12/18/2012Originator : SS

Checked : IP

WINCH BASE CALCULATION

1. 20 T Winch Skid

The pulling-in of the winch will produce a maximum liad of 20 T on the winch skid

σy 250N

mm2⋅:=

MPa 1 106× Pa=

kN 1 103× N=

1.1 Support Member

Length of member•

L 2380 mm⋅:=

Distance along member to load pos•

Loadposition 1190 mm⋅:=

Load applied to beam•

W 20000 1.3⋅ 9.81⋅ N⋅:=

W 2.551 105× N=

Allowable stress in bending•

As per AS3990 - 1993 Clause 5.2

Fb 0.66 σy⋅:= Fb 1.65 108× Pa=

Allowable stress in shear•

As per AS3990 - 1993 Clause 5.2

Fv 0.45 σy⋅:= Fv 1.125 108× Pa=

Calculation points over beam length•

x 0mm 1mm, 2380mm..:=

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12/18/2012Originator : SS

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Second moment of area for W12 x 65 member•

Iyy 174 in4⋅:=

Cross sectional area•

Aw 19.1 in2⋅:=

Section Width•

bf 12 in⋅:=

Beam reactions assuming a fixed-fixed analysis•

RAW( ) L Loadposition−( )⋅ L Loadposition−( )⋅ L 2 Loadposition⋅+( )⋅

L3( ):=

RA 1.275 105× N=

RB W RA−:=

RB 1.275 105× N=

End moments assuming a fixed-fixed analysis•

MA W−Loadposition

L2⋅ L Loadposition−( )2⋅:= MA 7.588− 104

× J=

MBW− Loadposition

2⋅

L2L Loadposition−( )⋅:= MB 7.588− 104

× J=

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12/18/2012Originator : SS

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Shear Force Diagram•

V x( ) if 0 x< Loadposition< RA, RA W− RB+ if Loadposition x< L< RA W−, RA W− RB+, ( )+, ( ):=

0 1 2

1− 105×

0

1 105×

V x( )

x

Shear Stress•

σv x( )V x( )Aw

:=

0 1 21− 108×

5− 107×

0

5 107×

1 108×

σv x( )

x

σv 1000mm( ) 1.035 107× Pa=

UCσv 1000mm( )

Fv0.092=:=

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12/18/2012Originator : SS

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Bending Moment Diagram•

M x( ) if 0 x≤ Loadposition≤ MA RA x⋅+, MA RA x⋅+ W x Loadposition−( )⋅−⎡⎣ ⎤⎦, ⎡⎣ ⎤⎦:=

0 1 21− 105×

5− 104×

0

5 104×

1 105×

M x( )

x

Bending Stress•

σb x( ) M x( )0.5bfIyy

⋅:=

0 1 22− 108×

1− 108×

0

1 108×

2 108×

σb x( )

x

σb 1190mm( ) 159.673 MPa⋅=

UCσb 1190mm( )

Fb0.968=:=

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12/18/2012Originator : SS

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1.2 Fastener Selection

Use M30 bolts Grade 8.8

Design details for fasteners as per AS3990 9.6

Design Load for 1 bolt if 16 bolt used (assuming only 8 bolts are loaded)•

Designload20000 1.3⋅ 9.81⋅ N⋅

4:=

Designload 6.377 104× N=

Bolt Diameter•

diabolt 30 mm⋅:=

Flange Thickness•

tt 16.3mm:=

Tensile strength of fastener•

Fuf 800 106⋅ Pa:=

Yield strength of fastener•

FYf 640 106⋅ Pa:=

Shank Area•

Ash 452mm2:=

Stress Area•

Bolt areas as per AS 1252

Ast 353mm2:=

Core Area•

Acore 324mm2:=

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12/18/2012Originator : SS

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Maximum permissable stress in tension, Fit (on the stress area)

Ftf min0.60 FYf⋅

0.45 Fuf⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

:=

Ftf 0.45 Fuf⋅:=

Ftf 3.6 108× Pa=

Consider base rotatiin about its forward bolts, creating tension in the aft bolts due to thevertical load form the upending moment of the winch

Perpendicular distance from base of winch to vertical wire load

ra 1.6m:=

Wwd 20000 9.81⋅ N:=

MA ra Wwd⋅:=

MA 3.139 105× J=

Lever arm length resisting rotation (distance between forward and aft bolts)•

rb 1.13m:=

Tensile bolt force shared by all aft bolts•

FbMArb

:=

Fb 2.778 105× N=

Tensile bolt force assuming tensile load carried by only 4 of the 8 aft bolts•

σtFb

8. Ast⋅:=

σt 9.837 107× Pa=

boltcheck if σt Ftf> "FAIL", "PASS", ( ):=

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12/18/2012Originator : SS

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boltcheck "PASS"=

Consider frame rotating about its aft bolts, creating tension in the forward bolts due to thehorizontal load on the base

Perpendicular distance from base of winch to horizontal winch wire load•rc 1.6m:=

Wwd 15000 9.81⋅ N:=

MB rc Wwd⋅:=

MB 2.354 105× J=

Level arm length resisting rotation (distance between forward and aft bolts)•

rb 1.13 m=

Tensile Bolt force shared by all aft bolts •

FbMBrb

:=

Fb 2.084 105× N=

Tensile Bolt force assuming tensile loadlts carried by only 4 of the 8 aft bolts•

σtFb

8 Ast⋅:=

σt 7.378 107× Pa=

boltcheck if σt Ftf> "FAIL", "PASS", ( ):=

boltcheck "PASS"=

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