Calculate the Maximum Rated Load

What is the maximum rated load?

Top View of scaffold

2” X 10” X 8’ (L.D.) Planks

6 “ 6 “

5’-0”

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”

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.

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?

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   -­‐-­‐-­‐  

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

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

6.25 #s per sq. inches 225 36 =

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 =

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)

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

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

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.

Indian Mill Corporation • Pin-Lok 2.0E Scaffold Plank

– Fb – 2900 psi • 3.95 #s / bf

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

Tipping Forces on Towers

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

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 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

Calculation for single point suspension

Calculation for single point suspension

CA≥3BW

C W

A B

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

Roof C

A

B

W

Calculation for single point suspension

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

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

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

Questions