THEORETICAL PART (10 p)

Transcription

1 Solutions to the written exam ( ) in Ship structures Basic course Solutions to the written exam in Ship structures Basic course THEORETICAL PART (10 p) Question 1 (.5 p) Somewhere between the top and the bottom of a beam subjected to bending loads is a surface in which longitudinal lines do not change in length. The neutral axis (NA) of the cross-section is the intersection between the neutral surface and the cross-section. On the neutral axis there are no longitudinal strains ( ε = 0 ) and hence no longitudinal stresses ( σ = 0 ). Question (1 p) a) A plane cross-section of a beam remains plane after bending (J. Bernoulli, 1694). Only normal stresses occur in bending and Hooke s law holds for the normal stress in every fibre of the beam (L. Navier, 186). b) The bending moments and the normal/axial force. Question 3 ( p) a) A safety factor is a value/constant which indicates how, for example, much load a structure is allowed to carry based on a limit value. Assume that yielding is this limit value. If the safety factor is set to one, yielding occurs when the applied load reaches the yield-limit-load. If it is set to two, only half of the applied load is allowed to make sure that yielding will not take place. A high value of the safety factor, is normally used in engineering design of structures which have a safe life design, i.e. failure is not an option. b) The general safety factor is usually defined as the actual strength divided by the required strength. c) Advantages: it is easy to use for the designer, simple and basic expressions, easy and quick to calculate, a safety factor may be divided into products of several safety factors on different levels in the calculation, hands-on i.e. influence of each design parameter is visible. d) Disadvantages: it is not a measure of probability of failure, i.e. it does not indicate the probability/risk of failure, it is not easy to decide what a proper value of the safety factor should be i.e. good engineering judgement is required. Question 4 (.5 p) The resultant of all shear stresses through thickness of a beam is defined as the shear flow, D. For a thin-walled beam, it can be assumed that the shear stress is uniformly distributed. Hence, the shear flow is D= τ b where b is the thickness of the beam. In the connection points, the sum of all shear flows should be equal to ero, i.e. D = 0. It can, for example, be used for calculation of beam shear stresses if the i thickness of the beams in a connection varies. 1 (7) WE _solutions

2 Solutions to the written exam ( ) in Ship structures Basic course Question 5 ( p) a) In probabilistic design, a safety margin, m, is defined as m = r s, where r is the resistance and s is the load effect. For example, the resistance may be a function of material properties and dimensions, while the load effect may be a function of applied and position of loads, density, and dimensions. Usually, both r and s vary in a random manner, and hence, m is also a random variable. b) The safety index is a geometrical interpretation of the shortest distance from the design point, which is defined by the expected values for all random variables in the equation of the safety margin, to the failure surface defined by the equation of the safety margin when it is equal to ero. In a transformed/normalised coordinate system, using normalised coordinates, the safety index, β, is the shortest distance from the origin to the failure surface expressed in normalised coordinates. c) The Cornell safety index, β C : The basic variables are assumed independent and Normal distributed. The safety margin should be a linear combination of the basic variables. The Hasofer-Lind safety index, β HL : We often assume that the basic variables are independent and Normal distributed, however, they may not be independent here. The safety margin can be a nonlinear function of basic variables. In the normalised coordinate system, the safety index must be found by iteration since there is no closed analytical solution to calculate it. (7) WE _solutions

3 Solutions to the written exam ( ) in Ship structures Basic course PROBLEM PART (40 p) Question 6 (10 p) The failure function is given as f ( x1, x) = 1+ x1+ x + x1x. The failure surface is then defined as 1+ x1+ x + x1x = 0, i.e. f( x1, x ) = 0. It is assumed that the basic variables x = ( X1, X) are independent and Normal distributed. Hence, the failure surface f( x ) = 0 is also Normal distributed. Since the failure is a nonlinear function of the basic variables, the Hasofer-Lind s definition of the safety index, β, should be used here. HL Normalise the basic variables: X1 µ X1 Z1 = X1 = Z1σ X + µ 1 X1 σ X1 X µ X Z = X = Zσ + µ σ X X X The normalised -coordinate system failure surface is given by: 1 + ( Zσ + µ ) + ( Z σ + µ ) + ( Zσ + µ ) ( Z σ + µ ) = 0, or 1 X1 X1 X X 1 X1 X1 X X (1 + µ + µ + µ µ ) + Z ( σ + µ σ ) + Z ( σ + µ σ ) + Z Z σ σ = 0 X1 X X1 X 1 X1 X X1 X X1 X 1 X1 X With numerical values, the normalised failure surface is: 1+ Z + Z + Z Z = 0, or Z1+ Z + Z1Z = The reliability index, β, and the design point, * = ( βα1, βα), are determined using the following four equations, see the UB1 literature: 1 β = ( α1+ α + βαα 1 ) 1 α1 = (1 + βα ) k 1 α = (1 + βα1 ) k k = ( α ) + ( α ) k = (1 + βα ) + (1 + βα ) 1 1 These equations must be solved by iteration. In the problem formulation, the start guess was given as [ β; α1; α ] = [3; 0.58;0.58]. Make a table and start with the iteration: Guess β α α (7) WE _solutions

4 Solutions to the written exam ( ) in Ship structures Basic course From the table it can be seen that after the fourth/fifth iteration, the safety index, β, has converged to βhl = β = HL The design point is therefore * = ( βα, βα ) = ( 1.0,0.0). 1 The probability of failure is P = Φ ( β ) = 1 Φ ( β ). The reliability that the structure will survive is P = 1 P = 1 (1 Φ ( β )) =Φ ( β ). r f f Table values for the Standardised Normal distribution gives Φ ( β ) =Φ (1.00) = Hence, P r = i.e. the probability that the structure will survive is 84%. ANSWER: βhl = β = 1.00 * = ( βα, βα ) = ( 1.0,0.0) 1 P = , i.e. 84% r 4 (7) WE _solutions

5 Solutions to the written exam ( ) in Ship structures Basic course Question 7 (10 p) Use St Venant torsion theory to calculate T x. No preventing of warping gives that τw = 0 τ = τsv. Use Vlasov torsion theory to calculate the relative displacement between A and B. ST literature Eq. (4.) on p. 57 gives: L Tx TxL ϕ = θ = ϕ = ϕ dx GK =. V GK 0 V The torsional constant of this cross-section is (ST literature Table 4. on p. 65) N KV = bt i i KV = m. i= 1 3 ϕgkv Hence, Tx = = { ϕ in radians! } =.95 knm. L The relative displacement is defined here as ua B= ua ub. Use ST literature Eq. (5.30) on p. 94 (no bending or axial loads): uxy (,, ) = ϕ ω( xy, ). St Venant torsion theory gives that ϕ is constant, and hence, ϕ = ϕ / L. The sectorial coordinates are given as ωa = ωb = 58.1 m. ϕ Hence, u A = ϕ ω A = ω A = { ϕ in radians! } = m. L Thus, ua B= ua ub = 0.07 m. ANSWER: T x =.95 knm ua B= ua ub = 0.07 m 5 (7) WE _solutions

6 Solutions to the written exam ( ) in Ship structures Basic course Question 8 (10 p) The ST literature Eq. (3.19) on p. 8: the Navier formula: N ymi ( y y + MI y) MI ( y + MI y ) σ = + A ( I I I ) y y We need to calculate M y and M to get an expression of σ in the coordinates y and. All other properties have been given in the problem. o M = Nξ =... = Nm. y o M = Nη =... = 754 Nm Hence, using given data, σ ( y, ) =... = y We are looking for the imum normal stress. Check values in proper corners of the cross-section the most critical one: ( y1, 1) = (1.16, 4.37) 10 m σ1( y1, 1) = 33.4 MPa. Therefore, σ = σ ( y, ) 33 MPa The normal strain in CG is calculated as ε xo N EA 4 = =... = ( M yi + MIy) The curvature, κ y = =... = m EII ( I ) y y -1 ( M yiy + MIy) -1 The curvature, κ = =... = m. EII ( I ) y y. ANSWER: σ = σ ( y, ) 33 MPa ( y1, 1) = (1.16, 4.37) 10 m 4 ε xo = κ y κ = m -1 = m -1 6 (7) WE _solutions

7 Solutions to the written exam ( ) in Ship structures Basic course Question 9 (10 p) Calculate where the shear centre of the cross-section is located using Table 1:34, case 5: o γ = c/ h= 3/1= 0.5 o β = b/ h= 8/1= 0.67 o ε =... = o e= ε h=... = m (see the figure in Table 1:34, case 5). MW ( x) Sω ( s) The warping shear stress is calculated as τw ( xs, ) = where s is a Iω t() s running coordinate around the cross-section. It is convenient here to put s = 0 at A. From the equation above we realise that we need to calculate the statical moment and the sectorial moment of inertia. The sectorial moment of inertia can be calculated using the table: o ω =... = 7.57 m ; ω =... = 0.43 m ; ω =... = 58.1 m o Iω = Kω =... = m. We are aiming for the imum warping shear stress and it is positioned here where the statical moment has its imum. In this case it is for S ω according to the table. 4 o S ω 1 =... =.83 m 4 o S ω =... = 3.66 m o τ = τ Sω =... = 5.65 MPa ANSWER: τ = τ Sω = 5.65 MPa 7 (7) WE _solutions