Civil Engineering Reference
In-Depth Information
Solution:
(i) Safe bearing capacity
Self-weight of foundation, Wf = 0.5 × 24 × 1.8 = 21.6
f
=
0.5
×
24
×
1.8
=
21.6 kN/m run
Weight of soil on top of foundation, W
s
=
0.25
×
20
×
1.8
=
9.0 kN/m run
Total weight of foundation
+
soil, W
=
21.6
+
9.0
=
30.6 kN/m run
P e
P W
×
+
50 0 4
50
.
×
+
P
Eccentricity of bearing pressure e
,
=
=
=
0.25 m
30 6
.
Since
e
≤
6
, the total force acts within the middle third of the foundation.
Effective width of footing, B
′
=
1.8
−
2
×
0.25
=
1.3 m
Footing is continuous, i.e. L
→
∞
; s
c
=
1.0.
1 0 4
0 75
1 3
.
.
=
d
c
= +
.
1 23
.
q
cN s d
−
γ
z
u net
c c
c
Safe bearing capacity per metre run
(
)
=
+ =
γ
z
+
γ
z
3
30 5 14 1 0 1 23 20 0 75
3
3
×
.
× ×
.
.
− ×
.
=
+ ×
20 0 75
.
73 2
.
kPa
=
Safe bearing load
=
73.2
×
B
′
=
95.2 kN/m run
(ii) Eurocode 7 GEO limit state
1.
Combination 1 (partial factor sets A1
+
M1
+
R1)
c
30
1
u
Design material property c
:
;
=
=
=
30
kPa
u d
γ
cu
Design actions:
Weight of foundation W W
,
= ×
γ
=
30 6 1 35
.
×
.
=
41 3
.
kN m run
/
d
G unfav
;
Applied line load P
,
= ×
P
γ
= ×
50 1 35
.
=
67 5
.
kN m run
/
d
G unfav
;
Effect of design actions:
Total vertical force F
,
=
41 3
.
+
67 5
.
=
108 8
.
kN m run
/
d
P e
P W
×
+
67 5 0 4
67 5
.
×
+
.
d
P
Eccentricity e
,
=
=
=
0 248
.
m
.
41 3
.
d
d
B
Since e
≤
6
,
the total force acts within the middle-third of
the foundation.
Effective width of footing B
,
′
=
1 8 2 0 248
.
.
1 3
.
m
− ×
=
Design resistance:
From before, N
c
=
5.14, N
q
=
1.0, N
γ
=
0, s
c
=
1.0.
Ultimate bearing capacity q
,
=
γ
30 5 14 1 20 0 75 1 0
169 2
c N s
+
zN
u
u d c c
;
q
= ×
.
× + ×
.
×
.
=
.
kPa