Calculate the Maximum Rated Load
What is the maximum rated load? 6
6
“
“
2 X 10 X 8 (L.D.) Planks ”
”
’
5 -0 ’
”
Top View of scaffold
What is the maximum rated load? 6
6
“
“
2 X 10 X 8 (L.D.) Planks ”
”
’
5 -0 ’
”
Top View of scaffold
What is the maximum rated load? 6
“
7 0 ’
6
“
7 X 5 (area)= 35 Square
”
’
’
Feet 2 X 12 X 8 (L.D.) Planks ”
”
’
35 X 25 (L.D.) = 875 5 -0 ’
”
Light Duty (L.D.) = 25#/sq.ft Medium Duty (M.D.) = 50#/sq.ft Heavy Duty (H.D.) = 75#/sq.ft.
pounds
Maximum Capacity
What is the maximum intended load? •
1-Bands of bricks = 400 #s each
400
•
Mortar = 150#s
150
•
Pan (mortar) = 100#s
100
•
1 worker = 250#s each
250 max. intended load
Will the light duty planks work? We calculated 875 # rate load for L.D.
900#s
What is the maximum intended load? 6
“
7 0 ’
6
“
7 X 5 (area)= 35 Square
”
’
’
Feet 2 X 12 X 8 (M.D.) Planks ”
”
35 X 50 = 1,750 pounds
’
5 -0 ’
Maximum Capacity
”
Will this work? Light Duty (L.D.) = 25#/sq.ft Medium Duty (M.D.) = 50#/sq.ft Heavy Duty (H.D.) = 75#/sq.ft.
900 pounds Intended Load
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What is the maximum intended load? 6
“
7 0 ’
6
“
7 X 5 (area)= 35 Square
”
’
’
Feet 2 X 12 X 8 (H.D.) Planks ”
”
’
35 X 75 = 2625 pounds 5 -0 ’
”
Light Duty (L.D.) = 25#/sq.ft Medium Duty (M.D.) = 50#/sq.ft Heavy Duty (H.D.) = 75#/sq.ft.
Maximum Capacity
1,800 pounds Intended Load
What is the weight per leg? We have a total of 900 #s.
900
4=
225 #s /leg (post)
What is the weight on the base plate? Each leg has a 6 X 6 base plate. ”
6 X 6 = ”
”
”
36 sq. inches 225 36 = 6.25 #s per sq. inches
What is the weight on the mud sill? Each base plate is on a 2 X 10 x 12 mud sill. ”
”
”
10 X 12 = ”
”
120 sq. inches 225
120 =
1.875 #s / sq. inch.
How important is it to have a base plate and a mud sill? From 900 #s intended load
To: 1.875 #s / sq. inch
Mud Sill • Typical mud sill: – 2 X 10 pad between 12 and 18 ”
”
”
”
This will be adequate for most scaffolds four levels or less in height and not heavily loaded
Mud Sill • Multiple levels (5 levels and up) may require a different size pad. • Calculate the weight imposed by the scaffold leg on the sill (leg load). • Divide that number by the square footage of the sill to determine the PSF imposed on soil. • Compressive force should be limited to 1000 PSF on soil (type C)
Mud Sill • As a conservative guideline base on a maximum 3,000 lb leg load: – 2 X10 X18 is adequate for Type A soil ”
”
”
– 18 square pad for Type B soil ”
– 3 X3 square pad for Type C soil ’
’
Don t use unstable objects, lose bricks, etc., as a sill. ’
Calculate a Board Foot
Calculate the board foot (BF)
(TXWXL) BF =
12
T = Thickness W = Width L = Length
1 X 12 X 12 = 144 cubic inches ”
”
”
A board foot is equal to 144 cubic inches of wood
Calculate the board foot (BF) For example: An 8-foot 2 X10 (nominal board )would contain ___ board feet. ”
”
(TXWXL) BF =
12
!1.5
– Thickness !9.25 = Width ”
(1.5 X 9.25 X 8 / 12)= 9.25 board feet
Nominal 2 X 10
Southern Pine Design Values
Table 4 Scaffold Plank1 2" and 3" thick, 8" and wider
Size
Grade
2" thick, 8" and wider MC!19%2
Dense Industrial 72 Scaffold Plank Dense Industrial 65 Scaffold Plank
3" thick, 8" and wider MC>19%
Dense Industrial 72 Scaffold Plank Dense Industrial 65 Scaffold Plank
Extreme Fiber Stress in Bending Fb Flat wise Use Only
Modulus of Elasticity E
3.7#s/b.f. (TOP GRADE)
2400 2200
1,800,000 1,800,000
2.7#s/b.f.
1800 1650
1,600,000 1,600,000
(1) Scaffold plank design values are for flat wise use only. They were calculated using ASTM D245 and D2555 standards and modified using procedures shown in "Calculating Apparent Reliability of Wood Scaffold Planks," as published by the Journal on Structural Safety, 2 (1984) 47-57, and updated in 1993. (2) For exposed conditions of use (where the moisture content in service may exceed 19%) the values shall be multiplied by: 0.85 for "F b" and 0.90 for "E".
Indian Mill Corporation • Pin-Lok 2.0E Scaffold Plank –Fb – 2900 psi • 3.95 #s / bf
Calculate the board foot (BF) For example: An 8-foot 2 X10 (nominal board )would contain ___ board feet. ”
”
(TXWXL) BF =
144
!1.5
– Thickness !9.25 = Width ”
(1.5 X 9.25 X 96 ) / 144 = 9.25 board feet
9.25 bf X 2.7 = 24.975 or 25 lbs.
PIN-LOK Scaffold Plank (1.5 X 9.25 X 120 ) / 144 = 11.56 board feet
11.56 bf X 3.9 = 45 lbs.
Nominal 2 X 10
Tipping Forces on Towers
Tipping Forces on Towers F = (W x Rm) / H F
F (Tipping Forces) H (Height) Rm (Resisting Moment) = " the width W (Weight of Tower) = 1,225 lbs 26 – 0 ’
”
Rm
F = (1,225 lbs. x 2.5 ft.) / 26 ft. F = 3,062.5 / 26 ft.. F = 117.7 lbs
5 -0 ’
”
Calculate Tipping Force
How much force must be exert to tip scaffold?
A 6 ft, 180 lb man is pushing an overhead beam while standing on a 450 lb wheeled scaffold which is 6 ft. wide and 18 ft. high. The coefficient of friction for the scaffold with the wheels locked is 0.88.
F1 d1 = F2 d2
How much force must be exert to tip scaffold?
A 6 ft, 180 lb man is pushing an overhead beam while standing on a 450 lb wheeled scaffold which is 6 180+450 ft. wide and 18 ft. high. The coefficient of friction for the scaffold with the wheels locked is F1 d1 0.88. (180+450) (3) F2 = 18 + 6
F2 d2
18 +6
F1 d 1 = F 2 d 2 F2 = F1 d1 / d2 F2 = (180+450)(3) / 18 + 6 F2 = 1890 / 24 F2 = 78.75
6 ft. 6/2
Calculation for single point suspension
Calculation for single point suspension CA 3BW
A
C
B
W
Calculation for single point suspension
C = counter weight
A C
B
A = distance from C to fulcrum ‘
W
B = distance from fulcrum to edge W = weight
’
Calculation for single point suspension B
A
W
C Roof
Calculation for single point suspension Case scenario # 1: W = 750#s
B
A = 10
A C
B=4 W
’
’
900 C = ____ C 10 ! 3 " 4 " 750 C 10/10 ! 3 " 4 " 750 10
CA 3BW
C ! 9000/10 C = 900
Calculation for single point suspension Case scenario # 2: W = 750#s
B
A = 15
A C
B=5 W
’
’
750 C = ____ C 15 ! 3 " 5 " 750 C 15/15 ! 3 " 5 " 750 15
CA 3BW
C ! 11,250/15 C = 750
Calculation for single point suspension Case scenario # 3: W = 1,250#s
B
A = 15
A C
B=3 W
’
’
750 C = ____ C 15 ! 3 " 3 " 1,250 C 15/15 ! 3 " 3 " 1,250 15
CA 3BW
C ! 11,250/15 C = 750
