Snow, Rain, and Ice Loads

Snow, Rain, and Ice Loads

The NCEES topic names all three hazards, so a snow-only review is incomplete. For a July 2026 exam, use ASCE 7-16 Chapters 7, 8, and 10 with IBC 2018 Sections 1608, 1611, and 1614.

Ground snow is not roof snow

Begin with ground snow load pg from the ASCE 7-16/IBC 2018 source or an approved site-specific study where required. The flat-roof snow load is

pf = 0.7 Ce Ct Is pg

where Ce represents exposure, Ct represents thermal condition, and Is is snow importance. These factors are not interchangeable: an exposed windswept roof, a continuously heated building, and a higher-risk occupancy affect different terms. For a sloped roof, apply the ASCE 7-16 roof-slope factor to obtain the balanced sloped-roof load, and check the prescribed minimum load where applicable.

Balanced load is only the baseline. Examine roof geometry for:

• unbalanced loading on sloped, curved, folded, or multilevel roofs;
• leeward drift at roof steps, projections, parapets, and adjacent taller structures;
• sliding snow onto a lower roof;
• local loads on roof equipment, pipes, and platforms; and
• rain-on-snow surcharge where the ASCE 7-16 criteria apply.

A drift is not spread uniformly across the entire roof unless the prescribed geometry says so. Treat the drift surcharge as its specified distribution and place the pattern to maximize the response being checked.

Worked snow pattern

Suppose pg = 30 psf, and the problem gives Ce = Ct = Is = 1.0. Then

pf = 0.7(1.0)(1.0)(1.0)(30) = 21 psf.

If the roof-slope factor provided for the roof is Cs = 0.90, the balanced sloped-roof snow load is

ps = 0.90(21) = 18.9 psf.

Consider a supporting line with a 20-ft tributary width. Per foot along the support, balanced snow is 18.9(20) = 378 lb/ft. Now assume the required ASCE 7-16 drift calculation has produced a triangular surcharge with 22-psf peak over an 8-ft drift width. The added line load is the triangle's area:

(1/2)(8 ft)(22 psf) = 88 lb/ft.

For the same support, the patterned total is 378 + 88 = 466 lb/ft. A complete solution also checks the balanced case, required unbalanced cases, local shears and moments near the drift, and the applicable load combination.

Rain: assume the primary drain is blocked

IBC 2018 requires each roof portion to support water that accumulates when its primary drainage is blocked, plus water above the secondary-drain inlet needed to produce the design flow. In US customary units, the rain pressure is

R = 5.2(ds + dh) in psf,

where ds is the static depth on the undeflected roof up to the secondary inlet and dh is the hydraulic head above that inlet at design flow, both in inches.

For example, if ds = 1.5 in. and dh = 1.0 in., then

R = 5.2(1.5 + 1.0) = 13.0 psf.

Over 200 ft², that is 13.0(200) = 2,600 lb, or 2.60 kip, on the undeflected-roof tributary area. Do not omit dh, and do not assume an operating primary drain eliminates the code case.

The pressure calculation is not the whole rain check. A flexible roof can deflect under water, creating a deeper basin that attracts more water, causing more deflection. IBC 2018 directs susceptible bays to the ASCE 7-16 ponding evaluation. Verify drainage geometry, stiffness, deflected behavior, secondary drainage, and controlled-drainage provisions as applicable.

Atmospheric ice is a separate structural condition

ASCE 7-16 Chapter 10 addresses atmospheric icing from freezing rain on ice-sensitive structures. Use the mapped nominal ice thickness and the standard's height, importance, and topographic provisions to obtain design ice thickness. Then determine both:

1. Di, the weight of ice attached to members, appurtenances, cables, or equipment; and
2. Wi, wind on the ice-covered structure using the enlarged projected dimensions and the Chapter 10 concurrent-wind procedure.

Do not use the full basic design wind event as though it automatically occurs with the design ice event. ASCE 7-16 supplies concurrent wind-on-ice requirements and dedicated strength and ASD ice combinations in Sections 2.3.3 and 2.4.3. Partial ice loading can also govern an asymmetric response.

As a geometry example, a circular member has bare diameter D = 4 in. and a code-derived radial design ice thickness t = 0.50 in. The iced diameter is D + 2t = 5 in. The annular ice area is

Ai = (π/4)(5^2 - 4^2) = 7.07 in² = 0.0491 ft².

Using the ASCE 7-16 glaze-ice unit weight of 56 pcf gives an added ice line weight of about 56(0.0491) = 2.75 lb/ft. The projected width for wind rises from 4 to 5 in., a 25% increase before applying the appropriate force coefficient and concurrent velocity pressure. Ignoring either added weight or increased wind area misses part of the ice condition.

Combination and review discipline

For ordinary roof gravity combinations, follow the exact ASCE 7-16/IBC 2018 treatment of S and R; do not automatically add full snow and full rain merely because both are water. Apply the separate rain-on-snow provision when triggered. For atmospheric ice, carry Di and Wi together through the dedicated ice combinations.

On the exam, finish with three questions: Did I convert the mapped hazard to the actual structural surface? Did I check the nonuniform or feedback case? Did I use the load combination written for this hazard in the 2016 standard? If all three answers are clear, snow, rain, and ice remain distinct and auditable.