《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

ExampleFindthemaximumshearstressintheT-sectionof a simplysupportedbeamas shown in Fig 1.400kN/m22mm4m305mm13mm6

Find the maximum shear stress in the T-section of a simply supported beam as shown in Fig 1. Example 4m 400kN/m 22mm 13mm 305mm 6

SolutionTaking moment about x-x(13×(305-22)+(22×230)x=13×283×163.5 +22×230×11657176.5：75.20mmx8739×13×2833 +13×283×(88.3)12×230×223 + 230×22×(64.2)12= 74.298×106mm4

13305 2222230x 13283163.5 2223011 x 75.20mm 8739 657176.5       6 4 2 3 2 3 74.298 10 230 22 230 22 64.2 12 1 13 283 13 283 88.3 12 1 mm NA                Taking moment about x-x. Solution 7

400kN/m4m400×4MaximumShearForce==800k2AyTxyLh53.2Ay=230×22×64.2+53.2×13x2=343249mm800x103x34.32×104X113×74.298×106=284.26N / mm228

4m 400kN/m Maximum Shear Force = 800kN 2 400 4     2 4 6 3 3 284.26 / 34.32 10 13 74.298 10 800 10 343249 2 53.2 230 22 64.2 53.2 13 . N mm mm Ay Ay b V x y x y                   8

CircularSectionLetus calculatenowtheshearingstressesalongthe lineppofthecircularcrosssectionInapplyingequation(1)tothecalculationoftheverticalcomponenttxy ofthese stresses,wemustfindthemomentofthesegmentofthecirclebelowthelineppwithrespecttotheZC1z-axisyiThe elemental area mn dA = 2/R2 - y2 .dly个ThemomentofthisstripaboutCzisydA.Thetotalmomentoftheentiresegment-y2 . y·dy=2(R? -福If we substitute this in basic equation (1) and take 2/R? - y? = bV(R?-y2)2)bV31zblzVRJR?-y?Txy.RThetotal shearingstress at point pis31zJR?-y?Itis seenthatthemaximumt occurswheny1=o iefortheneutral axis of the cross sectionTR44VR?4V4rATmax9TR23元R433

Circular Section Z C p m R y p m dy y1 Let us calculate now the shearing stresses along the line pp of the circular cross section. In applying equation (1) to the calculation of the vertical component xy of these stresses, we must find the moment of the segment of the circle below the line pp with respect to the z-axis. The elemental area mn dA 2 R y .dy 2 2   The moment of this strip about CZ is ydA. The total moment of the entire segment   2 3 2 1 2 2 2 3 2 2 1 R y y dy R y R y        If we substitute this in basic equation (1) and take R  y  b 2 1 2 2     Z Z x y I b V R y R y bI V 3 3 2 1 2 2 1 2        The total shearing stress at point p is Z xy I VR R y R y R 3 . 2 1 2 2 1 2       It is seen that the maximum  occurs when y1 = 0 ie for the neutral axis of the cross section. A V R V x R VR R I Z       3 4 3 4 4 3 4 4 2 2 max 4     9

In the case of a circular section, the maximum shearing stress is 33% larger than the averagevalueobtainedbydividingtheshearingforcebythecross-sectional area.I-beamsAssumptionsShearingstressesareparalleltothesharingforceVTheyareuniformlydistributed over thethicknessb1of theweb.Letusfindtheshearingstressatpointpp.ThenthemomentoftheshadedportionofthecrosssectionwithrespecttotheneutralaxisZIf we substitute this in the basic equation (1),we obtainh4b,110

In the case of a circular section, the maximum shearing stress is 33% larger than the average value obtained by dividing the shearing force by the cross-sectional area. h h1 b y p p V Z b1 y1 I-beams Assumptions Shearing stresses are parallel to the sharing force V They are uniformly distributed over the thickness b1 of the web. Let us find the shearing stress at point pp. Then the moment of the shaded portion of the cross section with respect to the neutral axis Z.                      2 1 2 1 1 2 1 2 2 1 2 4 4 2 4 y b h h b h ydA h y If we substitute this in the basic equation (1), we obtain                             2 1 2 1 1 2 1 2 1 2 4 4 2 4 y b h h b h b I V Z x y  10