《船舶与海洋工程结构风险评估》课程教学课件（讲稿）Lecture 8 & 9 Hull Girder Strength I

ASRAnetStructural Reliability & Risk Assessment4-8 July 2016Wuhan, ChinaLecture 11: Hull Girder Strength - IProfessor Purnendu K. DasB.E.,M.E., PhD, CEng,CMarEng, FRINA, FIStructE,FIMarEST

Structural Reliability & Risk Assessment 4 – 8 July 2016 Wuhan, China Lecture 11: Hull Girder Strength – I Professor Purnendu K. Das B.E., M.E., PhD, CEng, CMarEng, FRINA, FIStructE, FIMarEST 1

EShear StressShears on orthogonal forces are complimentaryV-dMdxdyMAMBdx6.X6Shears on orthogonal forces are complimentary2

dx dM V  Shears on orthogonal forces are complimentary. Shear Stress x y x y xy yx x xy dy dx MA MB y yx Shears on orthogonal forces are complimentary. 2

Derivation of EguationM+dMYn1dALetusconsideralengthdxof abeamsubjectedtobendingmomentMandM+dMinthecross sectionmnandm,n,respectivelyThenormal forceactingon anarea dAof the sidenppnMy.dAordAIzThesumofallforcesonthesidenppnMyrh/2dAIzyy3

Derivation of Equation yx dx M m M+dM m n n1 Z b m m n n p y1 p y y dA h max y x y p p1 Let us consider a length dx of a beam subjected to bending moment M and M+dM in the cross section mn and m1 n1 respectively. The normal force acting on an area dA of the side nppn dA I My x dA    The sum of all forces on the side nppn dA I My Z h y . / 2 1  3

Sumof normalforcesactingonsidenP1P,n,is(M +dM)yh/2dAIzVThe unbalanced force is balanced by a shear stress acting along the plane of the cutForce due to shearing stresses Tyx*TyxbdxZX=0(M +dM)yrh/2Mybdx=1Tyx片12dM1rh/2ydAFdxb.1Vh/2ydAor<(1)bl4

Sum of normal forces acting on side n1 p1 p1 n1 is dA I M dM y Z h y . / 2 ( ) 1   The unbalanced force is balanced by a shear stress acting along the plane of the cut. Force due to shearing stresses yx. 1 1 1 1 / 2 / 2 / 2 / 2 0 ( ) . . . yx h h yx y y Z Z h yx y Z h xy yx y Z b dx X M dM y My b dx dA dA I I dM I ydA dx b I V or ydA bI                    (1) 4

Theintegral representsmoment oftheshadedportionofthecrosssectionwithrespecttotheneutralaxisForarectangularsectiondA=bdyandoydy=21Maximum value of txy occurs when y, = O ie for points on the neutral axisVh?(txy)max:81zSince I z=bh123.VV)maxbh2bhq=t.b= shearflow5

The integral represents moment of the shaded portion of the cross section with respect to the neutral axis. For a rectangular section                    2 1 2 2 1 / 2 2 4 2 1 2 4 y h I V y b h and by dy dA bdy Z x y h y  Maximum value of xy occurs when y1 = 0 ie for points on the neutral axis q b shearflow bh V bh V bh Since I I Vh x y Z Z x y        . 1.5 2 3 ( )max 12 8 ( )max 3 2    5