Question 4.8: Determine the design snow loads for the public utility facil......
Determine the design snow loads for the public utility facility depicted in Figure 4.16. The facility is required to remain operational during an emergency.
| DESIGN DATA | |
| Ground snow load, p_g : | 60 psf |
| Terrain category: | D (flat unobstructed area near water) |
| Occupancy: | Essential facility |
| Thermal condition: | Unheated structure |
| Roof exposure condition: | Fully exposed |
| Roof surface: | Rubber membrane |
1. Determine ground snow load p_g.
From the design data, the ground snow load p_g is equal to 60 psf.
2. Determine flat roof snow load p_{\, f} by Eq. 7-1.
Use Flowchart 4.1 to determine p_{\, f} .
a. Determine exposure factor C_e from Table 7-2.
From the design data, the terrain category is D and the roof exposure is fully exposed. Therefore, C_e = 0.8 from Table 7-2.
b. Determine thermal factor C_t from Table 7-3.
From the design data, the structure is unheated. Thus, C_t = 1.2 from Table 7-3.
c. Determine the importance factor I from Table 7-4.
From IBC Table 1604.5, the Occupancy Category is IV for this essential facility. Thus, I = 1.2 from Table 7-4.
Therefore,
p_{f}=0.7C_{e}C_{t}I p_{g}\quad=0.7\times0.8\times1.2\times1.2\times60=48.4 psf
Check if the minimum snow load requirements are applicable:
Minimum values of p_f in accordance with 7.3 apply to curved roofs where the vertical angle from the eaves to the crown is less than 10 degrees. Since that slope in this example is equal to 12 degrees, minimum roof snow loads do not apply.
3. Determine sloped roof snow load p_s from Eq. 7-1.
Use Flowchart 4.2 to determine roof slope factor C_s .
a. Determine thermal factor C_t from Table 7-3.
From item 2 above, thermal factor C_t = 1.2 .
b. Determine if the roof is warm or cold.
Since C_t = 1.2 the roof is defined as a cold roof in accordance with 7.4.2.
c. Determine if the roof is unobstructed or not and if the roof is slippery or not.
There are no obstructions on the roof that inhibit the snow from sliding off the eaves.^{13} Also, the roof surface is a rubber membrane. According to 7.4, rubber membranes are considered to be slippery.
Since this roof is unobstructed and slippery, use the dashed line in Figure 7-2c to determine C_s .
For the tangent slope of 25 degrees at the eave, the roof slope factor C_s is determined by the equation in C7.4 for cold roofs with C_t = 1.2:
C_{S}=1.0-{\frac{(\mathrm{slope-15^{\circ}})}{55^{\circ}}}=1.0-{\frac{(25^{\circ}-15^{\circ})}{55^{\circ}}}=0.82Therefore, p_s = C_s p_{\, f }= 0.82 × 48.4 = 39.7 psf, which is the balanced snow load at the eaves.
Away from the eaves, the roof slope factor C_s is equal to 1.0 where the tangent roof slope is less than or equal to 15 degrees (see dashed line in Figure 7-2c). This occurs at distances of approximately 20.7 ft from the eaves at both ends of the roof.
Therefore, in the center portion of the roof, p_s = C_sp_f = 1.0 × 48.4 = 48.4 psf.
The balanced snow load is depicted in Figure 4.17, which is based on Case 1 in Figure 7-3 for slope at eaves less than 30 degrees.
4. Consider unbalanced snow loads.
Flowchart 4.5 is used to determine if unbalanced loads on this curved roof need to be considered or not.
Since the slope of the straight line from the eaves to the crown is greater than 10 degrees and is less than 60 degrees, unbalanced snow loads must be considered (7.6.2).
Unbalanced loads for this roof are given in Case 1 of Figure 7-3. No snow loads are applied on the windward side. On the leeward side, the snow load is equal to 0.5 p_f = 0.5 × 48.4 = 24.2 psf at the crown and 2 p_fC_s / C_e = 2 × 48.4 × 0.82 / 0.8 = 99.2 psf at the eaves where C_s is based on the slope at the eaves.
The unbalanced snow loads are shown in Figure 4.17.
^{13} In general, large vent pipes, snow guards, parapet walls and large rooftop equipment are a few common examples of obstructions that could prevent snow from sliding off the roof. Ice dams and icicles along eaves can also possibly inhibit snow from sliding off of two types of warm roofs, which are described in 7.4.5.
\, Table 7.2 Flood Loads
| Load Type | Equation No(s). |
\qquad\qquad\qquad\qquad\qquad\qquadEquation*\qquad\qquad\qquad\qquad\qquad\qquad | ||
| Hydrostatic | Lateral** | — |
F_{s t a}=\gamma_{w}d_{s}^{2}w/2 |
|
| Vertical upward | — |
F_{b o u y}=\gamma_{w}\times V{o}l u m e |
||
| Hydrodynamic | V ≤ 10 ft/sec | 5-1 |
F_{d y n}=\gamma_{w}d_{s}d_{h}w where d_h = (aV^2 / 2g) |
|
| V > 10 ft/sec | — |
F_{d y n}=a ρ V^{2}d_{s}w/2 |
||
| Wave | Breaking wave loads on vertical pilings and columns |
5-4،5-2،5-3 |
F_{D}=\gamma_{w}C_{D}D H_{b}^{2}\;/\,2 where H_b = 0.78d_s d_s = 0.65(BFE −G) |
|
| Breaking wave loads on vertical walls |
Space behind the vertical wall is dry |
5-5, 5-6 |
P_{\mathrm{max}}=C_{p}\gamma_{w}d_{s}+1.2\gamma_{w}d_{s} F_{t}=1.1C_{p}\gamma_{w}d_{s}^{2}+2.4\gamma_{w}d_{s}^{2} |
|
| Free water exists behind the vertical wall |
5-5, 5-7 |
P_{\mathrm{max}}=C_{p}\gamma_{w}d_{s}+1.2\gamma_{w}d_{s} F_{t}=1.1C_{p}\gamma_{w}d_{s}^{2}+1.9\gamma_{w}d_{s}^{2} |
||
| Breaking wave loads on nonvertical walls |
5-8 |
F_{nν}=F_{t}\sin^{2}\alpha. |
||
| Breaking wave loads from obliquely incident waves |
5-9 | F_{oi}=F_{t}\sin^{2}\alpha. | ||
| mpact | Normal | C5-3 |
F=\frac{\pi W V_{b}C_{I}C_{O}C_{D}C_{B}R_{\mathrm{max}}}{2g(\Delta t)} |
|
| Special | C5-4 | F = C_D{\rm ρ}AV^2 / 2 | ||
* γ_w = unit weight of water, which is equal to 62.4 pcf for fresh water and 64.0 pcf for salt water.
V = water velocity in ft/sec; see ASCE/SEI C5.4.3 for equations on how to determine V.
Additional information on these equations is given in ASCE/SEI Chapter 5 and in this section.
** In communities that participate in the NFIP, it is required that buildings in V Zones be elevated above the BFE on an open foundation; thus, hydrostatic loads are not applicable. It is also required that the foundation walls of buildings in A Zones be equipped with openings that allow flood water to enter so that internal and external hydrostatic pressure will equalize.
