Rigging and Lifting Handbook Sample

Whereas, if the load is divided into simple / symmetrical shapes, approximate measurements taken and basic calculations used i.e. load weight = volume x density (reference weight calculation in the following section), a greater degree of accuracy can be achieved.

However, even with this method, if the load weight is estimated, a suitable factor, for example 25%, should be added to compensate for the inaccuracy in the estimation.

16.2 Weight Calculation

To calculate load weight use the following formula:

Load Weight (LW) = Volume (VVOL) x Density (DDEN)

Note | The use of formulae in this handbook does not produce competence. Formulae should be used to enhance and refresh existing knowledge. Only authorised personnel with the appropriate engineering qualifications should produce, review or approve engineering calculations. To enhance efficiency, go to www.rigex.com for online calculation tools.

When using formulae:

• When two letters are together, e.g. ‘ab’ this means ‘a’ times ‘b’

• a/b means ‘a’ is divided by ‘b’• a2 means ‘a’ times ‘a’• a3 means ‘a’ times ‘a’ times by ‘a’• π = 3.142• Always write out the formula first• Ensure that all units are the same• Write calculations out line by line and keep work tidy

1 lb. 0.45 kg

2.2 lb. 1 kg

2000 lb.1

Short ton907 kg

2204 lb. 1000 kg 1 tonne

2240 lb.1

Long ton1016 kg

225 lb. 1 kN

38 People. Process. Performance.

16.3 Conversion Table

16.4 Density Table

Imperial Metric

1 inch 25.4 mm 2.54 cm 0.0254 m

1 foot 305 mm 30.5 cm 0.305 m

1 yard 914.4 mm 91.44 cm 0.9144 m

Material lb./ft³ kg/m³

Aluminium 165 2640

Brass 545 8730

Concrete 150 2400

Copper 557 8930

Lead 708 11340

Printing Paper 50 800

Stainless Steel 500 8000

Steel 490 7850

Water 62 1000

Water (sea) 64 1025

Wood (pine average) 36 580

Note | The values in the previous table are given as guidance only (1 kg/m³ = 0.0624 lb./ft³ = 0.000036127 lb./in³).

1 lb. 0.45 kg

2.2 lb. 1 kg

2000 lb.1

Short ton907 kg

2204 lb. 1000 kg 1 tonne

2240 lb.1

Long ton1016 kg

225 lb. 1 kN

www.rigex.com | [email protected] 39

Rigging and Lifting Operations Handbook

16.5 Volume Formulae

Cube

L³

Cuboid or Plate

L x W x H

Cylinder

πR² x H

L

W

L

L

L

H

H

R


Pipe

(πR² - πr²) x H

Cone

(πR²H) ÷ 3

Sphere

(4πR³) ÷ 3

R

H

H

Rr

R



17.0 Centre of Gravity

Where the CoG is located, is just as important as knowing the weight of the load.

The CoG is located where the total weight of an object is assumed to be concentrated.

For symmetrical loads the CoG will be located in the geometric centre.

For asymmetrical (not symmetrical) loads, that can be divided into symmetrical shapes, two of the simplest methods that can be used to find the CoG are to draw the load to scale and use inverse proportionality or use moments around an axis / datum line.

On the following pages:

• Example 1 uses a scale drawing and inverse proportionality • Example 2 uses moments around an axis / datum line


Example 1 - Scale Drawing and Inverse Proportionality

1. Draw the load (front elevation) to scale.

2. Determine the total weight of the load: 30 ton

3. Divide the load into symmetrical shapes and label: A and B

4. Determine the weight of each shape: A = 10 ton, B = 20 ton

5. Locate the CoG for each shape.

6. Draw a connecting line between each CoG.

7. Shape ‘A’ represents one third (1/3) of the total weight, shape ‘B’ two thirds (2/3).

8. Use these proportions inversely as distances on the connecting line to identify the location of the actual / combined CoG.

• Shape ‘A’ weight = 1/3, distance of CoG from ‘A’ = 2/3

• Shape ‘B’ weight = 2/3, distance of CoG from ‘B’ = 1/3

Note | If you have a load that can be divided into three symmetrical shapes, first determine the combined CoG for two and then use this information with the third shape.

10

20

20

10

10

10

A

B

10

20

20

10

10

10

10

20

20

10

10

10

10

20

20

10

10

10



Example 2 - Moments Around an Axis / Datum Line

For asymmetrical loads, that can be divided into symmetrical shapes, the location of a load's CoG can be determined by using moments around an axis / datum line. A moment, used in this way, is the distance a weight is from an axis / datum line, times weight:

• Moment = Distance x Weight

1. Draw the load (front elevation) to scale.

2. Determine the total weight of the load: 7 ton

3. Divide the load into symmetrical shapes and label: A, B and C

4. Determine the weight of each shape: A = 4 ton, B = 1 ton and C = 2 ton

10

10

30

20

40

10

10

A B C

10

10

30

20

40

10

10


5. Locate and indicate the CoG for each for shape.

6. Draw a datum line on the ‘Y’ axis. This is used to determine the distance along the ‘X’ axis where the combined / actual CoG will be located.

7. Determine the distance that each CoG is located from the datum line: A = 5, B = 15 and C = 25

A B C

10

10

30

20

40

10

10

A

Dat

um L

ine

B C

515

25



8. Produce the following table.

9. Enter the weight of each symmetrical shape and its distance from the ‘Y’ axis / datum line into the table. Next, multiply each to determine the moment.

10. Divide the total moment by the total weight to produce the distance from the ‘Y’ axis / datum line where the combined / actual CoG is located.

• 85 ÷ 7 = 12.1

Shape WeightDistance

from Datum

Weight x Distance (Moment)

A

B

C

Total: Total:

Shape WeightDistance

from Datum

Weight x Distance (Moment)

A 4 5 20

B 1 15 15

C 2 25 50

Total: 7 Total: 85

A

Dat

um L

ine

B C

12.1


21.0 Vertical Reaction Force

By determining the Vertical Reaction Force (VRF), before a load is lifted, personnel will ensure that the force to be applied is less then the lifting point’s capacity.

This information can then be used to determine the tension applied to lifting appliances and accessories.

21.1 Single Point Lifts

For vertical single point lifts the VRF is the same as the weight of the load, tension applied to the appliance and the accessories.

Load applied to Hoist Ring, Shackle, Link and Hoist = 200kg

Load Weight = 200kg



21.2 Two Point Lifts

To determine the VRF on each lifting point, use the following formulae:

• VRF1 = (TW x DDIST2) ÷ TD• VRF2 = (TW x DDIST1) ÷ TD

• TW: Total Weight• DDIST1: Distance between VRF1 and the CoG• DDIST2: Distance between VRF2 and the CoG• TD: Total Distance between VRF1 and VRF 2

Example

VRF1 = (TW x DDIST2) ÷ TD VRF2 = (TW x DDIST1) ÷ TD (600 x 600) ÷ 1400 (600 x 800) ÷ 1400 360,000 ÷ 1400 480,000 ÷ 1400VRF1 = 257 kg VRF1 = 343 kg

Note | Ensure that VRF1 and VRF2 add up to the Total Weight.

TD = 1400mm

Total Weight = 600 kg VRF 2VRF 1

DDIST1 = 800mm DDIST2 = 600mm


21.3 Three and Four Point Lifts

If the lifting points are all at the same height, uniformly and symmetrically distributed around the CoG then the VRF in each will be identical e.g. in a three point lift each VRF will be one third of the total weight.

If there is no symmetry, they are described as statically indeterminate and should be calculated by structural analysis. However, in some four point lift configurations if there is symmetry along at least one axis the VRFs can be manually calculated using the following formulae:

• VRF1 = VRF2 = (TW x DDIST2) ÷ 2TD• VRF3 = VRF4 = (TW x DDIST1) ÷ 2TD

VRF 1

VRF 4VRF 3

VRF 2



28.6 Wire Rope Slings

You will find that right-angled triangles are widely used when calculating sling tensions. It is therefore essential that you understand Pythagoras, Triangle Proportions and Trigonometry.

28.6.1 Pythagoras

If you know the lengths of two sides of a right-angled triangle, you can find the length of the third side, by using the following formula:

• a² + b² = c²

Note | ‘a’ and ‘b’ are usually the shortest sides, ‘c' is usually the longest.

Example

a² + b² = c²

• 3²+ 4² = c²• 9 + 16 = c²• c² = 25• c = 5

b = 4

a = 3

c = ?

c

a

+ =b b²

c²

a²

a²b² c²



28.6.2 Triangle Proportions

A simpler method of working out the length of a triangle’s sides can be used if the triangle has the correct proportions or angles.

If all the angles within a triangle are 60°, the length of each side will be the same (1 : 1 : 1).

If the angles within a triangle are 90°, 45° and 45°, the length of two sides will be the same, with the longest side 1.414 (√2) times the length of either of the two sides (1 : 1 : √2).

If the angles within a triangle are 90°, 60° and 30°, the length of the longest side will be twice the length of the shortest, with the other side 1.732 (√3) times the length of the shortest side (1 : 2 : √3).

√2 (1.414)

1

1

45˚

45˚

√3

1

2

30˚

60˚

1

1 1

60˚

60˚ 60˚


28.6.3 Trigonometry

Pythagoras and triangle proportions are used to determine the length of a triangle’s sides. Trigonometric formulae / ratios (Sine, Cosine and Tangent) are used to determine angles and the length of sides in right-angled triangles.

1. First name each side of the triangle in relation to the angle (θ).

• Hypotenuse (HHy) is the longest side• Opposite (OOpp) is the side opposite the angle• Adjacent (AAdj) is the side next to the angle

2. Next determine what information you have, what information you need and which of the following formulae / ratios to use.

• Sine θ (SSin) = Opposite (OOpp) ÷ Hypotenuse (HHy)• Cosine θ (CCos) = Adjacent (AAdj) ÷ Hypotenuse (HHy)• Tangent θ (TTan) = Opposite (OOpp) ÷ Adjacent (AAdj)

A common way to remember the formulae / ratios is to sound them out phonetically; SOH CAH TOA.

θAAdj

Adjacent

Hypotenuse

Op

po

site H HyOOpp



Example

What is the length of side X?

1. Names the sides.

2. Identify what information you have.

3. Identify what information you need?

4. Choose the formula / ratio that contains both the information you have and the information you need.

You have: the angle (Sin, Cos or Tan) Length of the OppositeYou need: Length of HypotenuseFormula: Sin θ = Opposite ÷ Hypotenuse Sin 31° = 4 ÷ X 0.515 = 4 ÷ X (transpose) X = 4 ÷ 0.515 X = 7.8

Note | Angles must be changed to a dimension ratio. Using a scientific calculator, press Sin, then 31 and then equals. Use inverse function to turn a dimension ratio back to an angle.

31˚

Adjacent

Hypotenuse

Op

po

site X

4

31˚

Adjacent

Hypotenuse

Op

po

site X

4

31˚

Adjacent

Hypotenuse

Op

po

site X

4

31˚

X4


28.6.4 Single Leg Wire Rope Slings

When wire rope is bent around, for example a pin, its capacity is reduced, and can be seriously damaged if bent around a sharp edge.

The following table shows D/d Ratios (D = diameter of sheave, pin or load, d = diameter of wire) and sling configuration efficiency, expressed as a percentage.

Note | To simplify operational use 50% (1:1), 75% (5:1) and 100% (25:1) are used to determine wire rope sling capacities. Specific values / efficiencies should be used out with these generalisations for engineered lifts by using the following formulae:

• When Ratio = 6 or less use: E = 100 – (50 ÷ √Ratio)• When Ratio = 6 or more use: E = 100 – (76 ÷ R0.73)

D/d

Rat

io 1

:1 =

50%

Effi

cien

cy

D/d

Rat

io 5

:1 =

77%

Effi

cien

cy

D/d

Rat

io 2

5:1

= 9

3% E

ffici

ency



28.6.8 Sling Tension Formulae

When lifting points and sling legs are at different heights and lengths to compensate for an offset CoG, Vertical Reaction Forces can be used with the following two formulae to determine sling tension:

• Sling Tension = (VRF x SL) ÷ SH (example 1)• Sling Tension = VRF ÷ Sin θ (example 2)

Example 1

Sling Tension 1 = (VRF1 x SL) ÷ SH (257 x 1592) ÷ 1389 409,144 ÷ 1389Sling Tension 1 = 295 kg

Sling Tension 2 = (VRF2 x SL) ÷ SH (343 x 1510) ÷ 1389 517,930 ÷ 1389Sling Tension 2 = 373 kg

VRF 2 = 343 kg

VRF 1 = 257 kg

Sling Tension 1 = 295 kg


Sling Length 2 = 1510 m

m

Sling

Heig

ht = 1389 m

mSlin

g Le

ngth

1 =

159

2 m

m



Example 2

Sling Tension 1 = VRF 1 ÷ Sin θ 257 ÷ Sin 60° 257 ÷ 0.866Sling Tension 1 = 297 kg

Sling Tension 2 = VRF 2 ÷ Sin θ 343 ÷ Sin 66° 343 ÷ 0.913Sling Tension 2 = 376 kg

VRF 2 = 343 kg

VRF 1 = 257 kg




Example 1

Cross-haul a load from Point A to Point B using two vertical rigging arrangements.

In this example, if the rigging does not exceed 45° the greatest tension i.e. the weight of the load, will be applied to the appliances and accessories at position A when the load is initially lifted, and at position B when the cross-haul is complete.

Note | Engineering input / calculation / approval is required if a vertical rigging arrangement is to exceed 45° and where more than two vertical rigging arrangements are connected to and used to lift a load.


Exa

mp

le 1

To d

eter

min

e th

e ac

tual

tens

ion

at a

ny p

oint

dur

ing

a cr

oss-

haul

usi

ng v

ertic

al r

iggi

ng, u

se th

e fo

llow

ing

form

ulae

:

Tens

ion

1 =

Lo

ad W

eig

ht x

S

L1 x

DD

IST2

(SH

1 x

DD

IST2)

+ (S

H2

x D

DIS

T1)

Tens

ion

2 =

Lo

ad W

eig

ht x

S

L2 x

DD

IST1

(SH

1 x

DD

IST2)

+ (S

H2

x D

DIS

T1)

SH2

SH1

SL1

SL2

DD

IST2

DD

IST1

Tens

ion

1

Tens

ion

2



Exa

mp

le 2

Tens

ion

1 =

Lo

ad W

eig

ht x

S

L1 x

DD

IST2

(SH

1 x

DD

IST2)

+ (S

H2

x D

DIS

T1)

Tens

ion

1 =

1 x

1

.4 x

0.7

5

(1

x 0

.75)

+ (0

.75

x 1)

Tens

ion

1 =

1 x

1.

05

0.7

5 +

0.7

5

Tens

ion

1 =

1 x

1.

05

1.5

Tens

ion

1 =

1 x

0.

7

Tens

ion

1 =

0

.7 t

onn

e

SH2 = 0.75m

SH1 = 1m

SL1 =

1.4

m

SL2 =

1.1

m

DD

IST2

= 0

.75m

DD

IST1

= 1

m

Tens

ion

1

Tens

ion

2


Tens

ion

in a

rig

ging

arr

ange

men

t can

als

o be

det

erm

ined

by

draw

ing

a sc

aled

vec

tor

diag

ram

. Thi

s is

whe

re a

rrow

s re

pres

ent t

ensi

on

(leng

th) a

nd d

irect

ion

(ang

le o

f ten

sion

).

Exa

mp

le 3

Sca

led

vect

or d

iagr

am u

sing

the

riggi

ng a

rran

gem

ent i

n ex

ampl

e 2.

1. D

raw

a v

ertic

al li

ne 1

0cm

long

, whi

ch w

ill re

pres

ent 1

to

nne

i.e. t

he w

eigh

t of t

he lo

ad2.

Fro

m th

e bo

ttom

of t

he v

ertic

al li

ne, d

raw

a li

ne a

t 45°

(th

is w

ill re

pres

ent t

he a

ngle

of t

ensi

on o

n th

e le

ft ha

nd

side

)3.

Fro

m th

e to

p of

the

vert

ical

line

, dra

w a

noth

er li

ne a

t 45

° un

til it

cro

sses

the

seco

nd li

ne (t

his

will

repr

esen

t th

e an

gle

of te

nsio

n on

the

right

han

d si

de)

4. M

easu

re th

e le

ngth

of t

he s

econ

d an

d th

ird li

nes.

Use

th

e sa

me

ratio

as

the

first

line

to d

eter

min

e te

nsio

n i.e

. 10

cm =

1 to

nne,

ther

efor

e 7

cm =

0.7

tonn

e

10cm = 1000kg

7cm

= 7

00kg

7cm

= 7

00kg



Exa

mp

le 4

Cro

ss-h

aul a

load

from

Poi

nt A

to P

oint

B u

sing

one

ver

tical

an

d on

e ho

rizon

tal r

iggi

ng a

rran

gem

ent.

In th

is e

xam

ple,

the

tens

ion

in th

e ve

rtic

al r

iggi

ng a

t 45°

w

ill be

app

roxi

mat

ely

1.5

times

the

wei

ght o

f the

load

. The

te

nsio

n in

the

leve

l hor

izon

tal r

iggi

ng w

ill be

the

sam

e as

the

wei

ght o

f the

load

.

No

te |

Eng

inee

ring

inpu

t / c

alcu

latio

n /

appr

oval

is re

quire

d if

a ve

rtic

al r

iggi

ng a

rran

gem

ent e

xcee

ds 4

5° a

nd /

or

the

horiz

onta

l rig

ging

exc

eeds

the

horiz

onta

l pla

ne.

H

ori

zont

al

Pla

ne


To d

eter

min

e th

e ac

tual

tens

ion

at a

ny p

oint

dur

ing

this

type

of c

ross

-hau

l a s

cale

d ve

ctor

dia

gram

can

be

use

d.

Exa

mp

le 5

Sca

led

vect

or d

iagr

am u

sing

the

riggi

ng a

rran

gem

ent

in e

xam

ple

4.

No

te |

Dur

ing

cros

s-ha

ulin

g op

erat

ions

, do

not s

tand

or

ope

rate

hoi

sts

in th

e po

tent

ial s

win

g pa

th o

f a lo

ad,

shou

ld th

e lif

ting

equi

pmen

t par

t.

10cm = 1 tonne

10cm

= 1

to

nne

14cm

= 1

.4 to

nne



Exa

mp

le 6

In th

is e

xam

ple

a sc

aled

vec

tor

diag

ram

illu

stra

tes

the

tens

ions

app

lied

to th

e ve

rtic

al a

nd h

oriz

onta

l rig

ging

arr

ange

men

ts a

fter

the

angl

es h

ave

been

in

crea

sed

by 1

0°.

10cm = 1 tonne

23.3

cm =

2.3

3 to

nne

19.4

cm =

1.9

4 to

nne


Rigging and Lifting Handbook Sample

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