Practical Design and the implications of the Blizzard of ‘96 BSE 2294 Animal Structures and Environment Dr. Susan Wood Gay & S. Christian Mariger Ph.D.

Practical sizing techniques for structural components. We have looked at some of the engineering approaches for analyzing stresses in wood structural components. In practice span tables are used to determine the size and quality required for wood structural components. Span tables are produced by calculating the maximum span (L) for a given dimension and grade of lumber with a given distributed load (W). When designing the structure the builder will simply use the table to determine dimension and grade required instead of calculating the actual stress and deflection.

Typical Span Table (Floor Joists) Table 1 Southern Pine Floor Joists (Maximum spans given in feet and inches inside to inside of bearings) Design Criteria: Deflection limited to span in inches divided by 360 (live load only) Strength based on 30, 40 or 50 psf live load plus 10 psf dead load. GradeLive L o a d Size (inches) and Spacing (inches on center) 2 x 62 x 82 x 102 x 12 12”oc16”oc24”oc12”oc16”oc24”oc12”oc16”oc24”oc12”oc16”oc24”oc No. 130 psf psf psf No. 230 psf psf psf No. 330 psf psf psf Note only the better grades of lumber No. 1 – No. 3 are included in the table!

Floor Joist Example: Determine the dimension and grade required for a floor joist spaced 24” on center, supporting a calculated live load of 47 lbs/ft 2 and spanning 14’- 0” between bearings.

Floor Joist Example: Determine the minimum dimension and grade required for a floor joist spaced 24” on center, supporting a calculated live load of 47 lbs/ft 2 and spanning 14’- 0” between bearings. –Look at the span table for floor joists, 47 psf is greater than 40 psf so look at the rows for 50 psf.

Floor Joist Example: Determine the minimum dimension and grade required for a floor joist spaced 24” on center, supporting a calculated live load of 47 lbs/ft 2 and spanning 14’- 0” between bearings. –Look at the span table for floor joists, 47 psf is greater than 40 psf so look at the rows for 50 psf. –Look at the columns for 24” on center. Work your way across the table from smallest to largest and up the table from lowest to highest grade.

Floor Joist Example: Determine the minimum dimension and grade required for a floor joist spaced 24” on center, supporting a calculated live load of 47 lbs/ft 2 and spanning 14’- 0” between bearings. –Look at the span table for floor joists, 47 psf is greater than 40 psf so look at the rows for 50 psf. –Look at the columns for 24” on center. Work your way across the table from smallest to largest and up the table from lowest to highest grade. –What is the smallest/lowest grade joist that can be used?

Answer A No. 2 Southern Pine 2” x 12” is the smallest lowest grade (most economical) joist that should be used to span 14’ with a live load of 47 lbs/ft 2.

Southern Pine Rafter Table Table 11 Rafters – No Finished Ceiling – Snow Load (C D = 1.15) 1 (Maximum spans given in feet and inches inside to inside of bearings) Design Criteria: Deflection limited to span in inches divided by 180 (live load only) Strength based on 30 or 40 psf live load plus 10 psf dead load. Grad e Live Load Size (inches) and Spacing (inches on center) 2 x 42 x 62 x 82 x 10 12”oc16”oc24”oc12”oc16”oc24”oc12”oc16”oc24”oc12”oc16”oc24”oc No. 130 psf psf No. 230 psf psf No. 330 psf psf (1) C D = duration of load factor see table A-3 for additional information on adjustment factors.

Rafter Example Given a closed 4/12 pitch gable roof heated high risk 20’ wide x 30’ long x 16’ wall height building in a windy unsheltered area near Richmond. The dead load for the roofing material is 9lbs – 8oz/ft 2. Determine the minimum size and grade lumber for common rafters spaced 24” on center for the building.

Rafter Example (Live/Environmental Loads) –P s = R (1.0) x Ce (0.8) x Is (1.0) x Cs (0.95) x Ct (1.0) x Pg (15) = 11.4 psf –Mid Roof Elevation = Eave height (16’) + ½ Gable Height (3.33’) = 17’-8” –(q) = x K z (1.08) x V 2 (80) 2 x I w (1.00) = 17.7 psf –P w = q (17.7) x G (0.85) x C p (0.7) = psf –Total Live/Environmental Load = 21.9 psf

Rafter Example (Finding the Rise) To find the rise for the mid roof elevation & the rafter span find the run (1/2 x width of building) so ½ x 20’ = a run of 10’ Then plug in the run as the denominator in a fraction and cross multiply with the pitch to find the unknown rise. 4/12 = n/10 (n (rise) = 3.33’ or 3’-4”)

Rafter Example (Rafter Span) c a b The length of the hypotenuse is the span a 2 + b 2 = c 2 c = rafter span a = run of the rafter b = rise of the rafter

Rafter Example (Span of the Rafter) Plug the rise and run into the Pythagorean formula for the hypotenuse to find the span. (10) 2 + (3.33) 2 = Solve for the square root of = 10.54’ or 10’ /8”

Rafter Example Refer to the rafter table for no-finished ceiling –Check the rows for 30 psf live load –Check the columns for 24” on center spacing –Work across the table from smallest to largest size –Work up the table from lowest to highest grade

Rafter Example (Answers) No. 3 2 x 10 No. 2 2 x 8 No. 1 2 x 6 How to choose (what is most important to you?) –Price –Ease of handling (weight)

Fastener Facts & Figures

Common Nail Lateral Loads Lateral Load (Pn) = KD 3/2 Pn =safe load in pounds per nail (assuming that the point penetrates ½ of it’s length into the second member for hardwoods and 2/3 of it’s length into the second member for softwoods) K =a constant depending on the type of wood D = diameter of the nail in inches

Common Nail Lateral Loads Number/WeightCalculated (Pn) lbs/nail Table Value (Pn) lbs/nail 6d d d d d d d d d

Common Nail Lateral Load Example Given a nailed joint between an S4S Southern Pine 2” x 8” and a 6” x 6” post; the joint contains 8 evenly spaced 20d common nails. –Does this joint meet the assumptions for P n ? –What is the maximum lateral load for the joint?

Common Nail Lateral Load Example Given a nailed joint between an S4S Southern Pine 2” x 8” and a 6” x 6” post; the joint contains 8 evenly spaced 20d common nails. –Does this joint meet the assumptions for P n ? A 20d common nail is 4” in length so it will penetrate the 1-1/2” 2” x 8” and about 2/3 of it’s length will be in the second member the 6” x 6” –What is the maximum lateral load for the joint? The max lateral load will be between 925 lbs and 1,112 lbs depending on the value you choose.

Common Nail Withdrawal Loads Withdrawal Load (P) = 1150 G 5/2 D P = load-pounds per inch of penetration (into the second member!) G = specific gravity of the wood D =diameter of the nail

Common Nail Withdrawal Loads NumberDiameter (in)Length (in)P lbs/in of penetration 6d d / d d / d / d d / d d / d

The Blizzard of 1996 affected much of the eastern US. January 6 – 8, 1996 January 6 – “Explosion” of moisture on satellite photos January 7 – Storm reached the Blue Ridge Mountains January 8 – Storm tapered off in NYC area in early morning Snowfall map of the Blizzard of January 6 – 8, 1996.

The Blizzard of ’96 snowfall totals for the Mid-Atlantic States.

Numerous agricultural structures collapsed due to the use of “reduced” or inadequate structural design loads. Barns Machinery sheds Poultry houses Snowfall map of the Blizzard of January 6 – 8, 1996.