Question

Calculate balanced and unbalanced snow loading for a heated garage/workshop located in the most northern part of Maine at the site of a 200 ft hight from the mean seal level. The area exhibits low wind and is populated with some trees. The structure is occupied by workers daily, has an asphalt gable roof with a pitch of 4 over 12 and is 44’ by 44’.

Answer 1

  

SOL: Information given in the problem:

height of site above the mean seal level = 200 ft

Slope of gable roof = 4 over 12 , Length = 44inch, width = 44inch

Location = northern part of Maine

Step 1 :The ground snow load from IBC Figure 1608.2 for northern part of Maine. is

pg = 50 psf.

Step 2: Determine the snow exposure factor, Ce, from IBC Table 1608.3.1 (ASCE 7 Table 7-2). Ce = 1.0 for exposure category B, partially exposed roof.

Step 3: Determine the thermal factor, Ct, from IBC Table 1608.3.2 (ASCE 7 Table-3). Ct = 1.1 for heated structures with ventilated roofs, R > 25.

Step 4: Determine the snow load importance factor, Is, from IBC Table 1604.5. For Category II ordinary structures, Is = 1.0. Step 5: Determine the Gable roof snow load, pf, in accordance with Equation 7-1.

pf = 0.7CeCtI spg = 0.7(1.0)(1.1)(1.0)(50) = 38.5 psf

Step 5: Determine the sloped roof snow load by adjusting the flat roof snow load by the slope factor, Cs, in accordance with Section 7.4. The slope of the roof in degrees, = arctan(4/12) = 18.43 degrees. Since Ct = 1.1, Cs is determined from Figure 7.2b. Cs = 1.0. The sloped roof snow load is, ps = 30.8 psf (balanced snow load).

Step 6: Unbalanced loading from the effects of wind loading on hip and gable roofs must be considered in accordance with Section 7.6.1. Unbalanced snow load need not be considered for >= 70 degrees or for   < (70/30 + 0.5) degrees. (70/30+ 0.5) =2.380 degrees. Since > 2.38 degrees and < 70 degrees, the unbalanced loading condition must be considered. (W=30) > 20 feet. The parameter is a function of L/W. L/W = 44/44 = 1.0,    = 0.33 + 0.167 (L/W) = 0.33 + 0.167 (1.0) = 0.497 The roof must be designed for an unbalanced snow load per Figure 7-5 as follows:

Windward load = pW = 0.3ps = 0.3(30.8) = 9 psf Leeward load = pL = 1.2(1 +   /2) ps / Ce = 1.2(1 + 0.49/2) (30.8)/(1.0) = 46.01 psf

The density of snow, , is given by Equation 7-4. = 0.13pg + 14 = 0.13(50) + 14 = 20.5 pcf The depth of unbalanced snow on the windward roof is d = 9/19.2 = 0.43 feet. The depth of unbalanced snow on the leeward roof is d =46.01/20.5 = 2.24 feet.