CE-REF.COM
ICE LOAD AND
CONCURRENT WIND LOAD
Design code: ASCE
7-02
Section 10.4:
Ice load due to freezing rain
Important
factors and equations:
1. ASCE 10.4.1:
Ice load shall be determined using the weight of glaze ice form on
all expose surfaces of the structural members
2. Cross section
of ice on structural shape, prismatic members, and other similar
shapes, shall be
A_{i }= p t_{d} (D_{c} +
t_{d})
Where
A_{i} is
cross section of ice, t_{d} is
thickness of ice, D_{c} is
equivalent diameter of structural shape as
3. For large
plate, and large three-Dimentional objects such as domes and sphere,
volume of ice shall be determined as
V_{i} = p t_{d} A_{s}
For flat plate A_{s} is
flat area of plate.
For domes and
spheres, As = p r^{2}.
r is radius of domes or spheres.
V_{i} can
be multiply by 0.8 for vertical plate, 0.6 for horizontal plate.
4. Ice density is
56 lb/ft^{3} minimum.
5. Normal ice
thickness, t, due to freezing rain at a height of 33 ft and the
concurrent wind speed is determined from Figure 10-2, 10-3, and 10-4
on ASCE 7-02.
6.
Adjustment factor for height is
f_{z} =
(z/33)^{0.1} for
0 to 900 ft., fz = 1.4 for 900 ft and higher.
7. Important
factor is listed in Table 10-1.
8. Topographic
Factor,
K_{zt} =
(1+K_{1}+K_{2}+K_{3})^{2} where
K_{1}, K_{2}, K_{3} are
determined from Figure 6-4 of ASCE 7-02.
9. Design ice
thickness is determined
as
t_{d} =
2 t I_{i} f_{z} (K_{zt})^{0.35}
Important
factors and equations:
Velociaty
pressure, q_{z} =
0,00256 K_{z} K_{zt} K_{d} V_{c}^{2} I_{i}.
K_{d} =
0.85 for using load combination, Kd = 1 for wind along.
K_{z} =
2.01 (z/z_{g})^{2/}^{a}
where z is height
above ground, z shall not be less than 15 ft. except that z shall
not be
less than 30 ft
for exposure B for low rise building and for component and cladding.
a and
z_{g} are taken
as follows:
Expsoure |
a |
z_{g }(ft) |
B |
7.0 |
1200 |
C |
9.5 |
900 |
D |
11.5 |
700 |
Design Procedure:
Location: New
York City, New York
Design data:
Height above
ground: 120 ft
Diameter of logo
sign: d = 10 ft
Requirement:
determine weight of ice on the logo and concurrent wind pressure.
Solution:
Design procedure
(ASCE 2-02 section 10.7) :
1. Detarmine
normal ice thickness from Figure 10-2: t = 1 in
Determine
concurrent wind speed: V_{c} =
50 mph
2. Determine
topographic factor from section 10.4.5 (Figure 6.4)
K_{1} =
0, K_{2} = 1, K_{3} =
1
Kzt = (1+K_{1} K_{2} K_{3})^{2} =
1
3. Determine
importance factor (Table 1.1 & 10.1), I_{i} =
1
4. Determine
height factor from section 10.4.3,
Z = 120 ft, fz =
(Z/33)0.1 = 1.14
5. Determine ice
thickness at roof from section 10.4.6
t_{d} =
2.0 t I_{i} f_{z} K_{zt} ^{0.35} =
2.28 in
6. Determine
weight of ice from section 10.4.1
Area
of ice of logo, A_{f} = p d^{2}/4
= 78.5 ft^{2}
Weight
of ice, W_{si} =
Af td x 56 pcf = 834 lb
7. Determine
velocity pressure
Kd
= 0.85, I = 1
Exposure
B, exposure coefficient
a =
9.5, Z_{g} = 900
ft, Z = 120 ft
K_{z} =
2.01 (Z/Z_{g}) ^{2/}^{a} =
1.32
8. Determine
velocity pressure
q_{z} =
0.00256 K_{z} K_{zt} K_{d} (V_{c})^{2} I
= 7.2 psf
9. Determine wind
force coefficient (Table 6.21)
C_{f} =
1.4
10. Gust effect
factor (Section. 6.5.8)
G
= 0.85
11. Determine
design wind pressure with ice load
p
= q_{z} G C_{f} =
8.5 psf
12. Concurrent
wind force on logo
A
= p (d
+ 2 t_{d})^{2}/4 = 84.6 ft^{2}.
F
= p A = 720.3 lbs