1. Induced Stress GLE/CEE 330 Lecture Notes Soil Mechanics
William J. Likos, Ph.D. Department of Civil and Environmental Engineering University of Wisconsin-Madison

2. Plane Strain and Axisymmetric Conditions
Strain in one direction is small or zero
(Induced) radial stress is same in every direction

3. Total Stress and Effective Stress
self-weight and external (induced) stress
stress carried by soil skeleton
s = total stress (sv and sh)
s’ = effective stress (s’v and s’h)
uw = pore water pressure (isotropic)
hydrostatic (no-flow) or flow cond.
TERZAGHI’S EFFECTIVE STRESS P sv sh t uw

4. Induced stress from fills of large areal extent, sv = Hfillg
Bridge Embankment
After Construction
Hfill
20’
Fill
g = 118 pcf
Before Construction 5’ 5’
Dry Sand
g = 110 pcf
Dry Sand
g = 110 pcf
10’
Sat Sand
g = 120 pcf
10’
Sat Sand
g = 120 pcf B B

5. Induced stress from foundation loads (small areal extent)
1) Point Load P
P = lb, kN, kips, etc. P
P/b = lb/ft, kN/m, etc.
2) Line Load P b A
Square P
3) Area Load
q = P/A
“Bearing Pressure”
q = psf, kPa, ksf, etc.
Circular
Area, A

6. Strip Footings & Spread Footings

7. Stress Bulbs
Induced stress dissipates with depth and with distance from center
Depth of induced stress ~2B

8. zA = zB g zc B A C
If foundation is flexible, it will settle more under the centerline because the induced stress is greater

9. Methods of Analysis Elastic Theory (Boussinesq’s Method, Newmark)
Numerical Solutions (Finite Element Methods, FEM)
Chart Solutions
Approximate Solutions (e.g., trapezoidal 2:1 rule)

10. Boussinesq’s Method Infinite elastic half space
See Coduto (1999) equations:
(Point Loads)
10.19 (Line Loads)
(Area Loads)
See Budhu (2007) equations:
(Point Loads, Displacement Dz, Dr)
(Line Loads)
(Strip Loads)
(Area Loads)

11. Boussinesq’s Method – Point Loads

13. Area Loads 2 General Solutions: Stress under center of area
Chart solutions
Stress under corner of area
Newmark method
Influence factor method
Often need to break up the problem (superposition) 1 2 3 4

14. Chart Solutions

15. Example Compute sz 10 m below edge of 25m diameter water tank
(mass of tank = 6.1 X 106 kg)

16. Newmark Method – Area Loads

17. Influence Factors (Budhu 5.11.6)
(Induced Beneath Corner of Rectangular Area, BXL)
Or use chart

18. Approximate Methods – Trapezoidal Rule (aka 1:2 Method)
Example:
Compute sz 10 ft below spread footing (3 ft X 3 ft) P = 10,000lb
Average Vertical Stress

19. Superposition Principle
See Budhu example 5.11
Point A – Break into 4 rectangles and multiply by 4
Point B – Break into 2 rectangles and multiply by 2
Point C – Break into 2 “fictitious” rectangles

20. Superposition Principle
475 kN
1.5 m
0.5 m
1 X 1.5 m
fictitious area
1.0 m I II
g = 17kN/m3
1.2m
0.5m A A
Compute sz at A (under corner)

22. Induced Stress (Boussinesq):
Geostatic Stress:
(Flat surface)

23. t s’ Superimposing… Effective Stress Element 846 160 213 160 213 43092 92
490 t
Effective Stress
Mohr Circle (x-z)
(490,213) s’
(846,-213)

24. t s’ What if we double load (P) to 180 kip? f’ c’ Mohr’s Circle grows!
Mohr-Coulomb Failure Envelope (Chap. 13) t f’
= c’ + s’tanf’
c’ = effective cohesion intercept
f’ = effective friction angle
(622,426)
(490,213) c’ s’
(846,-213)
(1166,-426)
Mohr’s Circle grows!
More shear stress placed on element
Failure if shear stress exceeds shear strength!!

25. Stress Under an Embankment
H = 40 ft
g = 110 pcf
B2 = 30 ft g H a1 a2 z
Still need to consider other side!

26. Lateral Stress on retaining wall with surface point load x = 1 m
P = 350 kN
3 m
???
g = 19.5 kN/m3
v = 0.3
K = 0.36
zw >>3m
1 m

27. Lateral Stress on retaining wall with surface point load x = 2 m
P = 350 kN
3 m
???
g = 19.5 kN/m3
v = 0.3
K = 0.36
zw >>3m
2 m

28. So how do we design the wall?.....
Bending Analysis
Shear Analysis
Overturning Analysis
Reinforcement (e.g., tie backs)
Effect of compaction?
Lightweight Fill?
P = 350 kN
???
g = 19.5 kN/m3
v = 0.3
K = 0.36
zw >>3m