Civil Engineering Reference
In-Depth Information
Width of slab
¼
250
=
6
¼
41
:
7cm
Width of haunch
¼
50
=
6
¼
8
:
3cm
A
1
¼
70
3
¼
210 cm
2
A
2
¼
174
1
4cm
2
A
3
¼
50
3
¼
150 cm
2
A
4
¼
8
:
6
¼
278
:
3
20
¼
166 cm
2
:
5cm
2
A
5
¼
41
:
7
25
¼
1042
:
9cm
2
To determine the centroid, we can take the first area moment around
the
x
1
-
x
1
axis, as follows:
A¼
1846
:
210
1
:
5 + 278
:
4
90 + 150
178
:
5 + 166
190 + 1042
:
5
212
:
5
y
c
¼
1846
:
9
¼
165
:
3cm
I
x
¼
70
3
3
8
2
6
174
3
3
2
½
=
12 + 210
163
:
+1
½
:
=
12 + 278
:
4
75
:
+50
3
3
2
2
3
20
3
7
2
½
=
12 + 150
13
:
+8
½
:
=
12 + 166
24
:
7
25
3
2
2
¼
10,425,383 cm
4
+41
½
:
=
12 + 1042
:
5
47
:
Shear flow at section
s
-
s
:
36
10
3
10
3
Q
Ed
S
ss
I
x
2193
:
166
24
½
:
7 + 1042
:
5
47
:
2
q ¼
¼
10,425,383
10
4
¼
1121
:
5N
=
mm
Maximum spacing between shear connectors in the longitudinal direction
s
235
f
y
r
235
275
ð
S
max
Þ¼
15
t
f
¼
15
30
¼
416mm
Force per two headed studs
¼S
1121
:
5N
Force per one headed stud
¼S
560
:
75N
Design resistance of headed studs can be calculated as follows:
8
f
u
pd
2
14
25
2
0
:
=
4
0
:
8
430
3
:
=
4
P
Rd
¼
¼
¼
135,088N
¼
135
:
1kN
g
v
1
:
25
a¼
1
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