chapter 1

CHAPTER 1 VERTICAL STRESSES DISTRIBUTION IN SOIL

LESSON OUTCOMES

At the end of this chapter, student should be able to:

a) Explain the meaning of total stress, effective stress and

overburden pressure.

b) Describe the concept of soil response to stress: elastic, plastic and

elasto-plastic.

c) Conduct empirical analysis and evaluate different stresses in soil

due to point load, line load, strip load, triangular strip load, uniformly

loaded circular area & uniformly loaded rectangular area using

Boussinesq Theory, Fadum’s and Newmark Chart.

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The Kissing Silos

What?

Why?

How?

REVISION : STRESS CONCEPT

In ECG303, you have learned about stresses in soil.

• Stress is defined as the intensity of loading per unit area.

You also should be able to differentiate in between total stress and

effective stress.

Total stress, σ Effective stress, σ’

The stress carried by the soil particles

and pore water, u in the soil void.

The stress carried by soil particles

alone.

σ=σ’+u σ’=σ-u

Short term analysis before pore water

dissipates.

Long term analysis.

Soil deformation, strength and stability

depend on effective stress because

water cannot resist shear stress.

Undrained condition. Drained condition.

REMEMBER : STRESS

CONCEPTWILL BE USED

THROUGHOUT THIS COURSE

OVERBURDEN PRESSURE • Pressure or stress imposed on the soil by the weight of overlying

material, also known as lithostatic pressure.

• The vertical stress imposed at point

x is σ=ϒz • If there is surcharge on the surface,

the total vertical stress at point x now is

σ=ϒz+q

Depth, z Unit weight of soil, ϒ (kN/m3)

x

Surcharge, q

Single layer

The total vertical stress imposed at point x is

σ=ϒ1z1 + ϒ2z2 + ϒ3z3

z1

ϒ3

x

ϒ1

ϒ2

z2

z3

Multiple layer

OVERBURDEN PRESSURE

• If the ground water is at the ground level, then total vertical

stress is σ=ϒsatz

• And, the pore water pressure at

point x is u=ϒwz

• Therefore, the effective vertical stress

is σ’= σ-u σ’= ϒsatz – ϒwz or σ’= ϒ’ z where ϒ’=(ϒsat –ϒw)

Saturated soil

Depth, z Saturated unit weight, ϒsat (kN/m3)

x

GWL

OVERBURDEN PRESSURE

• If the ground water is located deeper in

the soil layer, then vertical stress is σ= ϒ1z1 + ϒsatz2

• And, the pore water pressure at point x is

u=ϒwz2

• Therefore, the effective vertical stress is at

point x is σ’= ϒ1z1 + ϒsatz2 – ϒwz2 or σ’= ϒ1z1 + ϒ’z2

where ϒ’=(ϒsat –ϒw)

Saturated soil

z1

ϒsat

x

GWL

z2

ϒ1

SOIL RESPONSE TO STRESS

Foundation soil fails in supporting the

load.

An engineer must

ensure that a

geotechnical structure

must not collapse

under exposed

loading→ have to

determine stress and

strain in soil due to

external load.


• In mechanics, material response can be demonstrate

using stress-strain curve.

a) Elastic –

recovers its original

configuration during

unloading.

b) Plastic –

permanent deformation

(irrecoverable)

c) Yield point –

beyond this point, the

material start to behave

plastically.


What is soil behaviour? • Soil is very complex material. In general, soil is behaving

as an elasto-plastic material.

• Elasto-plastic material undergoes both elastic and

plastic deformation during loading.

• At small strains, soil behave like

an elastic material and thereafter

like an elastoplastic material


WHY it is IMPORTANT????????

• Calculate settlement • Proposed suitable foundation system • Remedial work to strengthen up the soil • Suggest for other construction site

STRESSES IN SOIL FROM

SURFACE LOAD.

• Point load • Vertical load transferred from

an electric power pole line.

• Line Load • Long brick wall

• Circular load • Water tannk


SURFACE LOAD.

• Strip load • Long road embankment

• Triangular load • Side of an embankment

• Rectangular load • Structural foundation


SURFACE LOAD.

What factors affecting the stress distribution in soil? 1 - Size and shape of foundation. 2 – Type of Load distribution. 3 - Contact pressure (surcharge load) 4 - Modulus of elasticity – Type of soil. 5 - Rigid boundary – stress increased is considered if the depth is less than 4B.

STRESSES CHANGES IN

SOIL FROM SURFACE LOAD.

Why need to learn? To know how surface stresses are distributed in the soil and resulting deformation.

Soil is elastic plastic material, so how to calculate stress?

To do stress analysis on soil, we must assumed that • soil is elastic and modulus of elasticity is constant. • Soil mass is homogeneous, isotropic. • Soil is semi-infinite (elastic half space)

By using this assumptions, we can apply Boussinesq’s Equation.

CHANGE IN VERTICAL

STRESSES UNDER POINT LOAD

Where Ip is the Influence Factor

Or alternatively, Ip can be obtained from Table 1.1.

Depth,

x

𝐼𝑃 =3

2𝜋

1

1 + 𝑥𝑧

2

52

Point load, P (kN) ∆𝜎𝑧=𝑃

𝑧2× 𝐼𝑃

Table 1.1 Influence Factor for Point Load

VERTICAL STRESSES

UNDER POINT LOAD

x/z x/z x/z

VERTICAL STRESSES UNDER

POINT LOAD

• Figure shows the distribution of pressure form point load (After Craig, 2001).


POINT LOAD - Example P1=1200 kN

P2=600 kN

x1=2m x2=6m

Load z x x/z Ip P Δσz (kN/m2)

P1 2 2 1 0.0844 1200 25.32

P2 2 6 3 0.0015 600 0.225

Vertical stress increased at Point A is 25.55 kN/m2

Solution (You can tabulated answer for simplicity)

z=2m

CHANGE IN VERTICAL STRESSES

UNDER LINE LOAD

(kN/m)

Where IL is the Influence Factor

𝐼𝐿 =2

𝜋

1

1 + 𝑥𝑧

2

12

Alternatively, IL can be obtained from Table 1.2.

∆𝜎𝑧=𝑃

𝑍× 𝐼𝐿

P (kN/m)

Table 1.2 Influence Factor, IL for Line Load

VERTICAL STRESSES

UNDER LINE LOAD

x/z x/z x/z


LINE LOAD -Example TWO (2) parallel line loads of 100 kN/m and 50 kN/m respectively and 2 m apart, act vertically on a horizontal soil surface. Calculate the vertical stress increase at depth of 2 m directly at the middle points between them.

Q1=100 kN/m Q2=50 kN/m

x1=1m x2=1m z=2m

Load z x x/z IL P Δσz (kN/m2)

Q1 2 1 0.5 0.2733 100 13.655

Q2 2 1 0.5 0.2733 50 6.83

Vertical stress increased is 20.485

Solution

Sketch out the situation

VERTICAL STRESSES

UNDER CIRCULAR LOAD

Where Ic is the Influence Factor

𝐼𝐶 = 1 −1

1 + 𝑎𝑧

2

32

Alternatively, Ic can be obtained from Table 1.3.

x

radius

∆𝜎𝑧= 𝑃 × 𝐼𝐶

P (kN/m2)

VERTICAL STRESSES

UNDER CIRCULAR LOAD Table 1.3 Influence Factor, Ic for Circular Load

x/a


CIRCULAR LOAD - Example

𝜎𝑧 = 𝑄𝐼𝐶

𝐼𝐶 = 1 −1

1 + 𝑎𝑧

2

32

A circular area with radius 1.6 m, induced a soil pressure at the surface of 100 kPa. Calculate the pressure at a depth of 3.2 m directly under the center of the circular area.

z x z/a x/a A B Ic (A+B) Δσz

(kN/m2)

3.2 0 2 0 0.106 0.179 0.285 28.5

Solution

Radius = a = 1.6m Top value

Bottom value


UNIFORM STRIP LOAD

Where IS is the Influence Factor

𝐼𝑠 =𝛼 + sin 𝛼 𝑐𝑜𝑠 𝛼 + 2𝛽

𝜋

Alternatively, Ip can be obtained from Table 1.4.

2b P

(kN/m2)

∆𝜎𝑧=𝑃

𝑍× 𝐼𝑆


UNIFORM STRIP LOAD

Table 1.4 Influence Factor, Is for Uniform Strip Load

x/b

z/b 0.10 0.20 0.40 0.60 0.80 1.00 1.25 1.50 2.00 3.00 5.00 10.00

0.00 1.000 1.000 1.000 1.000 1.000 0.500 0.000 0.000 0.000 0.000 0.000 0.000

0.20 0.997 0.996 0.992 0.979 0.979 0.500 0.059 0.011 0.002 0.000 0.000 0.000

0.40 0.977 0.973 0.955 0.906 0.773 0.498 0.178 0.059 0.011 0.001 0.000 0.000

0.60 0.937 0.928 0.896 0.825 0.691 0.495 0.258 0.120 0.030 0.004 0.000 0.000

0.80 0.881 0.869 0.829 0.755 0.638 0.489 0.305 0.173 0.056 0.010 0.001 0.000

1.00 0.818 0.805 0.766 0.696 0.598 0.480 0.332 0.214 0.084 0.017 0.002 0.000

1.20 0.755 0.743 0.707 0.646 0.564 0.466 0.347 0.243 0.111 0.026 0.004 0.000

1.40 0.696 0.685 0.653 0.602 0.534 0.455 0.354 0.263 0.135 0.037 0.005 0.000

1.60 0.642 0.633 0.605 0.562 0.515 0.440 0.356 0.276 0.155 0.048 0.008 0.000

1.80 0.593 0.585 0.563 0.526 0.497 0.425 0.353 0.284 0.172 0.060 0.010 0.000

2.00 0.550 0.543 0.524 0.494 0.455 0.409 0.348 0.288 0.185 0.071 0.013 0.001

2.50 0.462 0.456 0.445 0.426 0.400 0.370 0.328 0.285 0.205 0.095 0.022 0.002

3.00 0.396 0.393 0.385 0.372 0.355 0.334 0.305 0.274 0.211 0.114 0.032 0.003

3.50 0.345 0.343 0.338 0.329 0.317 0.302 0.281 0.258 0.210 0.127 0.042 0.004

4.00 0.306 0.304 0.301 0.294 0.285 0.275 0.259 0.242 0.205 0.134 0.051 0.006

5.00 0.248 0.247 0.245 0.242 0.237 0.231 0.222 0.212 0.188 0.139 0.065 0.010

6.00 0.208 0.208 0.207 0.205 0.202 0.198 0.192 0.186 0.171 0.136 0.075 0.015

8.00 0.158 0.157 0.157 0.156 0.155 0.153 0.150 0.147 0.140 0.122 0.083 0.025

10.00 0.126 0.126 0.126 0.126 0.125 0.124 0.123 0.121 0.117 0.107 0.082 0.032

15.00 0.085 0.085 0.085 0.084 0.084 0.084 0.083 0.083 0.087 0.078 0.069 0.041

20.00 0.054 0.064 0.064 0.063 0.063 0.063 0.063 0.063 0.062 0.061 0.056 0.041

50.00 0.025 100.00 0.013


TRIANGLE LOAD

(kN/m2)

Where IT is the Influence Factor

Alternatively, Ip can be obtained from Table 1.5.

𝐼𝑇 =

𝑥𝛽𝛼 −

12sin 2𝛽

𝜋

∆𝜎𝑧= 𝑃 × 𝐼𝑇


TRIANGLE LOAD

x/c

z/c -2.00 -1.50 -1.00 -0.50 0.00 0.20 0.40 0.60 0.80 1.00 1.25 1.50 2.00

0.00 0.000 0.000 0.000 0.000 0.000 0.200 0.400 0.600 0.800 0.500 0.000 0.000 0.000

0.20 0.000 0.000 0.000 0.002 0.061 0.209 0.395 0.577 0.697 0.437 0.050 0.009 0.001

0.40 0.000 0.001 0.003 0.013 0.110 0.227 0.372 0.497 0.527 0.379 0.136 0.042 0.007

0.60 0.001 0.003 0.008 0.031 0.140 0.023 0.334 0.409 0.414 0.328 0.177 0.080 0.018

0.80 0.003 0.006 0.016 0.049 0.155 0.225 0.294 0.339 0.337 0.285 0.187 0.106 0.032

1.00 0.005 0.011 0.025 0.064 0.159 0.211 0.258 0.286 0.283 0.250 0.184 0.121 0.046

1.20 0.008 0.016 0.034 0.075 0.157 0.195 0.227 0.245 0.243 0.221 0.175 0.126 0.057

1.40 0.011 0.021 0.041 0.083 0.151 0.179 0.202 0.215 0.213 0.197 0.165 0.127 0.066

1.60 0.015 0.026 0.048 0.087 0.143 0.165 0.182 0.190 0.189 0.178 0.154 0.124 0.072

1.80 0.018 0.031 0.053 0.089 0.135 0.152 0.164 0.171 0.170 0.161 0.143 0.120 0.076

2.00 0.021 0.035 0.057 0.089 0.127 0.140 0.150 0.155 0.015 0.148 0.134 0.115 0.078

2.50 0.028 0.042 0.062 0.086 0.110 0.117 0.122 0.125 0.124 0.121 0.113 0.103 0.078

3.00 0.033 0.046 0.062 0.080 0.095 0.100 0.103 0.105 0.104 0.102 0.095 0.091 0.074

3.50 0.037 0.048 0.060 0.073 0.084 0.087 0.089 0.090 0.090 0.089 0.085 0.081 0.069

4.00 0.038 0.048 0.058 0.067 0.075 0.077 0.078 0.079 0.079 0.076 0.076 0.073 0.064

5.00 0.039 0.045 0.051 0.057 0.061 0.062 0.063 0.063 0.063 0.062 0.062 0.060 0.055

6.00 0.037 0.041 0.046 0.049 0.052 0.052 0.530 0.053 0.053 0.052 0.052 0.051 0.048

8.00 0.032 0.035 0.037 0.038 0.039 0.039 0.040 0.040 0.040 0.039 0.039 0.039 0.038

10.00 0.028 0.029 0.030 0.031 0.032 0.032 0.032 0.032 0.032 0.032 0.032 0.031 0.031

15.00 0.020 0.020 0.021 0.021 0.021 0.021 0.021 0.021 0.021 0.021 0.021 0.021 0.021

20.00 0.015 0.016 0.016 0.016 0.016 0.016 0.016 0.016 0.016 0.016 0.016 0.016 0.016

50.00 0.006 0.006 0.006 0.006 0.006 0.006 0.006 0.006 0.006 0.006 0.006 0.006 0.006

Table 1.5 Influence Factor, Is for Triangle Load

VERTICAL STRESSES UNDER TRIANGLE

& STRIP LOAD - Example

A 6 m high embankment with unit weight of 20 kN/m3 is as shown below. Determine the increase in vertical stress at depth of 3.2 m at points A upon completion of the embankment.

6 m 8 m

c z x z/c x/c IT σz (kN/m2)

8 3.2 14 0.4 1.75 0.0245 120(0.0245) =2.94

Triangular load Embankment loading, 𝑞 = 𝛾𝐻 = 20 6 = 120𝑘𝑁/𝑚2

By referring to table 1.5, no value 1.75 at row x/c. So, need to interpolate IT. Interpolation technique:

2.00 1.50

0.007 0.042 IT

1.75

Refer row z/c=0.4

First, write down the equation 2.00 − 1.50

0.007 − 0.042=2.00 − 1.75

0.007 − 𝐼𝑇

Then rearrange, (2-1.5)(0.007-IT) = (2-1.75)(0.007-0.042)

3.5x10-3 – 0.5IT = -8.75x10-3

IT = 0.0245

?

x/c

z/c -2.00 -1.50 -1.00 …….... 1.25 1.50 2.00

0.20 0.000 0.000 0.000 …….... 0.050 0.009 0.001

0.40 0.000 0.001 0.003 …….... 0.136 0.042 0.007

0.60 0.001 0.003 0.008 …….... 0.177 0.080 0.018

1.75

?

2.00 1.50

0.007 0.042 IT

1.75

Triangle load

Transfer the value into this diagram

b z x z/b x/b IS σz (kN/m2)

6 3.2 3 0.53 0.5 0.885 120(0.885)=106.2

Strip load Embankment loading, 𝑞 = 𝛾𝐻 = 20 6 = 120𝑘𝑁/𝑚2

By referring to table 1.4, no value 0.5 at row x/b and no value of 0.53 at row z/b. So, need to interpolate IS three 5(3) times. Interpolation 1: (refer to z/b=0.4 and x/b between 0.4 &0.6)


0.906 − 0.955=

0.6 − 0.5

0.906 − 𝑎

Then rearrange, (0.6-0.4)(0.906-a) = (0.6-0.5)(0.906-0.955)

0.1812– 0.2a = -4.9x10-3

a = 0.9305

0.6 0.4

0.906 0.955 a

0.5

Refer row z/b=0.6

x/b

z/b 0.10 0.20 0.40 0.60 0.80

0.00 1.000 1.000 1.000 1.000 1.000

0.20 0.997 0.996 0.992 0.979 0.979

0.40 0.977 0.973 0.955 0.906 0.773

0.60 0.937 0.928 0.896 0.825 0.691

0.80 0.881 0.869 0.829 0.755 0.638

Strip load

0.53

0.5

a Interpolation 1

Interpolation 3

Interpolation 2 b

c

Interpolation 2: (refer to z/b=0.6 and x/b between 0.4 &0.6)

0.6 0.4

0.825 0.896 b

0.5

Refer row z/b=0.6


0.825 − 0.896=

0.6 − 0.5

0.825 − 𝑏

Then rearrange, (0.6-0.4)(0.825-b) = (0.6-0.5)(0.825-.896)

0.165– 0.2b = -7.1x10-3

b = 0.8605 Interpolation 3: (refer to z/b between 0.4&0.6 and and x/b = 0.5)

0.6 0.4

0.8605 0.9305 IS

0.53

Refer row z/b=0.6


0.8605 − 0.9305=

0.6 − 0.53

0.8605 − 𝐼𝑆

Then rearrange,

(0.6-0.4)(0.8605-IS) = (0.6-0.53)(0.8605-0.9305)

0.1721– 0.2IS = -4.9x10-3

IS = 0.885

Total vertical stress induced in the soil mass due to surface loading is : = Triangular + Strip = 2.94 + 106.2 = 109.14 kN/m2


RECTANGULAR LOAD

Where IR is the Influence Factor

Alternatively, IR can be obtained from Table 1.6 or by Using Fadum’s Chart.

𝐼𝑅 =1

4𝜋

2𝑚𝑛 𝑚2 + 𝑛2 + 1

𝑚2 + 𝑛2 +𝑚2𝑛2 + 1

𝑚2 + 𝑛2 + 2

𝑚2 + 𝑛2 + 1

+ tan−12𝑚𝑛 𝑚2 + 𝑛2 + 1

𝑚2 + 𝑛2 −𝑚2𝑛2 + 1

Where m=Longer span, L/z n=Shorter span, B/z (L or B can be interchangeable)

∆𝜎𝑧= 𝑃 × 𝐼𝑅


RECTANGULAR LOAD

Table 1.6 Influence Factor, IR for Rectangular Load

L/z

B/z 0.7 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.4 2.0 3.0 5.0

0.1 0.0047 0.0092 0.0132 0.0168 0.0198 0.0222 0.0242 0.0258 0.0270 0.0279 0.0301 0.0311 0.0315 0.0316 0.0316

0.2 0.0092 0.0179 0.0259 0.0328 0.0387 0.0435 0.0474 0.0504 0.0528 0.0547 0.0589 0.0610 0.0620 0.0620 0.0620

0.3 0.0132 0.0259 0.0374 0.0474 0.0560 0.0630 0.0686 0.0731 0.0766 0.0794 0.0856 0.0887 0.0898 0.0901 0.0902

0.4 0.0168 0.0328 0.0474 0.0602 0.0711 0.0801 0.0873 0.0931 0.0977 0.1013 0.1094 0.1134 0.1150 0.1154 0.1154

0.5 0.0198 0.0387 0.0560 0.0711 0.0840 0.0947 0.1034 0.1104 0.1158 0.1202 0.1300 0.1350 0.1368 0.1374 0.1375

0.6 0.0222 0.0435 0.0629 0.0801 0.0947 0.1069 0.1168 0.1247 0.1310 0.1361 0.1475 0.1533 0.1555 0.1561 0.1562

0.7 0.0240 0.0474 0.0686 0.0873 0.1034 0.1168 0.1277 0.1365 0.1436 0.1491 0.1620 0.1686 0.1711 0.1719 0.1720

0.8 0.0258 0.0504 0.0731 0.0931 0.1104 0.1247 0.1365 0.1461 0.1537 0.1598 0.1739 0.1812 0.1841 0.1849 0.1850

0.9 0.0270 0.0528 0.0766 0.0977 0.1158 0.1311 0.1436 0.1537 0.1619 0.1684 0.1836 0.1915 0.1947 0.1956 0.1958

1.0 0.0279 0.0547 0.0794 0.1013 0.1202 0.1361 0.1491 0.1598 0.1684 0.1752 0.1914 0.1999 0.2034 0.2044 0.2046

1.4 0.0301 0.0589 0.0856 0.1094 0.1300 0.1475 0.1620 0.1739 0.1836 0.1914 0.2102 0.2206 0.2250 0.2263 O.2266

2.0 0.0311 0.0610 0.0887 0.1134 0.1350 0.1533 0.1686 0.1812 0.1915 0.1999 0.2206 0.2325 0.2378 0.2395 0.2399

3.0 0.0315 0.0618 0.0898 0.1150 0.1368 0.1555 0.1711 0.1841 0.1947 0.2034 0.2250 O.2378 0.2420 0.2461 0.2465

5.0 0.0316 0.0620 0.0901 0.1154 0.1374 0.1561 0.1719 0.1849 0.1956 0.2044 0.2263 0.2395 0.2461 0.2486 0.2491

0.0316 0.0620 0.0902 0.1154 0.1375 0.1562 0.1720 0.1850 0.1958 0.2046 0.2266 0.2399 0.2465 0.2492 0.2500


RECTANGULAR LOAD

FADUM’S CHART

• The influence values referred to this chart is strictly for corner of a rectangular loaded foundation only.

• For points other than the corner, value of IR may be obtained by superpositions of rectangles


RECTANGULAR LOAD

SUPERPOSITIONS OF RECTANGLES

(a) Ir = Ir directly from table 1.6 or Fadum’s Chart (b) Ir = Ir AFIE + FBGI + GCHI +HDEI (c) Ir = Ir AEFD + EBCF (d) Ir = Ir GIAE – GHBE – GIDF +GHCF


RECTANGULAR LOAD - Example

Find the vertical stress at z = 2.5 m for a rectangular loading of q = 100kPa as shown below points R, S, T & U.

(Solid line denotes rectangular loading area)

Loaded area

Unloaded area



Point L B z L/z (m)

B/z (n)

IR σz

(kN/m2)

R

2 1.5 2.5 0.8 0.6 0.128 25.6

S

3 1 2.5 1.2 0.4 0.108 (0.108)(100) x 2 =21.6

T 4 1

1 1

2.5 2.5

1.6 0.4

0.4 0.4

0.112 0.06

(0.112-0.06)(100)

x 2 = 10.4

1.5=B

2=L

1=B

4=L

3=L

1=B

Consider half space

Consider half space

Where m=Longer span, L/z n=Shorter span, B/z (L or B can be interchangeable)

Point L B z L/z (m)

B/z (n)

IR σz

(kN/m2)

U (refer

dimension from question)

Box 1 4 3 2.5 1.6 1.2 0.205 20.5

Box 2

3 1 2.5 1.2 0.4 0.108 10.8

Box 3 4 1 2.5 1.6 0.4 0.112 11.2

Box 4 1 1 2.5 0.4 0.4 0.06 6



3

1

4 1

1 1

3

4

3

1 2

4



U

The vertical stress increment at point U is : = [box 1] – [box 2]-[box 3]+[box4] = 20.5 – 10.8 – 11.2 + 6 = 4.5 kN/m2

3

1 2

4


IRREGULAR SHAPE AREA

Value of n and I can be calculated from

Newmark’s Influence Chart

∆𝜎𝑧= 𝑄nI

Where Q= load per unit area n= number of influence units I=Influence value

The vertical stress are:


IRREGULAR SHAPE AREA - Example

Find the vertical stress at point U at a depth of 10 m with q= 100 kPa using Newmark’s Influence Chart.

Newmark’s Influence Chart

1. By using the depth scale at the bottom right of the chart, scaled out the foundation and draw with point U at the center of the chart.,

i.e. length AB=32mm depth = 10 m as given so, the scaled is 1 m=3.2mm if 4m= 12.8mm if 6m= 19.2mm if 10m =32mm if 1m= 3.2mm 2. Calculate the numbers of segment

covered by the foundation. For example 46. 3. Use the stress equation

U


IRREGULAR SHAPE AREA - Example

𝜎𝑧 = 𝑄nI 𝜎𝑧 = 100(46)(0.005)

σz= 23 kN/m2

Given z=10m

SUMMARY OF THE TOPIC VERTICAL STRESS

a) Explain the total stress and effective stress analysis.

b) Solve total stress and effective stress problems due to overburden pressure.

c) Analyze the empirical analysis for -point load, -line load, -strip load, -circular load -rectangular load by -using Boussinesq theory, -using Fadum’s Chart -using Newmark’s Chart.

- able to define x, a, b, c, L, B parameter

- know how to read table for influence factor

- able to do interpolation

HAPPY REVISING. DO WELL IN your

ASSESSMENTS!!!

chapter 1

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