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Correction of calculation example in ECCS book.
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4.5 BEAMS RESTRAINED BY SHEETING
_____ 361
- check at the intermediate support 2
,
1.22 0.033 12 2 18.435Ed
b Rd
VV
= = <⋅
– OK
Due to the fact that the shear force VEd < 0.5 Vb,Rd no reduction due to shear force need be done.
Stability for the uplift loading case
Gross properties of the free flange (a 1/5th part of the purlin web is considered (§§4.5.3.2, Figure 4.55)
At the span
22
200 66 21.2 1.46 185.71mm5 5fzhA b c t⎛ ⎞ ⎛ ⎞= + + = + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
28.12mmGy = ; 2 66 28.12 37.88mmGb y− = − =
( ) ( ) ( )
33 322
222
2 2
/ 512 12 12 5
2
fz G
G G
t bh t c t hI y t
b y b t b y c t
⋅⋅ ⋅ ⎛ ⎞= + + + +⎜ ⎟⎝ ⎠
⎛ ⎞+ − + − ⋅⎜ ⎟⎝ ⎠
4127881.05mmfzI =
3
2
3375.2mmfzfz
G
IW
b y= =
− Free flange identification
26.24mmfzfz
fz
Ii
A= =
Lateral bending
Lateral load
0
0
1101451.60 97.89 0.1633307851 200
/ 0.163 37 / 2000.163 0.185 0.022
= = ⋅ =
= − = − == − = −
yz sh
y
h h
I gkI h
k k a h
, 0.022 ( 0.38) 0.00836N/mmh d h dq k q= = − ⋅ − =
b
c
h
b 1
2
g s
shear centre
b
c
h
b
h/5
1
2
yG