Building Problem Solutions
Building Problem Solutions
Beam Design - Basic
By John F Mann, PE
BEAM DESIGN EXAMPLE
WOOD BEAM DESIGN REFERENCES
I-BEAM TYPE WOOD JOISTS
BEAM DESIGN SOFTWARE
This discussion is intended to provide an overview of basic requirements for general beam design, for any material. Example calculations are for design of basic wood beams.
A beam is most often understood to be an element that supports load over space. However, the key characteristic is that the beam resists load primarily in bending (flexure) and shear, perpendicular to the length of the beam.
In contrast, a column resists load primarily in compression parallel to the length of the column.
Most often we see horizontal beams. Yet, beams can be sloped, as for a gable roof, or even vertical, to resist wind. Some beams are also columns, such as wall studs.
Many sources of information are available to allow for selection of beams, including floor joists. However, except for very short headers over small windows and doors, beam design should be performed by qualified designers only. Proper beam design requires consideration of many factors (such as lateral bracing) that are often easy to overlook.
See other pages on this site (such as "Structural Ridge Beam", "Design To Minimize Deflection" and "Wood Framing For Tile Flooring") for more detailed information about beam design.
Several beams in view.
Yellow manufactured-wood beam supports floor joists (right side), ceiling joists (left side) and wall above, which also support floor joists for attic floor.
The following information is required for beam design;
- Values, type (uniform, concentrated,varying) and locations of design loads to be supported
- Number and locations of supports along beam which defines span lengths between supports
- Type of supports (hinge or "simple", "fixed", continuous)
- Allowable dimensions (depth or height, and width or thickness)
- Spacing between lateral bracing of compression edge or flange
- Design properties of beam (shear, bending, bearing)
- Design capacity of supports, which may include connection hardware (such as hanger)
- Allowable deflection
- Design factors that may be applicable, as specified by building code or other governing design code.
For a beam that must support floor joists only, and a rectangular floor layout, calculation of design loading is relatively straightforward. However, complications often occur, such as bearing walls supporting upper floors or heavy equipment not included in general uniform design loads.
Calculation of design loads can be complicated, especially when a beam on a lower floor must support loads from other beams, some of which may be from upper floors.
For general building design, the governing building code specifies "minimum" design loads. For some conditions, special design codes are applicable. For conditions not addressed by any code, careful study of expected and unexpected events must be performed to develop adequate design loads.
Any beam must have adequate strength, to resist failure, and stiffness, to prevent excessive movement (deflection). Stiffness is the inverse of flexibility; greater stiffness results in less flexibility.
In general, strength and stiffness of a beam is influenced much more by the depth (height) than by width (thickness).
For a beam with rectangular cross section, bending (flexural) strength is a function of the square (second power) of the depth (height). A similar, though more complicated, relation applies for I-beam shapes.
With all other factors the same, a rectangular beam with depth of 12 inches has a bending strength that is 4 times that of a beam with 6-inch depth.
Bending strength of the beam is dependent on the "section modulus".
For a beam with cross section that is symmetrical about the horizontal axis (of the cross section), section modulus is the same for compression and tension bending stress. For non-symmetrical cross section, section modulus is different for compression and tension flexural stress.
Large & long one-piece "specialty" wood beam (Longleaf Pine) supporting floor joists and column load on floor above.
Bending stress (compression and tension) acts parallel to length of the beam. At any section of a beam, bending stress varies through the depth. The general formula for bending stress is;
Bending stress, fb = M c / I
M = Bending moment at the section
c = distance (vertically) from neutral axis
I = Moment of inertia at the section
Stress is zero at the neutral axis (c = 0), which is at mid-height for a rectangular beam or any beam with section that is symmetrical about a horizontal axis. At any section, maximum bending stress (compression and tension) occurs at the upper and lower edges.
At any section of the beam, maximum bending stress (compression and tension) is equal to moment (M) divided by section modulus (S). For a beam with two simple (hinge) supports (one at each end), and for load acting towards the beam, maximum compression stress occurs at the loaded edge or face. Maximum tension stress occurs at the opposite edge or face. For load acting away from the beam, the sense of maximum stresses are of course reversed.
General formula for section modulus is;
S = I / cmax
I = moment of inertia at section
cmax = distance from neutral axis to upper edge or lower edge
As noted above, there are generally two section modulus values at any cross section; one for top edge of beam (S-top; compression flexural stress for simple beam) and one for bottom edge of beam (S-bottom; tension flexural stress for simple beam). However, for a beam with symmetrical cross section, each section modulus value is the same (S-top = S-bottom).
For a rectangular beam, section modulus is calculated as;
S = B H^2 / 6
where; B = width
H = depth (parallel to direction of load)
Width (B) and depth (H) must of course be in the same units (generally inches). In the US, units for section modulus are generally calculated as inches to the third power (cubed). Outside the US, SI units would generally be used, such as centimeters to the third power.
For a built-up beam consisting of two 2x10s (9-1/4 inch deep by 3 inches thick), section modulus is calculated as;
S = [ 3" x (9.25")^2 ] / 6 = 42.8 in^3
For a beam supporting uniform load (w) along the entire length, maximum bending moment (at midspan) is calculated as;
Moment, M = wL^2 / 8
Units for uniform load w are force per distance, such as pounds per foot (often noted as PLF or pounds per linear foot).
Reaction force, R (acting in opposite direction to loading) occurs at each support;
Reaction, R = wL/2
Moment varies with square of span length (L), such that required moment strength increases very quickly as the span length increases. For example, increasing span length from 10 feet to 12 feet results in a 44% increase in maximum moment (for the same uniform load).
Units for moment are often calculated initially in foot-pounds (ft-lbs). However, for calculation of bending stress (M/S) the length units for moment must be the same as for section modulus (generally inches). Therefore, moment in foot-pounds often must be converted into inch-pounds. A common mistake is to use inconsistent length units when calculating moment.
The resulting value from dividing moment (M) by section modulus (S) will most often be in pounds per square inch (psi).
Span length is clear span plus the lesser of; (1) Half width of support, such as wall, at each end, or (2) Half depth of beam, at each end.
Calculated bending stress must be compared to allowable bending stress, which is typically defined by the building code. In general, the building code will reference another governing code for each particular material (wood, steel, concrete, aluminum).
For design of wood beams, the governing design code is National Design Specification For Wood Construction (NDS), published by the American Wood Council. The NDS code includes detailed provisons that govern design.
Basic allowable bending stress for typical wood species and grade of sawn lumber (such as Douglas Fir, Hem Fir, Spruce Pine Fir) is in the range of 850 psi to 950 psi. However, there are numerous adjustment factors that may be applicable which will most often (though not always) allow for increase of the basic allowable stress.
For a continuous beam with more than two supports, bending stress reversal occurs over and near interior supports; tension stress at loaded face (for loading acting towards the beam) and compression stress at opposite face. This condition also occurs for the cantilevered segment of any beam.
A beam must also resist "shear", which is most often visualized as the beam being cut through, as if cut with a saw. However, shear is generally much more complicated. For wood beams, "horizontal" shear (parallel to wood grain) is most important. For reinforced concrete beams, "diagonal" shear governs.
For practical design of wood beams, shear stress caused by design shear force (at a specific location along the beam) is most often compared to allowable shear stress, as defined by the governing code. For reinforced concrete and steel beams, design shear force (at any location along beam) is also compared to design shear capacity (at that same location or section).
Design shear force to be resisted may be modified to account for the complexity of shear behavior for the particular material.
For a uniformly loaded beam, shear force varies at a constant rate along length of beam, with the mathematical sign (positive or negative) changing from one end to the other. Shear is zero at midspan.
Shear force changes abruptly at concentrated ("point") loads.
For a simply-supported beam with two supports, basic design (maximum) shear force is the "reaction" force that occurs at each end-support. This may be reduced slightly to account for beneficial effects of compression near the support, in accordance with code provisions. However, for practical design, the full reaction force is most often used.
Average shear stress is simply design shear force divided by area of the beam "web", which is generally a part of the cross section with long dimension parallel to direction of force. For wood beams and concrete beams with rectangular cross section, the web is the same as the cross section. However, for I-beam shapes (wood, steel, prestressed concrete), flanges are generally not included in the web.
For a wood beam with rectangular cross section, design shear stress is the average shear stress times 1.5 (3/2), which accounts for the actual distribution of shear stress within the cross section, which varies from zero at a free edge to maximum at the neutral axis. Multiplication of basic shear stress times the 1.5 factor is often overlooked.
An essential but often overlooked requirement for any beam, especially a long beam, is adequate lateral bracing. Without proper bracing, the beam will fail by buckling sideways long before failure in bending or shear. Such buckling is similar to how an unbraced compression member (column) fails.
The compression edge (flange for steel I-beam) must be braced against lateral movement. Typically this can be accomplished by connecting framing members (joists) or decking. However, a positive connection must be specified by the designer, not merely assumed.
Allowable design stress in bending is reduced when spacing of lateral bracing exceeds the maximum allowed for use of full allowable design stress. However, at some spacing value, allowable stress is zero. Calculation of allowable bending stress for various spacing of lateral bracing is performed using standard formulas specified by governing design codes.
Careful consideration of lateral bracing must be made for design of beams subject to reversal of loading, such as for a roof beam that must resist wind uplift. The lower, tension edge for gravity (downward) loading then becomes the compression edge for upward loading. However, since there is no roof deck available to brace the lower edge, some other method of lateral bracing must be provided, or the beam must have adequate strength without lateral bracing.
Temporary bracing must often be installed during construction. Such bracing may be necessary for lateral bracing of compression edge. Bracing may also be required to prevent rollover of one or more beams.
A key requirement, also overlooked, is that the bracing itself must also be braced by some substantial building element. Entire roof and floor systems have collapsed during construction due to failure of builder to properly brace each line of lateral bracing.
Beam Design Example
A rectangular wood beam is to be designed to support first floor joists only (no bearing walls or other loads). The beam is to be simply supported at each end on support walls (2x4s). Clear span between stud walls is 14 feet. Span length is then 14.3 feet.
Joist span lengths are 10 feet on one side and 16 feet on the other side. Uniform design loads are 10 psf dead load and 40 psf live load. Total design load is then calculated;
Design load on beam = (10 psf + 40 psf) (10 feet / 2 + 16 feet / 2)
= 50 psf x 13 feet
= 650 PLF
Weight of the beam must also be included. For initial design, assume 20 PLF beam weight.
In general, strength is checked for total load. Deflection is typically checked for live load only and for total load.
Forces are first calculated for total load;
Reaction (each end), R = 670 PLF (14.3 ft / 2) = 4,791 lbs
Moment (midspan), M = 670 PLF (14.3 ft)^2 / 8 = 17,126 ft-lbs
Moment is converted to 205, 513 inch-lbs.
No adjustment factors are applicable.
For allowable bending stress of 900 psi, required section modulus is calculated as;
S required = 205,513 in-lbs / 900 psi = 228 in^3
Required depth (h) can be calculated for a given thickness using basic algebra. In practice, section modulus of standard beam sizes are checked to determine required size. For a built-up wood beam, using 2x12s, section modulus is calculated for a beam of 2-2x12s and 4-2x12s;
Section modulus (2-2x12s), S = 3 in (11.25 in)^2 / 6 = 63 in^3
This is clearly much less than required. Even 6-2x12s do not have adequate strength. Therefore, an LVL (laminated veneer lumber) beam is considered.
For a standard Microllam LVL beam, allowable bending stress is 2600 psi. Small adjustment might be applicable as function of beam depth (per manufacturer formula) however the basic allowable is used for this example.
S required = 205,513 in-lbs / 2600 psi = 79 in^3
Section modulus is first calculated for 3-1/2 inch thick beam, with 11-1/4 inch depth (same as 2x12);
S =3.5 in (11.25in)^2 / 6 = 74 in^3 < 79 in^3 Not Adequate
Beam depth or thickness (or both) can be increased to increase section modulus. For any specific case, one or the other may be limited by functional requirements. Beam thickness may have to be increased to limit bearing stresses, especially for relatively short beams or when there is a large point load near a support.
Increasing beam depth (if allowable) is most effective based on material quantity. Section modulus is then calculated for standard LVL depth of 11-7/8 inches;
S = 3.5 in (11.875 in)^2 / 6 = 82 in^3 OK
Shear stress is calculated for maximum shear force. For a simply supported beam, maximum shear, which occurs at end of beam, is equal to the end reaction force. A small reduction can be calculated if necessary, as discussed below. Maximum shear stress is shear force divided by cross sectional area of the beam (at the location of shear force), multiplied by 1.5 as previously noted;
Shear stress, v = 1.5 (4,791 lbs / (11.875 in x 3.5 in) )
= 173 psi
Allowable shear stress for Microllam LVL is 285 psi, so that the proposed beam has adequate shear strength.
If necessary, for beam with uniform load, and supported from underneath, design shear force can be taken as the shear that occurs a distance "d" from face of support, with "d" being depth of beam. This standard code provision reduces design shear by an amount equal to the uniform load times the sum of; (1) Distance from face of support to end of beam (per design), and (2) Depth of beam. Of course length units for load and each distance must be consistent.
Live load deflection is most often the key deflection value to be checked. See "Design To Minimize Deflection" for detailed discussion.
For the LVL beam (11-7/8 x 3-1/2), moment of inertia (I) is 488 in^4. Modulus of elasticity is 1900 ksi or 1,900,000 psi. Midspan deflection of 0.48 inches is equal to Span / 360, which happens to also be 0.48 inches.
Of course, for many applications, there may be a more conservative deflection limit. In that case, a deeper beam is most likely required, unless the difference (between calculated and allowable values) is relatively small, such that increasing beam thickness might work.
Wood Beam Design References
"Beam Design Formulas With Shear And Moment Diagrams", published by the American Wood Council (AWC), is a useful reference for those with basic understanding of beam design.
APA publishes a design guide for glued-laminated beams ("glulam") which have similar properties to LVL beams. More importantly, the guide includes general discussion that applies to all wood beams.
Note that, as with almost all such design guides, tables are applicable for uniform load only. Also, additional calculations must be performed to complete the design, such as bearing stresses at supports (for beam and support) and lateral stability.
I-Beam Type Wood Joists
Manufacturers of I-beam type wood joists publish design guidelines with detailed tables listing required joist sizes for various loading and span conditions. Although very useful for simple uniform loading, the tables must not be used for more complex loading conditions without careful analysis.
Beam Design Software
Numerous software programs are available for beam design. However, as for any program, quality of results is dependent on quality of input. Overlooking just one important factor will produce defective results.
Check back for beam design software.