3. Longitudinal strength calculation

3.1 Nature of Ship Structural Reactions The reactions of structural components of the ship hull to external loads are usually measured by either stresses or deflections. Strength Structural performance criteria and the associated analyses involving stresses are referred to under the general term of strength. Stiffness Deflection-based considerations are referred to under the term stiffness.

Response patterns of typical ship structures Primary, secondary and tertiary structure

Primary response is the response of the entire hull when bending and twisting as a beam, under the external longitudinal distribution of vertical, lateral and twisting loads.

Secondary response comprises the stress and deflection of a single panel of stiffened plating, e.g., the panel of bottom structure contained between two adjacent transverse bulkheads. The loading of the panel is normal to its plane and the boundaries of the secondary panel are usually formed by other secondary panels (side shell and bulkheads) .

Tertiary response describes the out-of-plane deflection and associated stress of an individual panel of plating. The loading is normal to the panel, and its boundaries are formed by the stiffeners of the secondary panel of which it is a part .

The results are combined in an appropriate manner to obtain the total response of the structure. 3.2 Primary Direct Stress Elementary Bernoulli-Euler beam theory is usually utilized in computing the component of primary stress or deflection due to vertical or lateral hull bending loads.

The longitudinal stress A transverse axis through the centroid is termed the neutral axis of the beam and is a location of zero stress and strain.

It is clear that the extreme stresses are found at the top or bottom of the beam where z takes on its numerically largest values. The quantity ， where is either of these extreme values, is termed the section modulus of the beam.

3.3 Calculation of Section Modulus In general, the following items may be included in the calculation of the section modulus, provided they are continuous or effectively developed. Deck plating (strength deck and other effective decks). Shell and inner-bottom plating. Deck and bottom girders

Longitudinal strength members Plating and longitudinal stiffeners of longitudinal bulkheads. All longitudinals of deck, sides, bottom and inner bottom. Continuous longitudinal hatch coamings Longitudinal strength members

where I is moment of inertia of the section about a line parallel to the base through the true neutral axis (center of gravity), is moment of inertia of the half-section about an assumed axis parallel to the true neutral axis, A is total half-section area of effective longitudinal strength members . Generally no deduction is made for rivet holes.

is distance from the assumed axis to the true axis. is vertical moment of inertia (about its own center of gravity) of each individual plate or shape effective for longitudinal strength. is area of each such plate or shape. is distance of the center of gravity of each such plate or shape from the assumed axis.

Beams composed of two or more materials of different moduli of elasticity, for example, steel and aluminum. Where: A is cross sectional area, E(z) is modulus of elasticity of an element of area dA located at distance z from the neutral axis..

The neutral axis is located at such height that In practice,

Example two Consider the simplified cross-section of a hull with a superstructure shown in Figure 2. The hull girder is made of steel and the superstructure of aluminum. The cross-sectional dimensions are shown in the Figure. Compute the stresses at top deck house and at the bottom shell respectively assuming only a vertical sagging moment of Nm considered. 16m 8m 4m 10m 10 16 20 Figure 2. A simplified hull section

Solution: Given Sectional area First moment of the sectional area

The distance from the neutral axis to baseline The moment of inertia of the section about the neutral axis

Another way: The moment of inertia of the section about an assumed axis parallel to the true neutral axis

The moment of inertia of the section about the neutral axis Stress at top deck house Stress at bottom shell

3.4 Calculation of reduction coefficients Basic concept  a i cr b 0.25b

The reduction coefficient is defined as the ratio of to A. Plating combating the longitudinal bending only

Plating combating both longitudinal bending and grillage bending Case 1 Case 2

3.5 Buckling of a longitudinal and a plate The longitudinal Ei cr cr a

Deck plating and bottom shell in longitudinal framing system Computational model: simply supported on four sides

Inner-bottom plating and bottom shell in transverse framing system K=1.5 if floor with every spacing K=1.25 if floor with every two spacing K=1.0 if floor with every three spacing

Deck plating in transverse framing system Euler stress of a plating with simply supported on four sides due to pure shear stress

3.6 Iterative calculation of longitudinal strength Revised sectional properties

Revised longitudinal bending stresses Revised sectional properties Revised longitudinal bending stresses

3.7 Stress superposition Inner-bottom plating 1 2 3 4 Figure 3 Transverse bulkhead Inner-bottom plating Bottom shell Floor Longitudinal girder

1---longitudinal bending stress, assumed to be the same between two adjacent transverse bulkheads. 2---grillage bending stress, assumed to be the same through the spacing. Stresses at end portions and in the middle of the grillage should be noticed. 3---stress due to longitudinal bending, stresses at ends and in the middle of the longitudinal should be noticed. 4---plating bending stress, stresses at middle points of the short sides and center should be noticed.

Example three A transversely framed ship balances on a hogging wave. The longitudinal bending stresses at bottom shell and inner-bottom plating are 100 MPa and 78 MPa respectively. Grillage bending stresses at the transverse bulkhead and middle of the grillage for the bottom shell are 112 MPa and 50 MPa respectively. For the inner-bottom plating, bending stresses of the grillage at the transverse bulkhead and middle of the grillage are 130 MPa and 67 MPa respectively. Panel bending stresses in the bottom shell at midpoint and boundary are 6 MPa and 3 MPa respectively while those in the inner-bottom plating at midpoint and boundary are 8 MPa and 4 MPa respectively. Evaluate resultant stresses at top surface of inner-bottom plating in Sections 1, 2, 3 and 4 as shown in Figure 4 and counterparts at lower surface of bottom shell. Note that above values of stresses are given in terms of their absolute values.

Solution: Stresses at top surface of inner-bottom plating Section 1: A=-78-130+4=-204 MPa Section 2: A=-78-130-8=-216 MPa Section 3: A=-78+67+4=-7 MPa Section 4: A=-78+67-8=-19 MPa

Solution: Stresses at lower surface of bottom shell Section 1: B=-100+112+3=15 MPa Section 2: B=-100+112-6=6 MPa Section 3: B=-100-50+3=-147 MPa Section 4: B=-100-50-6=-156 MPa

3.8 Calculation of ultimate bending moment The ultimate bending moment is the one corresponding to the case of the tensile stress of the furthermost member from the neural axis reaching the yield strength of the material or the one corresponding to the case of the compressive stress of the furthermost member from the neural axis reaching the critical buckling stress of the member under consideration.

Case 1: extreme tensile stress at top reaches the yield strength of the material (Hogging) Case 2: extreme tensile stress at bottom reaches the yield strength of the material (Sagging) Case 3: extreme compressive stress at top reaches the critical buckling stress of the member (Sagging) Case 4: extreme compressive stress at bottom reaches the critical buckling stress of the member (Hogging) The ultimate bending moment will be the minimum of bending moments of above-mentioned cases.

General speaking, the ratio of the ultimate bending moment to the bending moment in standard calculation is more than 2.0 for the central part of the ship girder and one is more than 1.5 for either the stern part or the fore part of the ship girder.

The factor n represents the ability of the ship girder combating the over loading. In practice, where Ws is the section modulus corresponding to ultimate bending moment while W is one corresponding to the bending moment from standard calculation.

3.9 Distribution of shear and transverse stress components In many parts of the ship, the longitudinal stress, , is the dominant component. There are, however, locations in which the shear component becomes important and under unusual circumstances the transverse component may, likewise, become important.

shear stress resultant or shear flow stresses in the longitudinal stress resultants shear stress resultant or shear flow stresses in the longitudinal and girth-wise directions shear stress plate thickness

s Nx N N Ns x Element of plate structure in deck or side shell, illustrating components of bending stress resultants

The static equilibrium conditions for the element of plate subject only to in-plane stress (i.e., no bending of the plate) are:

In these expressions, s, is the girth-wise coordinate measured on the surface of the section from the x-axis as shown in the Figure.

Here, , the constant of integration, is equal to the value of the shear flow at the origin of integration, s=0. By proper choice of the origin, can often be set equal to zero.

The quantity is the first moment about the neutral axis of the cross sectional area of the plating between the origin at the centerline and the plating between the origin at the centerline and the variable location designated by s.

Free body diagram of plate-frame joint

If a longitudinal frame or girder that carries longitudinal stress is attached to the plate, as shown in above figure, there will be a discontinuity in the shear flow, , at the frame corresponding to a jump in . The total force in the frame, , is obtained by integrating this expression over the cross-section area, , of the frame.

where is the moment of the section area of the frame about the ship neutral axis. Equilibrium of forces in the x-direction requires that

or, the stepwise change in shear flows at the frame will be given by The moment of the stiffener cross section may be written as, where: is sectional area of frame i is distance from neutral axis to centroid of

Therefore Shear flow and girth-wise stress around a rectangular ship cross section

The transverse stress, , or equivalently, the stress resultant, , may be found by integration of the following Equation,

Example four A 50m long box-shaped barge in still fresh water is loaded with gravel as shown in Figure 3. The interior of the barge is uniformly loaded up to the deck. The combined weight of the barge’s internal load and structure is 60 kN/m. Once the level of the deck is reached, the load varies linearly from zero at the two ends to a maximum load of 360 kN/m at amidships. The simplified cross-section of the hull is shown in Figure 4 and the thickness of the plating is constant throughout and is equal to 12mm. a) Compute the shear stress at neutral axis of the section.

Figure 3 A loaded box-shaped barge 360 kN/m WL Deck Structure +Gravel=60 kN/m 25 m Figure 3 A loaded box-shaped barge 8000 1500 Figure 4 Simplified section of the barge 4200

Weight distribution along ship length Buoyancy distribution along ship length

Load distribution along ship length Maximum shear forces q(x)=0 x=12.5m, x=37.5m

Sectional area First moment of the sectional area about assumed neutral axis

The distance from the neutral axis to baseline The moment of inertia of the section about the neutral axis

First moment about the neutral axis of the cross sectional area of the plating between the origin and the plating between the origin and neutral axis Shear stress at the neutral axis

Selection of the sections to be checked Seven to nine sections may be required to be checked for their longitudinal strength Central parts of the ship Sections with large openings Sections with changes in framing system, materials Ends of superstructures Sections in which the bending moments are large due to special weight distribution Sections in which the shear forces are large …

3.10 Deflection of the ship girder Deflection due to bending Boundary conditions: y(0)=0, y(L)=0

Approximate calculation Deflection at midsection due to bending =0.08~0.10

Deflection due to shearing