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Load on Derricks

Stresses or Forces Acting on a Derrick

There are many components to the forces that act on a derrick when lifting a weight, but the four main ones are the following:

� The stress on the derrick head block (P1)� The thrust on the derrick (P2)� The stress on span block (P3)� The stress on the heel block (P4)

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The stresses on a derrick can be found by using the parallelogram method, which is explained in the Mathematics, Sciences and Conversions chapter. Fundamentally, when two or more forces act on the same point at an angle to each other, there is a Resultant Force which combines these forces.

Applying the Parallelogram

Vector Method to a Derrick

Assume a derrick has a single weight, W, hanging on its head, as shown in the figure below. A force acts downward equal to W load, which is represented by vector AB. Another force on the span acts on the derrick head to hold it, which is vector AD. From these two vectors representing the two forces acting on a derrick, we can use the parallelogram method to obtain the resultant force, which acts as a thrust on the derrick, by the following steps:

1. From B, draw a line parallel to AD, which is the span of the derrick, and cut the derrick at C.

2. Connect CD, which should be parallel to AB.

AC should be the resultant force of vector AB and AD. So AC is the thrust on the derrick.

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The figure above shows the components that create the four main resultant forces on the derrick. These resultant forces can be obtained by using the parallelogram method as follows:

1. Draw a diagram of the mast, derrick and span as per their dimensions and chosen scale; the scale doesn’t have to be the same as the scale used for the vectors in the parallelogram;

2. Determine the resultant stress on the derrick head block;

3. Determine the thrust on the derrick;

4. Determine the stress on the span block;

5. Determine the stress on heel block.

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Procedure to draw Diagram of Derrick

with Span and Derrick

1. Choose a scale for the diagram and draw a vertical line that represents the mast;

2. From the bottom end of the mast, use a compass to draw the arc with radius equalling the length of the derrick;

3. From the top end of the mast, use a compass to draw the arc with radius equalling the length of the span. The intersection of these two arcs is the derrick head.

Procedure to obtain Stress on the Derrick Head Block

1. Choose a scale for the vector in a parallelogram;

2. Calculate the tension on the hauling part by using the formula:

( )+=

W 10 nS

10P

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3. From the derrick head, lay down the vector (AD) along the derrick, which represents tension on the hauling part as per chosen scale;

4. From the derrick head, lay downward vertically a vector (AB), which represents the weight being lifted;

5. Complete the parallelogram; the stress on the derrick is the resultant vector AC.

In parallelogram ABCD

AB : Weight being liftedAD : Tension on hauling partAC : Stress on the derrick head block AC = AB + AD

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Procedure to obtain the Thrust on the

Derrick

Obtain the thrust on a derrick as follows:

1. From C, draw CE parallel to the topping lift span;

2. AE should be the thrust on the derrick.

If the weight of the derrick and the hauling parts is taken into consideration, then from C, extending CG equals the total weight of derrick and hauling part. Then the thrust becomes AH, instead of AF, and the tension on the topping lift span becomes AI, instead of AE.

In the parallelogram ACFEAF : Tension on topping lift spanAE : Thrust on derrick AE = AC + AF

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Procedure to obtain the Stress on a Span

Block

1. From X, draw vector XV along the lifting span, with magnitude equalling AF, the tension on the topping lift span;

2. Draw vector XW; if there is a single block rigged at the top of the mast to pull the lifting span, then the magnitude equals XV. If a tackle is rigged, then XW equals the tension on the hauling part;

3. Complete the parallelogram; the stress on the span block derrick is the resultant vector XY.

In parallelogram XVYWXV : Tension in topping lift spanXW : Stress in the downhaul of the topping liftXY : Stress on the span block XY = XV + XW

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Procedure to obtain the Stress on the

Heel Block

1. From K, draw vector KM along the derrick with KM=AD;

2. From K, draw vector KL pointing toward the winch, with magnitude equalling the pulling power of the winch;

3. Complete the parallelogram; the stress on the heel block is the resultant vector KN.

In parallelogram KLNMKM : stress in cargo runner acting in the direction of the derrickKL : stress in cargo runner acting in the direction of the winchKN : stress on the heel block KN = KM + KL

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Example A derrick 15 m long has a span 10 m which is attached to the mast at 12 m height above the heel. A weight of 5 tonnes is lifted by a gun tackle using disadvantage with the runner arranged through the derrick head, then along the derrick, then led to the winch, which is 2 m below the heel block and 2 m forward of the heel:

AB 5tonnes=

( ) ( )( )

W 10 P 5 10 2AD 3tonnes

10P 10 2+ +

= = =

From triangle XAK:( )( )2 2 2AX KX AK 2 KX AK cosXKA= + −

Therefore:

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( )( ) ( )( )2 2 2 2 2 2

1 1KX AK AX 12 15 10XKA cos cos 41.72 KX AK 2 12 15

− − + − + −

= = = °

Similarly:

( )( ) ( )( )2 2 2 2 2 2

1 1AX AK KX 10 15 12XAK cos cos 52.92 AX AK 2 10 15

− − + − + −

= = = °

KXA 180 XKA XAK 180 41.7 52.9 85.4= °− − = ° − ° − ° = °

Stress on derrick head block

DAB XKA 41.7 (KX AB)= = ° ll

From parallelogram ABCD:

ADC 180 DAB180 41.7 138.3 (Ajacent angles in parallelogram)

= °−= ° − ° = °

From triangle ADC:

( )( )( )( )

( )( )

2 2 2

2 2

2 2

AC AD DC 2 AD DC cos ADC

AC AD DC 2 AD DC cos ADC

3 5 2 3 5 cos138.3

7.5tonnes

= + −

= + −

= + − °

=

Thrust on the derrick From triangle DEC:

SinDEC SinDCEDC DE

=

DCE AXK (AX EC and KX DC)= ll ll

DCSinDEC 5sin85.4DE 6.3SinDEC sin52.9

AE AD DE 3 6.3 9.3tonnes

°= = =

°

= + = + =

Stress on span block

SinDAC SinADCDC AC

=

DC sin ADCsinDACAC

×∴ =

1 1DC sin ADC 5 sin138.3DAC sin sin 26.3AC 7.5

− −× × ° = = = °

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DAC FEA 26.3 (EF AC)= = ° ll

SinFEA SinFAEFA FE

=

FE sinFEA 7.5 sin26.3FA 4.2tonnessinFAE sin52.9× × °

= = =°

XV XW FA 4.2tonnes= = =

XVY 180 WXV 180 85.4 94.6= °− = ° − = °

From triangle XVY

( )( )( )( )( )( )

2 2 2

2 2

2 2

XY XV VY 2 XV VY cosXVY

XY XV VY 2 XV VY cosXVY

4.2 4.2 2 4.2 4.2 cos94.6

6.2tonnes

= + −

= + −

= + − °

=

Thrust on heel block KO OL 2m OKL 45= = ∴ = °

MKL 180 OKL XKA 180 45 41.7 93.3= °− − = ° − ° − ° = °

KMN 180 MKL 180 93.3 86.7= °− = ° − ° = °

From triangle KMN

( )( )( )( )( )( )

2 2 2

2 2

2 2

KN KM MN 2 KM MN cosKMN

KN KM MN 2 KM MN cosKMN

4.2 4.2 2 4.2 4.2 cos86.7

5.8tonnes

= + −

= + −

= + − °

=