5. Torsional strength calculation
5.1 Torsional loads acting on a ship hull
As the ship moves through a seaway
encountering waves from directions other
than directly ahead or astern, it will
experience lateral bending loads and
twisting moments in addition to the vertical
loads.
The twisting or torsional loads will require
some special consideration. The response
of the ship to the overall hull twisting
loading should be considered a primary
response.
H
 The distributed twisting moment per unit
length due to distributed vertical loading
c1 ( x)  v( x)  e
H
 Twisting or torsional moment at location x
due to vertical loading
x
T1 ( x)   c1 ( x)dx
0
H
 The distributed twisting moment per unit length
due to distributed horizontal loading
c2 ( x)  h( x)  e1
 The distributed twisting moment per unit length
due to distributed horizontal loading
c2 ( x)  h( x)  e1
Twisting or torsional moment at location x
due to horizontal loading
x
T2 ( x)   c2 ( x)dx
0
Total twisting moment
T ( x)  T1 ( x)  T2 ( x)
dT ( x)
c
dx
Similar to
 M ( x )  x V ( x ) dx
0


dM ( x )
 V ( x) 
dx

Twisting moment diagram
Note: Either rolling or loading/unloading
may also cause twisting moments.
5.2 Torsional effects
5.2.1 Twist of a closed single cell thinwalled prismatic section
Consider a closed, single cell, thin-walled
prismatic section subject only to a twisting
moment, M T , which is constant along the
length as shown in the following figure.
Twist of closed prismatic tube
Consider equilibrium of forces in the xdirection for the element dxds of the
tube wall as shown in the above figure.
t11dx  t2 2 dx  0
The shear flow, N  t , is therefore seen
to be constant around the section.
The magnitude of the moment, M T , may
be computed by integrating the moment
of the elementary force arising from this
shear flow about any convenient axis.
MT 
 rNds  N  rds  2 N 
The symbol  indicates that the integral is
taken entirely around the section and,
therefore,  is the area enclosed by the midthickness line of the tubular cross section
The constant shear flow is then related to
the applied twisting moment by
N  M T / 2
Consider the deformation of the element
dsdx which results from this shear. Let u, v
be the displacements in the axial and
tangential directions respectively of a point
on the surface of the tube.
v
d
r
x
dx
u v


s x


G
E
G
2(1  v)
v
d
r
x
dx
u v


s x


G
E
G
2(1  v)
u 
d
 r
s G
dx

d
1 N
d
u ( s )   ds   r
ds  u0   ds 
rds  u0

G
dx
G t
dx
where
u0 is a constant of integration.
The quantity, u(s), termed the warp, is
seen to be the longitudinal displacement
of a point on the cell wall, which results
from the shear distortion of the material
due to twist. If the section is circular the
rotation will take place without warping,
but for other shapes the warping will be
nonzero and will vary around the
perimeter of the section.
For a closed section, the differential warp
must be zero if the integral is evaluated
around the entire section. This is expressed
by
1
G

N
d
ds 
t
dx
 rds  0
Noting that N is constant around the section and
recalling that the second integral was previously
represented by 2 ,
d
N

dx 2G
MT
d

2
dx 4 G
4 2
Set J 
ds
 t

ds
t

ds
t
d M T

dx GJ
5.2.2 Twist of an open single cell thinwalled section
Basic assumptions
• The shear deformation at the midthickness of the wall being equal to
zero
• The cross sections rotating without
distortion of their shape.
The relative torsional stiffness of closed and
open sections may be visualized by the
following figure.
The closed tube will be able to resist a much
greater torque per unit angular deflection
than the open tube.
Twist of open and closed tubes
 The case of the open tube without
longitudinal restraint
The only resistance to torsion in the case
of the open tube without longitudinal
restraint is provided by the twisting
resistance of the thin material of which
the tube is composed. The resistance to
twist of the entirely open section is given
by the St. Venant torsion equation,
d
M T 1  GJ
dx
d
dx
where
is twist angle per unit length,
G is shear modulus of the material, J is
torsional constant of the section. For a
1 S 3
thin-walled open section, J  3 0 t ds
 The case of the open tube with
longitudinal restraint
If warping resistance is present, another
component of torsional resistance is
developed through the shear stresses
that result from this warping restraint.
In ship structures, warping resistance
comes from four sources:
 The closed sections of the structure
between hatch openings.
 The closed ends of the ship.
 Double wall transverse bulkheads.
 Closed, torsionally stiff parts of the
cross section (longitudinal torsion
tubes or boxes, including double
bottom).
For an open section, we may assume that
u v


0
s x
d
u (s)  
dx
d
0 r (s)ds  u0  2 (s) dx  u0
1 s
 ( s )   r ( s )ds
2 0
u0
s
the sectorial area
the warping displacement at
the origin of the coordinate s.
The average warp, which is, given by the
integral of u(s) around the entire section
periphery, S, divided by S,
1 S
1 S
d
u   u ( s )ds    u0 ds 
S 0
S 0
dx

S
0

2 ( s )ds 

S
Set S  2S0  0 2(s)ds
S w d
u  u0 
S dx
S
S  2S0   2(s)ds
0
S  2S0 The first sectorial moment
with respect to the origin of s.
There will in general, be one or more points
on the contour which, if used as origins for
the S-integration, will result in a zero value
for 0 or S . These points are referred to as
sectorial centroids.
To measure the warp, u(s), from the
plane of the mean warp,
d
u ( s)  u  u 
(20  2 ( s))
dx
If the origin of s were chosen as a
sectorial centroid, the term containing
0 would vanish.
The x-strain is,
u
d
x 
 2 2 (0   ( s))
x
dx
2
Neglecting the transverse (Poisson)
effect, the x-stress is
d
 x  E x  2 E 2 (0   ( s))
dx
2
N x N

0
x
s
N s N

0
s
x
N x
[ x  t ]
N
d 2


 2 Et ( s )
(0   ( s))
2
s
x
x
dx
d
N ( s)  N 0  2 E 3
dx
3

s
0
t ( s)[(0   ( s)]ds
d
N ( s)  N 0  2 E 3
dx
3

s
0
t ( s)[(0   ( s)]ds
If the origin of the s-integration is now
taken on a free edge of the contour, the
value of N0 is zero.
d
N ( s)  2 E 3
dx
3

s
0
t ( s)[(0   ( s)]ds
The twisting moment on the end of the
section is obtained by integration of the
moment of N(s) around the entire contour,
S
T2 ( x)   N (s)r (s)ds
0
d
N ( s)  2 E 3
dx
3

s
0
t ( s)[(0   ( s)]ds
The twisting moment due to restrained
warping
d
T2 ( x)   E 3
dx
3
S

S
0
d
4(0   ( s)) t ( s)ds   E 3
dx
  4 (0   ( s))2 t ( s)ds
0
3
2
the warping constant
of the section.
If T ( x)  T1  T2 is the total twisting
moment at station x the differential
equation of twist, taking into
consideration unrestrained warping
and restrained warping effects, is
d
d
E 3  GJ
 T ( x)
dx
dx
3
Shear stress in multi-cell section
Shear flow due to the shear force
For single thin-walled closed section,
u v


s x
v
d
r
x
dx
u    ds  0
When only a pure vertical loading acting at the bending
center of the section, the twist of the section d / dx ,
must be zero, and only the bending displacement appears.
u    ds  0

1
G
N
 
G Gt

N
ds  0
t
V ( x) s
V ( x)
N ( s) 
tzds  N 0 
m( s)  N 0

I 0
I
 N1  N0
1
G
ds 1
 ( N1  N0 ) t  G

ds N 0
N1 
t
G

ds
0
t
1
G
ds 1
 ( N1  N0 ) t  G

ds N 0
N1 
t
G

ds
0
t
The unknown constant of integration N 0 can be obtained.
ds
N
 1t
N0  
ds
 t
Assume that the closed tubular section subject to a
vertical loading without twist is transformed into an open
section by an imaginary longitudinal slit, and the edge of
this slit is taken as the origin of the s-coordinate.
Warp associated with N1 and N0
1
G

ds N 0
N1 
t
G

ds
0
t
The first term is the warping displacement
caused by the statically determinate shear flow
N1,
while the second term is the warp due to the
constant shear flow N 0
Apply a similar procedure to the multicell section
Shear flow in multiple-bulkhead tanker section
We first imagine each cell to be cut with longitudinal slits at points a, b, c. Let Ni ( s) be the shear
flow in cell i obtained with the origin of s
located at the slit in that cell. Let Ni 0 be the
constant of integration for cell i.
The relative warp at slit a due to N1 ( s) is given by
1
u1a 
G

1
ds
N1 ( s )
t
For cell 1, the additional warp at slit a, resulting
from the constant shear flows acting on the
boundaries of cell 1, is
1
ds 1
ds
u0 a  N 01   N 02 
1 2 t
G
t
G
1
The second integral in this expression is
evaluated only over the bulkhead dividing cells 1
and 2 and is negative since the constant shear
flows of the two cells oppose each other in the
bulkhead.
We now require that the total warp at slit a
must vanish and this is given by the condition

1
ds
ds
ds
N1 ( s )  N 01 
 N 02 
0
1 2 t
t
t
1
Similar equations may be written for the
remaining cells.
For cell 1:

1
ds
ds
ds
N1 ( s )  N 01 
 N 02 
0
1 2 t
t
t
1
For the middle cell, 2:

2
ds
ds
ds
ds
N 2 ( s )  N 02 
 N 01 
 N 03 
0
2 1 t
2 3 t
t
t
2
For cell 3:
ds
ds
ds
3 N3 (s) t  N03 3 t  N02 32 t  0
Note also that the moment term m(s) in
the expressions for N1 , N2 , and N3 must
include the moment of the area of
longitudinal stiffeners as well as plating,
while the integrals in above three
Equations include only the plating.
Example
Consider a tanker section,
Calculate the shear flow
in the section.
Assume:
Shear force: V
Moment of inertia: I
D B/2
t A  tB  1/ 2tC
1. calculate
m( s)
obtain neutral axis
A  D *(t A  2tB  t C )  5Dt A
B  ( Dt A )* D  2 Dt B *( D / 2)  2 D t A
2
e  B / A  2D / 5
z0  2D / 5
imagine the cell to be cut with longitudinal slits
near points 3
Path： 3-4-5
3-4：
S1
m3 4   ( z1  s1 )t B ds1
0
 ( z1s1  s12 / 2)t B
m4  0.1D 2t B
4-5：
S2
m45  m4   ( z0 )tC ds2
0
 0.1D 2t B  z0 s2tC
m5  0.3D 2t B
Path： 6-5
S3
m65   ( z0 )tC ds3  0.8DtB s3
0
m5  0.4 D 2t B
Path： 1-2
S5
m12   ( z1 )t Ads5  0.6Dt A s5
0
m2  0.3D 2t B
Path： 3-2-5
3-2：
S4
m32   ( z1 )t Ads4  0.6Dt A s4
0
m2  0.3D 2t B
2-5:
S6
m25  m2   ( z1  s6 )t Ads6
0
 0.6 D t B  (0.6 Ds6  0.5s )t A
2
m5  0.7 D 2t B
2
6
2. calculate 
Ni ( s)
ds
t (s)
Clockwise
D
3-4：  ( z1s1  s12 / 2)ds1  0.4D3 / 3
0
2
0.1
D
t

z
s
t
3
B
0
2
C
4-5: 
ds2  0.025D
0
tC
2
2
D [0.6 D t  (0.6 Ds  0.5s )t ]
B
6
6 A
5-2: 
ds6  2.2 D3 / 3
0
tB
D /2
2-3:


D /2
0
0.6Ds4ds4  0.3D / 4
3
N (s)
3V
3V
ds  (0.4 / 3  0.025  2.2 / 3  0.3 / 4) D
 0.7 D
t ( s)
I
I
3. calculate


1
ds
t (s)
D ds
D /2 ds
D ds
D /2 ds
ds
6
1
2
4




0 t
0
0 t
0
t
t
tA
B
C
B
D
D
 (1  1/ 4  1  1/ 2)  2.75
tB
tB
4. Calculate constant shear flow N 0
k
Ni
1
ds
N0i  ds   Ni , j 
 
ds  0
t
t cell i t
j 1
cell i
i, j
Single cell
N1
1
N0  ds   
ds
t
t
cell
cell
m( s )
ds

V cell t
V
N0  
 (0.255D 2t B )
1
I
I
cell t ds
5. Calculate total shear flow
V
2
N  N1  N 0  [m( s )  0.255D t B ]
I
shear flow in longitudinal bulkhead
V
2
N  [m( s ) 25  0.255 D t B ]
I
shear flow at neutral axis
S6
m25  m2   ( z1  s6 )t Ads6
0
 0.6 D 2t B  (0.6 Ds6  0.5s62 )t A
s6  3D / 5
V
N    [0.6 D 2t B  (0.36 D 2  0.18 D 2 )t B ]  0.255 D 2t B 
I
V
  (0.525D 2t B )
I