Technical Specifications
Weyerhaeuser Edge
™
and Edge Gold
™
Specier’s Guide OSB-4000
|
February 2018
5
CALCULATING UNIFORM LOADS
The equations above are based on one-way “beam” action. They are provided to help develop allowable uniform loads based on moment, shear, and deflection as applied
to one-, two-, and three-span conditions. Loads derived from the equations provided are assumed to be applied over full-size panels in normal sheathing applications. The
following definitions apply:
Δ
......deflection (in.)
EI ......design bending stiffness capacity (lb-in.
2
/ft)
F
b
S ......design moment capacity (lb-in./ft)
F
s
(Ib/Q) ...design shear capacity (lb/ft)
L ......span (in.)
L
M
......span, center-to-center of supports, used for moment calculation (in.)
L
V
......clear span, used for shear calculation (in.)
L
Δ
......clear span plus SW, used for deflection calculations (in.)
R ...... denominator of chosen deflection limit. Example: deflection limit =
L/360 then R = 360
Find the maximum allowable uniform load (psf) for 24" o.c. span-rated flooring
over 16" on-center joists.
Assumptions
■
24" o.c. span-rated flooring
– Full 4'x8' panel
– Strength axis perpendicular to joists
– Use 3-span equations
■
Joist Spacing = 16" o.c.
■
Joist Width = 1.5"
■
Deflection = L/360
Locate panel design values for moment, shear, and stiffness on page 3.
Moment capacity (primary) = F
b
S = 770 lb-in./ft of width
Shear capacity (in-the-plane) = F
s
(Ib/Q) = 250 lb/ft of width
Stiffness = EI = 300,000 lb-in.
2
/ft of width
1 Calculate Allowable Uniform Load Based on Moment Capacity
W
M =
120F
b
S/L
2
M
Calculate appropriate span for moment (center-to-center), L
M
= 16"
Using: F
b
S = 770 lb-in./ft and L
M
= 16"
W
M
= 120 x 770/16
2
W
M
= 361 psf
2 Calculate Allowable Uniform Load Based on Shear Capacity
W
V = 20F
S
(
I
b/Q)/L
v
Calculate appropriate span for shear (clear span), L
V
= 16"-1.5" = 14.5"
Using: F
S
(Ib/Q) = 250 lb and L
V
= 14.5"
W
V
= 20 x 250/14.5
W
V
= 345 psf
3 Calculate Allowable Uniform Load Based on Deflection
W
Δ
= L
M
1743 EI/L
4
Δ
R
SW = 0.25 (from above)
Calculate appropriate span for deflection (clear span + SW),
L
Δ
= 14.5" + 0.25" = 14.75"
Using: L
M
= 16", R = 360, and
EI = 300,000 lb-in
2
/ft
W
Δ
= (16 x 1743 x 300,000)/(14.75
4
x 360)
W
Δ
= 491 psf
4 Compare Calculated Allowable Uniform Loads
Calculated allowable uniform loads based on strength:
W
M
= 361 psf
W
V
= 345 psf
W
V
controls
Calculated allowable uniform load based on deflection:
W
Δ
= 491 psf
Example Problem
L
W
W
LL
W
L
L
L
One-Span Equations
Uniform load equations based on:
Uniform load equations based on:
Uniform load equations based on:
Three-Span Equations
Two-Span Equations
Moment
Capacity
Shear
Capacity
Deflection
WM=
96F
b
S
L
2
M
WV =
24F
s
(Ib/Q)
L
V
WΔ=
L
M
921.6 EI
L
4
Δ
R
Moment
Capacity
Shear
Capacity
Deflection
WM=
120F
b
S
L
2
M
WV =
20F
s
(Ib/Q)
L
V
WΔ=
L
M
1743 EI
L
4
Δ
R
Moment
Capacity
Shear
Capacity
Deflection
WM=
96F
b
S
L
2
M
WV =
19.2F
s
(Ib/Q)
L
V
WΔ=
L
M
2220 EI
L
4
Δ
R
SW ....support width factor:
– 0.25 for 2x nominal lumber
– 0.625 for 4x nominal lumber
– For additional information refer to the current Manual for Engineered
Wood Construction
W ....uniform load (psf)
W
M
....uniform load based on moment capacity (psf)
W
V
....uniform load based on shear capacity (psf)
W
Δ
....uniform load based on deflection (psf)