Dead Loads

EGR 437: The Class: Loads: Gravity Loads: Dead Loads

Gravity Loads

Dead Loads, D

Dead loads are long-term stationary forces that include the self weight of the structure and the weight of any permanent equipment.
How much do various building materials weigh?
- For example see Appendices A and B in Design and Wood Structures by D. Breyer.
- These material weights are given in lb/ft² (also written as psf).
- How do you use this information?
Consider the following flat roof:
1. Compute the dead load carried by a typical 2 X 10 beam and draw its proper load diagram.
  1. Consider the weight of the basic roof.
    - The basic roofing loads are uniformly applied over the entire roof:
      
      Material psf
      
      3 ply with gravel 5.5
      
      5/8" plywood 5/8 (3.0) = 1.9
      
      4" loose insulation 4 (.5) = 2.0
      
      å 9.4 psf
  2. How much of this uniform load does the typical 2 X 10 carry?
    - It is a function of the member's tributary area.
    - The area assumed to load any given member is called tributary area (t.a.)
      - t.a.= span X tributary width
      - Where tributary width is often equal to the on center, (o.c.) spacing.
    - For a typical 2 X 10, the tributary area = 20 X 8 = 160 ft²
    - For an exterior 2 X 10, tributary area = 20 X 4 = 80 ft²
    - Model the 2 X 10 with basic roof load as:
      
      with w = uniformly distributed line load = unit psf load X tributary width.
      For this example then, w = 9.4 lb/ft² (8ft) = 75.2 lb/ft.
  3. But is this the only dead load carried by this beam?
    - Need to also consider the 2X10's self weight and the weight of the 2 X 4 members framing into the beam.
    - There are a number of ways to calculate a member's self weight, but all methods:
      - Are a function of wood species, moisture content, and dimensions.
      - And, you must be careful to properly track units.
    - For this example problem, self-weight can be found as:
      - Assume that the 2X10's are dry Douglas Fir-Larch, (DF-L), members with a specific weight of » 35 lb/ft³.
      - 2" nominal Þ 1.5" actual.
      - 10" nominal Þ 9.25" actual.
    - Calculate the dead weight of 2 X 4 at 2' o.c. spanning into the supporting 2X10 members.
      - Assume DF-L again
      - One 2 X 4 member weighs:
        
        4" nominal Þ 3.5" actual.
      - Each 2X4 spans 8' with a 2X10 supporting each end of the 2X4:
        Þ 1/2 its weight is supported by each 2X10 or
        Þ 1/2 (1.28 lb/ft) (8ft) = 5.12 lbs.
      - Each typical 2X10 supports 9-2X4's per side:
        Þ the total load on the 2 X 10 is 2 (9) (5.12 lb) = 92.16 lbs.
        Þ distribute this over 20' span yielding 92.1 lbs / 20 ft = 4.61 lb/ft.
        But there is an easier way to get the 2X4 dead weight effect on a typical 2X10 member:
  4. Therefore the total dead weight supported by a typical 2X10 is the summation of the basic roof (and ceiling, if directly attached to underside of roof members), self-weight, and self-weight of any supporting members.

Send Email to Deb Larson at Debra.Larson@nau.edu

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