Dome at Burning Man 2005

Strengths of Materials (Part 2)
May 21, 2005


Contents
Summary
Introduction
Tension in 2x4s
Compression in 2x4s
Bending of 2x4s
Everything Together


Summary

The 2x4s are the dome's main support, and so here's an entire (and pretty long) page devoted to them. Perhaps I should summarize the results here:

  1. The 2x4s are sufficiently strong to support the dome in a 60 MPH wind.
  2. Holes in the 2x4s must be set back 1 to 2 inches from the end of the wood member (the exact amount depends on the width of the 2x4 after its end is angled).

Still coming along? Well, buckle up!

You sure? Last chance . . . here we go.

Introduction

Each of the dome's 2x4 members is in compression or in tension, depending on the loads that the structure faces. We can analyze both of these conditions. All members need to resist some amount of bending. This condition can be analyzed, too.

Remember that in allowable-stress design, actual stresses are compared to allowable stresses to determine whether a design is safe.

Tension in 2x4s

2x4s in tension are described by the allowable tensile stress, FT. The allowable tensile strength of a select structural hemlock fir 2x4 is 925 psi. Just how much tensile load can such a 2x4 withstand?

Area = 1.5 * 3.5 = 5.25 in2

Pallowable = FT Area = 925 Area

Pallowable = 4856 pounds

So the tensile strength of a 2x4 is outrageously strong. It's the bolt hole that is the weakest part. Tearing through a bolt hole in a 2x4 requires shearing two planes, one on each side of the hole. Resistance to this shearing is quantified by the allowable shear stress, FV, which for all grades of hemlock fir is 75 psi.

Remember that in Part 1 the shear load of a bolt was calculated to be 437 pounds. It makes no sense to require the bolt hole in the 2x4 to be stronger than the bolt, and so below I calculate the backset of the hole in the 2x4 for a tensile load of 437 pounds.

P = 437 pounds

FV = 75 psi

Area = P / FV = 437 / 75 Area = 5.83 in2

Since tearing through a hole requires failure of the shear planes on both sides of the hole, the shear area of the hole backset is just half of the value calculated above.

Shear-Plane Area = 2.91 in2

For a long bolt going through the entire 3.5" height of a 2x4 (5-plate attachment), the backset can be calculated from the shear-plane area.

H = height = 3.5

backset = Shear-Plane Area / H = 2.91 / 3.5

backset = 0.84 in

(Note: I'm assuming that both ends of the bolt are supported, which is not true with the plate attachment used here. So I really should increase the backset to compensate.)

For a bolt going through the end of 2x4 that is cut at an angle (3-plate attachment), the backset is a function of the height remaining after the cut. For a conservative measurement, I'll use the thickness of the narrowest part of the cut, ignoring the thin wedge of material that results from the cut angle. 

  height of cut 2x4   required hole backset
         3.5 in               0.84 in
         3.0                  0.97
         2.5                  1.67
         2.0                  1.46
         1.5                  1.94

The result is that the hole in the end of each 2x4 should be backset by 1 to 2 inches in order for the 2x4 to be as strong as the bolt that attaches it to the joint plate.

Compression in 2x4s

A 2x4 in compression is considered a column, which is a configuration that we can analyze.

Allowable compressive stress is calculated below.

Fc = 405 psi
CD = 1.15
CF = 1.15
Fc* = CD CF Fc
Fc* = 536 psi

The dimensions of an 8-foot 2x4 are

d1 = 1.5 in
d2 = 3.5 in
L = 96 in

Assume that the attachment holes in the 2x4s are 2 inches back from the ends. Then the apparent length of the column, between these supports, is 4 inches shorter.

L' = 92 in

The elastic modulus of select structural hemlock fir 2x4s is 1.6*106.

The allowable compressive stress, FcE, is

FcE = 0.3 E / (L' / d1)2

FcE = 0.3 * 1.6*106 / (92 / 1.5)2

FcE = 128 psi

Then the allowable compressive load is calculated here.

ratio = FcE / Fc* = 128 / 536 = 0.24

Cp = really hairy equation = 0.23

Fc' = Cp Fc* = 0.23 * 128 = 121 psi

P = Fc' d1 d2 = 121 * 1.5 * 3.5

P = 633 pounds

Yeah! So the 2x4 is strong enough in compression.

A final note on the column analysis. An 8-foot 2x4 violates the "slenderness ratio" for columns, which recommends an L/d ratio of 50 maximum. The L/d of a 92-inch 2x4 is 61. I'm just going to ignore this for now. A possible solution is to brace the 2x4s, if I can convince myself that it's really necessary. But given that the capacity of the column exceeds the requirements by almost 45%, I don't think I'm going to worry too much.

Bending in 2x4s

The bending analysis is incredibly complicated, so I'm just going to summarize the calculations here.

  1. Calculate the total wind load on a member. I considered one of the 8' members that is facing the wind. It has the largest wind load P = 143 pounds.
  2. Maximum shear is half of 143 pounds: Vmax = 71.5 pounds.
  3. Calculating the maximum moment is more complicated because the load is distributed non-uniformly (the shade fabric is in triangular panels, and so the load at the member ends is zero, but the load in the middle is large). After doing lots of math, Mmax = 2288 inch-pounds.
  4. Allowable bending stress is calculated by multipling the base allowable bending stress Fb = 1400 psi by two bending adjustment factors. One factor, CD = 1.15, compensates for the short load duration (less than two months). The other factor, CF = 1.5, adjusts the base bending stress for the size of a 2x4. Finally, after multiplying everything together, Fb' = 2415 psi.
  5. The required section modulus is calculated at the point of maximum stress. It is the ratio of the maximum moment and the allowable bending stress. Sreq = 0.95 in3.
  6. The section modulus of a 2x4 is 3.06 in3, which is three times the required value, and so the 2x4 will withstand the wind load.
 Everything Together

All loading on a 2x4 must be combined simultaneously. The worst case is a 2x4 that is in compression and is withstanding a wind load. There is an equation for this. It specifies that the total stress in a member must be less than total allowable stress.

The equation, below, is a real beauty. 

         fc2         fb
         --- + --------------  <=  1.0
         Fc'2            fc
               Fb' ( 1 - --- )
                        FcE

I calculated all of its values: 

         fc  =    71 psi
         Fc' =   121 psi
         fb  =   702 psi
         Fb' = 2,415 psi
         FcE =   128 psi

I obtained the result that even the longest 2x4 will withstand the wind load. The analysis assumes that holes in the 70% shade cloth decrease the wind load by 6% (the amount necessary to meet the equation's condition).

I assumed that either end of the 2x4s could rotate, since only a single bolt holds each end, but if two bolts attach each end, or if the joint plates otherwise prevent 2x4 rotation, then in the analysis I get to decrease the 2x4 length by 35% (or increase the member thickness). This observation suggests that joint plates that prevent rotation allow narrower members while maintaining the same strength. It even helps put the slenderness ratio into the right region.

Conclusion

Using 2x4s for this structure will be okay in a 60-MPH wind.

Wow! You made it to the end. Congratulations!

-- Kerry
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