Theoretical strength assessment of unstiffened bilge shell plating and some considerations on rule prescriptions

Abstract

In ship hull structures, bilge part is often composed of unstiffened radiused plating, for which the harmonized CSR stipulates required scantlings, locations of the longitudinals adjacent to the bilge radius, and so on. In recent years, under the increased demand for rational structural design, larger radius than ever is sought, expecting reduction of hull steel weight and welding lengths, thus contributing to the reduction of production cost of structures. However, structural problems arising from the larger bilge radius and associated structural arrangement around the bilge shell are not yet sufficiently identified. In these circumstances, the authors developed theoretical formulae, assuming radiused plating connected to continuous stiffened flat plating with regular stiffener spacing. The results of the theoretical calculations are compared with the results of finite element analysis, and it was found that the derived theoretical formulae well explain the complicated phenomena of the curved shell plating connected to the flat stiffened panels. Utilizing the derived theoretical formulae, parametric studies were carried out with regard to the radius of the bilge shell plating, the distance between the position where the curvature of the bilge plate starts and the adjacent longitudinal, and so on. As a result of the calculations, it was found that large bilge radius exceeding current usual practice is feasible, on the condition that the location of the longitudinals is well controlled to reduce bending stresses on the shell plate, and the buckling strength is satisfied. In such case, the stipulation in the harmonized CSR is not always rational, and the authors propose modified structural design methodologies around the unstiffened bilge shell plating.

Appendices

Appendix 1: Theoretical formulation of curved shell plating under uniform load

General solution of the curved shell plate under uniformly distributed load as shown in Fig. 34 can be derived as follows [4], where nomenclature is as shown in the figure and in Chapter 2.

Fig. 34

Curved shell plate with uniformly distributed load

First, we consider the infinitesimal element as shown in Fig. 35. From the equilibrium of forces to radial direction,

Fig. 35

Equilibrium of forces and moments

$$\frac{{{\text{d}}F}}{{{\text{d}}\theta }} + q + pR = 0$$

(49)

From the equilibrium of forces to circumferential direction,

$$\frac{{{\text{d}}q}}{{{\text{d}}\theta }} - F = 0$$

(50)

From the equilibrium of moments around the center of the infinitesimal element,

$$\frac{{{\text{d}}M}}{{{\text{d}}\theta }} - FR = 0$$

(51)

From the stress–strain relationship, axial force q and bending moment M can be obtained as follows:

$$q = \frac{{Et_{p} }}{R}\left( {\frac{{{\text{d}}v}}{{{\text{d}}\theta }} - w} \right)$$

(52)

$$M = - \frac{{EI_{p} }}{{R^{2} }}\left( {\frac{{{\text{d}}^{2} w}}{{{\text{d}}\theta^{2} }} + w} \right)$$

(53)

From Eqs. 49–51 and 53, we can obtain the differential equation of deflection of the curved shell as follows:

$$\frac{{{\text{d}}^{5} w}}{{{\text{d}}\theta^{5} }} + 2\frac{{{\text{d}}^{3} w}}{{{\text{d}}\theta^{3} }} + \frac{{{\text{d}}w}}{{{\text{d}}\theta }} = 0$$

(54)

Taking account of the symmetry about the center of the shell, we can obtain the following general solution.

$$w = C_{1} + C_{2} \cos \theta + C_{3} \theta \sin \theta$$

(55)

Then, we can obtain slope, bending moment, shearing force and axial force as follows:

$$\phi = \frac{{{\text{d}}w}}{{{\text{d}}x}} = \frac{{{\text{d}}\theta }}{{{\text{d}}x}} \times \frac{{{\text{d}}w}}{{{\text{d}}\theta }} = - \frac{1}{R}C_{2} \sin \theta + \frac{1}{R}C_{3} \left( {\sin \theta + \theta \cos \theta } \right)$$

(56)

$$M = - \frac{{EI_{p} }}{{R^{2} }}\left( {C_{1} + 2C_{3} \cos \theta } \right)$$

(57)

$$F = \frac{{2EI_{p} }}{{R^{3} }}C_{3} \sin \theta$$

(58)

$$q = - \frac{{2EI_{p} }}{{R^{3} }}C_{3} \cos \theta - pR$$

(59)

Substituting Eqs. 55 and 59 into Eq. 52,

$$\frac{{{\text{d}}v}}{{{\text{d}}\theta }} = C_{1} - \frac{{pR^{2} }}{{Et_{p} }} + \left( {C_{2} - 2\lambda_{p} C_{3} } \right)\cos \theta + C_{3} \theta \sin \theta$$

(60)

where

$$\lambda_{p} = \frac{{I_{p} }}{{t_{p} R^{2} }}$$

(61)

Then, integrating this equation, circumferential displacement, v, is obtained as follows:

$$v = \left( {C_{1} - \frac{{pR^{2} }}{{Et_{p} }}} \right)\theta + \left( {C_{2} - 2\lambda_{p} C{}_{3}} \right)\sin \theta + C_{3} \left( { - \theta \cos \theta + \sin \theta } \right)$$

(62)

Appendix 2: Rotational spring constant of infinite continuous beam

In this APPENDIX, spring constant for rotation of an infinite continuous beam is derived. Let us consider a beam as shown in Fig. 36, where point B is supported with rotational spring of spring constant $K_{B} = {{k_{B} EI_{p} } \mathord{\left/ {\vphantom {{k_{B} EI_{p} } l}} \right. \kern-0pt} l}$, and bending moment M _A is applied at point A.

Fig. 36

Beam with rotational spring support at one end and moment applied at the other end

Differential equations of the elastic deflection is obtained as follows.

$$EI_{p} \frac{{{\text{d}}^{2} w}}{{{\text{d}}x^{2} }} = - M = - M_{A} + R_{A} l - R_{A} x$$

(63)

$$EI_{p} \frac{{{\text{d}}w}}{{{\text{d}}x}} = - \frac{{R_{A} }}{2}x^{2} + \left( {R_{A} l - M_{A} } \right)x + C_{1}$$

(64)

$$EI_{p} w = - \frac{{R_{A} }}{6}x^{3} + \frac{{R_{A} l - M_{A} }}{2}x^{2} + C_{1} x + C_{2}$$

(65)

Boundary conditions are

$$w = 0$$

(66)

$$M = - K_{B} \frac{{{\text{d}}w}}{{{\text{d}}x}} = - k_{B} \frac{{EI_{p} }}{l}\frac{{{\text{d}}w}}{{{\text{d}}x}}$$

(67)

at $x = 0$ and

$$w = 0$$

(68)

at $x = l$. Solving Eqs. 63–65 using these boundary conditions, we can obtain the integral constant and R _A as follows.

$$C_{1} = \frac{{M_{A} l}}{{2\left( {k_{B} + 3} \right)}}$$

(69)

$$C_{2} = 0$$

(70)

$$R_{A} = \frac{{3\left( {k_{B} + 2} \right)M_{A} }}{{2\left( {k_{B} + 3} \right)l}}$$

(71)

Substituting Eqs. 69 and 71 into Eqs. 63 and 64, we can obtain the bending moment M and the slope $\phi$ as follows.

$$M = M_{A} \left[ {\frac{{3\left( {k_{B} + 2} \right)}}{{2\left( {k_{B} + 3} \right)}}\left( {\frac{x}{l}} \right) - \frac{{k_{B} }}{{2\left( {k_{B} + 3} \right)}}} \right]$$

(72)

$$\phi = \frac{{M_{A} l}}{{EI_{p} }}\left[ { - \frac{{3\left( {k_{B} + 2} \right)}}{{4\left( {k_{B} + 3} \right)}}\left( {\frac{x}{l}} \right)^{2} + \frac{{k_{B} }}{{2\left( {k_{B} + 3} \right)}}\left( {\frac{x}{l}} \right) + \frac{1}{{2\left( {k_{B} + 3} \right)}}} \right]$$

(73)

Then, to derive the spring constant at point A, substituting $x = l$ into Eq. 73,

$$\phi = - \frac{{M_{A} l}}{{EI_{p} }} \times \frac{{k_{B} + 4}}{{4\left( {k_{B} + 3} \right)}}$$

(74)

Thus, spring constant k _A is obtained as follows.

$$k_{A} = - \frac{l}{{EI_{p} }} \times \frac{{M_{A} }}{\phi } = \frac{{4\left( {k_{B} + 3} \right)}}{{k_{B} + 4}}$$

(75)

When the beam is an infinite continuous beam, substituting $k_{A} = k_{B}$ into the recurrence formula, Eq. 75, and $k_{A} = 3.464$ is finally obtained. In this case, the bending moment at point B is obtained as follows, substituting $x = 0$ and $k_{A} = 3.464$ to Eq. 72.

$$M = - 0.268M_{A}$$

(76)