06_Ship_strength

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Oct 30, 2023

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NE224 Spring 2023 Instructor: An Wang 7 Ship Strength 7.1 Loads on the ship structure Static loads: —Weights —Buoyancy forces —Internal/external force on hull/liquid tank surfaces —Stresses caused by thermal effects. —... Dynamic loads: — Wave induced forces. — Inertial forces due to ship acceleration. —Slamming forces. —Periodic loading due to forced vibration from rotating machinery. —Sloshing forces for containers partially filled with liquid. —... Occasional and unusual operational loads: —Ice breaking loads. —Loads caused by explosion, missile blast, air craft landing, etc. —Loads during launching, drydocking and berthing. —Loads due to grounding, collision. —... Figure 7.1: Longitudinal bending. 7.2 Structural arrangements of ships (Framing systems) 7.2.1 Structural element Stiffened plating : fundamental structural element. 1

NE224 Spring 2023 Instructor: An Wang Figure 7.2: An example of stiffened plating. Considerations for structural design: — Structural efficiency: have the least weight for a given strength. Unless the arrange- ment with the least weight costs more. — Cost of materials and fabrication. — Structural continuity. Avoid stress concentrations. — Utilization of space. Framing system of a ship can be characterized by the relative number, size and spacing of its transverse stiffeners compared to the number, size and spacing of its longitudinal stiffeners. 7.2.2 Transverse framing system Transverse framing system consists of many small, closely spaced transverse stiffeners and fewer larger, widely spaced longitudinal stiffeners, see figure 7.3. 2

NE224 Spring 2023 Instructor: An Wang Figure 7.3: Transverse framing. Figure 7.4: An example of transverse framing system. Transverse framing system originate from wooden ships and dominated during the early years for steel ships as well, when ship structural design mostly depends on empirical proce- dures and past experience. 3

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NE224 Spring 2023 Instructor: An Wang Most stiffeners in the transverse framing system do not contribute to longitudinal strength (resistance to longitudinal bending). Therefore, transverse framing system is not optimal for structural efficiency. 7.2.3 Longitudinal framing system Longitudinal framing system consists of many small, closely spaced longitudinals sup- porting the plating directly and being supported in turn by a few large, widely spaced longitudinals, see figure 7.5. Figure 7.5: Longitudinal framing. 4

NE224 Spring 2023 Instructor: An Wang Figure 7.6: Examples of longitudinal framing system. Longitudinal framing system becomes more popular after structural analysis were in- creasing applied to ship structural design. It is especially popular in oil tankers and bulk carriers. Advantages of longitudinal framing system compared with transverse framing system: 1) Longitudinal framing system has better structural efficiency than transverse fram- ing system (longitudinally framed ship has better longitudinal strength to a transversely framed ship of equal size and structural weight, thus permitting the carriage of more cargo or payload). 2) Longitudinal framing system is more resistant to plate buckling (especially for deck plate and bottom plate). 3) Longitudinal framing system can easily adapt a graduated size configuration at different depth to accommodate the varying pressure on shell plating or liquid container plating at different depth. 5

NE224 Spring 2023 Instructor: An Wang Drawbacks of longitudinal framing system compared with transverse framing system: 1) The deep webs penetrate into cargo space and interfere with packed cargo . 2) Difficult in construction at the bow and stern , where the longitudinal stiffeners are too close together. Therefore, transverse framing is usually used at the bow and stern of a longitudinally framed ship. 7.2.4 Combination framing system Figure 7.7: Combination framing. 6

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NE224 Spring 2023 Instructor: An Wang Figure 7.8: An example of combination framing system. Most ships are built with a combination framing system. For example, in figure 7.7, the longitudinal framing is used in the bottom and decks (why?—need more longitudinal strength and resistance to plate buckling). The transverse framing is used in the sides (why?—preclude the need for deep web that will inhibit efficient cargo space). 7.3 Ship structural materials 7.3.1 Requirement of ship hull material 1) Availability and cost. 2) Uniformity. Need controllable and repeatable manufacturing as well as careful quality control of the manufacturing process for the material. 3) Ease of fabrication. a) Easy and cheap to be formed into different shapes. b) The material properties remains unchanged during the fabrication process. c) Joints between structural members should be as strong as the materials being joined(for example, welding technique: electric arc welding to achieve full-strength joints). 4) Ease of maintenance. 5) Strength vs weight. 6) Resistance to distortion under load. 7

NE224 Spring 2023 Instructor: An Wang 7.3.2 Materials used in ship structure 1. Steels 2. Aluminum 3. Composites 7.3.3 Mechanical properties of steel The properties of ship steel that relate to its strength are called its mechanical properties, which can be determined by tensile test. The external load will induce internal reactions within the material and the reaction will cause the material to deform. Stress is the internal force per unit area (unit: Pa): σ = P A , where σ is the stress, P is the load on the stressed member and A is the load-carrying area of the stressed member. Principal stresses: —Tensile stress —Compression stress —Shear stress Figure 7.9: Stress three principal stresses. The deformation of the material per unit of original length is called a strain. Strain is a non-dimensional quantity and can be written as ϵ = δ L , where ϵ is the strain, δ is the deformation and L is the original length. Note: Both stress and strain are tensors. 8

NE224 Spring 2023 Instructor: An Wang Tensile test: tensile loading is applied to the specimen and increases from zero to the load required to break the specimen. During the loading process, the stress ( σ ) and strain ( ϵ ) are measured continuously. A plot of the stress and strain during the test is made and shown in figure 7.10. Figure 7.10: Stress-strain curve. Proportional limit: below the proportional limit, the stress is a linear function of strain (Hooke’s law). The ratio of stress to strain is called modulus of elasticity, or Young’s modulus E = σ ϵ , where E is the modulus of elasticity (unit: Pa), σ is the stress (below the proportional limit), ϵ is the strain corresponding to the above stress. The modulus of elasticity is the measure of stiffness of a material (capacity to resist deformation under a given load below proportional limit). For different steel, the proportional limit varies, however, the modulus of elasticity ( E ) is nearly the same for different steel. Elastic limit: Below the elastic limit, the deformation of the material is recoverable when the load (or stress) is removed. The material is able to return to its original dimension. This behavior is called elasticity . 9

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NE224 Spring 2023 Instructor: An Wang For steel, elastic limit is very close to proportional limit. For practically purpose, the two limits can be considered identical. Yield point: After yield point, the strain (deformation) increases rapidly without in- crease in stress (called yield). The stress at yield point is called yield stress . The deformation beyond the elastic limit is not recoverable after the load is removed and is called plastic deformation . Yield overcome: when the strain is beyond the yield point, the stress needs to increase again to create further increase in strain. Ultimate stress: The maximum value of stress. The ultimate stress is the measure of the strength of a material. After the ultimate stress is reached, necking down (reduction in cross section area) occurs as the deformation further increases. Rupture: Elongation: the percentage of the amount of stretch to the original length of the material at the point of failure, denoted as % δ and defined by % δ = L f − L 0 L 0 × 100% , where L f is the final length of the specimen at rupture and L 0 is the original length of the specimen. The elongation is the measure of material’s ductility. If very small elongation: brittle fracture , if large elongation: ductile fracture . Toughness: toughness is the area under the stress-strain curve. It is the ability of a material to absob energy and plastically deform without fracturing. High tensile strength and large elongation (ductility) result in better toughness. 7.4 Longitudinal strength (shear force and bending moment) 7.4.1 Shear force and bending moment in beams Shear force and bending moment are internal force and moment at a structure’s cross section of interest. Sign convention: — Shear force: for the surface on the right side of an element, down is positive; for surface on the left side of an element, up is positive. In other words, positive shear force tends to turn the element clockwise, no matter on which side, see figure 7.11. 10

NE224 Spring 2023 Instructor: An Wang Figure 7.11: Direction of positive shear force. — Bending Moment: For the surface on the right side of an element, counterclockwise is positive; for surface on the left side of an element, clockwise is positive. In other words, the positive moment tends to compress the top surface and stretch the lower surface of the element, see figure 7.12. Figure 7.12: Direction of positive moment. Determine the (internal) shear force and moment in a beam: Example: a simply supported beam (one end hinge, one end roller) under a concentrated force (see figure 7.13). — 1. Reaction force at each end: R 1 and R 2 . According to force balance and moment balance: R 1 + R 2 − P = 0 R 1 · a − R 2 · b = 0 11

NE224 Spring 2023 Instructor: An Wang Solve the above equation to obtain the reaction force at each end: R 1 = b L P R 2 = a L P — 2. The internal shear force and moment along the beam, V ( x ) and M ( x ). We can make a cut at a given position ( x ) along the beam, either the left element or the right element can be chosen for the analysis, as long as the sign convention is followed. Let’s use the element on the left of the cut. For 0 < x < a , R 1 , V ( x ) and M ( x ) are the forces and moment on the element. According to force and moment balance: V ( x ) = R 1 = b L P M ( x ) = R 1 x = bx L P For a < x < L , R 1 , P , V ( x ) and M ( x ) are the forces and moment on the element. V ( x ) = R 1 − P = − a L P M ( x ) = R 1 x − P ( x − a ) = a ( L − x ) L P — 3. The shear force and bending moment diagram. Relationship between shear force and bending moment: M ( x ) = Z V ( x ) dx or V ( x ) = dM ( x ) dx 12

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NE224 Spring 2023 Instructor: An Wang a b P R 1 R 2 x V(x) M(x) x R 1 V(x) M(x) P a R 1 M(x) V(x) L Figure 7.13: Example of a simply supported beam under concentrated force P . 7.4.2 Longitudinal bending of a ship as a floating beam The longitudinal bending of a ship can be treated as the bending of a beam under distributed loads, i.e., the buoyancy per unit length b ( x ) and weight per unit length w ( x ). The buoyancy per unit length b ( x ) is a function of the shape of hull along x . The weight per unit length w ( x ) is a function of the ship weight distribution along x . Load diagram: 13

NE224 Spring 2023 Instructor: An Wang Figure 7.14: Example load diagram of distributed buoyancy and weight. — 1. The area enclosed by the weight curve w ( x ) is equal to the displacement of the ship. — 2. The longitudinal position of the centroid of the weight curve is at the longitudinal center of gravity of the ship, LCG. — 3. The area enclosed by the buoyancy curve is equal to the displacement of the ship, expressed as the total buoyancy force. — 4. The longitudinal position of the centroid of the buoyancy curve is at the longitudinal center of buoyancy of the ship, LCB. We can use q ( x ) to represent the net distributed load for a ship and q ( x ) = b ( x ) − w ( x ) (7.1) . The load diagram of q ( x ) is the superposition of the weight diagram and the buoyancy diagrams. In order to find the internal shear force and bending moment, make a cut at a given location x along the ship and use the element on the left. Within the element, we introduce a local coordinate s , where 0 ≤ s ≤ x . The shear shear force at x can be calculated by (see figure 7.15) V ( x ) = Z x 0 ( b ( s ) − w ( s )) ds = Z x 0 q ( s ) ds (7.2) and the bending moment at x can be calculated by M ( x ) = Z x 0 ( x − s ) q ( s ) ds (7.3) 14

NE224 Spring 2023 Instructor: An Wang V(x) M(x) x s x q(s) q(x) Figure 7.15: Shear force and bending moment of an element of a beam under distributed load. Therefore, the following relationship can be obtained: q ( x ) = dV ( x ) dx (7.4) and V ( x ) = dM ( x ) dx (7.5) Therefore, we have the following conclusions: — 1. The slope of the shear force diagram at any point is equal to the net load (per unit length) at that point. — 2. The slope of the bending moment diagram at any point is equal to the shear force at that point. — 3. The maximum value of shear occurs at that point in the length of the ship at which the load curve is zero, i.e., where the load curve crosses its axis. — 4. The maximum value of bending moment occurs at that point in the length of the ship at which the shear curve is zero, i.e., where the shear curve crosses its axis. Complication introduced by ship form and loading In real sea state, the load (magnitude, direction) distribution along the ship varies from time to time. Shown in figure 7.16, depending on the distribution of water surface elevation along the ship, the distribution of the hydrostatic pressure varies from time to time. In addition, the weight distribution of the ship also changes from time to time due to load- ing/discharging, consumption of fuel and fresh water, change in blast water, etc. 15

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NE224 Spring 2023 Instructor: An Wang Sagging: the hull girder deflects downward at amidship. The deck is under compression and the bottom is under tension. Hogging: the hull girder deflects upward at amidship. The deck is under tension and the bottom is under compression. The Sagging-Hogging transition often occurs during head sea or following sea condi- tions with wave length on the same order of the length of the ship. It is estimated that during a 20-year ship life time, a typical ship undergoes on the order of 100 million bending reversals. Figure 7.16: Hull girder bending in waves. An example of stress variation over time during a voyage is shown in figure 7.17. 16

NE224 Spring 2023 Instructor: An Wang Figure 7.17: Stress variation over time during voyage (Lewis 1988). Note: Even though the total weight equals to the total buoyancy for a ship at equi- librium, due to the nonuniform distribution of the hydrostatic force on the hull, as well as the nonuniform weight distribution along the ship, the net load is not always zero along the ship. Therefore, at a longitudinal location x along the ship, the ship is subject to shear force and bending moment. Example: A rectangular barge of 250-ft length, 35-ft beam and 20-ft depth floats in seawater at a draft of 2 ft when it is empty. The light ship weight may be assumed to be uniformly distributed over the length of the barge. It has 5 holds, each 50 ft long. The barge floats in seawater loaded with cargo as shown in the figure below. Cargo weights within each hold are to be assumed uniformly distributed over the length of the hold. Calculate and plot diagrams of the distributions of weight, buoyancy, load, shear and bending moment. Determine the values of the curves at each bulkhead an at their maximum points. 17

NE224 Spring 2023 Instructor: An Wang Solution: Light ship weight (displacement when the ship is empty): W 0 = ∆ 0 = ρ w gLBT = 250 ft × 35 ft × 2 ft 35 ton / ft 3 = 500 ton , where ρ w g = 1 / 35 ton/ft 3 is the specific weight of sea water. The light sheep weight per unit length (ft) is w 0 = W 0 L = 500 ton 250 ft = 2 ton / ft The weight per unit length of the cargo: Holds 1 and 5: w 1 , 5 = 400 / 50 = 8 ton/ft Holds 2 and 4: w 2 , 4 = 700 / 50 = 14 ton/ft Holds 3: w 3 = 800 / 50 = 16 ton/ft Therefore, the total weight distribution along the ship is: Holds 1 and 5: w = w 0 + w 1 , 5 = 2 + 8 = 10 ton/ft Holds 2 and 4: w = w 0 + w 2 , 4 = 2 + 14 = 16 ton/ft Holds 3: w = w 0 + w 3 = 2 + 16 = 18 ton/ft The total buoyancy after the cargo is loaded: ∆ = W 0 + 2 × W 1 , 5 + 2 × W 2 , 4 + W 3 = 500 + 2 × 400 + 2 × 700 + 800 = 3500 ton The distribution of the buoyancy is uniform. Therefore, the buoyancy per unit length: b = ∆ L = 3500 ton 250 ft = 14 ton / ft The net load distribution q = b − w : Holds 1 and 5: q = 14 − 10 = 4 ton/ft Holds 2 and 4: q = 14 − 16 = − 2 ton/ft Holds 3: q = 14 − 18 = − 4 ton/ft 18

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NE224 Spring 2023 Instructor: An Wang Method 1: The diagram of the weight, buoyancy and the net load distribution is shown in the following figure. First make the diagram of net load distribution q ( x ) = b ( x ) − w ( x ). Use the diagram of q ( x ) to determine the shear force diagram V ( x ). Then use the diagram of V ( x ) to determine the moment diagram M ( x ), according to the following. At any x , the shear force, V ( x ), is the cumulative area under q ( s ) diagram for 0 ≤ s ≤ x . Area under horizontal axis is considered as negative area. At any x , the moment, M ( x ), is the cumulative area under V ( s ) diagram for 0 ≤ s ≤ x . Area under horizontal axis is considered as negative area. The results are shown in the following figure. *************** Do not present the following method in class ****************** Method 2: We can also use equations 7.2 and 7.3 to calculate the shear force and moment distri- 19

NE224 Spring 2023 Instructor: An Wang bution in the above example. Even though equations 7.2 and 7.3 are general for all types of loading, as you will see, it is sometimes mathematically tedious, not recommended for simple distributions of q ( x ) like the one in the above example. Hold 1: For 0 ≤ s ≤ x ≤ 50: q ( s ) = 4. Use equation 7.2 to calculate shear force: V ( x ) = Z x 0 q ( s ) ds = Z x 0 4 ds = 4 x At the bulkhead x = 50, V (50) = 4 × 50 = 200. Use equation 7.3 to calculate the moment: M ( x ) = Z x 0 ( x − s ) q ( s ) ds = 4 Z x 0 ( x − s ) ds = 4 xs − s 2 2 x 0 = 4 x 2 − x 2 2 = 2 x 2 At the bulkhead x = 50, M (50) = 2 × 50 2 = 5000. Hold 2: For 50 ≤ s ≤ x ≤ 100: q ( s ) = − 2. Use equation 7.2 to calculate shear force: V ( x ) = Z x 0 q ( s ) ds = Z 50 0 q ( s ) ds + Z x 50 q ( s ) ds = Z 50 0 4 ds + Z x 50 ( − 2) ds = 4(50 − 0) + ( − 2)( x − 50) = − 2 x + 300 At the bulkhead x = 100, V (100) = − 2 × 100 + 300 = 100. Use equation 7.3 to calculate the moment: M ( x ) = Z x 0 ( x − s ) q ( s ) ds = Z 50 0 ( x − s ) q ( s ) ds + Z x 50 ( x − s ) q ( s ) ds = 4 Z 50 0 ( x − s ) ds + ( − 2) Z x 50 ( x − s ) ds = 4 xs − s 2 2 50 0 − 2 xs − s 2 2 x 50 = 4 (50 x − 1250 − 0 + 0) − 2 x 2 − x 2 2 − 50 x + 1250 = − x 2 + 300 x − 7500 At the bulkhead x = 100, M (100) = − 100 2 + 300 × 100 − 7500 = 12500. Hold 3: 20

NE224 Spring 2023 Instructor: An Wang For 100 ≤ s ≤ x ≤ 150: q ( s ) = − 4. Use equation 7.2 to calculate shear force: V ( x ) = Z x 0 q ( s ) ds = Z 50 0 q ( s ) ds + Z 100 50 q ( s ) ds + Z x 100 q ( s ) ds = Z 50 0 4 ds + Z 100 50 ( − 2) ds + Z x 100 ( − 4) ds = 4(50 − 0) + ( − 2)(100 − 50) + ( − 4)( x − 100) = − 4 x + 500 At the bulkhead x = 150, V (150) = − 4 × 150 + 500 = − 100. Use equation 7.3 to calculate the moment: M ( x ) = Z x 0 ( x − s ) q ( s ) ds = Z 50 0 ( x − s ) q ( s ) ds + Z 100 50 ( x − s ) q ( s ) ds + Z x 100 ( x − s ) q ( s ) ds = 4 Z 50 0 ( x − s ) ds + ( − 2) Z 100 50 ( x − s ) ds + ( − 4) Z x 100 ( x − s ) ds = 4 xs − s 2 2 50 0 − 2 xs − s 2 2 100 50 − 4 xs − s 2 2 x 100 = 4 (50 x − 1250 − 0 + 0) − 2 (100 x − 5000 − 50 x + 1250) − 4 x 2 − x 2 2 − 100 x + 5000 = − 2 x 2 + 500 x − 17500 At the bulkhead x = 150, M (150) = − 2 × 150 2 + 500 × 150 − 17500 = 12500. Maximum value occurs at x = 500 / 2 / 2 = 125, and the maximum value is M (125) = − 2 × 125 2 + 500 × 125 − 17500 = 13750. Similarly, one can continue to calculate hold 4 and hold 5 ....... 7.5 Longitudinal strength (structural response) 7.5.1 Stresses caused by longitudinal bending The bending moment at a longitudinal position x will cause stress at the cross section. The stress at the cross section at distance z from the neutral axis of the cross section: σ z = Mz I (7.6) , where σ z is the tensile or compressive stress at point z . M is the bending moment at the cross section under consideration. z is the vertical distance to the neutral axis. 21

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NE224 Spring 2023 Instructor: An Wang I is the moment of inertial of the cross section of the ship (beam) about its neutral axis . On the neutral axis (plane) , there is no compression or stretching of the material. The neutral axis of a beam is located at the centroid of the beam’s cross section. Figure 7.18: Neutral axis of a beam and the stress at the cross section. The maximum stress appears at either the top or bottom of the beam. If we denote the distances from the top and bottom of the cross section to the neutral axis as c top and c bot , respectively, the stresses at the top and bottom are: σ top = Mc top I (7.7) σ bot = Mc bot I (7.8) Sectional modulus of the beam: Z = I c (7.9) The sectional modulus Z is a geometrical property of the beam cross section and can be determined independently by the beam’s dimensions. The sectional modulus is a measure of the beam’s ability to resist bending. Beam with greater Z has smaller maximum stress. 22

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NE224 Spring 2023 Instructor: An Wang With the definition of the sectional modulus, the stresses at the top and bottom can be written as σ top = M Z top (7.10) σ bot = M Z bot (7.11) 7.5.2 Calculation of sectional modulus The maximum bending moment often occurs near amidship. Therefore, the section modulus near the middle portion of the ship must be sufficient so that the ship can withstand the maximum bending moment. Only longitudinal structural member continuous through transverse bulkheads should be included in the calculation of sectional modulus. Structural members which should be included in the calculation of sectional modulus: — 1. Deck plating outboard of hatches. — 2. Plating of side shell, bottom shell, inner bottom, and longitudinal bulkheads. — 3. Center vertical keel, side girders (keelsons), and deck girders. — 4. Longitudinals on deck, sides, bottom, inner bottom and longitudinal bulkheads. Note: In the following, assuming all the cross sectional shapes of a ship are symmetric about the ship’s centerline plane. Therefore, only calculation of half of the sectional shape is needed. Procedures: 1. Choose the axis of moment. The baseline is recommended. 2. Determine: — a) cross section area of each shape, a n , in 2 or cm 2 . — b) vertical distance of the center of gravity of each shape from the assumed axis, d n , ft or m. — c) moment of inertia of each shape about the horizontal axis passing through the centroid of the shape, i n . The i n of horizontal plates may be omitted from the calculations because they are negligibly small. 3. Calculate: — a) The first moment of each shape about assumed axis, a n d n . — b) The second moments of each shape about assumed axis, a n d 2 n . 23

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NE224 Spring 2023 Instructor: An Wang 4. Calculate: — a) The total sectional area: A n = 2 ∑ n a n . — b) Total first moment of the section about the assumed axis: 2 ∑ n a n d n . — c) Total moment of inertia of the shape about the assumed axis: I n = 2 ∑ n ( i n + a n d 2 n ) (parallel axis theorem). 5. Calculate the vertical distance from the assumed axis to the true neutral axis of the whole cross section. D g = ∑ n a n d n ∑ n a n 6. Calculate, using parallel axis theorem, the moment of inertia of the whole ship section about its true neutral axis: I = I n − A n D 2 g 7. Calculate the top and bottom values of c and Z . Z top = I c top and Z bot = I c bot Example: The midship section including only the longitudinal continuous structure of a rectangular barge of L = 150 ft, B = 30 ft, D = 15 ft, is shown in the figure below. All plating is 1/2-inch thick. Determine the location of the neutral axis, the total moment of inertia of the midship section (both sides), and the top and bottom section moduli. Determine also the stresses in the deck and bottom plating when the barge is subjected to a hogging bending moment of 10,000 ft-ton. Solution: 24

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NE224 Spring 2023 Instructor: An Wang The moment of inertia of the horizontal platings are so small and is neglected. Only need to calculate the moment of inertia of the side shell and central keel. The only half of the central keel thickness should be included for the half-section. Use the following table for the calculation. Item Scantlings a n d n a n d n a n d 2 n h i n in 2 ft in 2 ft in 2 ft 2 ft in 2 ft 2 Main deck 5 × 12 × 0.5 30 15.0 450.0 6750.0 Side shell 15 × 12 × 0.5 90 7.5 675 5062.5 15 1688 Tank top 15 × 12 × 0.5 90 3.0 270 810 Central keel 1 2 (3 × 12 × 0.5) 9 1.5 13.5 20.3 3 6.8 Bottom shell 15 × 12 × 0.5 90 0 0 0 Half section totals 309 1408.5 12642.8 1695 The total area of the cross section: A n = 2 X n a n = 309 × 2 = 618 in 2 The vertical distance from the baseline to the neutral axis of the cross section: D g = ∑ n a n d n ∑ n a n = 1408 . 5 309 = 4 . 56 in The moment of inertia of the cross section about baseline: I n = 2 X n ( i n + a n d 2 n ) = 2 × (1695 + 12642 . 8) = 28676 in 2 ft 2 The moment of inertia of the cross section about its neutral axis: I = I n − A n D 2 g = 28676 − 618 × 4 . 56 2 = 15826 in 2 ft 2 The distance from the top of cross section to neutral axis: c top = D − D g = 15 − 4 . 56 = 10 . 44 ft The distance from the bottom of cross section to neutral axis: c bot = D g = 4 . 56 ft Sectional modulus for the top: Z top = I c top = 15826 10 . 44 = 1516 in 2 ft 25

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NE224 Spring 2023 Instructor: An Wang Sectional modulus for the bottom: Z bot = I c bot = 15826 4 . 56 = 3471 in 2 ft Tensile stress in the deck σ top = M Z top = 10000 1516 = 6 . 60 ton / in 2 = 14784 lb / in 2 Compressive stress in the bottom plating σ bot = M Z bot = 10000 3471 = 2 . 88 ton / in 2 = 6451 lb / in 2 7.5.3 Permissible bending stresses and required sectional modulus 1. Bending moment amidship ( M t ): Total maximum bending moment: M t = M sw + M w , where M sw is the still-water bending moment and M w is the wave-induced bending moment. 2. Nominal permissible bending stress ( f p ): typically smaller than the yield strength of steel by a safety factor (for example, 2). 3. Required section modulus (SM) SM = M t f p 26

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