Military institute of science & Technology
Course Code-NAME-300
Ship Design Project & Presentation
Topic-Shafting Arrangement
Submitted By Submitted to
Name- MD. Shihab Name- SM Ikhtiar Mahmud
Imrul kayes Ibne Rahim
ID- 201924025, Asst.Professor
201924033 NAME,MIST
Batch-NAME-07 Course-Ship Design Project &
Presentation
Level-3
Date of submission 2.1.2022
Shafting Arrangement Scantling
Rulebook Followed-BV(2013)
Principal Particulars
Engine Speed (RPM) 1032 RPM
Gear Ratio 2.5
Shaft length, L(m) 6.44m
Max Tensile Stress, (MPa) 600 Mpa
Shear modulus of elasticity, G (Pa) 82 Gpa
Engine Power(1x in KW) 1324 kW
Calculation
1. Propeller RPM
We know,
1032
Propeller RPM, n= 2.5
=412.8 ≅ 413
2. Torque
Engine Power(KW )× 9.5488 1032× 9.5488
Torque,T = Propeller RPM
=
412.8
=30.626 Nm
3. Shaft Diameter
Data From Rulebook-
[From Rules for the Classification of ship 2013, BV Rule]
The diameter of the propeller shaft d P shall not be less than that
calculated from the following formula:
√
P 1 560
d P=F P ∙ k p 3 ∙ ∙
( )
n d uP 4
RmP +160 [ mm ]
1−
d vP
Where,
F P=¿ factor for the type of propulsion installation.
¿ 100 for-
Turbine
I.C. Engine
Electric Propulsion Engine
Installations
k P=¿factor dependent on shaft design features
¿ 1.26 for-
propeller shafts where the propeller is keyed on the
propeller shaft.
P=¿rated power of the main engine, [ kW ]
¿ 1324 kW
n=¿rated speed of propeller shaft,[ rpm ]
¿ 412.8
d uP=¿diameter of internal longitudinal shaft bore,[mm]
d vP =¿ outside shaft diameter,[mm]
Consideration-
If d uP ≤ d vP it may be taken that:-
1− ( )
d uP 4
d vP
=1.0
RmP =¿tensile strength of the shaft material[N /mm2 ]
2
¿ 600[ N /mm ]
Hence we have-
√
P 1 560
d P=F P ∙ k p 3 ∙ ∙
( )
n d
4
RmP +160 [ mm ]
1− uP
d vP
¿ 100 ∙1.26
√
1324 1
3
∙ ∙
560
412.8 1 600+160
¿ 167.833 mm ≅ 168 mm
[ mm ]
∴ The diameter of the propeller shaft
d P=168 [ mm ]
4. Polar Moment of Inertia
4
π ∙D
Polar moment of Inertia, J= 32 P [ m4 ]
π ∙(0.1678)4 4
¿ [m ]
32
¿ 0.00007789[m¿¿ 4 ]¿
5. Twisting Angle
TL 30.626 × 6.4 −5
Twisting Angle,θ= GJ = 82 ×109 ×0.00007789 =3.087 × 10 [radius]
¿ 3.087 ×10−5 ×57.2958 [ degree]
¿ 0.001769 °
Calculation Summary
1.Propeller RPM, n 1032 RPM
2.Torque,T 30.626 Nm
3. Propeller shaft dia d P 168 mm
4
4. Polar moment of Inertia, J 0.00007789 m
5. Twisting Angle,θ 0.001769 °