A Beginner's Guide to ASCE 7-10

Chapter 3 - D: Dead Loads

© 2012, T. Bartlett Quimby

BGASCE7-10 Section 3.2

Typical Unit Area Dead Load Calculations

Last Revised: 11/04/2014

Computing the unit dead load for a region of surface area generally starts by identifying the region of a roof plan, floor plan, or elevation where the unit load is needed then looking at a typical section of that area to see how it is constructed.   Once the components of the system have been identified, a weight is computed for each item.  The total unit dead load is the sum of the component weights plus a "miscellaneous" factor to account of minor items not included specifically in the calculation.

A Floor System Example

Given the floor section shown in Figure 3.2.1, determine the unit dead load for the region that has this construction.  The floor system is in an office building that is likely to see reconfiguration of interior partitions over the course of it's life.

Figure 3.2.1
Floor System Section

Determine the weights of the various components:

• Carpet & Pad:  There is a wide variety of carpet & pad out there with a wide variety of weights.  In this case, since the carpet and pad weight is likely to be small with respect to some of the other components, we can be a little conservative without making a significant impact on the design.  Lets use 3 psf for the carpet and pad.
• Concrete on Deck:  The decking manufacturer provides weight tables for their product.  In this case, the weight is 35 psf.
• Steel Joists:  A 36LH07 steel joist weighs approximately 16 lbs per lineal foot according to the Steel Joist Institute literature.  The tributary width of the joists is 5' 10", so the weight per floor area equals 16 lb/ft / 5.8333 ft = 2.74 psf
• Lightweight Suspended Ceiling:  According to one reference, a metal suspension system with tiles weighs about 1.8 psf.  We'll use that.
• Partition Loading: This is an office occupancy so we'll use a partition load instead of computing the actual wall dead load.  Note that this can be included in live load instead of dead load because of the uncertainty of the value, however we are choosing a conservative value for the load so feel comfortable leaving it as a dead load.  This also means that the partition load is already included in the dead load when later computing effective seismic weight.
• Misc. Loading:  We expect that there will be misc. light fixtures in the ceiling, wiring, and minor plumbing, so we will select a value between 1 psf and 2 psf that will give us an integer value for the total load.

Table 3.2.1 summarizes the calculation:

Table 3.2.1
Floor System Dead Load

Note that this calculation does not include the weight of any girders or other structure that is in the floor system.  To determine the total weight of the system, you need to add in the other items that are a part of it.

A Roof System Example

Figure 3.2.2 shows a roof section for wood framed residential roof system.  Include a reroofing allowance when computing the design dead load.

Figure 3.2.2
Wood Framed Roof System Section 

Determine the weights of the various components.  Remember that we want to have the loads expressed in terms of weight per square foot of horizontal area so we have to convert sloped surfaces.  This is a simple trigonometry problem as shown in Figure 3.2.3.  The correction factor equal 1/cos q.  For our particular problem, the slope angle equals the arctangent of (8/12) or 33.7 degrees.  The correction factor becomes 1.20.

Figure 3.2.3
Slope Correction

• Asphalt Shingles:  The shingles weigh 2.5 lbs per square foot of surface area.  We need to convert this to weight on a horizontal projected area, so the shingle weight becomes 2.5 psf times 1.2 or 3.0 psf.  This includes the roofing felt that underlies the shingles.
• Reroofing Allowance:  This equals the original roofing weight of  3.0 psf.
• 1/2" Wood Sheathing:  Most wood products weigh about 35 lb per cubic foot, so the weight of 1/2" of wood weighs about 35 pcf times the ratio 0.5"/12", resulting in a weight of approximately 1.5 psf.
• Trusses:  As the trusses have probably not been designed yet (these are often done by the truss manufacturer while the project is under construction) we need to make a reasonable approximation.  Experience would suggest that the top and bottom chords are made from 2x6 material and the web members from 2x4 material.  The top chord and web members are sloped and require slope correction while the bottom chord does not.  In this case we will estimate the weight by adding the top chord weight (2x6 times slope correction), the web members (2x4 times same slope correction), and the bottom chord together to get a unit weight of 3.26 psf.
• Insulation: Insulation comes in various weights.  You need to consult with the architect on this or make a conservative estimate.  One source puts batt insulation at 0.1 psf/in to 0.2 psf/in.  We will use the larger to get a unit weight of 2.4 psf.  Note that there is no slope correction to be applied here since the insulation is already on a horizontal surface.
• Gypsum Wall Board (GWB) & Vapor Barrier:  Normally there is no addition for the vapor barrier because it's weight is so small compared to the miscellaneous value.  The GWB (or sheetrock) is heavy.  One source put 5/8" thick GWB at 2.8 psf so that is what we will use.
• Miscellaneous:   This roof system also has wind bracing at the end trusses, blocking at the supports, and minor wiring.  We will use a value between 1.5 and 2.5 to account for this extra stuff.

Table 3.2.2 summarizes the calculation:

Table 3.2.2
Roof System Dead Load

If you look closely at the right side of the roof section shown in Figure 3.2.2, you will notice that the roof framing is different.  This means that the average unit dead load computed here really only applies to where these trusses are.  A separate calculation is required for roof areas that don't match the one for which we did the calculation.

Interior Wall Example

Figure 3.2.4 shows a wall section for light gage steel framed wall system. 

Figure 3.2.4
Typical Interior Wall Section

Determine the weights of the various components:

• GWB:  The wall has two layers of 1/2" GWB, each weighing 2.2 psf.  The total weight of GWB on the wall is 4.4 psf. 
• Steel Studs:  The manufacturers website says that the cross sectional area away from cutouts equals 0.513 in² for this stud.   Since steel weighs about 492 pcf, this equates to 1.75 plf for the stud.  This is conservative since it does not account for the cutouts.  The studs are placed at 16 inches O.C. (on center), giving us a weight of 1.31 lbs per square foot of wall.
• Miscellaneous:  The wall is likely to see minor weight from electrical wiring, but not much else.  We will chose a miscellaneous load that is between 1 psf and 2 psf.

Note that we have not accounted for door opening or windows.  These items usually weight near or less per square foot of wall than the main part of the wall.  Using the average value computed here is usually close enough without being overly conservative.  If you have opening with out doors or windows, then subtract those areas from your wall weight calculation.

You should consult the architectural drawings for the project to see if there are any other wall coverings being added, such as paneling or sound board.  If there are other items, these must be added in as well.

Table 3.2.3 summarizes the calculation:

Table 3.2.3
Interior Wall Dead Load