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:
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
1
Search WWH ::




Custom Search