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Baøi giaûng A. Prof. Dr Chaâu Ngoïc AÅn. ÖÙNG SUAÁT TRONG NEÀN ÑAÁT. Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn. ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG. Haït theùp. nöôùc. Ñaát baõo hoøa nöôùc. Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn. - PowerPoint PPT Presentation
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Baøi giaûng A. Prof. Dr Chaâu Ngoïc AÅn
ÖÙNG SUAÁT TRONG NEÀN ÑAÁT
ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG
Ñaát baõo hoøa nöôùc
Haït theùp
nöôùc
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG
Ñaát baõo hoøa nöôùc
Haït theùp
nöôùc
* bình beân traùi theâm vaøo treân maët lôùp ñaát caùc haït theùp taïo moät aùp löïc p, maãu ñaát bò luùn xuoáng. AÙp löïc p coù aûnh höôûng leân öùng suaát khung neân laø öùng suaát höõu hieäu, kyù hieäu laø ’* bình beân phaûi theâm nöôùc treân maët ñeå taïo aùp löïc p, maãu ñaát khoâng luùn xuoáng vì nöôùc theâm vaøo thoâng vôùi nöôùc trong loã roãng taùc ñoäng leân ñaùy bình chöùa, p do coät nöôùc khoâng aûnh höôûng leân khung haït (trung hoøa).
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
ÖÙng suaát taïi moät ñieåm trong neàn ñaát goàm “öùng suaát giöõa caùc haït” hay öùng suaát höõu hieäu ’ vaø aùp löïc nöôùc trong loã roãng u, theo ñònh ñeà Terzaghi
= ’ + u
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
TÍNH ÖÙNG SUAÁT TRONG ÑAÁT NEÀN DO TROÏNG LÖÔÏNG BAÛN THAÂN
ÖÙng suaát toång do troïng löôïng baûn thaân ñaát theo phöông thaúng ñöùng kyù hieäu laø bt hay v taïi moät ñieåm baát kyø trong ñaát caùch maët ñaát moät chieàu saâu baèng H, coù theå tính nhö laø troïng löôïng khoái ñaát beân treân truyeàn xuoáng.
H
zbtv dzz0
, )(
n
iizbt h1
,
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
vôùi K0 laø heä soá aùp löïc ngang ôû traïng thaùi tónh cuûa ñaát coá keát thöôøng (ñaëc ñieåm coá keát thöôøng vaø coá keát tröôùc seõ ñöôïc phaân tích trong caùc chöông sau). Khoaûng nöûa theá kyû tröôùc, heä soá aùp löïc ngang ñöôïc vay möôïn töø lyù thuyeát ñaøn hoài vôùi kyù hieäu laø vaø coù daïng : trong ñoù laø heä soá Poisson.Vôùi toång keát töø raát nhieàu keát quaû thí nghieäm vaø ño ñaïc giaùn tieáp, Jaky ñaõ ñöa ra moät coâng thöùc ñeå tính heä soá aùp löïc ngang ôû traïng thaùi tónh (cuûa ñaát coá keát thöôøng) nhö sau :K0 = 1 - sin’Vôùi ’ laø goùc ma saùt trong ñieàu kieän caét thoaùt nöôùc (seõ phaân tích roõ trong chöông choáng caét). Coâng thöùc cuûa Jaky phuø hôïp cho ñaát rôøi hoaëc ñaát loaïi caùt.Neáu goùc ma saùt ’= 350 thì K0 = 1 – sin350 = 0,426Ñoái vôùi ñaát dính hoaëc ñaát loaïi seùt coá keát thöôøng, Alpan ñeà nghò moät coâng thöùc thöïc nghieäm.
K0 = 0,19 + 0,233logIP Neáu moät maãu seùt coù chæ soá deûo IP = 20, thì heä soá aùp löïc ngang ôû traïng thaùi tónh K0 theo coâng thöùc Alpan laø :
K0 = 0,19 + 0,233log20 = 0,493
10K
'0
'vh K
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
Seùt, c2
Seùt, c1
Caùùt, s2
Caùùt, s1
V
hWhS1
hC1
hS2
hC2 ’Vu
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
Seùt, C1
Seùt, C2
Caùùt, S2
Caùùt, S1
hW1
hW2
hS1
hC1
hS2
hC2
BS
BC
z
’u
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
Nguyeân lyù ño aùp löïc nöôùc loã roãngtrong ñaát
W
W
Soil loaded by an applied weight W
Soil loaded by water weighing W
W
W
Soil loaded by an applied weight W
Soil loaded by water weighing W
Compression No deformation
v Vertical StressVertical Force
Cross Sectional Area (1)
Definition of Total and Effective Stress
v Vertical StressVertical Force
Cross Sectional Area
v v wu'
(1)
(2)
Definition of Total and Effective Stress
Effective vertical stress
v Vertical StressVertical Force
Cross Sectional Area
v v wu'
(1)
(2)
Definition of Total and Effective Stress
Effective vertical stress
vuwv́
Case (a) W
A0 W
A
v Vertical StressVertical Force
Cross Sectional Area
v v wu'
(1)
(2)
Definition of Total and Effective Stress
Effective vertical stress
vuwv́
Case (a) W
A0 W
A
Case (b) W
A
W
A0
Fig 2 Two Pieces of Rock in Contact
N
T
Water pressure uw
Effective Stress
N
T
Water pressure uw
Effective Force
N N U (3a)
Effective Stress
U = uw ( A - Ac )
N
T
Water pressure uw
Effective Force
N N U (3a)
Effective Stress
U = uw ( A - Ac )
Frictional Failure
T N
N
T
Water pressure uw
Effective Force
N N U (3a)
Effective Stress
U = uw ( A - Ac )
Frictional Failure
T N
Failure in terms of stress
v (3b)
T
A
N
A
U
A
Layer 1
Layer 2
Layer 3
d1
d2
d3
bulk 1
bulk 2
bulk 3
Fig 3 Soil Profile
Surcharge q
v
z
Calculation of Effective Stress
d1
d2
q
v
z
A
Plan
Elevation
Calculation of Total Vertical Stress
z
Force on base = Force on top + Weight of soil
d1
d2
q
v
z
A
Plan
Elevation
Calculation of Total Vertical Stress
z
Force on base = Force on top + Weight of soil
A v = A q + A 1 d1 + A 2 d2 +
A 3 ( z - d1 - d2 )
d1
d2
q
v
z
A
Plan
Elevation
(4)
Calculation of Total Vertical Stress
z
Force on base = Force on top + Weight of soil
A v = A q + A 1 d1 + A 2 d2 +
A 3 ( z - d1 - d2 )
v = q + 1 d1 + 2 d2 +3 ( z - d1 - d2 )
Fig 4 Soil with a static water table
Water table
H
P u P Hw w( ) (5)
Calculation of pore water pressure
Fig 4 Soil with a static water table
Water table
H
P u P Hw w( ) (5)
• The water table is the level of the water surface in a borehole.
Calculation of pore water pressure
Fig 4 Soil with a static water table
Water table
H
P u P Hw w( ) (5)
• The water table is the level of the water surface in a borehole.
• It is the level at which the pore water pressure uw = 0
Calculation of pore water pressure
Dry
Saturated
2 m
3m
Fig 5 Soil Stratigraphy
b u l k d r y
b u lk s a t
Step 1: Draw ground profile showing soil stratigraphy and water table
Example: determining the effective stress
Distribution by Volume
Solid
VoidsVv=e Vs
= 0.7m3
Vs= 1m3
Step 2: Calculation of relevant bulk unit weights
Example
Distribution by Volume
Solid
VoidsVv=e Vs
= 0.7m3
Distribution by weight for the dry soil
Vs= 1m3
W V G
k N
k N
s s s w
1 2 7 9 8
2 6 4 6
. .
.
W = 0
Step 2: Calculation of relevant bulk unit weights
Example
Distribution by Volume
Solid
VoidsVv=e Vs
= 0.7m3
Distribution by weight for the dry soil
Distribution by weightfor the saturated soil
Vs= 1m3
W V kN
kN
kN
w v w
0 7 9 8
6 86
. .
.
W V G
k N
k N
s s s w
1 2 7 9 8
2 6 4 6
. .
.
W V G
kN
kN
s s s w
1 2 7 9 8
26 46
. .
.
W = 0
Step 2: Calculation of relevant bulk unit weights
Example
Distribution by Volume
Solid
VoidsVv=e Vs
= 0.7m3
Distribution by weight for the dry soil
Distribution by weightfor the saturated soil
Vs= 1m3
W V kN
kN
kN
w v w
0 7 9 8
6 86
. .
.
W V G
k N
k N
s s s w
1 2 7 9 8
2 6 4 6
. .
.
W V G
kN
kN
s s s w
1 2 7 9 8
26 46
. .
.
Ww=0
Step 2: Calculation of relevant bulk unit weights
Example
drys wkN
mkN m
G
e
26 46
1701556
133.
.. /
Distribution by Volume
Solid
VoidsVv=e Vs
= 0.7m3
Distribution by weight for the dry soil
Distribution by weightfor the saturated soil
Vs= 1m3
W V kN
kN
kN
w v w
0 7 9 8
6 86
. .
.
W V G
k N
k N
s s s w
1 2 7 9 8
2 6 4 6
. .
.
W V G
kN
kN
s s s w
1 2 7 9 8
26 46
. .
.
Ww=0
drys w
satw s
kN
mkN m
G
e
kN
mkN m
G e
e
26 46
1701556
1
26 46 686
17019 60
1
33
33
.
.. /
( . . )
.. /
( )
Step 2: Calculation of relevant bulk unit weights
Example
2 m
3m
v kPa kN m 15 56 2 19 60 3 89 92 2. . . ( / )
Step 3 Calculate total stress
Example
2 m
3m
v kPa kN m 15 56 2 19 60 3 89 92 2. . . ( / )
Step 3 Calculate total stress
Example
u kPaw 3 9 8 29 40. .
Step 4 Calculate pore water pressure
2 m
3m
v kPa kN m 15 56 2 19 60 3 89 92 2. . . ( / )
Step 3 Calculate total stress
Example
u kPaw 3 9 8 29 40. .
Step 4 Calculate pore water pressure
Step 5 Calculate effective stress
v v wu kPa89 92 29 40 60 52. . .
0 50 100 150
0m
2m
4m
6m
8m
kPa
pore waterpressure Effective
stress
TotalStress (5m)
Depth
Vertical stress and pore pressure variation
Fig 7 Definition of Stress Components
x
y
z
xz
yz
zz
yy
xy
zy
xx
yx
zx
z
x
Stresses acting on a soil element
Effective stress relations for general stress states
xx xx w yz yz
yy yy w zx zx
zz zz w xy xy
u
u
u
;
;
;(10)
Principle of Effective Stress
Example
Clay
Rock aquifer
Initial GWL
z
Lowered GWL3 m1 m
Example
Clay
Rock aquifer
Initial GWL Lowered GWL
v bulk z bulk z
u w (z - 1) w (z - 3)
v´ bulk - w )z + w bulk - w )z + 3 w
Initial GWL
z
Lowered GWL3 m1 m
• Effective stress increases - soil compresses - ground surface settles
• Effective stress decreases- soil swells - ground surface heaves. The following problems may then occur
• surface flooding
• flooding of basements built when GWL lowered
• uplift of buildings
• failure of retaining structures
• failures due to reductions in bearing capacity
Example
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
ÖÙNG SUAÁT TRONG NEÀN ÑAÁT DO TAÛI NGOAØI
Slide 16 of 36
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
c1
P
a1 a
c
R
d R
Döôùi taùc duïng cuûa P ñieåm M chuyeån vò moät ñoaïn S theo phöông baùn kính R. M caøng xa O thì S caøng nhoû. Maët khaùc, vôùi R = const, goùc caøng lôùn thì S cuõng caøng nhoû. Xuaát phaùt töø nhaän xeùt ñoù, ta coù theå vieát bieåu thöùc S coù daïng : R
AScos
Tröông töï, taïi M1 caùch M moät ñoaïn dR, coù chuyeån vò S1
dRRAS
cos1
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
Bieán daïng töông ñoái R cuûa ñoaïn dR laø:
cos.
cos
dR
S - S2
1
dRRR
A
dRR
A
R
A
dRR
Boû qua R.dR vì raát nhoû so vôùi R2
cos2R
AR
Theo giaû thuyeát quan heä giöõa öùng suaát vaø bieán daïng laø tuyeán tính do ñoù öùng suaát xuyeân taâm R gaây neân bieán daïng R ñöôïc xaùc ñònh nhö sau
Trò soá A.B coù theå xaùc ñònh döïa theo ñieàu caân baèng tónh hoïc. Xeùt ñieàu kieän caân baèng tónh hoïc cuûa baùn caàu (O; R)
cos2R
ABR
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
2
0
.cos
dFP R
Trong ñoù: dF – dieän tích maët ñai troøn caa1c1
dF = 2(Rsin)(Rd)
2
0
22
2
0
sin2coscos.cos
dRR
ABdFP R
dBAP sincos2..2
0
2 BAP ...3
2 P
AB .2
3
cos.2
32R
PR
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
5
3
2
3
R
zPz
23
2
3
22
5
2
)(
)2(
)(
.
3
21
2
3
zRR
zRx
zRR
zzRR
R
zxPx
23
2
3
22
5
2
)(
)2(
)(
.
3
21
2
3
zRR
zRy
zRR
zzRR
R
zyPy
5
2.
.2
.3
R
zyPzy
5
2.
.2
.3
R
zxPzx
235 )(
).2(..
3
.21..
.2
.3
zRR
zRyx
R
zyxPxy
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
RR
z
E
Pw
1)1(2
..2
)1(3
2
)(
).21(.
..2
)1(3 zRR
x
R
zx
E
Pu
)(
).21(.
..2
)1(3 zRR
y
R
zy
E
Pv
Chuyeån vò theo chieàu caùc truïc :
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
yA
x
R1
O
z
M(x,y,z)r
h
h
z
R2
P
72
3**
52
**2**
51
3*
32
*
31
*
30)5(3)43(3
3)21()21(
)1(8
R
Zhz
R
hzZhZz
R
Z
R
Z
R
ZPz
52
2**
32
2**
31
2*
2
2
1
)(62))(43(
)43()1(843
)1(16
R
Zhz
R
zhZ
R
Z
RRG
Pw
Baøi toaùn Mindlin
Slide 17 of 36
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
ÑÖÔØNG LÖÏC HAY LÖÏC ÑÖÔØNG THAÚNGÖÙng suaát do aûnh höôûng cuûa löïc daïng ñöôøng thaúng phaân boá ñeàu (kM/m) nhö: ñöôøng raây; töôøng chòu löïc trong neàn ñaát, … ñöôïc Flamant phaùt trieån töø baøi toaùn Boussinesq (1892) baèng caùch chia ñöôøng löïc thaønh voâ soá löïc taùc ñoäng pdy leân moät ñoaïn thaät ngaén dy, aùp duïng coâng thöùc Boussinesq cho löïc nhoû naøy roài tích phaân leân caû chieàu daøi taùc ñoäng löïc ñeå coù ñöôïc caùc coâng thöùc sau:
x
y
O
z y
z
p
dy
pdy
M
R
dy
zyx
zpz
25
222
3
.2
.3
222
3.2
zx
zpz
222
2 ..2
zx
zxpx
222
2..2
zx
zxpzx
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Taûi phaân boá ñeàu treân dieän tích baêng (Flamant)
Taûi phaân boá treân dieän tích baêng laø daïng raát thöôøng gaëp trong neàn moùng coâng trình nhö: moùng baêng, ñöôøng, ñeâ,… Khaûo saùt moät ñoaïn dx trong phaïm vi töø -b/2 ñeán +b/2, giaù trò taûi töông öùng laø pdx töông töï moät ñöôøng löïc, tính öùng suaát dz do ñöôøng löïc pdx gaây ra vaø ñoåi bieán soá sang goùc nhìn töø M veà ñaùy moùng.
o
M
1
2
b
d
z
r
x
dx
P
z
x
p
M
3
222
3
cos..
..2..2
r
dxp
zx
zdxpd z
cos
zr
tgzx .
2cos
.dzdx
dp
d z .cos..2 2
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
2211z 2sin2
1)(2sin
2
1P
1
2
1
2
.cos.21.cos.2 2
dP
dP
z
2211 2sin
2
1)(2sin
2
1
Px
12xzzx 2cos2cos.2
P
Trò soá 2 laáy vôùi daáu döông khi ñieåm M naèm ngoaøi phaïm vi hai ñöôøng thaúng ñi qua hai meùp cuûa taûi troïng
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
Tröôøng hôïp ñôn giaûn nhaát laø ñoái vôùi caùc ñieåm naèm treân maët chöùa Oz (ñi qua truïc taâm taûi troïng). Vì tính ñoái xöùng cho neân :
1 = 2 =
2sin21 p
z
2sin2P
3x
02cos2cos.2
P12xzzx
2
Slide 20 of 36
Slide 21 of 36
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Taûi phaân boá ñeàu treân dieän tích chöõ nhaät
b
l x
y d
d
p
O
ÖÙng suaát thaúng ñöùng Z cuûa ñieåm naèm treân truïc thaúng ñöùng ñi qua taâm dieän chòu taûi, ôû ñoä saâu z:
pk
zlbzlzb
zlbzlb
zlbz
lbarctg
pz 022
12
122
122
1
221
2111
221
21
11 .22
1
1
1
12
5222
3
)(.2
...3b
b
l
l
z
zyx
zddp
2
ll1
21
bb
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
ÖÙng suaát thaúng ñöùng z do taûi phaân boá ñeàu treân dieän chòu taûi chöõ nhaät, doïc truïc thaúng ñöùng beân döôùi ñieåm goùc dieän chòu taûi.
pklbzlbz
zlbblz
zlb
zlb
lbzlbz
zlbblzpgz
222222
2/12221
222
222
222222
2/3222
)(
)(2tan
2.
)(
)(2
4
pInmnm
nmmn
nm
nm
nmnm
nmmnpz
1
12tan
1
2
1
12
4 2222
221
22
22
2222
22
z
bm
Hoaëc
z
ln
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn PHÖÔNG PHAÙP ÑIEÅM GOÙC.
A B
CD
M
A B
CD
M
A1
B1
C1
D1
öùng suaát (z) cuûa dieän chöùa taûi ABCD döôùi ñieåm M laø toång caùc öùng suaát goùc M cuûa caùc dieän MA1AD1; MB1BA1; MC1CB1; MC1DD1.Z,M = p[kg(MA1AD1) + kg(MB1BA1) + kg(MC1CB1) + kg(MC1DD1)]p laø aùp löïc phaân boá ñeàu treân dieän chòu taûi ABCD.
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
A B
CD
M
A B
CD
M
B2
C2
D2 A2
öùng suaát (z) cuûa taïi M do dieän chöùa taûi ABCD taùc ñoäng laø toång ñaïi soá caùc öùng suaát goùc M cuûa caùc dieän MA2CC2; MD2DC2; MD2AB2; MA2BB2. Trong ñoù chæ coù dieän tích ABCD laø chöùa taûi phaân boá ñeàu p, phaàn dieän tích coøn laïi khoâng coù taûi. Z,M = p[kg(MD2AB2) - kg(MD2DC2) - kg(MA2BB2) + kg(MA2CC2)]p laø aùp löïc phaân boá ñeàu treân dieän chòu taûi ABCD.
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
TÍNH ÖÙNG SUAÁT THEO PHÖÔNG PHAÙP THAÙP LAN TOÛA
)2)(2(
ztglztgb
Qz
))(( zlzb
Qz
Ñôn giaûn hôn thaùp lan toûa coù ñoä doác 2:1 öùng suaát z ôû ñaùy thaùp lan toûa coù daïng:
Slide 18 of 36
Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn
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Ñoái vôùi ñaát caùt thöôøng coù caáu truùc haït vaø ôû ba traïng thaùi rôøi, chaët trung bình vaø chaët. Duø ôû traïng thaùi naøo, dieän tích tieáp xuùc giöõa caùc haït heát söùc nhoû so vôùi dieän tích xung quanh cuûa caùc haït.
Caùc dieåm tieáp xuùc haït trong caáu truùc haït (ñaát caùt)Nhö tyû leä dieän tích tieáp xuùc thöôøng gaëp cuûa caùt laø 0,03%, khi taùc duïng moät öùng suaát phaùp 100kPa thöøng öùng suaát thöïc taùc ñoäng leân dieän tích tieáp xuùc laø 330 Mpa. Vaø vôùi aùp löïc naøy thì nöôùc lieân keát taïi caùc ñieåm tieáp xuùc seõ bò ñaåy khoûi voû nöôùc, do vaäy coù ñieåm tieáp xuùc raát toát giöõa caùc haït.
Stresses in a Soil Mass
How are stresses produced?
• Geostatic stresses– total stress– effective stress– pore water pressure
• Additional stresses– surface loads
• foundation
• embankment
• vehicle
Effects of stresses
• Geostatic stresses– soil compresses or consolidates based on
geostatic stress levels
• Additional stresses– produces additional strain in soil which causes
settlement under point of load.
Normal and Shear Stresseson a Plane
• Use methods learned in Mechanics of Materials
• The normal stress on any plane is
• The shear stress on any plane is
n y x
2 y x
2cos2 xy sin2
n y x
2sin 2 xy cos 2
Normal and Shear Stresseson a Plane
• Use Mohr’s circle method to graphically depict stresses: orientation, and magnitude.
Normal and Shear Stresseson a Plane
• The major principal stress is
• The minor principal stress is
n 1 y x
2
y x
2
2
xy2
n 3 y x
2
y x
2
2
xy2
Pole Method of Finding Stresses on a Plane
• Also called “Origin of Planes”
• Draw a line from a known point on the Mohr’s circle parallel to the plane on which the state of stress acts.
• The point of intersection of this line with the Mohr’s circle is called the pole.
• To find the state of stress on any other plane, draw a line parallel to the plane of interest through the pole. State of stress is the intersection of this line with Mohr’s circle.
Basics of Surface Loads
• Categorize into groups based on areal extent– infinite extent (fills, surface surcharge)– finite extent
• point load• line load• strip load• linearly increasing load• uniformly loaded circular area• rectangularly loaded area
Basics of Surface Loads
• Load produces stress and strain– Stress and strain occur in all directions– Commonly focus only on vertical stress
increase
• Analysis based on elastic theory– isotropic, homogeneous material– linear elastic behavior (spring-like)– material comprises a half-space
Stress Caused by a Point Load
• First solved by Boussinesq (1883)
• Stress is maximum nearest applied load and diminishes at distances away.
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Stress Caused by Line Load
• Line load of magnitude q/unit length acts on half-space surface
• Vertical stress increase is:
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Superposition
• This principle works because this system is linear elastic, isotropic, and homogeneous.
• This procedure greatly simplifies subsurface stress analysis.
• Merely add up each separate component.
• Tham khảo thêm trong sách Cơ học đất