: Erddruck aktiv und passiv, analytisch
Aufgabenstellung
Im Zuge der Baumaßnahmen ist die Startgrube für einen horizontalen Rohrvortrieb herzustellen.
Zur sinnvollen Auswahl einer geeigneten Presse ist die maximal aufnehmbare Stützkraft des
Erdwiderlagers zu berechnen.
p=20 kN/m²
Schicht 1
γ = 18.0 kN/m 3
ϕ = 35°, c = 0 kN/m 2
δ a = 2/3 ϕ, δ p
=−2/3ϕ
3,0 m
Schicht 2
γ =
18.0
20.0
kN/m
kN/m 30°, 0 3
ϕ = 25°, c = 5
kN/m
kN/m 2/3 ϕ
2
δ a = 2/3 ϕ, δ p =-2/3ϕ
3,0 m
1 Eingangsgrößen, Erddruckbeiwerte
Die Erddruckbeiwerte werdem mit den Tab. 6.2 - 6.7 aus dem Lehrbuch ermittelt. Zwischenwerte
sind linear zu interpolieren.
Nr. φ δ a α β δ p K agh K aph K ach K pgh K pph K pch
1 35 23.333 0 0 -23.333 0.224 0.224 0.813 7.262 6.564 6.834
2 25 16.667 0 0 -16.667 0.346 0.346 1.043 3.557 3.398 4.33
2 Aktiver Erddruck
2.1 horizontaler aktiver Erddruck infolge Eigengewicht
erddruck horizontal=Vertikalspannung mal Erddruckbeiwert
e agh1,o = h 0 · γ 1 · K agh1 = 0 · 18 · 0.224 = 0
e agh1,u = h 1 · γ 1 · K agh1 = 3 · 18 · 0.224 = 12.096
e agh2,o = h 1 · γ 1 · K agh2 = 3 · 18 · 0.346 = 18.684
e agh2,u = e agh2,o + h 2 · γ 2 · K agh2 = 18.684 + 3 · 20 · 0.346 = 39.444
2.2 horizontaler Erddruck infolge Kohösion
e ach1 = 0
e ach2 = c 2 · K ach2 = 5 · 1.043 = 5.215
1

2.3 Superposition Erddruck infolge Eigengewicht und Kohäsion
e agch1,o = e agh1,o − e ach1 = 0 − 0 = 0
e agch1,u = e agh1,u − e ach1 = 12.096 − 0 = 12.096
e agch2,o = e agh2,o − e ach2 = 18.684 − 5.215 = 13.469
e agch2,u = e agh2,u − e ach2 = 39.444 − 5.215 = 34.229
2.4 Mindesterddruck
e ah2,omin = γ 1 · h 1 · K ah,min2 = 18 · 3 · 0.179 = 9.666
e ah2,umin = e ah2,omin + γ 2 · h 2 · K ah,min2 = 9.666 + 20 · 3 · 0.179 = 20.406
Ist hier nicht maßgebend!
2.5 horizontaler Erddruck infolge unbegrenzter Auflast
e ahp1 = p · K aph1 = 20 · 0.224 = 4.48
e ahp2 = p · K aph2 = 20 · 0.346 = 6.92
2.6 Superposition mit Auflast
e ah1,o = e ahp1 = 4.48
e ah1,u = e agh1,u + e ahp1 = 12.096 + 4.48 = 16.576
e ah2,o = e agch2,o + e ahp2 = 13.469 + 6.92 = 20.389
e ah2,u = e agch2,u + e ahp2 = 34.229 + 6.92 = 41.149
2.7 Berechnung der Hebelarme und der Erddruckkräfte
E ah1 = (e ah1,o + e ah1,u ) · h 1
2
h ea1 = e ah1,o · 2 + e ah1,u
3 · (e ah1,o + e ah1,u ) · h 1 + h 2 =
E ah2 = (e ah2,o + e ah2,u ) · h 2
2
h ea2 = e ah2,o · 2 + e ah2,u
3 · (e ah2,o + e ah2,u ) · h 2 =
2.8 Resultierender aktiver Erddruck
(4.48 + 16.576) · 3
= = 31.584
2
4.48 · 2 + 16.576
3 · (4.48 + 16.576) · 3 + 3 = 4.213
(20.389 + 41.149) · 3
= = 92.307
2
20.389 · 2 + 41.149
3 · (20.389 + 41.149) · 3 = 1.331
E ah = E ah1 + E ah2 = 31.584 + 92.307 = 123.891
E av = E ah1 · tan (α ◦ 1 + δ ◦ a1) + E ah2 · tan (α ◦ 2 + δ ◦ a2)
= 31.584 · tan (0 ◦ + 23.333 ◦ ) + 92.307 · tan (0 ◦ + 16.667 ◦ ) = 41.259
√
E a = E 2 ah + E 2 av = √ 123.891 2 + 41.259 2 = 130.581
h a = E ah1 · h ea1 + E ah2 · h ea2 31.584 · 4.213 + 92.307 · 1.331
= = 2.066
E ah 123.891
3 passiver Erddruck
3.1 Schicht 1
2

Erddruck aus Bodeneigengewicht
e pgh1,o = 0 · γ 1 · K pgh1 = 0 · 18 · 7.262 = 0
e pgh1,u = h 1 · γ 1 · K pgh1 = 3 · 18 · 7.262 = 392.148
Erddruck aus Auflast
e pph1 = p · K pph1 = 20 · 6.564 = 131.28
3.2 Schicht2
Erddruck aus Bodeneigengewicht
e pgh2,o = e pgh1,u
· K pgh2 = 392.148 · 3.557 = 192.078
K pgh1 7.262
e pgh2,u = e pgh2,o + h 2 · γ 2 · K pgh2 = 192.078 + 3 · 20 · 3.557 = 405.498
Erddruck aus Auflast
e pph2 = p · K pph2 = 20 · 3.398 = 67.96
Erddruck aus Kohäsionsanteil
e pch2 = c 2 · K pch2 = 5 · 4.33 = 21.65
3.3 Superposition der Erddruckspannungen
e ph1,o = e pgh1,o + e pph1 = 0 + 131.28 = 131.28
e ph1,u = e pgh1,u + e pph1 = 392.148 + 131.28 = 523.428
e ph2,o = e pgh2,o + e pph2 + e pch2 = 192.078 + 67.96 + 21.65 = 281.688
e ph2,u = e pgh2,u + e pph2 + e pch2 = 405.498 + 67.96 + 21.65 = 495.108
3.4 Resultierender passiver Erddruck
E ph1 = e ph1,o + e ph1,u
2
E ph2 = e ph2,o + e ph2,u
2
h ep1 = e ph1,o · 2 + e ph1,u
3 · (e ph1,o + e ph1,u ) · h 1 + h 2 =
h ep2 = e ph2,o · 2 + e ph2,u
3 · (e ph2,o + e ph2,u ) · h 2 =
131.28 + 523.428
· h 1 = · 3 = 982.062
2
281.688 + 495.108
· h 2 = · 3 = 1165.194
2
131.28 · 2 + 523.428
3 · (131.28 + 523.428) · 3 + 3 = 4.201
281.688 · 2 + 495.108
3 · (281.688 + 495.108) · 3 = 1.363
E ph = E ph1 + E ph2 = 982.062 + 1165.194 = 2147.256
( )
( )
E pv = E ph1 · tan α1 ◦ + δp1
◦ + E ph2 · tan α2 ◦ + δp2
◦
= 982.062 · tan (0 ◦ + −23.333 ◦ ) + 1165.194 · tan (0 ◦ + −16.667 ◦ )
= −772.456
√
E p = E 2 ph + E 2 pv = √ 2147.256 2 + −772.456 2 = 2281.972
h p = E ph1 · h ep1 + E ph2 · h ep2 982.062 · 4.201 + 1165.194 · 1.363
= = 2.661
E ph 2147.256
3