calculate rated load
TRANSCRIPT
![Page 1: Calculate Rated Load](https://reader034.vdocuments.site/reader034/viewer/2022042604/577cc7991a28aba711a170d2/html5/thumbnails/1.jpg)
Calculate the Maximum Rated Load
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What is the maximum rated load?
Top View of scaffold
2” X 10” X 8’ (L.D.) Planks
6 “ 6 “
5’-0”
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What is the maximum rated load?
7’ X 5’ (area)= 35 Square Feet
35 X 25 (L.D.) = 875 pounds
Maximum Capacity
2” X 12” X 8’ (L.D.) Planks
6 “ 6 “ 7’ 0”
Light Duty (L.D.) = 25#/sq.ft Medium Duty (M.D.) = 50#/sq.ft Heavy Duty (H.D.) = 75#/sq.ft.
5’-0”
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What is the maximum intended load?
• 1-Bands of bricks = 400 #s each • Mortar = 150#s • Pan (mortar) = 100#s • 1 worker = 250#s each
400
150
100
250
max. intended load 900#s
Will the light duty planks work?
We calculated 875 # rate load for L.D.
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What is the maximum intended load?
7’ X 5’ (area)= 35 Square Feet
35 X 50 = 1,750 pounds
Maximum Capacity
2” X 12” X 8’ (M.D.) Planks
6 “ 6 “ 7’ 0”
Light Duty (L.D.) = 25#/sq.ft Medium Duty (M.D.) = 50#/sq.ft Heavy Duty (H.D.) = 75#/sq.ft.
5’-0”
900 pounds Intended Load
Will this work?
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Allowable Spans
Maximum Intended Load
Maximum Permissible Span
Using Full Thickness Lumber
Maximum Permissible Span Using Nominal
Thickness Lumber
25 lbs./square foot 10 feet 8 feet
50 lbs./square foot 8 feet 6 feet
75 lbs./square foot 6 feet -‐-‐-‐
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What is the maximum intended load?
7’ X 5’ (area)= 35 Square Feet
35 X 75 = 2625 pounds
Maximum Capacity
2” X 12” X 8’ (H.D.) Planks
6 “ 6 “ 7’ 0”
Light Duty (L.D.) = 25#/sq.ft Medium Duty (M.D.) = 50#/sq.ft Heavy Duty (H.D.) = 75#/sq.ft.
5’-0”
1,800 pounds Intended Load
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What is the weight per leg?
We have a total of 900 #s.
900 4 = 225 #s /leg (post)
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What is the weight on the base plate?
Each leg has a 6” X 6” base plate.
6” X 6” = 36 sq. inches
6.25 #s per sq. inches 225 36 =
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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
1.875 #s / sq. inch.
225 120 =
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How important is it to have a base plate and a mud sill?
From 900 #s intended load
To: 1.875 #s / sq. inch
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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
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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)
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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.
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Calculate a Board Foot
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Calculate the board foot (BF)
1” X 12” X 12” = 144 cubic inches
BF = ( T X W X L )
12
T = Thickness W = Width L = Length
A board foot is equal to 144 cubic inches of wood
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Calculate the board foot (BF)
(1.5 X 9.25 X 8 / 12)= 9.25 board feet
For example: An 8-foot 2”X10” (nominal board )would contain ___ board feet. Ø 1.5 – Thickness Ø 9.25” = Width
Nominal 2” X 10”
BF = ( T X W X L )
12
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Table 4 Scaffold Plank1 2" and 3" thick, 8" and wider
Size Grade
Extreme Fiber Stress
in Bending Fb Flat wise Use Only
Modulus of Elasticity E
2" thick, 8" and wider
MC≤19%2
Dense Industrial 72 Scaffold Plank
Dense Industrial 65 Scaffold Plank
2400 2200
1,800,000 1,800,000
3" thick, 8" and wider
MC>19%
Dense Industrial 72 Scaffold Plank
Dense Industrial 65 Scaffold Plank
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 "Fb" and 0.90 for "E".
Southern Pine Design Values
3.7#s/b.f. (TOP GRADE)
2.7#s/b.f.
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Indian Mill Corporation • Pin-Lok 2.0E Scaffold Plank
– Fb – 2900 psi • 3.95 #s / bf
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Calculate the board foot (BF)
(1.5 X 9.25 X 96 ) / 144” = 9.25 board feet
For example: An 8-foot 2”X10” (nominal board )would contain ___ board feet.
Ø 1.5 – Thickness Ø 9.25” = Width
Nominal 2” X 10”
BF = ( T X W X L )
144
9.25 bf X 2.7 = 24.975 or 25 lbs.
11.56 bf X 3.9 = 45 lbs.
(1.5 X 9.25 X 120 ) / 144” = 11.56 board feet
PIN-LOK Scaffold Plank
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Tipping Forces on Towers
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Tipping Forces on Towers
Rm
F (Tipping Forces) H (Height) Rm (Resisting Moment) = ½ the width W (Weight of Tower) = 1,225 lbs
F = (1,225 lbs. x 2.5 ft.) / 26 ft.
F = 3,062.5 / 26 ft..
F = 117.7 lbs
26’ – 0”
F
5’-0”
F = (W x Rm) / H
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Calculate Tipping Force
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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
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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.
F2 d2
18 +6 180+450
6 ft. F1 d1
F2 = (180+450) (3)
18 + 6
6/2
F1 d1 = F2 d2
F2 = F1 d1 / d2
F2 = (180+450)(3) / 18 + 6
F2 = 1890 / 24
F2 = 78.75
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Calculation for single point suspension
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Calculation for single point suspension
CA≥3BW
C W
A B
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C = counter weight A = distance from ‘C’ to
fulcrum B = distance from
fulcrum to edge W = weight
C W
A B
Calculation for single point suspension
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Roof C
A
B
W
Calculation for single point suspension
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Case scenario # 1: W = 750#s A = 10’ B = 4’ C = ____
C
AB
W
Calculation for single point suspension
C 10 ≥ 3 ⋅ 4 ⋅ 750
C 10/10 ≥ 3 ⋅ 4 ⋅ 750 10
C ≥ 9000/10
C = 900
900
CA≥3BW
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Case scenario # 2: W = 750#s A = 15’ B = 5’ C = ____
C
AB
W
Calculation for single point suspension
C 15 ≥ 3 ⋅ 5 ⋅ 750
C 15/15 ≥ 3 ⋅ 5 ⋅ 750 15
C ≥ 11,250/15
C = 750
750
CA≥3BW
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Case scenario # 3: W = 1,250#s A = 15’ B = 3’ C = ____
C
AB
W
Calculation for single point suspension
C 15 ≥ 3 ⋅ 3 ⋅ 1,250
C 15/15 ≥ 3 ⋅ 3 ⋅ 1,250 15
C ≥ 11,250/15
C = 750
750
CA≥3BW
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Questions