Mechanics of Materials and Ship Structures

Euler buckling of columns

Euler buckling is the sudden lateral instability of a long, slender, initially straight column under a pure axial compressive load. The column fails by elastic buckling rather than by material crushing. The critical load is P_{cr} = \pi^2 EI / L_e^2, where E is Young's modulus, I is the least second moment of area, and L_e is the effective length. Buckling occurs about the axis of least stiffness, so the least I is always used.

The effective length is L_e = KL, where K depends on the end conditions. For pinned ends K = 1. For a column fixed at both ends K = 0.5, so L_e = L/2 and the critical load is four times that of a pinned column of the same length. For a fixed-free cantilever K = 2, so the critical load is one quarter of the pinned value. For a hinged-hinged solid circular column, I = \pi d^4/64 is substituted into the same formula.

The theory assumes a perfectly straight, elastic column whose length is very large compared with its cross-sectional dimensions. The direct compressive stress is neglected relative to the bending stress caused by lateral deflection. The load is purely axial with no transverse loading, so the failure mode is elastic buckling and not yielding.

• effective length — the length between points of inflection of the buckled shape, not the physical length. Example: a pinned column of length 3 m with EI = 2 MN\cdotm^2 has P_{cr} = \pi^2(2)/3^2 \approx 2.19 MN.
• unit consistency — with E in N/mm^2 and lengths in mm, the result is in N, so convert to kN before comparing values.

Midship section neutral axis and bending

For a hull girder under pure vertical bending, the neutral axis passes through the centroid of the midship section, and the longitudinal bending stress is \sigma = My/I. The stress is zero at the neutral axis and largest at the points farthest from it. If the distance to the deck is greater than the distance to the bottom, the deck carries the larger bending stress.

For a symmetric section whose deck and bottom flanges are identical, the neutral axis lies at mid-depth. In a simplified calculation the self second moment of area of the flange plating is ignored, so the inertia is dominated by the flange areas multiplied by the square of their distance from the neutral axis. The bending stress is then found from \sigma = My/I with y measured to the outer fibre, excluding the plate thickness.

The vertical shear flow in a thin-walled section is q = VQ/I, where Q is the first moment of area of the part of the section beyond the point considered. The shear flow is largest at the neutral axis, where Q is largest, and vanishes at free edges.

When an isolated local member such as a bottom longitudinal is attached to the plating, its own centroid differs from the ship's neutral axis. The offset is found by the area-weighted centroid of the composite member, and the distance to the ship's neutral axis is then the height of the ship's neutral axis above the reference minus that centroid.

• neutral axis — the locus of zero longitudinal stress under pure bending.
• first moment of area — Q = \int y\,dA about the neutral axis, used in shear flow.
• section modulus — Z = I/y_{max}, the ratio giving the maximum stress for a given moment.

Principal stresses in plane stress

In plane stress the principal stresses are the extreme normal stresses, \sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}. The principal plane is the plane on which the shear stress is zero, so only normal stresses act on it.

The orientation of the principal plane from the x-axis is given by \tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}. The angle \varphi between the x-direction and the direction of \sigma_1 is the principal angle, and its absolute value is taken from this relation.

The maximum in-plane shear stress is \tau_{max} = \dfrac{\sigma_1-\sigma_2}{2}, acting on planes at 45^\circ to the principal planes, and its value is half the difference of the principal stresses.

For a state of direct stresses \sigma_x and \sigma_y with no shear, the stresses on a plane at angle \theta are \sigma_\theta = \dfrac{\sigma_x+\sigma_y}{2} + \dfrac{\sigma_x-\sigma_y}{2}\cos 2\theta and \tau_\theta = \dfrac{\sigma_x-\sigma_y}{2}\sin 2\theta. On the 45^\circ plane the normal stress is the average of \sigma_x and \sigma_y, and the shear stress is half their difference.

• principal plane — a plane with zero shear stress.
• maximum in-plane shear — \tau_{max} = (\sigma_1-\sigma_2)/2, not \sigma_1 divided by 2.
• average stress — (\sigma_x+\sigma_y)/2, which is the centre of Mohr's circle.

Longitudinal strength of ship hull girder

The ship hull is treated as a beam whose longitudinal strength is governed by the distribution of weight and buoyancy along its length. In a wave, the buoyancy curve shows the immersed sectional area along the length. A wave crest amidships raises the water level at midship, so the buoyancy peaks there. This produces hogging, with the deck in tension and the bottom in compression. A wave trough amidships gives a buoyancy dip at midship and sagging, with the deck in compression and the bottom in tension.

A freely floating body has no external support, so its weight and buoyancy form a self-equilibrating system. For longitudinal strength analysis it is modelled as a free-free beam. At both free ends the shear force and bending moment are zero, because no external force or moment acts there.

For a hollow box girder of breadth B and depth D with thin walls, the second moment of area is I = \dfrac{BD^3 - bd^3}{12}, where b and d are the internal dimensions. The maximum bending stress at the outer fibre is \sigma = M(D/2)/I.

When extra cargo is added to the ends of a freely floating barge, the total load is balanced by additional buoyancy, and the loads are distributed along the length. The maximum bending moment occurs where the shear force is zero. The bending moment is found by taking moments of the forces on one side of that section.

• hogging — hull bends with the middle higher than the ends, deck in tension.
• sagging — hull bends with the middle lower than the ends, deck in compression.
• free-free beam — a beam with no supports, where shear and moment vanish at both ends.

Shear force and bending moment diagrams

The shear force and bending moment are related by dM/dx = V, and the load intensity is related by dV/dx = -w. Therefore the moment diagram is linear where no load acts, and a jump in shear force corresponds to a concentrated load.

A concentrated load at the free end of an overhang gives a linear moment diagram, with the moment zero at the free end and rising to its peak at the support. The reaction at the far support depends on the load position, and it causes the moment to change slope at the support.

A concentrated couple M acting at a point does not change the shear force, but it produces a jump in bending moment of magnitude M at that point. For a simply supported beam with a couple at midspan, the reactions are R_A = R_B = M/L and the bending moment at distance x < L/2 from the support is (M/L)x. At the quarter point this gives M/4.

The correct diagram for a couple at midspan rises linearly from zero at one support to the value M/2 just left of the couple, jumps by M at the couple, and then falls linearly to zero at the other support. Sign convention matters, and in this convention hogging is taken as positive.

• jump — a sudden change in bending moment caused by a concentrated couple.
• kink — a change of slope in the moment diagram caused by a concentrated load.
• hogging positive — a convention used in some question papers, opposite to the usual sagging-positive convention.

Strain energy in beams

The strain energy stored in a beam in bending is U = \int_0^L \dfrac{M^2}{2EI}\,dx. It is therefore proportional to the square of the moment, inversely proportional to the flexural rigidity EI, and depends on Young's modulus through EI.

For a simply supported beam under a uniformly distributed load w over span L, the strain energy is U = \dfrac{w^2L^5}{240EI}. Doubling the load quadruples the energy, and increasing the cross-sectional second moment of area reduces the energy.

Castigliano's theorem states that the partial derivative of the strain energy with respect to a load gives the displacement at the point of application of that load, \Delta = \partial U/\partial P. This is the method based directly on the strain energy principle. The conjugate beam, slope-deflection and moment distribution methods instead use analogy, displacement equations, or iterative distribution.

• strain energy — the energy stored in elastic deformation, always positive.
• flexural rigidity — the product EI, which governs bending deformation.
• Castigliano's method — a strain energy method for deflections, not for equilibrium of the whole structure.

Torsion of solid circular shaft

For a solid circular shaft in pure torsion, the shear stress at radius r is \tau = Tr/J, where J = \pi d^4/32 is the polar second moment of area. The maximum shear stress at the outer surface is \tau_{max} = \dfrac{16T}{\pi d^3}. It depends only on the applied torque and the diameter, and it does not depend on the length or the shear modulus.

The angle of twist for a shaft or girder with a torque at the free end and the other end rigidly fixed is \theta = \dfrac{TL}{GJ}, where G is the shear modulus and J is the polar moment of inertia. The twist is directly proportional to T and L and inversely proportional to GJ.

• polar moment of inertia — J, the torsional analogue of I for a circular section.
• shear modulus — G = E/[2(1+\nu)] for an isotropic material.
• angle of twist — \theta in radians, which must be multiplied by consistent units of T, L, G and J.

Offshore platform types

Offshore platforms are classified by how they are supported and what operations they serve. A jackup is a mobile unit with independent legs that are lowered onto the seabed for operation and raised for transit. It is used for drilling in relatively shallow water, and it is the platform type associated only with drilling operations.

A jacket is a fixed steel lattice structure, used to support production facilities and drilling equipment. A tension leg platform is a compliant floating structure held by tensioned tendons, and a SPAR is a deep-draft floating structure. Both TLPs and SPARs support production as well as drilling, so they are not restricted to drilling.

• jackup — mobile, legs on the seabed, drilling use.
• jacket — fixed, lattice steel base.
• tension leg platform — floating, tensioned tendons.
• SPAR — deep-draft, floating, production-oriented.

Plane strain assumptions

Plane strain applies to a long body in which the displacement in one direction, here z, is zero and the strain components involving z vanish. The out-of-plane normal strain is zero, \varepsilon_z = 0, and this is the necessary condition for plane strain in the x-y plane.

In plane strain the in-plane stresses \sigma_x and \sigma_y are generally nonzero, and the out-of-plane normal stress is not zero. For an elastic material it equals \sigma_z = \nu(\sigma_x+\sigma_y). Plane stress is the opposite case, where \sigma_z = 0 and the out-of-plane strain is not zero.