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CIVL 232 Design of Structural Concrete
Solution to Assignment 1
1. Solution
(a) Two principal types of limited state:
Ultimate limit state
:
The whole structure or its components should not
collapse, overturn or buckle when subjected to design loads.
Serviceability limit state
: The structure should not become unfit for use due to
excessive deflection, cracking, vibration and so on.
(b)
γ
m
takes the account of:
i.
Uncertainties in strength
ii.
Uncertainties in the accuracy of the methods of analysis
γ
f
takes the account of:
i.
Possible increase in loads
ii.
Inaccurate assessment of the effects of loads
iii. Unforeseen stress redistribution
iv. Constructional inaccuracies
2. Solution
(a) The weight of foundation at
A
can be determined by considering the following
load arrangement:
Take moment about
B
,
kN
1
.
69
0
2
192
1
2
2
.
46
4
2
4
25
=
=
×
−
×
×
−
×
+
×
×
A
R
A
kN/m
25
0
.
1
=
k
G
kN/m
2
.
46
6
.
1
4
.
1
=
+
k
k
Q
G
kN
192
6
.
1
=
k
Q
4 m
2 m
A
B
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(b)
Loading case 1
: 1.4G
k
+ 1.6 Q
k
on span AB and 1.0 G
k
on span BC
The reaction force at A = 79.9 kN (upward)
The reaction force at B = 154.9 kN (upward)
The maximum moment location =
m
73
.
1
2
.
46
9
.
79
=
Consider the free body from A to the 1.73 m from A,
kNm
11
.
69
0
2
73
.
1
2
.
46
73
.
1
9
.
79
max
max
2
=
=
+
×
+
×
−
M
M
Loading case 2
: 1.4G
k
+ 1.6 Q
k
on span AB and 1.0 G
k
on span BC
The reaction force at A = 69 kN (downward)
The reaction force at B = 64.5 kN (upward)
4m
2m
A
B
1.4
1.6
46.2 kN/m
kk
GQ
+=
1.0
25 kN/m
k
G
=
Loading case 1
m
kN
G
k
/
25
1
=
⋅
m
kN
Q
G
k
k
/
2
.
46
6
.
1
4
.
1
=
+
kN
Q
k
192
6
.
1
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.
 Spring '06
 JSKUANG

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