Wood Beam Span Calculator

Created by Kenneth Alambra
Reviewed by Steven Wooding
Based on research by
American Wood Council National Design Specification for Wood Construction (2018)See 1 more source
American Wood Council National Design Specification for Wood Construction Supplement (2018)
Last updated: Oct 04, 2022

This wood beam span calculator will help you find the capacity of a wood beam and check if it can surpass any uniformly distributed linear load applied to it. In this wood beam calculator, we'll perform wood beam deflection calculations, consider a wood beam's adjusted allowable design values, and compare them to the actual bending and shear stresses it has to support.

Keep on reading to explore:

  • The importance of wood beam calculations;
  • How to perform wood beam deflection calculations;
  • How to calculate the actual bending stress due to loading and the adjusted allowable bending stress in a wood beam; and
  • How to calculate the actual shear stress from the applied linear loading and the adjusted allowable shear stress on a wood beam.

The importance of wood beam calculations

When choosing what size of lumber to use as a beam, we must consider various factors so we won't have a wood beam that can pose a danger to us. We want to choose the size of lumber that can support the beam load we need to apply to it and can handle the effects of humidity and moisture, extreme temperature, bending, and shearing (to name a few).

Other than the size of the beam, we also have a wide range of selection of wood species and commercial grade. Each wood species and grade has its own set of stiffness or design values, including bending stress, shear stress, tension and compression stresses, and modulus of elasticity. We then adjust these design values to consider the long-term environmental and thermal effects mentioned above and see if the wood beam can still support the loading we anticipate it to carry. Performing these calculations will help us choose the beam size and species that can support our anticipated loading and handle some unforeseen additional loading and natural weakening of lumber over time.

This wood beam span calculator will focus on the first three parameters we typically test when designing a wood beam. These parameters are the beam's allowable deflection, bending stress, and shear stress.

We'll get all the necessary data from the National Design Specifications (NDS®) Supplement: Design Values for Wood Construction 2018 Edition and follow the adjustment guidelines we need from the National Design Specification (NDS®) for Wood Construction 2018 Edition prepared by the American Wood Council (AWC).

Then, we'll calculate the resulting deflection, bending stress, and shear stress due to the loading on our beam and compare them to the adjusted design values of our chosen wood beam.

Checking the actual and allowable deflection of a wood beam

To find the deflection of a wood beam with the typical rectangular cross-section, we use this formula:

δ=5×w×L4384×E×I,\small \delta = \frac{5\times w\times L^4}{384\times E\times I},

where:

  • δ\deltaDeflection (in inches) at the midspan of the wood beam due to the loading applied;
  • wwUniformly distributed linear load applied to the beam in pound-force per inch (lbf/in)\small(\text{lbf}/\text{in});
  • LLBeam span or unbraced length of the beam in inches;
  • EEModulus of elasticity of the wood species used in pounds per square inch (psi)(\small\text{psi}); and
  • IIArea moment of inertia of the beam's cross-section in inches to the fourth power (in4)(\small\text{in}^4).

We can get the modulus of elasticity (E) of any common American wood species from the NDS Supplement Table 4A Reference Design Values for Visually Graded Dimension Lumber. Here are some of the wood species' moduli of elasticity from that table:

Species

Modulus of Elasticity, E (×106 psi)

Select struct.

No. 1

No. 2

No.3

Stud

Const.

Standard

Utility

Alaska Cedar

1.4

1.3

1.2

1.1

1.1

1.2

1.1

1.0

Alaska Spruce

1.6

1.5

1.4

1.3

1.3

1.3

1.2

1.1

Alaska Yellow Cedar

1.5

1.4

1.3

1.2

1.2

1.3

1.1

1.1

Beech-Birch-Hickory

1.7

1.6

1.5

1.3

1.3

1.4

1.3

1.2

Coast Sitka Spruce

1.7

1.5

1.5

1.4

1.4

1.4

1.3

1.2

Douglas Fir-Larch

1.9

1.7

1.6

1.4

1.4

1.5

1.4

1.3

Douglas Fir-Larch (North)

1.9

1.8

1.6

1.4

1.4

1.5

1.4

1.3

Douglas Fir-South

1.4

1.3

1.2

1.1

1.1

1.2

1.1

1.0

Eastern Hemlock-Balsam Fir

1.2

1.1

1.1

0.9

0.9

1.0

0.9

0.8

Eastern White Pine

1.2

1.1

1.1

0.9

0.9

1.0

0.9

0.8

Hem-Fir

1.6

1.5

1.5

1.3

1.2

1.3

1.2

1.1

Hem-Fir (North)

1.7

1.7

1.6

1.4

1.4

1.5

1.4

1.3

Mixed Maple

1.3

1.2

1.1

1.0

1.0

1.1

1.0

0.9

Mixed Oak

1.1

1.0

0.9

0.8

0.8

0.9

0.8

0.8

Mixed Southern Pine

1.6

1.5

1.4

1.2

1.2

1.3

1.2

1.1

Northern Red Oak

1.4

1.4

1.3

1.2

1.2

1.2

1.1

1.0

Northern White Cedar

0.8

0.7

0.7

0.6

0.6

0.7

0.6

0.6

Norway Spruce (North)

1.5

1.3

1.3

1.2

1.2

1.2

1.1

1.1

Red Maple

1.7

1.6

1.5

1.3

1.3

1.4

1.3

1.2

Red Oak

1.4

1.3

1.2

1.1

1.1

1.2

1.1

1.0

Redwood

1.4

1.3

1.2

1.1

0.9

0.9

0.9

0.8

Southern Pine

1.8

1.6

1.4

1.3

1.3

1.4

1.2

1.2

Spruce-Pine-Fir

1.5

1.4

1.4

1.2

1.2

1.3

1.2

1.1

Spruce-Pine-Fir (South)

1.3

1.2

1.1

1.0

1.0

1.0

0.9

0.9

Western Cedars

1.1

1.0

1.0

0.9

0.9

0.9

0.8

0.8

White Oak

1.1

1.0

0.9

0.8

0.8

0.9

0.8

0.8

Yellow Cedar

1.6

1.4

1.4

1.2

1.2

1.3

1.2

1.1

💡 You can learn more about the concept of the modulus of elasticity in our young's modulus calculator.

On the other hand, we calculate the area moment of inertia (I) of our beam's cross-section using this formula:

I=b×d312,\small I = \frac{b\times d^3}{12},

where:

  • IIArea moment of inertia in inches to the fourth power (in4\small\text{in}^4);
  • bbActual base width or thickness of the lumber in inches; and
  • ddActual height of the lumber in inches.

Note that we use the actual dimensions of the lumber for the calculation of II instead of the lumber's nominal dimensions. We usually reduce the nominal size by half an inch to find the actual dimension of the lumber. That means a 2" × 10" piece of lumber has an actual thickness and height of 1.5 inches and 9.5 inches, respectively.

For example, we want to find the deflection of a 2" × 10" select structural Douglas Fir Larch beam that spans 8 feet (or 96 inches) in length, and we anticipate it to carry a uniform linear load of 240 pounds per foot (or 20 pounds per inches).

From the table, we can see that a select structural grade Douglas Fir Larch has a modulus of elasticity of 1.9×106 psi\small 1.9 \times 10^6\ \text{psi} or 1,900,000 lb/in2\small 1,\hspace{-0.04cm}900,\hspace{-0.04cm}000\ \text{lb}/\text{in}^2.

On the other hand, by calculating its area moment of inertia, we get:

I=b×d312=1.5 in×(9.5 in)312=1.5 in×857.375 in312=107.171875 in4107.17 in4\small \begin{align*} I &= \frac{b\times d^3}{12}\\[1.0em] &= \frac{1.5\ \text{in}\times (9.5\ \text{in})^3}{12}\\[1.0em] &= \frac{1.5\ \text{in}\times 857.375\ \text{in}^3}{12}\\[1.0em] &= 107.171875\ \text{in}^4\\[0.5em] &\approx 107.17\ \text{in}^4 \end{align*}

Since we have all the values we need to calculate the deflection due to the applied loading, we can already substitute them into our deflection formula, as shown below:

δ=5×w×L4384×E×I=5×20lbin×(96 in)4384×1,900,000lbin2×107.17 in4=0.10862 in\small \begin{align*} \delta &= \frac{5\times w\times L^4}{384\times E\times I}\\[1.0em] &= \frac{5\times 20\, \tfrac{\text{lb}}{\text{in}}\times (96\ \text{in})^4}{384\times 1,\hspace{-0.04cm}900,\hspace{-0.04cm}000\, \tfrac{\text{lb}}{\text{in}^2}\times 107.17\ \text{in}^4}\\[1.5em] &= 0.10862\ \text{in} \end{align*}

Our next step is to check if this deflection is less than the allowable deflection for our given beam span. For that, we use the deflection criteria provided by the 2012 International Building Code, stating that beams under a combination of dead load (permanent load) and live load (loads that can vary in time) should at most have an allowable deflection equal to the span of the beam divided by 240\small 240.

Solving for the maximum allowable deflection, δmax\small\delta_\text{max}, of the beam in our example, we have:

δmax=L240=96 in240=0.4 in\small \begin{align*} \delta_\text{max} &= \frac{L}{240}\\[1.0em] &= \frac{96\ \text{in}}{240}\\[1.0em] &= 0.4\ \text{in} \end{align*}

Now that we're done with our wood beam deflection calculation, we compare the two values we got. Since our beam's deflection due to loading is less than the maximum allowable, we can say that our beam passed the deflection check, and we can now proceed to the bending and shear stress calculations. ✔

🙋 For other purposes, we can also divide our beam span by 360\small 360, 480\small 480, 600\small 600, or 720\small 720 to consider smaller allowable deflections. Choosing a larger value is perfect if you're unsure if your beam will have to support more load in the future.

If the beam's deflection due to loading is greater than the maximum allowable deflection, the beam will fail. For such cases, we should choose either a different wood species, commercial grade, or a larger beam size, then perform the wood beam deflection calculation again.

✅ You can visit our beam deflection calculator if you want to learn more about beam deflection. 🙂

Checking the adjusted and allowable bending stress of a wood beam

Now that we've checked the deflection of our beam, we can now check for its bending stress. For that, we first need to calculate the required or actual moment, MM, our beam experiences due to the loading applied (in this case, a uniformly distributed linear loading), as shown below:

M=w×L28=20 lbin×(96 in)28=20 lbin×9216 in28=23,040 lbfin\small \begin{align*} M &= \frac{w\times L^2}{8}\\[1.0em] &= \frac{20\ \tfrac{\text{lb}}{\text{in}} \times (96\ \text{in})^2}{8}\\[1.0em] &= \frac{20\ \tfrac{\text{lb}}{\text{in}} \times 9216\ \text{in}^2}{8}\\[1.0em] &= 23,\hspace{-0.04cm}040\ \text{lbf}\cdot\text{in} \end{align*}

Next, we can determine the actual bending stress due to this bending moment with this equation:

fb=MS,\small f_\text{b} = \frac{M}{S},

where:

  • fbf_\text{b}Required or actual bending stress in pounds per square inch (psi\small\text{psi} or lb/in2\small\text{lb}/\text{in}^2);
  • MMBending moment due to loading in pound-force inches (lbfin\small\text{lbf}\cdot\text{in}); and
  • SSSection modulus of the beam in inches cubed (in3\small \text{in}^3).

We can calculate the section modulus of our beam using this equation:

S=b×d26=1.5 in×(9.5 in)26=1.5 in×90.25 in26=22.5625 in322.563 in3\small \begin{align*} S &= \frac{b\times d^2}{6}\\[1.0em] &= \frac{1.5\ \text{in}\times (9.5\ \text{in})^2}{6}\\[1.0em] &= \frac{1.5\ \text{in}\times 90.25\ \text{in}^2}{6}\\[1.0em] &= 22.5625\ \text{in}^3\\[0.5em] &\approx 22.563\ \text{in}^3 \end{align*}

🔎 Check out our section modulus calculator if you want to learn how to calculate the section modulus of cross-sections other than rectangular cross-sections.

Solving for the actual bending stress, we have:

fb=MS=23,040 lbfin22.5625 in3=1,021.163435 lbin21,021.2 psi\small \begin{align*} f_\text{b} &= \frac{M}{S}\\[1.0em] &= \frac{23,\hspace{-0.04cm}040\ \text{lbf}\cdot\text{in}}{22.5625\ \text{in}^3}\\[1.0em] &= 1,\hspace{-0.04cm}021.163435\ \tfrac{\text{lb}}{\text{in}^2}\\[0.5em] &\approx 1,\hspace{-0.04cm}021.2\ \text{psi} \end{align*}

Like in the deflection check, we must compare the actual bending stress with the beam's (adjusted) allowable bending stress design value. Looking at NDS Supplement Table 4A, we can see that the bending stress design value (Fb)(F_\text{b}) of select structural grade Douglas Fir Larch is 1,500 psi\small 1,\hspace{-0.04cm}500\ \text{psi}. However, we need to adjust that value to consider the different factors that can affect the bending stress capacity of any wood beams. To determine the adjusted allowable bending stress in wood beams, which we denote as FbF_\text{b}', we must multiply (FbF_\text{b}) by the following factors:

  • CDC_\text{D}Duration factor;
  • CMC_\text{M}Wet service factor;
  • CtC_\text{t}Temperature factor;
  • CLC_\text{L}Beam stability factor;
  • CFC_\text{F}Size factor;
  • CfuC_\text{fu}Flat use factor;
  • CiC_\text{i}Incising factor; and
  • CrC_\text{r}Repetitive member factor.

We won't anymore discuss each one of these factors. Some of these factors have values depending on which adjusted design values we want (e.g., Ci=0.80\small C_\text{i}=0.80 for Fb\small F_\text{b} and Fv\small F_\text{v}, but Ci=0.95\small C_\text{i}=0.95 for Emin\small E_\text{min}). Same with for CMC_\text{M}:

* If Fb×CF ≤ 1,150 psi, we use CM = 1.0

Design value

CM

Fb

0.85*

Fv

0.97

E and Emin

0.9

Some of them require separate calculations before we can use them. In contrast, some have values in a tabulated format where we choose the multiplier constant that fits our beam. For example for CDC_\text{D}:

Load duration

CD

Typical design loads

Permanent

0.90

Dead load

Ten years

1.00

Occupancy live load

Two months

1.15

Snow load

Seven days

1.25

Construction load

Ten minutes

1.60

Wind/earthquake load

Impact

2.00

Impact load

Please refer to the NDS for Wood Construction for the other adjustment factors.

For our example, let's say upon following the NDS, our CtotalC_\text{total} or the product of all the factors we need to multiply is 0.711\small 0.711. We can now calculate our beam's adjusted allowable bending stress design value to be:

Fb=Fb×Ctotal=1,500 psi×0.7111,066.4 psi\small \begin{align*} F_\text{b}' &= F_\text{b} \times C_\text{total}\\[0.5em] &= 1,\hspace{-0.04cm}500\ \text{psi} \times 0.711\\[0.5em] &\approx 1,\hspace{-0.04cm}066.4\ \text{psi} \end{align*}

Since Fb=1,066.4 psi\small F_\text{b}' = 1,\hspace{-0.04cm}066.4\ \text{psi} is greater than fb=1,021.2 psi\small f_\text{b} = 1,\hspace{-0.04cm}021.2\ \text{psi}, our beam also passed the bending stress check. ✔

Checking the actual and allowable shear stress of a wood beam

For the last parameter we want to check in this calculator, we'll find the actual shear stress due to the beam loading and compare that to our beam's adjusted shear stress design value. First, we must calculate the required or actual shear, which we denote as VV, our beam needs to overcome using this equation:

V=w×L2\small V = \frac{w\times L}{2}

Then we determine its corresponding actual shear stress, fvf_\text{v}, on our beam by dividing the actual shear by the cross-sectional area, AA, of our beam, as shown below:

fv=VAf_\text{v} = \frac{V}{A}

We can also combine these two equations, together with A=b×dA = b\times d to directly calculate the shear stress using the shear stress formula, as shown here:

fv=w×L2×b×df_\text{v} = \frac{w\times L}{2\times b\times d}

Substituting the values we have in our example, we then have:

fv=w×L2×b×d=20 lbin×96 in2×1.5 in×9.5 in=1920 lb2×14.25 in2=67.36842105 lbin2=67.37 psi\small \begin{align*} f_\text{v} &= \frac{w\times L}{2\times b\times d}\\[1.0em] &= \frac{20\ \tfrac{\text{lb}}{\text{in}}\times 96\ \text{in}}{2\times 1.5\ \text{in}\times 9.5\ \text{in}}\\[1.0em] &= \frac{1920\ \text{lb}}{2\times 14.25\ \text{in}^2}\\[1.2em] &= 67.36842105\ \tfrac{\text{lb}}{\text{in}^2}\\[0.5em] &= 67.37\ \text{psi} \end{align*}

Like with the bending stress, we also need to compare our actual shear stress with our beam's adjusted shear stress design value, which we denote as FvF_\text{v}'. Using the same NDS Supplement Table 4A, we can find the shear stress design value or FvF_\text{v} of select structural grade Douglas Fir Larch to be equal to 180 psi\small 180\ \text{psi}. To calculate FvF_\text{v}', we have to use the following adjustment factors:

  • CDC_\text{D}Duration factor;
  • CMC_\text{M}Wet service factor;
  • CtC_\text{t}Temperature factor; and
  • CiC_\text{i}Incising factor.

Let's say we're using CD = 1.0\small C_\text{D = 1.0} (for considering 10 years duration of the structure), CM=0.97C_\text{M} = 0.97 (as shown in the table of CM values for Fv in the previous section of this text), Ct=1.0C_\text{t} = 1.0, and Ci=0.8C_\text{i} = 0.8. We now calculate the adjusted shear stress design value as follows:

Fv=Fv ⁣× ⁣CD ⁣× ⁣CM ⁣× ⁣Ct ⁣× ⁣Ci=180 psi ⁣× ⁣1.0 ⁣× ⁣0.97 ⁣× ⁣1.0 ⁣× ⁣0.8=139.68 psi\small \begin{align*} F_\text{v}' &= F_\text{v}\!\times\! C_\text{D}\!\times\! C_\text{M}\!\times\! C_\text{t}\!\times\! C_\text{i}\\[0.5em] &= 180\ \text{psi}\!\times\! 1.0 \!\times\! 0.97 \!\times\! 1.0 \!\times\! 0.8\\[0.5em] &= 139.68\ \text{psi} \end{align*}

Fortunately, our beam's actual shear stress is less than the adjusted shear stress design value. ✔

If, again, our beam did not pass this test, we have to repeat the calculation using a larger size beam or a stiffer wood species and grade. Imagine how tiresome that would be to do it manually. This is where our wood beam span calculator comes in super handy. Let's learn how to use this wood beam calculator in the next section of this text.