How to Understand Wood Properties Used in Design

Working in the truss and EWP industry, you see a lot of terms and numbers that deal with the design properties of wood such and moment, shear, Fb and Modulus of Elasticity (E) to name a few.

With the design software that is available, you are able to plug-in the parameters for the design and hit run.  The program will tell you if the design works.  All the analysis is done for you and you move on to the next task.  But this present two problems:

What exactly do some of the terms mean and how do they relate to the design of a truss or beam?

And do we know how to factor these ourselves?

E – Modulus of Elasticity

As the name suggests, the modulus of elasticity is defined as a number that measures an object or substance’s resistance to being deformed elastically (i.e., non-permanently) when a force is applied to it.

To determine this number you use Hooke’s Law, which is defined as:

hookes-law

The higher the “E” value, the stiffer the material.

Example:

Hem-fir #2 has an E value of 1,300,000 psi

Hem Fir Select Structural has an E value of 1,600,000 psi.

F– Extreme Fiber Bending

As wood bends the outermost (extreme) fibers are compressed along the top edge of the piece and stretched along the bottom edge.  Fb indicates the design strength for these fibers.

fiber-bending

The higher the Fb, the stronger the wood.

Hem-fir # 2 has Fb of 850 psi

Hem-fir Select Structural has Fb of 1,400 psi

Fv – Allowable Shear Parallel to Grain

Shear forces result when wood fibers want to move past each other.  This is demonstrated in the illustration below.  The planks would represent wood fibers in a member.

shear-parallel-to-grain

The shear value will be the same for different grades of the same species.

Fc perp – Compression Perpendicular to Grain

This is an important value that will determine if the bearing width is sufficient for the reaction at the end of the member. Wood will begin to crush if the load is not distributed across a large enough area.  Note that the values for compression parallel to the grain are significantly higher.

Compression-Perpendicular-to-Grain

Hem-fir # 2 has Fc perp of 405 psi

Hem-fir #2 has Fc parallel to grain of 1,400 psi.  Big difference.

This can come in handy when designing trusses that have an end vertical.  You can run the end vertical down to the bearing and use the higher parallel to grain value to help with an overstressed bearing.

 

Beam Design Example:

So how do these values play into the design of a wood member?  Let’s look at an example of a simply supported beam with a uniform load.

lvl-beam

For this example will assume a 50 plf load (w) across the beam.  We’ll look at a single ply SYP 2×10 #1 x 10’-0” (L) beam and see if it works.

What we know about a SYP 2×10 #1:

2x10

E = 1,600,000 psi

Fb = 1,050 psi

Fv = 175 psi

Fc perp = 565 psi

Section Modulus – S  = 21.39 in3   (bd2 /6)

Moment of Inertia – I = 98.93 in (bd3 /12)

The values above can be found in the NDS Wood Design Spec

We will use L/360 for the max. deflection criteria.

Assume a 3.5” bearing.

First, figure the allowable Moment, Shear, Deflection and bearing width:

Mallow = Fb*S = (1,050*21.39)/12in. = 1,871.63 ft-#

Vallow   =( Fv*Area)/1.5 = (175*1.5*9.25)/1.5 = 1,618.75 #

Deflallow = L / 360 = 120 in. /360 = 0.334 in.

Bearing width = Fc perp * bearing area = 565*(1.5×3.5) = 2,966.25 #

Second, figure the max. Moment, Shear and Deflection in the beam to see if beam is acceptable:

Mmax = wL2 / 8 = (50(102 ))/8 = 625 ft-#           < 1,871.63   OK

Vmax  = wL/2 =( 50(10))/2 = 250 #  < 1,618.75    OK

Defl. = (5wL4 ) / (384*EI) = .071 in.   <  0.334   OK

Reaction = wL/2 = 250#    <  2,966.25   OK

So we can see that the member works.  This is a very simple example and there are a number of additional factors that can be used when designing a member such as load duration, size, wet service, and repetitive member factors that can be applied in many cases.  These were not used for this example.

This is the type of analysis that goes on behind the scenes when you hit the “design” button.  The equations used are only for a simply supported beam with a uniform load.  Things get much more involved when you have start adding

  • Partial uniform loads
  • Point loads
  • Drift loads
  • Trapezoidal loads
  • Wind loads
  • Etc., etc.

I hope this information is useful and gives a better understanding of what goes on behind the scenes when designing wood members. Was this information helpful to you? Look forward to hearing from you.

Bill Hoover – Design Professional

Gould Design, Inc.