《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

Applicationtoshipstructure16

R V Application to ship structure 16

ExampleproblemAmidshipsectionofacargoshipisshowninFig1and issubjectedtoaverticalshearforceof20,oo0kNCalculateanddrawtheshearstressdistributioninthesectionabovetheneutralaxis,N-A.Ifaverticalbendingmomentof6x1o5kNmalsoactsonthesection,calculatetheequivalentstressoeinthetopdeckusingthevonMisesequation:2 +3t2Ce=1Owherebandarethebendingandshearstressesrespectively.Themoment ofinertiaofthe sectionabouttheneutralaxis(N-A)is49.5m4andtheneutralaxisis5.27mfromthebase,asshowninFig1.7m18mm4mFig 1C14m2mm12mm7m+5.27R1.4m15mm12mm16mm14m17

A midship section of a cargo ship is shown in Fig 1 and is subjected to a vertical shear force of 20,000 kN. Calculate and draw the shear stress distribution in the section above the neutral axis, N-A. If a vertical bending moment of 6 x 105 kNm also acts on the section, calculate the equivalent stress e in the top deck using the von Mises equation: Example problem 2 2     3 e b where b and  are the bending and shear stresses respectively. The moment of inertia of the section about the neutral axis (N-A) is 49.5 m4 and the neutral axis is 5.27m from the base, as shown in Fig 1. 7m 7m 18mm 16mm 15mm 16mm 12mm 12mm 14m 14m 1.4m N A C D B A 5.27 m C 12mm 4m Fig 1 17

SolutionV=20.000kNV20,0004.95×1012=4.04×10-7N/mm41=TA=OVAy=4.04×10-7x7000×18×8730(TB)BA18T107=4.04×10-7×109.99x=24.68N/mm218107(tB)Bc =4.04×10-7 ×109.99×27.77N/mm216109.99×107+4000×16×6730(tc)cB = 4.04×10-7 x16=4.04×10-7x153.06x107=38.65N/mm216D = 4.04×10-7×7000×12×4730t)12=4.04x10-739.73x107=13.38N /mm21218

Solution V = 20,000 kN A = 0   2 7 7 7 24.68 / 18 10 4.04 10 109.99 18 7000 18 8730 4.04 10 N mm t Ay I V B BA                  2 7 7 27.77 / 16 10 B BC  4.0410 109.99  N mm     2 7 7 7 7 38.65 / 16 153.06 10 4.04 10 16 109.99 10 4000 16 6730 4.04 10 N mm C CB                          7 7 2 7 10 13.38 / 12 39.73 4.04 10 12 7000 12 4730 4.04 10 N mm C CD              𝑉 𝐼 = 20,000 4.95 × 1012 = 4.04 × 10−7𝑁/𝑚𝑚4 18

(153.06×107+39.73×107=4.04×10-7T16192.79=4.04×10-7,×107=48.68N/mm2164730192.79×107+4730×16x2(tc)Nc.= 4.04×10-716210.68= 4.04×10-7 x=53.195N/mm16ob at topdeckM8730·y=6×105×10°×1034.95×10131=105.82N/mm20=Jo, +3t2=/105.822+3×24.682=114.09N/mm2Theshearstressdistributionisshown inthefigure19

  7 7 2 7 7 7 10 48.68 / 16 192.79 4.04 10 16 153.06 10 39.73 10 4.04 10 N mm C CN                         7 2 7 7 53.195 / 16 210.68 4.04 10 16 2 4730 192.79 10 4730 16 4.04 10 N mm C NC                           b at top deck 5 3 3 13 2 8730 6 10 10 10 4.95 10 = 105.82N/mm M        y   2 2 2 2 2 3 105.82 3 24.68 = 114.09N/mm    e b      The shear stress distribution is shown in the figure. 19