REFERENCE EQUATIONS AND EXAMPLES

Reference Equations

1) Safe column load psi = P/A = E(0.3)/(L/d)^2 < f. Doug Fir lumber E = 1,210,000,
f = 1250. Check supporting bearing.

2) The maximum bending moment M in foot-pounds due to uniform load “w” pounds per foot is M= (1/8)wl^2, where l is the span in feet (simple support).

3) If the load W is concentrated at mid span, M =(1/4) (W l)

4) The stress in psi due to bending moment is f = 12 M / S

5) S is a “shape factor” depending on the dimensions of the beam. For a rectangular beam, S = b d^2/ 6 with b = the beam width and d = the beam depth, in inches.

2x4 S = (1.5x3.5x3.5)/6 = 3.06, M capacity = 3.06 x 1250 = 3825 in-lbs or 319 ft lbs
2x6 S = (1.5x5.5x5.5)/6 = 7.56, M capacity = 7.56 x 1250 = 9450 in-lbs or 787 ft lbs
2x8 S = (1.5x7.5x7.5)/6 = 14.1, M capacity = 14.1 x 1250 = 17578 in-lbs or 1465 ft lb

Handbooks give the value of S for non-rectangular shapes.

Examples:

1) You are remodeling a building and need to temporarily support the end of a purlin that is holding up a section of asphalt shingle roof that has three layers of shingles. You need to shore up one end of the purlin at 15 ft. height. The purlins are spaced six feet on center, and span 22 feet. What is the load on the purlin, and what is the size of post needed? Assume the total existing roof dead load is 15 psf.

Total roof load = 6 x 22 x 15 = 1980 lbs

Load on each end of the purlin = half = 990 lbs, use a double 2 x 4 post in T shape.

2) It is planned to add another layer of shingles onto the existing roof of example one. The new shingles will add 5 psf dead load. Is the existing purlin with S = 50 adequate? Assume the allowable f = 1250 psi.

Existing distributed load along the purlin = 1980/22 = 90 plf

Existing maximum bending moment = 1/8 (90 x 22 x 22) = 5445 ft-lbs

Existing bending stress = 12 x 5445/50 = 1307 psi, which is greater than allowable. The purlin appears to be a bit over-stressed under current loading. Existing shingles must be removed.

3) Scaffolding is needed along side a building, at 14 foot height. You want to use 2 each, 2x12x10’ planks, spanning 8 feet. Assume a concentrated live load at mid span of 300 lbs. per plank. Is this a safe span for these planks? What is the best 14 foot column to use? What should be used as a cross beam to support the planks?

S value per plank (actual size 1.5x11) is bdd/6 = (11x1.5x1.5)/6 = 4.125

M due to concentrated load W is (¼) 300x 8 = 600 ft-lbs = 600x12 =7200 inch-pounds

Ignoring dead load, bending stress f is M/S = 7200/4.125 = 1745 psi. This is greater than allowable, which will lead to excessive deflection and “springiness”. The planks can be stiffened by fastening a 2x4 “T” underneath along the centerline. Reducing the load or span also are options.

Each scaffold bay will be supported by four corner posts. The live load at each corner post will be ¼ of 600 pounds, or 150 pounds per single bay, or twice that for posts at double bays, i.e. 300 pounds. Notice from the column table that if a single 2x post is used, it must be braced at half-height, because L/d exceeds 50. A double 2x4 post will support this load securely.

The cross beam to support 4 planks at a double bay must carry 600 lbs live load, plus the weight of the planks themselves. At 3 pounds per FBM, each 2x12 plank weighs roughly 10x2x1x3 = 60 lbs. Therefore total load per beam is 720 lbs, assumed uniformly distributed over a span of (say) 4 feet. M = 1/8wl^2 = 1/8 (720/4)16 = 360 ft-lbs. A 2x6 is adequate. Shear at each beam end will be half 720 lbs = 360 lbs, requiring at least a four 16 penny nail connection to each post. Nail in a random pattern, with no more than one nail per growth ring. Avoid nailing close to the top edge of the beam.