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. 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

H Twisting or torsional moment at location x due to vertical loading

H The distributed twisting moment per unit length due to distributed horizontal loading

Twisting or torsional moment at location x due to horizontal loading Total twisting moment

Note: Either rolling or loading/unloading may also cause twisting moments. Similar to Twisting moment diagram

5.2 Torsional effects

5.2.1 Twist of a closed single cell thin- walled prismatic section Consider a closed, single cell, thin-walled prismatic section subject only to a twisting moment,, which is constant along the length as shown in the following figure.

Twist of closed prismatic tube

Consider equilibrium of forces in the x- direction for the element dxds of the tube wall as shown in the above figure. The shear flow,, is therefore seen to be constant around the section.

, The magnitude of the moment,, may be computed by integrating the moment of the elementary force arising from this shear flow about any convenient axis. The symbol indicates that the integral is taken entirely around the section and, therefore, is the area enclosed by the mid- thickness line of the tubular cross section

The constant shear flow is then related to the applied twisting moment by 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.

where 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 Noting that N is constant around the section and recalling that the second integral was previously represented by,

Set

5.2.2 Twist of an open single cell thin- walled section Basic assumptions The shear deformation at the mid- thickness of the wall being equal to zero The cross sections rotating without distortion of their shape.

The closed tube will be able to resist a much greater torque per unit angular deflection than the open tube. The relative torsional stiffness of closed and open sections may be visualized by the following figure.

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,

where is twist angle per unit length, G is shear modulus of the material, J is torsional constant of the section. For a thin-walled open section,.

  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 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, Set

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 or. These points are referred to as sectorial centroids.

To measure the warp, u(s), from the plane of the mean warp, If the origin of s were chosen as a sectorial centroid, the term containing would vanish.

Neglecting the transverse (Poisson) effect, the x-stress is The x-strain is,

If the origin of the s-integration is now taken on a free edge of the contour, the value of N0 is zero.

The twisting moment on the end of the section is obtained by integration of the moment of N(s) around the entire contour,

The twisting moment due to restrained warping the warping constant of the section.

If is the total twisting moment at station x the differential equation of twist, taking into consideration unrestrained warping and restrained warping effects, is

Shear stress in multi-cell section   Shear flow due to the shear force For single thin-walled closed section, When only a pure vertical loading acting at the bending center of the section, the twist of the section, must be zero, and only the bending displacement appears.

The unknown constant of integration can be obtained.

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 N 1 and N 0

The first term is the warping displacement caused by the statically determinate shear flow N 1, while the second term is the warp due to the constant shear flow

Apply a similar procedure to the multicell section Shear flow in multiple-bulkhead tanker section

We first imagine each cell to be cut with longitu- dinal slits at points a, b, c. Let be the shear flow in cell i obtained with the origin of s located at the slit in that cell. Let be the constant of integration for cell i. The relative warp at slit a due to is given by

For cell 1, the additional warp at slit a, resulting from the constant shear flows acting on the boundaries of cell 1, is 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 Similar equations may be written for the remaining cells.

For cell 1: For the middle cell, 2: For cell 3:

Note also that the moment term m(s) in the expressions for 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

1. calculate obtain neutral axis

imagine the cell to be cut with longitudinal slits near points 3

Path ： ： 4-5 ：

Path ： 6-5 Path ： 1-2

Path ： ： 2-5:

2. calculate Clockwise

3-4 ： 4-5: 5-2: 2-3:

3. calculate

4. Calculate constant shear flow Single cell

5. Calculate total shear flow shear flow in longitudinal bulkhead

neutral axis shear flow at neutral axis