CE 366 SHALLOW FOUNDATIONS - Adnan Menderes...
Transcript of CE 366 SHALLOW FOUNDATIONS - Adnan Menderes...
CE 366 – SHALLOW FOUNDATIONS
P.1) CANTILEVER FOOTING
Question:
Given: Q1 = 1500 kN, Q2 = 2500 kN, qall = 200 kN/m2
Ignore the weight of footings and find dimensions B and B2 of a cantilever footing for a uniform
soil pressure distribution. Draw shear and bending moment distributions.
B
B2 B
Q1=1500 kN
L=8m
Q2=2500 kN
0.8 m 0.8 m
B1=2.0m
Solution:
B1=2m B
B2
Strap
Locate ΣQ = Q1 + Q2
B b=1500 × 8 / 4000=3
For B1=2m
L=8 m
Q1 = 1500 kN Q1+Q2 =4000kN Q = 2500 kN
0.4m
a = 8.0 + 0.4 – 1.0 – 3.0
a = 4.40m
R1= (4000 × 3) / 7.4 = 1622kN
R2 = 4000 – 1622 = 2378 kN
Determine B2,
qall = Q / (2 × B2)
200 = 1620 / (2 × B2) B2 = 4 m
1.0
200 kN/m2
Q1=1500 kN Q2-R2=122 kN
0.6m
R1=1622 kN
122kN
OR
Without considering resultant (Q1 +Q2)
Moment w.r.t Q2 or (R2);
Q1x8- R1x7.4=0 R1=1622 kN
From force equilibrium;
ΣFvertical =0
1500+25000-1622-R2 =0
R2=2378 kN
Determine
qall = Q / (2
200 = 1622 / (2 B2) B2 ≈ 4
200 = 2378 / B2 B ≈ 3.45
0.8 m 0.8 m
FBD
V(kN)
M(kN.m)
1500 kN
900 kN.m
P.2) TRAPEZOIDAL FOOTING
Question:
Determine B1 and B2 of a trapezoidal footing for a uniform soil pressure of 300 kN/m2. (γconc =
24kN/m3)
M1=600kN-m
Q1=3000kN
M2=800kN-m
Q2=2000kN M3=1200kN-m
Q3=4000kN
1
2 6 6 2
B1 B2
Centroid of the base
ΣQ
x
M1=600kN-m
Q1=3000kN
M2=800kN-m
Q2=2000kN M3=1200kN-m
Q3=4000kN
1
Solution:
After finding the location of resultant force, you can decide whether the wider side of
the trapezoid will be at the left or right side.
2.0 m 6.0 m 6.0 m 2.0 m
x Centroid of the base
ΣQ=9000 kN
B1 B2
General formulation for finding coordinates of a centroid;
ΣQ=9000kN
Weight of footing= 16x24x(B1+B2)/2 = 192(B1+B2)
Area of footing = 8(B1+B2)
ΣFv=0 9000+192(B1+B2)=8(B1+B2)x300
B1+B2 = 4.07m ………(1)
ΣM=0 (moment about centroid of the base)
3000(14-x)+600+2000(8-x)+800-4000(x-2)-1200+(wght of ftg)x0=(base pressure)x0
x = 7.36m
x=7.36=
………..(2)
From (1) and (2) B1= 1.55m ; B2 = 2.52m
Note: You can also take moment about the left or right side but keep in mind that weight and base
pressure will also have moment about left or right side.
B =
16
2
2
6
6
b =
6
P.3) MAT FOUNDATION
Question:
A mat foundation rests on a sand deposit whose allowable bearing value is 150 kN/m2
. Column
loads are given in the figure. The thickness of the mat is 2.0 m (concrete = 24 kN/m3). Calculate
base pressures assuming that the lines passing through the centroid of the mat and parallel to the
sides are principal axes. Find the base pressure distribution beneath the base and check whether the
mat foundation given is safe?
l = 12 m
1000 kN 2000 kN 3000 kN
EXISTING
STRUCTURE
2500 kN 3500 kN
4000 kN
3500 kN
2500 kN
1000 kN 2000 kN 3000 kN 2000 kN 1000 kN
2 m 6 m 6 m 6 m
6 m 2 m
L = 28 m
2 m
2 m
•
Solution:
Area of foundation = 28 × 16 – 12 × 6 = 376 m2
Total vertical load = Σ V = Column loads + Weight of mat=31000+(376) × 24 × 2 = 49048 kN
* Center of gravity (CG) of mat:
28 × 16 × 14 − 6 × 12 × (16 + 6) = 12.47
376
m from left
28 × 16 × 8 − 6 × 12 × (3 + 10) = 7.04
376
m from bottom
16 6
• 3
12.47 CG •
10 7.04
* Location of ΣQ :
Take moment about the left side:
= (1 / 49048) . [ 2 × (1000+2500+1000) + 8 × (2000+3500+2000) + 14 × (3000 +
+ 4000 + 3000) + 20 × (3500+2000) + 26 × (2500+1000) + 376 × 2 × 24 × 12.47 ]
= 12.95 m from left
Take moment about bottom side :
= (1 / 49048) . [ 2 × (1000+2000+3000+2000+1000) + 8 × (2500 + 3500 + 4000 +
+ 3500 + 2500) + 14 × (1000+2000+3000) + 376 × 2 × 24 × 7.04 ]
= 7.28 m from bottom
Centroid of the
16mx28m rectangular
Centroid of the 12mx6m
cut section due to
existing structure
Eccentricity :
e1 = 12.95 – 12.47 = 0.48 m
e2 = 7.28 – 7.04 = 0.24 m
1
8.96
2 • 2
7.04
12.47 15.53
1
M1
28 m
M2
16 m
M1 about 1-1 axis:
M1 = ΣQ ⋅ e1 = 49048 ⋅ (0.48) = 23543 kN.m
M2 about 2-2 axis:
M2 = ΣQ ⋅ e2 = 49048 ⋅ (0.24) = 11772 kN.m
⎡ B ⋅ L3 2 ⎤ ⎡b ⋅ l 2 ⎤
I
1−1 = ⎢
12 + B ⋅ L ⋅ (D
1 ) ⎢ − ⎢
12 + b ⋅ l ⋅ (d
1 ) ⎢ =
⎣ ⎦ ⎣ ⎦
⎡ × 3
⎤ ⎡ 3 ⎤ 4
= 16 28
+ 16 × 28 × (14 − 12.47) 2
⎢ − ⎢ 6 ×12
+ 6 ×12 × (22 − 12.47) 2
⎢ = 22915 m
⎢ 12 ⎦ ⎣ 12 ⎦
⎡ L ⋅ B 3 2 ⎤ ⎡ l ⋅ b 2 ⎤
I
2−2 = ⎢
12 + B ⋅ L ⋅ (D
2 ) ⎢ − ⎢
12 + b ⋅ l ⋅ (d
2 ) ⎢ =
⎣ ⎦ ⎣ ⎦
= ⎡ 28 ×16
3
+ 16 × 28 × (8 − 7.04) 2 ⎤ −
⎡12 × 6 + 12 × 6 × (13 − 7.04) 2
⎢ = 7197 m4
⎢ 12 ⎢ ⎢
⎦ ⎣ ⎦
e1=0.48m
e2=0.24m
Note: In soil mechanics compression is taken as positive (+)
q = ΣQ ±
M 1 ⋅ y1 ± M 2 ⋅ y2
Area I1−1 I
2−2
q = 49048
± 23543 ⋅ y
1 ± 11772 ⋅ y
2 = 130.4 ± 1.03 y
± 1.64 y
376 22915 7197 1 2
q A
= 130.4 ± 1.03 y1 ± 1.64 y
2 = 130.4 + 1.03 ⋅ (3.53) + 1.64 ⋅ (2.96) = 138.9 kN/m
2
qB
= 130.4 + 1.03 ⋅ (3.53) +1.64 ⋅ (8.96) = 148.7
kN/m2
qC
= 130.4 −1.03 ⋅ (12.47) + 1.64 ⋅ (8.96) = 132.2
kN/m2
qD
= 130.4 −1.03 ⋅ (12.47) −1.64 ⋅ (7.04) = 106
qE = 134.9 kN/m2
qF = 151.3 kN/m2
kN/m2
ΣQ = 49048 kN 1
qC = 132.2kN/m2
C B qB = 148.7 kN/m2
qA = 138.9 kN/m2
A e2 = 0.24
m
CG
F qF = 151.3 kN/m2
(MAX)
2 2
D
qD = 106 kN/m
2
(MIN)
e1 = 0.48 m
E
qE = 134.9 kN/m2
1
Since at all critical points stress values are almost < qall = 150 kN/m2
given mat foundation is
safe.
P.4) COMBINED FOOTING ANALYZED BY RIGID METHOD
Question:
A rectangular combined footing which supports three columns is to be constructed on a sandy clay
layer whose allowable bearing value (base pressure) is 84 kN/m2.The thickness of the concrete
footing is 0.85m. There is a 1.20m thick soil fill having same unit weight with sandy clay on the
footing. Unit weight of the sandy clay and the concrete are 20 kN/m3 and 24 kN/m3 respectively.
Analyze the footing by rigid method and plot shear and moment diagrams.
800kN 600kN 500kN
1.2
0.85
2m 6m 6m 2m
B=?
Solution:
Neglecting column weights;
ΣQnet=800+600+500+(24-20)x16xBx0.85 = 1900+54.4B
e = ΣM
ΣV
= 800 x6 − 500 x6
= 1900 + 54.4B
1800
1900 + 54.4B
qmax =
ΣV
BxL (1 +
6e ) = 84kN / m 2
L
84 = 1900 + 54.4B
(1 + Bx16
6x1800 ) (1900 + 54.4B) x16 ⇒ B = 2.0m
Use B=2.0m qmax = 84 kN/m2
; qmin = 41.6 kN/m2
Downward uniform pressure = 0.85 × (24-20) = 3.4 kN/m2
These are the diagrams related to forces and moments acting on the foundation. Explanations are
at the next page;
800kN 600kN 500kN
2m 6m 6m 2m
3.4
84
3.4
41.6
1.01
a A b
4.2
B e
7.7
C g
84– 3.4
≅ 80.5
c
75.2
d
59.4 43.5 h
f
41.6 - 3.4 = 38.2
311.4 319.2 336
+ j m
_
Shear
(shear diagram is
488.6
-507
280.8
-444
164
parabolic in fact,
but we linearized
it for simplicity)
_ -99.4
+
Moment
315 167
V = 80.5 + 75.2
x 2 x 2 = 311.4 kN A
2
area (abcd) B = 2
VA’ = 311.4 – 800 = – 488.6 kN
To find the centroid of a trapezoid on the horizontal axis;
x x = L 2B1+ B2
x: distance from shorter dimension
B1 B2 3 B1+ B2
L
x (abcd) = 2 2x80.5 + 75.2
=1.01m
3 80.5 + 75.2
MA = 311.4 x 1.01 = 315 kN.m
VBA = 80.5 + 59.4
x 8 x 2 − 800 = 319.2 kN 2
VBC = 319.2 – 600 = -280.8 kN
x (aefc) = 8 2x80.5 + 59.4
= 4.2 m
3 80.5 + 59.4
M = ⎡80.5 + 59.4
x 8 x 2⎤
x 4.2 − 800 x 6 = − 99.4 kN.m
B ⎢ ⎢ ⎣ ⎦
VCB = 80.5 + 43.5
x14 x 2 − 800 − 600 = 336 kN 2
VC = 336 – 500 = -164 kN
Check the end point; 80.5 + 38.2
x16 x 2 − 800 − 600 − 500 = 0 OK
2
x (aghc) = 14 2x80.5 + 43.5
= 7.7 m
3 80.5 + 43.5
M = ⎡80.5 + 43.5
x 14 x 2⎤
x 7.7 − 800 x12 − 600 x 6 =167 kN.m
C ⎢ ⎢ ⎣ ⎦
Slope of V b/w x = 2 and x = 8 m Æ (488.6+319.2)/6=134.6 kN/m
Equation of V b/w x = 2 and x = 8 m Æ V(x) = -488.6 + 134.6 (x) for V(x)=0 Æ x=3.6 m
Base pressure @ x = 2 + 3.6 = 5.6 m Æ 65.6 kN/m2
Similarly @ x = 10.7 m V = 0
Base pressure @ x =10.7 m Æ 52.2 kN/m2
maximum points of the moment diagram;
a A b j
65.6
B e m C g
52.2 h
f n
c d k
j and m are the points where shear forces are equal to zero (i.e. moment is max)
the distance between point b and point j
x (ajkc) = (2 + 3.6) 2x80.5 + 65.6
= 2.9 m
3 80.5 + 65.6
M = ⎡80.5 + 65.6
x (2 + 3.6) x 2⎤
x 2.9 − 800 x 3.6 = − 507 kN.m
AB ⎢ ⎢ ⎣ ⎦
x (amnc) = 5.73 m
MBC = -444 kN.m