HW5_solutions

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Oct 30, 2023

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NE 224 Spring 2023 Homework 05 (Due 5/5 (Friday), Canvas upload) 1. (Self-study example 6-1 in Zubaly starting on p.178 and solve the following problem) A rectangular barge 350 ft long and 40 ft wide floats on an even keel at a draft of 15 ft. An empty ( µ = 95%, µ s = 100%) forward end compartment 30 ft long has been bilged. Determine the forward and aft drafts that result. Solution: The volume of displacement of the barge: ∇ = LBT = 350 ft × 40 ft × 15 ft = 210000 ft 3 1

The waterplane area before flooding: A w = LB = 350 ft × 40 ft = 14000 ft 2 The lost waterplane area: a = µ s a s = µ s lB = 100% × 30 ft × 40 ft = 1200 ft 2 The waterplane area after flooding: A ′ w = A w − a = 14000 ft 2 − 1200 ft 2 = 12800 ft 2 Volume of the lost buoyancy: v = µlBT = 95% × 30 ft × 40 ft × 15 ft = 17100 ft 3 First, consider the parallel sinkage of the barge caused by flooding. The increase of draft by parallel sinkage: δT p = v A ′ w = 17100 ft 3 12800 ft 2 = 1 . 336 ft It is considered that the centroid of lost buoyancy is shifted to the centroid of the re- gained buoyancy, resulting in a moment that cause the barge to trim. The longitudinal positions relative to FP of the lost buoyancy ( X l ) and regained buoyancy ( X r ) should satisfy: X l × v + X r × ( ∇ − v ) = L 2 × ∇ Since X l = l 2 , we can use the equation above to find the distance of the centroid of the regained buoyancy to FP, X r : X r = L 2 × ∇ − l 2 × v ∇ − v = 350 / 2 ft × 210000 ft 3 − 30 / 2 ft × 17100 ft 3 210000 ft 3 − 17100 ft 3 = 189 . 184 ft Therefore, the tangent of the angle of trim caused by the shifting of volume from lost buoyancy to regained buoyancy is: tan θ = t L = w ( X r − X l ) ∆ GM L = ρgv ( X r − X l ) ρg ∇ GM L = v ( X r − X l ) ∇ GM L The trim is t = v ( X r − X l ) ∇ GM L L ≈ v ( X r − X l ) ∇ BM L L = v ( X r − X l ) ∇ I L ∇ L = v ( X r − X l ) I ′ L L 2

Need to find I ′ L (moment of inertia of the remaining waterplane about its centroid). The centroid of the remaining water plane should be above the centroid of the regained volume at X r = 189 . 184 ft from FP. I ′ L = " I L + A w X r − L 2 2 # − µ s " i L + a X r − l 2 2 # = " BL 3 12 + A w X r − L 2 2 # − µ s " Bl 3 12 + a X r − l 2 2 # = " (40 ft)(350 ft) 3 12 + (14000 ft 2 ) 189 . 184 ft − 350 ft 2 2 # − 100% " (40 ft)(30 ft) 3 12 + (1200 ft 2 ) 189 . 184 ft − 30 2 2 # = 109235189 . 7 ft 4 Therefore, the trim t = v ( X r − X l ) I ′ L L = 17100 ft 3 (189 . 184 ft − 15 ft) 109235189 . 7 ft 4 (350 ft) = 9 . 544 ft Draft at FP: T F = T + δT p + F L t = T + δT p + X r L t = 15 ft + 1 . 336 ft + 189 . 184 ft 350 ft (9 . 544 ft) = 21 . 50 ft Draft at AP: T A = T + δT p − A L t = T + δT p − L − X r L t = 15 ft + 1 . 336 ft − 350 ft − 189 . 184 ft 350 ft (9 . 544 ft) = 11 . 95 ft 3

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2. A ship of ∆ = 10 , 000 tons of displacement and L = 350 ft long has a maximum hogging bending moment (ft-ton) of ∆ L/ 30. The depth of the midship section is 37 ft and the neutral axis is 19.1 ft above the keel. The moment of inertia of the midship section is 4000 ft 4 . Calculate the maximum tensile and compressive stresses and state where each occurs (deck or bottom). Solution: The top sectional modulus Z top = I c top = 4000 ft 4 37 ft − 19 . 1 ft = 223 . 464 ft 3 Maximum stress (tensile) at the top: σ top = M Z top = ∆ L/ 30 Z top = 10000 ton × 350 ft / 30 223 . 464 ft 3 = 522 . 083 ton / ft 2 = 8121 . 285 psi The bottom sectional modulus Z bot = I c bot = 4000 ft 4 19 . 1 ft = 209 . 424 ft 3 Maximum stress (compressive) at the top: σ bot = M Z bot = ∆ L/ 30 Z bot = 10000 ton × 350 ft / 30 209 . 424 ft 3 = 557 . 084 ton / ft 2 = 8665 . 744 psi 4

3. A rectangular barge is 128 ft long, 30 ft wide and 18 ft deep, and it weighs 544 tons when empty. The barge light weight may be assumed uniformly distributed over its length. It is divided into four holds of equal length. Cargo is loaded into each hold as follows, stowed uniformly and level in each hold: Hold #1 192 tons Hold #2 224 tons Hold #3 272 tons Hold #4 176 tons 1) Construct weight, buoyancy, load, shear and bending moment diagrams for the loaded barge, and calculate the values of each at the ends, at each bulkhead, and at their maximum points. 2) Calculate the minimum required section modulus (in 2 -ft) of the barge such that in the loaded condition the bending stress does not exceed 15,000 psi. (1 ton (long ton) = 2240 lb, 1 ft = 12 in). Solution: 1) The light ship weight per unit length: w 0 = W 0 L = 544 ton 128 ft = 4 . 25 ton / ft Length of each hold: l = L/ 4 = 128 / 4 = 32 ft Weight per unit length for each hold: Hold 1: w 1 = w 0 + 192 ton / 32 ft = 4 . 25 + 6 = 10 . 25 ton / ft. Hold 2: w 2 = w 0 + 224 ton / 32 ft = 4 . 25 + 7 = 11 . 25 ton / ft. Hold 3: w 3 = w 0 + 272 ton / 32 ft = 4 . 25 + 8 . 5 = 12 . 75 ton / ft. Hold 4: w 4 = w 0 + 176 ton / 32 ft = 4 . 25 + 5 . 5 = 9 . 75 ton / ft. Total buoyancy B = 544 + 192 + 224 + 272 + 176 = 1408 ton Buoyancy per unit length: b = B L = 1408 128 = 11 ton / ft The net load distribution q ( x ) = b ( x ) − w ( x ): 5

Hold 1: q = 11 − 10 . 25 = 0 . 75 ton / ft. Hold 2: q = 11 − 11 . 25 = − 0 . 25 ton / ft. Hold 3: q = 11 − 12 . 75 = − 1 . 75 ton / ft. Hold 4: q = 11 − 9 . 75 = 1 . 25 ton / ft. Draw the distribution of q ( x ). See below. Calculate the area under q ( x ) to get V ( x ) and then calculate the area under V ( x ) to get M ( x ). The maximum shear force is -40 N and the maximum moment is 1097 ton · ft. 0 20 40 60 80 100 120 10 11 12 0 20 40 60 80 100 120 -2 0 2 0 20 40 60 80 100 120 -40 -20 0 20 0 20 40 60 80 100 120 0 500 1000 6

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2) Maximum moment M m = 1097 ton · ft. Required sectional modulus: Z = M m σ = 1097 ton · ft 15000 lb / in 2 = 1097 ton · ft 15000 / 2240 ton / in 2 = 163 . 82 in 2 ft 7

4. Calculate the top and bottom section moduli (in 2 -ft) of the structural section shown below. Plating dimensions and thickness are as labeled. If the steel has a yield stress of 34,000 psi, what safety factor based on yield will this structure have when it is subjected to a bending moment of 12,000 ft-tons? (1 ton (long ton) = 2240 lb, 1 ft = 12 in). Solution: Item Scantlings a n d n a n d n a n d 2 n h i n in × in in 2 ft in 2 ft in 2 ft 2 ft in 2 ft 2 Main deck 120 × 0.75 90 10.0 900.0 9000 Side shell 120 × 0.25 30 5 150 750 10 250 Quarter keel 12 × 0.75 9 0.5 4.5 2.25 0.5 0.1875 Central keel 1 2 (12 × 0 . 75) 4.5 0.5 2.25 1.125 0.5 0.09375 Bottom plating 180 × 0 . 5 90 0 0 0 Half section totals 223.5 1056.75 9753.375 250.2813 The total area of the cross section: A n = 2 X n a n = 223 . 5 × 2 = 447 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 = 1056 . 75 223 . 5 = 4 . 728 ft The moment of inertia of the cross section about baseline: I n = 2 X n ( i n + a n d 2 n ) = 2 × (250 . 2813 + 9753 . 375) = 20007 . 31 in 2 ft 2 8

The moment of inertia of the cross section about its neutral axis: I = I n − A n D 2 g = 20007 . 31 − 447 × 4 . 728 2 = 10014 . 29 in 2 ft 2 The distance from the top of cross section to neutral axis: c top = D − D g = 10 − 4 . 728 = 5 . 272 ft The distance from the bottom of cross section to neutral axis: c bot = D g = 4 . 728 ft Sectional modulus for the top: Z top = I c top = 10014 . 29 5 . 272 = 1899 . 591 in 2 ft Sectional modulus for the bottom: Z bot = I c bot = 10014 . 29 4 . 728 = 2117 . 997 in 2 ft Tensile stress in the deck σ top = M Z top = 12000 ton · ft 1899 . 591 in 2 ft = 6 . 317 ton / in 2 = 14150 . 41 lb / in 2 Compressive stress in the bottom plating σ bot = M Z bot = 12000 ton · ft 2117 . 997 in 2 ft = 5 . 666 ton / in 2 = 12691 . 24 lb / in 2 Safety factor: σ yield /σ top = 34000 / 14150 . 41 = 2 . 403. 9

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