Some information contained in it may be outdated. Understanding how loads are transferred through a structure and act on structural members is the first step to sizing headers and beams. Most builders automatically choose double -2 x 8 or -2 x 10 headers to frame windows and doors in every house they build. These headers work to support most residential loads and coincidentally keep the window tops to a uniform height. A neat solution, but is this an efficient and cost effective use of material?

The same is true for beams like structural ridge beams and center girders.

## Dead Load vs Live Load

Too often builders gang together 2-inch dimension lumber to support roof and floor loads without considering other options. Parallam, Timberstrand, Laminated Veneer Lumber and Anthony Power Beam are examples of alternative materials that provide builders with some exciting choices. In this 2-part series we will review how sawn lumber and these engineered materials measure up as headers and beams. Part I will show you how to trace structural loads to headers and beams.

The job of headers and beams is a simple one. They transfer loads from above to the foundation below through a network of structural elements. The idea behind sizing headers and beams is straight-forward: Add together all live loads and dead loads that act on the member and then choose a material that will resist the load.

However, the process for sizing these structural elements can be complicated if you are not an engineer. Here is a simplified approach that will help you specify the appropriate material for many applications. The first step is the same for sawn- and engineered wood materials: add up all the loads acting on a header or beam and then translate this load into terms of how much load each lineal foot of header or beam will feel. In beam-speak you say: this header must carry X-pounds per lineal foot.

This translation is the key to any structural sizing problem. Armed with this information you can determine the minimum size, span or strength of the beam credit julio. Engineered wood components are sized using span tables that match various spans to pounds per foot of beam.

For sawn-lumber you must perform mathematical calculations. Loads are considered to be either distributed or point loads. A layer of sand spread evenly over a surface is an example of a pure distributed load. Each square foot of the surface feels the same load. Live and dead loads listed in the building code for roofs and floors are approximations of distributed loads.

### Load Limits on the Roof of a Building

Point loads occur when a weight is imposed on one spot in a structure, like a column. The load is not shared equally by the supporting structure. Analysis of point loading is best left to engineers. We will consider only distributed loads. This will enable us to size beams for most common applications. Figure 1. Assume that all are located in the same climate, but have different loading paths because of the way they are built.Bartlett Quimby.

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.

Given the floor section shown in Figure 3. The floor system is in an office building that is likely to see reconfiguration of interior partitions over the course of it's life.

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.

Figure 3. Include a reroofing allowance when computing the design dead load. 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. The correction factor becomes 1. If you look closely at the right side of the roof section shown in Figure 3.

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.

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. Homework Problems. Report Errors or Make Suggestions. Section 3. 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. In this case, the weight is 35 psf.

We'll use that.

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Promoting, selling, recruiting, coursework and thesis posting is forbidden. Students Click Here. Related Projects. I am working on a project where we are using wood trusses 2x6 for top and bottom chord, 2x4 for web.

I am trying to figure out a psf from the weight of the truss but am hitting a road block. I have attached a picture of my attempt. Any help here would be much appreciated. I use a 44pcf for wood to be conservative. Check with a local supplier. It has many great tables supplying all kinds of info like this. See if you can get yourself a copy. I think it is Alpeng. The truss will weight about 6. I got this out of tables in "Wood Engineering and Construction" 3 rd Ed.

Faherty and williamson. I don't design roof trusses but I use 4psf when considering gravity loads on framed walls and 2psf when design tie-down. As for the calculations, calculate your total length of 2x6 and total length of 2x4, convert these to total weights, multiply by 1. Red Flag This Post Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

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This includes challenging trends on contractors and distributors like new refrigerants, growing automation, complex sensors and monitoring, green initiatives and a technician shortage.A roof must be designed to support its own weight, plus any temporary loads that will be placed on it.

Roofs are under a lot of pressure. In order to stay intact and in place, a roof must be able to resist loads, both permanent and temporary, that are pushing both downward and upward on it. The limit of the load a given roof can support is determined by its specific design, but typical roofs are expected to withstand a few commonly anticipated loads, such as the weight of people working on it.

The dead load on a roof is the weight of the roof structure itself, along with any permanently attached materials or structures on the roof, so it must be designed, first of all, to support itself.

The dead load of a typical asphalt-shingled, wood-framed roof is about 15 pounds per square foot. The load increases with the use of heavier roofing material. A clay-tiled roof may have a dead load of as much as 27 psf.

The live load on a roof is the weight of any temporary objects on the roof. Where snow isn't a problem, the live load can come from people working on the roof and any equipment they take on to the roof with them. The roof must be able to support the sum of its dead load and any anticipated live load, so the roof has to be designed with a load limit that takes into account both of these loads. A typical roof is expected to support a live load of 20 psf; this minimum live load is in addition to the dead load that the roof must bear.

When wind hits the exterior wall of a building, the wind's energy disperses upward and downward along the wall. The upward movement of the wind exerts an uplift load on the roof, and the roof must be able to resist this uplift.

## Understanding Loads and Using Span Tables

A typical uplift load limit assumes a maximum wind speed of about 90 miles per hour and expects a load of about 20 psf. Most of this load will be resisted by the roof's downward-pushing dead load. Load limit calculations assume loads are pushing downward uniformly on the roof's horizontal surface area. On a steeply sloped roof, more weight is pushing downward on a relatively smaller horizontal surface, so the roof's load limits must be adjusted to take this difference into account.

On a roof with a slope greater than 4 to 12, the live load limit is typically adjusted downward from 20 psf to 15 psf to allow for the relatively greater dead load on the steeper roof. To determine which roof framing members are appropriate for a building you're designing, you'll first have to determine the loads, both dead and live, that the kind of roof you want to build will have to bear.

You should begin by consulting the minimum load-bearing requirements in your local building code. The code book will give you minimum load limits and a deflection limit, which is a measurement of how much the rafter is allowed to bend under its load. Then you'll determine the span of the roof's rafters, which is a measurement of the horizontal distance from the inside surface of the ridge board to the inside surface of the wall that supports the rafter.

Using these numbers in consultation with span tables, which are available from sources such as the American Wood Council, you can determine which lumber dimensions and wood species are suitable for your rafters. Evan Gillespie grew up working in his family's hardware and home-improvement business and is an experienced gardener.

He has been writing on home, garden and design topics since Skip to main content.

### How to Calculate Floor Load Capacity

Home Guides Home Home Improvement Troubleshoot, Fix and Repair A roof must be designed to support its own weight, plus any temporary loads that will be placed on it. Dead Loads The dead load on a roof is the weight of the roof structure itself, along with any permanently attached materials or structures on the roof, so it must be designed, first of all, to support itself.

Live Loads The live load on a roof is the weight of any temporary objects on the roof. Uplift Load When wind hits the exterior wall of a building, the wind's energy disperses upward and downward along the wall.

Slope Correction Load limit calculations assume loads are pushing downward uniformly on the roof's horizontal surface area. Calculating Load To determine which roof framing members are appropriate for a building you're designing, you'll first have to determine the loads, both dead and live, that the kind of roof you want to build will have to bear. About the Author Evan Gillespie grew up working in his family's hardware and home-improvement business and is an experienced gardener.

Customer Service Newsroom Contacts.Introduction on Dead Load vs Live Load :. Generally, structural loads including forces, deformations or accelerations applied to a structure can be divided into two main category: 1- Dead load, and 2- Live load. Dead loads are static forces that are relatively constant for an extended time; usually the weight of materials plus immovable fixtures such as carpet, roof and etc. Building materials are not dead loads until constructed in permanent position.

Roof and floor live loads are produced during maintenance by workers, equipment and materials, and during the life of the structure by movable objects, such as planters and people.

For live load not exceeding psf, not in passenger garages, not in group A occupancies of IBC, supporting more than square ft 2Live load can be reduced by the following percent.

R is percent of reduction shall not be more than 40 percent for horzontal member, and 60 percent for vertical member, and not exeeding.

Maximum Crane wheel load shall be the sum of weight of bridge, crane load capacity, and weight of trolley with trolley in a position that produces maximum load.

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No files available for this topic.Bartlett Quimby. ASCE 3. This is generally considered to mean that anything that is a fixed part of the structure is a dead load.

One test that generally works is that if it can be moved without cutting it loose or detaching it from the structure then it is not a dead load. Items that can be considered to be dead load include construction materials that make up the building beams, columns, floor systems, ceiling systems, wall systems, doors, windows, floor coverings, wall coverings, cabinets, and the like and permanently attached equipment such as heating and ventilating systems, electrical trays, piping, etc.

Items that are not considered to be dead load include such things as movable shelving, desks, chairs, beds, chests, books, copiers, stored items, or anything else that can or may be moved around during the life of the structure. One feature of dead loads is that they are the weights of the final structure. This creates a bit of a dilemma for the design engineer. The engineer must know the weights of the structure in order to design it, but the engineer also needs the final structure to accurately define the weights!

At the beginning of the design process, the framing and other structural elements are all unknown, however the weight is needed to determine the internal forces for the members being designed.

The solution to this dilemma often involves a few iterations where an educated guess is made as to where the design will end up, compute the dead loads based on this estimate, select members based on the estimated loads, recompute the dead loads, then continue the cycle until member sizes don't change.

Most floor, roof, and wall systems have fairly uniform density and their weights can be expressed in terms of weight per unit area.

With dead loads expressed in this manner, tributary area concepts can be used to determine the forces exerted on the supporting members. To compute the average unit weight i. Generally, this method works if all items in the system they are either spread over the surface of the area like carpet on a floor system Items that are not spread over the surface and not uniformly spaced over the entire area are generally not included in the unit weight calculation but are treated as individual loads in addition to the unit load.

The exceptions to this are minor electrical and mechanical items such as wiring and plumbing whose exact location is unknown during design, as well as added density at the connections to supporting structures.

To account for these items an additional "miscellaneous" load may be applied. The magnitude of the miscellaneous load generally varies depending on what you expect the extent of these items to be.

For typical cases, values between 1 and 2 psf are typical. Unit dead loads for roof areas that are sloped are commonly expressed in terms of weight per unit area of horizontal projection. This is done because most load computations are done in terms of plan view unit areas. The use of weight per unit horizontal projected area requires a slope adjustment as the weight of a sloped surface expressed in terms of weight per unit area of horizontal projection is larger than the weight per unit area along the surface.

Consequently the slope adjustment is commonly done only for slopes greater than You will see this calculation in section 3.

ASCE section 4. The seismic requirements ASCE Certain types of roofing systems allow for a new roof to be placed over the old when it is time to replace the roofing. This is the case with asphalt shingles. When determining the weights to be used for roofing, many engineers include a reroofing allowance for the added weight of the second layer of roofing material.

If in doubt, consult with the architect on the likelihood of the selected roof system requiring a reroofing allowance. Given that area dead load calculations, there are a number of sources that present tables of unit weights expressed in terms of weight of per unit area. The following are a couple that I have used in the past:. You will find, when you compare sources, there is not complete agreement in the estimates that each source puts together.

This is to be expected. As with using any tables, you need to apply a little engineering judgment, taking into account your actual job conditions.

When in doubt it is probably best to be mildly conservative with your estimate. Being too conservative may have big ramification in structures with large dead loads.

Given the inexact nature of dead load computations, it is common to round dead loads to integer values of no more than three significant figures. The dead load of linear members such as beam, columns and large pipes are generally expressed in terms of weight per unit length.Because figuring out floor joist load capacity requires a thorough understanding of the structural properties of wood, as well as building code requirements, it's a job best left to a structural engineer when it comes to determining the specifics of a building's design.

Understanding the basics of the calculations, however, can help you to understand the load limits of your existing floor. Floors must be able to support two different kinds of weight loads. The dead load on the floor is the weight of the floor structure itself and anything else that is permanently attached to the floor.

The live load is the weight of furnishings, people and anything else that the floor needs to support, but which isn't permanently attached. The dead load on a floor is determined by the materials used in the floor's construction.

A typical wood-frame floor covered with carpet or vinyl flooring has a dead load of about 8 pounds per square foot; if there's wall-board covered ceiling suspended from the underside of that floor, the dead load increases to about 10 pounds per square foot.

Heavier flooring materials increase the dead load even more. Local building codes specify the minimum live load that floors must be able to bear.

The International Residential Code, on which most local building codes are based, requires that floors in non-sleeping rooms must support a minimum live load of 40 pounds per square foot, and floors in sleeping rooms must be able to handle a live load of 30 pounds per square foot.

Codes also specify how much floors are allowed to bend under load, a measurement called a deflection limit. Span tables and tables that show design limits for particular lumber types allow you to determine whether a given floor design will meet code and design requirements. For a given joist size and spacing, the tables indicate the strength value, called the Fb valueand stiffness, called the E valueof the joist.

Architects and engineers use these tables to determine the required size and spacing of joists as they design buildings, but you can use them to work backwards and calculate the load capacity of an existing floor. First, determine the size, the spacing and span, and the species and lumber grade of your floor joists.

Look for a stamp on the joist that indicates the lumber's species and grade. Use a design value table to find the Fb value for your floor joists. As an example, consider a room with a floor area of 10 feet by 11 feet, 2 inches built with No. A design value table shows that the joists have an Fb value of 1, and an E value of 1, Next, consult a span table to cross reference the spacing and span to find the required Fb value for that table's load limits.

In the example, the joists have a spacing of 16 inches and a span of 11 feet, 2 inches. In this case, your joists are adequate to support a 30 psf live load and 10 psf dead load.

**Types of loads act on a structure - dead load - live load - wind load - CIVIL ENGINEERING**

Use span tables for progressively heavier loads until you find the limits of your floor. That means that the existing joists are not adequate to support a 40 psf live load.

After you've determined the load limit of your joists, you can use that figure to determine the total acceptable load for the room or building in question. In the example, the floor area of the room is approximately square feet. With an evenly distributed live load of 30 psf, which the tables show the floor is able to support, the total weight on the floor would be about 3, pounds.

Increasing the total weight on the floor to 4, pounds, however, results in a live load of 40 psf, which is beyond the floor's load capacity. Joseph West has been writing about engineering, agriculture and religion since He is actively involved in the science and practice of sustainable agriculture and now writes primarily on these topics.

He completed his copy-editing certificate in and holds a Bachelor of Science degree from the University of California-San Diego. Hunker may earn compensation through affiliate links in this story. Share this article. Joseph West. Show Comments.

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