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Part 2

This document contains calculations to determine: 1) the depth of an anchored sheet pile wall, which is found to be 2.10 meters, 2) the length of the anchor, which using AutoCAD is found to be 7.55 meters, 3) the required area of the anchor if the allowable stress is 200 MPa, which is found to be 5.83 x 104 square meters.

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10 views3 pages

Part 2

This document contains calculations to determine: 1) the depth of an anchored sheet pile wall, which is found to be 2.10 meters, 2) the length of the anchor, which using AutoCAD is found to be 7.55 meters, 3) the required area of the anchor if the allowable stress is 200 MPa, which is found to be 5.83 x 104 square meters.

Uploaded by

amazing world
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Part 2

Task1

Ka = 1−sin 30 / 1+sin 30 = 0.33


Kp = 1 Ka = 3.00
Ϭa1 = 10*0.33= 3.3 kn/m2
Ϭa2 = 10+(20*(5+d))*0.33= 6.6d+43 kn/m2
Ea1 = 3.3*(5+d)
Ya1 = 5+d/2

Ea2 = 3.3d2 + 38d


Ya2 = 5+d/3
Ϭp = 60d

Ep = 30d2
Yp = d/3

For equilibrium, Σ moment (point A) = zero


Ea1*Ya1 + Ea2*Ya2 = Ep*Yp
(16.5+3.3d)*(2.5+0.5d) + (3.3d2 +38d+107.5)*(1.67+0.33d) = (30d2 )*(0.33d)

 d = 5.43m
 D = 1.2*5.43 = 6.51 ≈ 6.55m
Task 2

1) Calculate the depth of the anchored sheet pile wall.


Ka = 1−sn 30 / 1+sin 30 = 0.33
Kp = 1 Ka = 3.00
Ϭa1 = 10*0.33= 3.3 kn/m2
Ϭa2 = (10+20*5)*0.33= 36.3 kn/m2
Ϭa3 = (110+20*d)*0.33= 36.3 + 6.6d kn/m2
Ϭp4 = 0*3= 0 kn/m2
Ϭp5 = 3*(20*d) = 60d kn/m2
Ea1 = 3.3*5 = 16.5 kn
Ya1 = 0.5 m
Ea2 = 0.5*33*5 = 82.5 kn
Ya2 = 1.33 m
Ea3 = 36.3d*d= 36.3d kn
Ya3 = 5+d/2 m
Ea4 = 6.6d*0.5*d = 3.3d2 kn
Ya4 = 3+2d/3 m

Ep5 = 0.5*60d2 = 30d2 kn


Yp5 = 3+2d/3 m

For equilibrium, Σ moment at the tie rod (point A) = zero


117.975 + 90.75d + 18.15d2 + 2.2d3 + 9.9d2 - 20d3 - 90d2 = 0
 d = 1.73 m
 D = 1.2*1.73= 2.07 ≈ 2.10 m

Σ Fx = 0  99 + 36.3*1.73 + 3.3∗ 1.732 – 30*1.732 = 81.88 kn


2) Calculate the length of the anchor.

Length of the anchor using AutoCAD with scale drawing = 7.55 m

3) Calculate the required area for the anchored if the allowable stress 200 Mpa
Σ Pa = 16.5 + 82.5 > 81.88 kn
Ϭz = 0.33(10+20*z) = 3.3 + 6.6z kn/m2
Ea1 = 3.3*z
Ea2 = 1.65z + 3.3Z2

Σ Fx = 0  3.3 + 6.6z + 1.65z + 3.3Z2 = 81.88


 Z= 3.78 m

Mmax = 81.88*1.78 – 3.3*3.78*0.11 – 1.65*3.78 +


3.3*3.782*0.52 = 116.61 kn.m/m’

 Z = 116.61 / 200 x 103 = 5.83 x 104 m3

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