1. Ãbungsaufgabe: Aktiver Erddruck analytisch
1. Ãbungsaufgabe: Aktiver Erddruck analytisch
1. Ãbungsaufgabe: Aktiver Erddruck analytisch
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Übungsbeispiel 1: <strong>Erddruck</strong> aktiv, <strong>analytisch</strong><br />
Aufgabenstellung<br />
Bestimmen Sie die Größe und Lage der resultierenden <strong>Erddruck</strong>kraft.<br />
b k<br />
0.5m<br />
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p=5.0 kN/m 2<br />
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h<br />
h f<br />
h L<br />
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b f<br />
b s<br />
Stützmauer Nr. 4<br />
3.2m<br />
0.8m<br />
3.0m<br />
0.3m<br />
γ Β = 24.0 kN/m 3<br />
<strong>1.</strong>0m<br />
<strong>1.</strong>3m<br />
Sand<br />
γ = 18.0 kN/m 3<br />
γ´ = 1<strong>1.</strong>0 kN/m 3<br />
ϕ = 40°<br />
c´=0.0 kN/m 2<br />
δ a = 2/3 ϕ<br />
δ p = −2/3 ϕ<br />
Ton<br />
γ = 20.0 kN/m 3<br />
γ´= 10.0 kN/m 3<br />
ϕ = 22°<br />
c´=7.0 kN/m 2<br />
δ a = 2/3 ϕ<br />
δ p = −2/3 ϕ<br />
1 Anfangsangaben<br />
<strong>1.</strong>1 Geometrie<br />
α = 0; β = 0 Neigung der Mauerrückseite bzw. Geländeneigung in ◦<br />
h = 4 m Höhe der Mauerrückseite<br />
b k = 0.5 m Breite der Krone<br />
b s = <strong>1.</strong>3 m Breite der Sohle<br />
b f = 0.3 m Breite Vorsprung<br />
h L = 3 m Höhe der Wand luftseitig<br />
h f = 0.8 m Höhe Fundament<br />
<strong>1.</strong>2 Materialkennwerte, Belastung<br />
γ B = 24 kN/m 3 Wichte des Betons<br />
p = 5 kN/m 2 Auflast<br />
1
<strong>1.</strong>3 Bodenkennwerte<br />
φ 1 = 40 ◦ Reibungswinkel Hinterfüllung<br />
γ 1 = 18 kN/m 3 Wichte der Hinterfüllung<br />
δ a1 = 2 3 · φ 1 = 2 · 40 = 26.667<br />
3<br />
δ p1 = − 2 3 · φ 1 = − 2 · 40 = −26.667<br />
3<br />
φ 2 = 22 ◦ Reibungswinkel Untergrund<br />
γ 2 = 10 kN/m 3 Wichte Untergrund<br />
c 2 = 7 kN/m 2 Kohäsion<br />
2 Berechnung des <strong>Erddruck</strong>s<br />
2.1 <strong>Erddruck</strong>beiwerte<br />
K agh1 = Kagh100 [α ◦ ; β ◦ ; δa1; ◦ φ ◦ 1] = Kagh100 [0 ◦ ; 0 ◦ ; 26.667 ◦ ; 40 ◦ ]<br />
= 0.179<br />
K aph1 = K agh1 = 0.179<br />
2.2 <strong>Erddruck</strong><br />
e agh1o = 0 kN/m 2 <strong>Erddruck</strong> infolge Eigengewicht - oben<br />
e agh1u = e agh1o + h · γ 1 · K agh1 = 0 + 4 · 18 · 0.179<br />
= 12.859 kN/m 2 <strong>Erddruck</strong> infolge Eigengewicht - unten<br />
e aph1 = p · K aph1 = 5 · 0.179<br />
= 0.893 kN/m 2 <strong>Erddruck</strong> infolge Auflast<br />
e ah1o = e aph1 = 0.893 kN/m 2 Gesamterddruck horizontal, oben<br />
e ah1u = e aph1 + e agh1u = 0.893 + 12.859<br />
= 13.751 kN/m 2 Gesamterddruck horizontal, unten<br />
3 <strong>Erddruck</strong>kräfte<br />
3.1 <strong>Erddruck</strong> infolge Eigengewicht<br />
E agh = e agh1o + e agh1u<br />
· h = 0 + 12.859 · 4 = 25.717 kN/m<br />
2<br />
2<br />
e agh1o·2+e agh1u 0·2+12.859<br />
3<br />
3<br />
h f,eag =<br />
· h =<br />
e agh1o + e agh1u 0 + 12.859 · 4 = <strong>1.</strong>333 m<br />
E agv = E agh · tan (α ◦ + δa1) ◦ = 25.717 · tan (0 ◦ + 26.667 ◦ )<br />
= 12.916 kN/m<br />
3.2 <strong>Erddruck</strong> infolge Auflast<br />
E aph = e aph1 · h = 0.893 · 4 = 3.572 kN/m<br />
h f,eap = h 2 = 4 2 = 2 m<br />
E agv = E agh · tan (α ◦ + δ ◦ a1) = 25.717 · tan (0 ◦ + 26.667 ◦ )<br />
2
= 12.916 kN/m<br />
E ah = e ah1o + e ah1u<br />
2<br />
0.893 + 13.751<br />
· h = · 4 = 29.289 kN/m<br />
2<br />
0.893 · 2 + 13.751<br />
· h =<br />
3 · (0.893 + 13.751) · 4 = <strong>1.</strong>415 m<br />
h f,ea1 = e ah1o · 2 + e ah1u<br />
3 · (e ah1o + e ah1u )<br />
E av = E ah · tan (α ◦ + δa1) ◦ = 29.289 · tan (0 ◦ + 26.667 ◦ ) = 14.709 kN/m<br />
h f,ea1 = h f,eag · E agh + h f,eap · E aph <strong>1.</strong>333 · 25.717 + 2 · 3.572<br />
= = <strong>1.</strong>415 m<br />
E agh + E aph 25.717 + 3.572<br />
3