Ship structural design

Ship structural design - 1. Revision Equation Chapter 1...

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1 1. Revision Equation Chapter 1 Section 1 For statically determinate structure, the components of the reaction at the supports can be calculated from: 0 0 0 M H V = = = (1.1) where, M is the bending moment, H is the horizontal component of the load or the reaction, and V is the vertical component of the load or the reaction. The rate of change of the bending moment M at any point is equal to the value of the shear force Q at the same point: dM Q dx M Qdx C = = (1.2) The rate of change of the shear force Q at any point is equal to the value of the load density p at the same point: dQ p dx = (1.3) Sign convention Figure 1 sign convention Sagging Hogging Positive bending Negative bending Positive shear Negative shear
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2 1.1. Horizontal cantilever 1.1.1. Single concentrated load The reactions at the support B are (Figure 2): VP = , 0 H = , . M PL = The internal stress effects at any section (s-s) at a distance x from the end A are: 0 Normal force Shear force Bending moment x x x N QP M Px = = − = − (1.4) Figure 2 1.1.2. Two concentrated loads The reactions at B are (Figure 3):
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3 ( ) ( ) 0 4.0 2.0 6.0 4 4.0 2 1.5 19 . N V ton M ton mt = =+= = += (1.5) The cantilever can be divided to three parts; DB, CD, and AC. AC has not any internal stress effects (no forces exist at the left of any section through this part). The internal stress effects through CD, at any section 1 s at a distance 1 x from C are: 1 4.0 4 x x Q Mx = − = − (1.6) The internal stress effects through DB, at any section 2 s at a distance 2 x from D are: ( ) 22 2 4.0 2.0 6.0 4 2.5 2 10 6 x x Q tons M xx x = −− = = + − = (1.7) Figure 3
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4 1.1.3. Uniformly distributed load The reactions at B are (Figure 4): ( ) 2 0.5 0.5 V pL M pL L pL = = = (1.8) The resultant of the loads at the left of a section (s-s) at a distance x from A is: x W px = (1.9) This force act at the middle of x , then the shear force and the bending moment at (s-s) are: 2 0.5 x x Q px M px = − = − (1.10) Figure 4 1.1.4. Linearly distributed load The resultant of the distributed load is (Figure 5): 0.5 W pL = (1.11)
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5 Which acts at /3 L from the end B. The reactions at B are: 2 0.5 6 V pL pL M = = (1.12) Load density x p at any section (s-s) at a distance x from the end A is: x x pp L  =   (1.13) The resultant of the loads at the left of the section (s-s) is: 2 0.5 2 xx px W xp L = = (1.14) which acts at a distance 3 x from the left of the section (s-s). So, the shear force and the bending moment at (s-s) will be: 2 3 2 36 px QW L x px MW L = −= = (1.15) Figure 5
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6 1.1.5. Examples Figure 6
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7 Figure 7
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8 Figure 8
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9 Figure 9 1.2. Simple beam 1.2.1. Single concentrated load The vertical reactions at the ends are (Figure 10): a b b VP L a L  =   = (1.16) The maximum bending moment will be: max ab MP L = (1.17)
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10 Figure 10 1.2.2. Uniformly distributed load The vertical reactions at the ends are (Figure 11): 2 ab L VVP  = =   (1.18) The shear force at any distance x from A will be: 2 x L Qp x = (1.19) The bending moment at any distance x from A will be: ( ) 0.5 x M px L x = (1.20) The maximum bending moment will be at the middle of the beam and equal: 2 max 8 pL M = (1.21)
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11 Figure 11 1.2.3. Symmetrical triangular load The total load on the beam is 0.5 pL and the reaction at each support equals 0.25
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This note was uploaded on 03/19/2012 for the course ECON 256 taught by Professor Lopez during the Spring '10 term at École Normale Supérieure.

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Ship structural design - 1. Revision Equation Chapter 1...

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