EBK PRINCIPLES OF GEOTECHNICAL ENGINEER
EBK PRINCIPLES OF GEOTECHNICAL ENGINEER
9th Edition
ISBN: 9781337517218
Author: SOBHAN
Publisher: VST
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Chapter 10, Problem 10.1P

(a)

To determine

Calculate the maximum and minimum principal stresses.

(a)

Expert Solution
Check Mark

Answer to Problem 10.1P

The maximum principal stress (σ1) is 200.72kN/m2_.

The minimum principal stress (σ3) is 116.28kN/m2_.

Explanation of Solution

Given information:

The normal stress along x axis (σx) is 172kN/m2.

The normal stress along y axis (σy) is 145kN/m2.

The shear stress along xy axis (τxy) is +40kN/m2.

Calculation:

Find the horizontal angle as follows:

θ=90°68°=22°

Calculate the maximum principal stress (σ1) using the relation.

σ1=σy+σx2+[σy+σx2]2+τxy2

Substitute 172kN/m2 for σx, 145kN/m2 for σy, and 40kN/m2 for τxy.

σ1=145+1722+[1451722]2+402=158.5+1,782.25=158.5+42.22=200.72kN/m2

Hence, the maximum principal stress (σ1) is 200.72kN/m2_.

Calculate the minimum principal stress (σ3) using the relation.

σ3=σy+σx2[σy+σx2]2+τxy2

Substitute 172kN/m2 for σx, 145kN/m2 for σy, and 40kN/m2 for τxy.

σ3=145+1722[1451722]2+402=158.51,782.25=158.542.22=116.28kN/m2

Hence, the minimum principal stress (σ3) is 116.28kN/m2_.

(b)

To determine

Calculate the normal and shear stresses on plane AB.

(b)

Expert Solution
Check Mark

Answer to Problem 10.1P

The normal stress on plane AB (σn) is 176.6kN/m2_.

The shear stress on plane AB (τn) is 38.15kN/m2_.

Explanation of Solution

Given information:

The normal stress along x axis (σx) is 172kN/m2.

The normal stress along y axis (σy) is 145kN/m2.

The shear stress along xy axis (τxy) is +40kN/m2.

The angle (θ) is 22°.

Calculation:

Calculate the normal stress (σn) using the relation.

σn=σy+σx2+σyσx2cos2θ+τxysin2θ

Substitute 172kN/m2 for σx, 145kN/m2 for σy, 40kN/m2 for τxy, and 22° for θ.

σn=145+1722+1451722cos(2×22°)+40sin(2×22°)=158.513.5cos44°+40sin44°=176.6kN/m2

Hence, the normal stress on plane AB (σn) is 176.6kN/m2_.

Calculate the shear stress (τn) using the relation.

τn=σyσx2sin2θτxycos2θ

Substitute 172kN/m2 for σx, 145kN/m2 for σy, 40kN/m2 for τxy, and 22° for θ.

τn=1451722sin(2×22°)40cos(2×22°)=13.5sin44°40cos44°=38.15kN/m2

Therefore, the shear stress on plane AB (τn) is 38.15kN/m2_.

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