型材剖面模数计算

剖面模数计算

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56.00 50.00 50.00 72.00 50.00 50.00 84.00 50.00 50.00 50.00 91.00 91.00 50.00 50.00 50.00 50.00 50.00 50.00 60.00 60.00 50.00 50.00 50.00 50.00 50.00 50.00 50.00 50.00 50.00 50.00 50.00 50.00 50.00
zn
h
5.73 12.68680168
1.165
0.926
2.72
2.04
3.57
1.95
4.42
1.86
5.27
1.77
6.12
1.64
6.97
1.49
L400× 120×12
×23
W=1624.
0.85
12.69
4
662.22 89 cm3
L400×
120×12
×23
W=1624.
0.85
11.43
135.01 156.35 143.10 176.66 198.08 222.24 228.33 254.82 286.55 315.72 288.72 335.45 358.61
395.82 398.97 450.07 497.58 541.77 68.72 99.58 111.09 139.40 155.61 188.68 206.01 252.54 275.21 324.37 348.50 407.54 436.64 517.49 548.87
T型材计算
T型材带板面积
轧制型材计算
L63×40×4 L63×40×5 L63×40×6 L63×40×7 L75×50×5 L75×50×6 L75×50×8 L75×50×10 L90×56×5 L90×56×6 L90×56×7 L90×56×8 L100×63×6 L100×63×7 L100×63×8 L100×63×10 L100×80×6 L100×80×7 L100×80×8 L100×80×10 L110×70×6 L110×70×7 L110×70×8 L110×70×10 L125×80×7 L125×80×8

项目六--6.2.3型材剖面设计实例(精)

试确定该船的实肋板尺寸。
176N / m m2， M 77kN m，N 71kN， Y 235N / m m2，
任务2 优化设计船舶型材剖面
项目六
船舶型材剖面设计
1、计算 W1 和 f 0
W1

M
437.5cm2，f 0
N 9.5cm2 0.85
船舶技术设计
项目六
船舶型材剖面设计 6.2.3 型材剖面设计实例
学习内容：某船用T型材的剖面优化设计学习目标：初步具有优化设计型材剖面的能力
任务2 优化设计船舶型材剖面
项目六
船舶型材剖面设计
例题：
已知条件：
88N / m m2，f 2 18cm2，t0 4m m，l 8m
式中： a1
W1 f ，a2 2 fh f
任务2 优化设计船舶型材剖面
项目六
船舶型材剖面设计
5、第二次近似决定m 因为：

2 f1 f 0.805 2 f2 f N 74.6n / m m2 0.85 f
0.424
则运用式（7-19），可得m=78.4
所以总稳定性可以得到保证。
任务2 优化设计船舶型材剖面
ቤተ መጻሕፍቲ ባይዱ
项目六
船舶型材剖面设计
拓展与思考
案例1：某船T型材主肋骨剖面的优化设计。
设规范对某船货舱主肋骨所要求的剖面模数 [W]=1200cm3 ，要求的剖面惯性矩 [I]=17640cm4 ，已知此货舱舷侧外板厚度 t2 =18mm，肋距 s=0.78m，主肋骨跨距 l=4.5m，试设计此主肋骨用T型材的剖面尺寸。
任务2 优化设计船舶型材剖面

第十六讲型材剖面设计概要

100 102 m 70 Y 102 A
因为Ｋ＝４，所以 2 3 3 W0 0.75m t0 235.2cm （４）第一次近似计算型材剖面尺寸 2 < 及 < W0 W1 2W0 f 0 < mt0 hopt m t0 28cm 则
2 f m t0 11.2cm2
船体强度与结构设计
初取腹板尺寸为３００×４，则
（７）确定面板尺寸
a2 (3a1 1) 0.25(6a1 1) f1 f 12.1cm2 3a2 1
由式（5.3.2）及（5.3.3）决定的n0在９～１８之间。由式（5.3.32）面板宽度为：
b1 2n0 f1 14.8 ~ 20.9cm
船体教研室吴春芳编 Email:qiuyuan_cat@
船体强度与结构设计
３）设计变量的类型
a.结构的剖面几何参数
b.结构的布局参数（如结构的型式，节点位置，构件间距等） c.结构所用材料的参数其中a是最常见也是最简单的一类结构优化设计
问题
明确一个标志：设计空间
船体教研室吴春芳编 Email:qiuyuan_cat@
船体教研室吴春芳编 Email:qiuyuan_cat@
船体强度与结构设计
3)可行域满足所有约束条件的结构设计方案是可以被应用的设计，称为可行域。所有可行设计点的集合称为可行域。构成可行域边界的约束曲面称为临界约束。
x1
0 = ) x h1(
h2(x )=0
0
可行域
0 = ) x ( h
f x min或 max
T
使目标函数并受到约束
h j X 0
g k X 0
j 1,2,3, J k 1,2,3, K

剖面模数计算

剖面模数计算剖面模数是指材料或构件在剖面方向上的刚度。

它是用来计算剖面弯曲刚度或剖面截面形状对剖面变形的影响程度的一个参数。

在工程设计中，剖面模数的计算常常涉及到材料力学性质和截面形状等多个因素。

下面将详细介绍剖面模数的计算方法，并给出一些相关参考内容。

一、剖面模数的定义：剖面模数是指材料或构件在剖面方向上的刚度，表示材料或构件抵抗剖面方向上变形的能力。

剖面模数的单位通常是mm^3。

二、剖面模数的计算方法：1.矩形截面：对于矩形截面，剖面模数的计算公式为：W = bh^2/6其中，W表示剖面模数，b和h分别表示矩形截面的宽度和高度。

2.圆形截面：对于圆形截面，剖面模数的计算公式为：W = πd^3/32其中，W表示剖面模数，d表示圆形截面的直径。

3.其他复杂截面：对于其他复杂的截面形状，可以通过将其分割成多个简单的几何体来计算剖面模数，然后将结果进行相加。

例如，可以将T 形截面分割成矩形和两个L形截面，然后计算每个部分的剖面模数，并将它们相加得到总的剖面模数。

三、相关参考内容：1.《结构计算手册》（程季男著）：该书对剖面模数的计算方法作了详细的介绍，涵盖了各种常见截面形状的剖面模数计算公式，同时还对剖面模数的应用进行了讲解。

2.《结构力学与结构设计》（刘妙德、吴少军著）：该书对剖面模数的计算方法进行了深入的研究，重点介绍了复杂截面形状的剖面模数计算方法，并给出了实际应用案例。

3.《钢结构设计手册》（沈大成、杨智、吕玉奇著）：该手册对剖面模数的计算方法进行了详细的介绍，包括各种常见钢结构截面形状的剖面模数计算公式，并对剖面模数的应用进行了实例分析。

4.《建筑结构》（许春明、王道生主编）：该书对剖面模数的计算方法进行了全面的介绍，包括各种常见剖面形状的剖面模数计算公式以及计算步骤，对于各类材料的剖面模数计算均有详细说明。

以上是关于剖面模数计算的相关参考内容，读者可以根据实际需求选择参考书籍或文献来进行学习和研究。

剖面模数计算方法

Allowable stress to ABS MODU 2001, part 3, charpter 2, section 1, item 3.3F=Fy/F.S., whereFy = 235 N/mm2 , or 34 ksiF.S. = 1.67 for axial or bending stress2.50 for shear stressHence, F = 140.7 N/mm2 , or 20.4 ksi for axial or bending stress94.0 N/mm2 , or 13.6 ksi for shear stress1. Bulkhead1.1 Wind pressure p = f V k2.c h.c s N/m2wheref = 0.611Vk = 100 knots = 51.44 m/sc s = 1.0c h = 1.1hence p = 1778.4 N/m2or 37.13 lbf/ft21.2 Bulkhead platingPlate panel maximum size (mm)4070 by 690Plate thickness, t (mm)8Bulkhead load to wind pressure p = 1778.4 N/m2or 37.13 lbf/ft2Stress due to lateral perpendicular load:σ = kpb2/t2 wherek = 0.741 for panel size ratio of 5.9 (4070/690)p =37.13lbf/ft2, or0.26 lbf/in2b =690 mmt =8mmHenceσ =1421 lbf/in2, or 1.42ksi OK3Shear stress at support,τ = RF max/A web = 4.49N/mm2, or0.7ksi OK2. Bottom2.1. bottom platingPlate panel maximum size (mm)2650 by 830Plate thickness, t (mm)8Deck load to MODU 2001, w920 kgf/m2, or 188 lbf/ft2Stress due to lateral perpendicular load:σ = kwb2/t2 wherek = 0.718 for panel size ratio of 3.19 (2650/830)w =188lbf/ft2, or 1.31 lbf/in2b =830mmt =8mmHenceσ =10090 lbf/in2, or10.1ksi OK33. APV' lower Supporting StructureAs per contract specification 2.22G, foundations for equipment shall be designed for combined staticand dynamic load of 1.5g vertical and 0.5g horizontal for roll and pitch.According to HYDRALIFT Drawing: T2820-D1157-G0040 APV's arrangement,per WORKING APV' average weight: 2750kg,add 10% variables: 3025kg is to be used in following calculation.3.1 check supporting plate panelThe supporting plate panel, which is supported at four sides, is considered conservatively as plate beam supported at two longer edges.Plate panel concentrated load maximum size (mm)1420 by 760Plate thickness, t (mm) =25.5Deck load to MODU 2001, w =920kgf/m2, or 188 lbf/ft2Max moment due to deck load q: M q =qL/8 =925N.mwhere L =0.76mMax reaction force due to deck loa R q=qL/2=4870NLoad Case 1 (LC1): Heave at 1.5gForce due to static and dynamic load:P = ma,wherem=3025kga=14.7m/s2 (1.5g)P=44467.5NHence,Q=2P = 88935NM1max=Ql1l2/L=16605N.mwhere L=0.76ml1=0.33ml2=0.43mR1max=Ql2/L=50318NForce due to pitch:P=ma,wherem=3025kga= 4.9m/s2 (0.5g)P pitch=14822.5NHence,Q2=2.755*P/5.76 = 7090NThe force acts on plate as a longitudinal tension, as illustrated in sketchLC3: Roll at 0.5g to starboardForce due to roll:P=ma,wherem=3025kga= 4.9m/s2 (0.5g)P=14822.5NHence,Q2=2.755*P/5.76 = 7090NThe force acts on plate as a transverse tension, as illustrated in sketchLC4: Heave at 1.5g, pitch at 0.5g to forward and roll at 0.5g to starboard (LC1+LC2+LC3)moment:BM max=M1max + Mq =17530N.mshear:RF max=R1max + Rq =55188Nlongitudinal tension:TF x =14179Ntransverse tension:TF y =14179Nplate beam modulus:SM=bt2/6 =154cm3where b =142cmt = 2.55cmplate beam area:A1 =bt =362cm2A2 =at =194cm2where a =76cmBending stress,σ = BM max/SM =113.91N/mm2, or16.5ksi OK Shear stress,τ = RF max/A1 = 1.52N/mm2, or0.2ksi OK Longitudinal tension stress:σx = TF x/A2 =0.73N/mm2, or0.1ksi OK Transverse tension stress:σy = TF y/A1 =0.39N/mm2, or0.1ksi OK3.2 Check supporting structurewhere L= 1.42mBM max = (q1+q2)L2/8 =1774kgf.mRFmax = (q1+q2)L/2 = 4997kgf3Bending stress ,σ = BM max/SM = 6.21N/mm2, or0.9ksi OK Shear stress ,τ = RF max/A1 = 6.81N/mm2, or 1.0ksi OKb. Beam A2-B2Similar to beam A1-B1, check beam A2-B2 stress is OK.R B2 =4964kgfc. Beam A3-B3Similar to beam A1-B1, check beam A3-B3 stress is OK.R B3 =2697kgfd. Beam A4-B4Similar to beam A1-B1, check beam A4-B4 stress is OK.R B4 =2482kgfe. Beam A5-B5Similar to beam A1-B1, check beam A5-B5 stress is OK.R B5 =4964kgff. Beam A6-B6Similar to beam A1-B1, check beam A6-B6 stress is OK.R B6 =4964kgfg. Beam A7-B7Similar to beam A1-B1, check beam A7-B7 stress is OK.R B7 =4964kgfh. Beam A8-B8Similar to beam A1-B1, check beam A8-B8 stress is OK.R B8 =4964kgfi. Beam A9-B9Similar to beam A1-B1, check beam A4-B4 stress is OK.R B9 =2482kgfj. Beam C1-D1Similar to beam A1-B1, check beam C1-D1 stress is OK.R C1 =4989kgfR D1 =4989kgfk. Beam C2-D2Similar to beam A1-B1, check beam C2-D2 stress is OK.R C2 =4957kgfR D2 =4957kgfl. Beam C3-D3Similar to beam A1-B1, check beam C2-D2 stress is OK.R C3 =2690kgfR D3 =2690kgf3.2.2 Check transverse girdersMax moment due to force R B1: M B1 = 0.76*1.985*R B1/2.745 =2746kgf.mMax moment due to force R B2: M B2 = 1.42*1.325*R B2/2.745 =3402kgf.mMax moment due to force R B3: M B3 = 2.08*0.665*R B3/2.745 =1359kgf.m Combined moment: BM max =6163kgf.mReaction force: R E1 = 1.985*R B1/2.745 + 1.325*R B2/2.745 + 0.665*R B3/2.745 =6663kgf Reaction force: R F1a = 0.76*R B1/2.745 + 1.42*R B2/2.745 + 2.08*R B3/2.745 =5995kgf hence,RF max =6663kgfBending stress ,σ = BM max/SM =24.00N/mm2, or 3.5ksi OK Shear stress ,τ = RF max/A WEB =8.17N/mm2, or 1.2ksi OKn. Beam E2-F2Similar to beam E1-E1, check beam E2-F2 stress is OK.Reaction force: R F2 =5984kgfDistributed load along the beam length due to bulkhead weight, q = 660kgf/mMax moment due to load q: M q =qL2/8 =622kgf.mMax moment due to force R D1: M D1 = 0.76*1.985*R D1/2.745 =2742kgf.mMax moment due to force R D2: M D2 = 1.42*1.325*R D2/2.745 =3398kgf.mMax moment due to force R D3: M D3 = 2.08*0.665*R D3/2.745 =1355kgf.mCombined moment: BM max =6774kgf.mReaction force: R E3 =7558kgfReaction force: R F3a =6890kgfhence,RF max =7558kgfBending stress ,σ = BM max/SM =26.38N/mm2, or 3.8ksi OK Shear stress ,τ = RF max/A WEB =9.27N/mm2, or 1.3ksi OKDeck load to MODU 2001, w = 920kgf/m2 or 188 lbf/ft2Distributed load along the beam length, q = 0.165*w =151.8kgf/mMax moment due to load q: M q =q*1.4452*(1+1.3/2.745)2/8 =86kgf.mMax moment due to force R B4: M B4 = 1.445*1.3*R B4/2.745 =1699kgf.mMax moment due to force R B5: M B5 = 2.105*0.64*R B5/2.745 =3402kgf.mCombined moment: BM max =4259kgf.mReaction force: R F1b =2424kgfReaction force: R =5146kgfthk(cm)width(cm)sectionarea(cm2)ctr.dist. toplt top(cm)d(cm)I0 (cm4)mom. ofinert.(cm4)SM(cm3)top flg 2.5516.542.075 1.27522.844135.0web1808042.5542666.748997.3btm flg0.816.513.282.950.732077.6 Combined135.27533.7125210.02520 Bending stress ,σ = BM max/SM =16.58N/mm2, or 2.4ksi OK Shear stress ,τ = RF max/A WEB = 6.31N/mm2, or0.9ksi OKDeck load to MODU 2001, w = 920kgf/m2 or 188 lbf/ft2Distributed load along the beam length, q1 = 0.165*w =151.8kgf/mDistributed load along the beam length due to bulkhead weight, q2 = 660kgf/m BM max = (q1+q2)L2/8 =765kgf.mRFmax = (q1+q2)L/2 = 1114kgfHence,R =1114kgfBending stress ,σ = BM max/SM = 2.98N/mm2, or0.4ksi OK Shear stress ,τ = RF max/A WEB = 1.37N/mm2, or0.2ksi OKr. Beam E5-F5Similar to beam F3-E5, check beam E5-F5 stress is OK.Reaction force: R E5b =1185kgfR F5 =1185kgfDeck load to MODU 2001, w = 920kgf/m2 or 188 lbf/ft2Distributed load along the beam length, q = 0.165*w =151.8kgf/mMax moment due to load q: M q =q*0.832*(1+2.66/3.49)2/8 =41kgf.mMax moment due to force R B6: M B6 = 0.68*2.81*R B6/3.49 =2718kgf.mMax moment due to force R B7: M B7 = 1.34*2.15*R B7/3.49 =4098kgf.mMax moment due to force R B8: M B8 = 2.0*1.49*R B8/3.49 =4239kgf.mMax moment due to force R B9: M B9 = 2.66*0.83*R B9/3.49 =1570kgf.mCombined moment: BM max =9829kgf.mReaction force: R E4b =9779kgfBending stress ,σ = BM max/SM =38.27N/mm2, or 5.6ksi OK Shear stress ,τ = RF max/A WEB =11.99N/mm2, or 1.7ksi OK3.2.3 Check longitudinal girdersDeck load to MODU 2001, w = 920kgf/m2 or 188 lbf/ft2Distributed load along the beam length, q = 0.3*w =276kgf/mMax moment due to load q: M q =q*3.5882/2 =1777kgf.mMax moment due to force R F1a +R F1b: M F1 = 0.938*(R F1a+R F1b) =7897kgf.mMax moment due to force R F2: M F2 = 2.193*R F2 =13123kgf.mMax moment due to force R F3a +R F3b: M F1 = 3.588*(R F3a+R F3b) =29041kgf.mCombined moment: BM max =51838kgf.mReaction force: R G1 = q*3.588 + RF1a + RF1b + RF2 + RF3a + RF3b=23397kgfBending stress ,σ = BM max/SM =167.46N/mm2, or24.3ksi OK Shear stress , 1.2τ = RF max/A WEB =65.58N/mm2, or9.5ksi OKDeck load to MODU 2001, w = 920kgf/m2 or 188 lbf/ft2Distributed load along the beam length, q1 = 0.3*w =276kgf/mLoad as Heave at 1.5gForce due to static and dynamic load:P = ma,wherem=3025kga=14.7m/s2 (1.5g)P=44468NHence,q2=2P/L = 6384kgf/mwhere L= 1.42mMax moment due to load q1: M q1 =q1*4.072/2 =2286kgf.mMax moment due to load q2: M q2 =q2*1.422/2 =6437kgf.mMax moment due to force R E4a +R E4b: M E4 = 1.42*(R E4a+R E4b) =21194kgf.mMax moment due to force R E5a +R E5b: M E4 = 4.07*(R E5a+R E5b) =9357kgf.mCombined moment: BM max =39273kgf.mReaction force: R G2 = q1*4.07 +q2*1.42 + R E4a + R E4b + R E5a + R E5b=27413kgf hence,RF =27413kgfBending stress ,σ = BM max/SM =65.74N/mm2, or9.5ksi OK Shear stress ,τ = RF max/A WEB =26.89N/mm2, or 3.9ksi OKv. Beam G3-F5Deck load to MODU 2001, w = 920kgf/m or 188 lbf/ft2Distributed load along the beam length, q1 = 0.165*w =151.8kgf/mDistributed load along the beam length due to bulkhead weight, q2 = 660kgf/m Max moment due to load q1: M q1 =q1*4.072/2 =1257kgf.mMax moment due to load q2: M q2 =q2*4.072/2 =5466kgf.mMax moment due to force R F4: M F4 = 1.42*R F4 =10964kgf.mMax moment due to force R F5: M F5 = 4.07*R F5 =4823kgf.mCombined moment: BM max =22510kgf.mBending stress ,σ = BM max/SM =62.18N/mm2, or9.0ksi OK Shear stress ,τ = RF max/A WEB =11.26N/mm2, or 1.6ksi OK4. APV' Upper Supporting Structure3.1 :P pitch =14822.5NQ1pitch =7733N Load due to a APV's Roll at 0.5g to starboard has calculated as 3.1 :P roll =14822.5NQ1roll =7733N 4.1 Check APV' end box mounting structure on forward transverse bulkhead4.1.1 Check stiffener' flange subjected to tensionAs per "Yield Line Analysis of Bolted Hanging Connections", AISC, Engineering Journal, Vol.14, No.3 1977, For hanger rods, the allowable working load is the smaller of following :P1 = F y t b2(2r)1/2(1+a/b)/LFP2 = F y t b2[r(1+a/b)]1/2/LFwhere F y=235N/mm2t b=13mmr= (F y-F b)/F y =0.401F b=140.7N/mm2a=50mmb=35.5mmLF = 1.7P1 =50388NP2 =22959Nhence,the allowable total force carried by flange[ P ]=22959Nmaximal load forced on stiffener L100x75x13 is P max = 1.5Q1roll = 11600 N < [ P ]OK!4.1.2 Check stiffener subjected to compressionR max =8522N9thk(cm)plt width/sect dep(cm)sectionarea(cm2)ctr.dist. toplt top(cm)d(cm)I0 (cm4)mom. ofinert.(cm4)SM(cm3)att plt0.85644.80.4 2.493.9section-7.515.46 5.9794.6359.7Combined60.26 1.8453.6704.3in3 Bending stress ,σ = BM max/SM =23.83N/mm2, or 3.5ksi OKR max=R F =8738Nthk(cm)plt width/sect dep(cm)sectionarea(cm2)ctr.dist. toplt top(cm)d(cm)I0 (cm4)mom. ofinert.(cm4)SM(cm3)att plt 1.2519.2240.625 3.1196.0section-7.521.06 6.6994.6314.4Combined45.06 3.5510.3965.9in3Bending stress ,σ = BM max/SM =22.61N/mm2, or 3.3ksi OK Shear stress ,τ = RF max/A1 = 4.15N/mm2, or0.6ksi OKC. Check beam L-MR max =11934Nthk(cm)width(cm)sectionarea(cm2)ctr.dist. toplt top(cm)d(cm)I0 (cm4)mom. ofinert.(cm4)SM(cm3)top flg00000.00.0web0.9 2.5 2.25 1.25 1.2 4.8btm flg0.97.5 6.75 2.950.5 1.7Combined9 2.5 6.530.2in3 Bending stress ,σ = BM max/SM =4145.20N/mm2, or601.6ksi OK Shear stress ,τ = RF max/A1 =53.04N/mm2, or7.7ksi OK4.2 Check APV' end box mounting structure on inboard longitudinal bulkheadAs per "Yield Line Analysis of Bolted Hanging Connections", AISC, Engineering Journal, Vol.14, No.3, 1977, For hanger rods, the allowable working load is the smaller of following :P1 = F y t b2(2r)1/2(1+a/b)/LFP2 = F y t b2[r(1+a/b)]1/2/LFwhere F y=235N/mm2t b=19mmr= (F y-F b)/F y =0.401F b=140.7N/mm2a=50mmb=35.5mmLF = 1.7hence,P1 =107634NP2 =49042Nhence,the allowable total force carried by flange[ P ]=49042Nmaximal concentrated load forced on girder T 811x12.5w P max = 3Q2roll = 23199 N < [ P ]OK!4.2.2 Check longitudinal girder' web stability under compression when roll to starboardAs per "Manual of STEEL CONSTRUCTION Allowable Stress Design", AISC,Slenderness ratio Kl/r =450> 200where K =2l =811mmr = 3.61mmAnd C c =(2*3.142E/F y)1/2 =130where E =200000MpaF y =235N/mm2here,Kl/r >C chence,the allowable stress F a = 12*3.142E/(23*(Kl/r)2 = 5.08N/mm2Compression total load forced on Girder' web section Q =12*Q2roll92796N web section area A=19625mm2RF max =92796Nthk(cm)width(cm)sectionarea(cm2)ctr.dist. toplt top(cm)d(cm)I0 (cm4)mom. ofinert.(cm4)SM(cm3)top flg 2.5547.3120.615 1.27565.474578.2web 1.2581.1101.37543.155563.784757.5btm flg 1.911.521.8584.6 6.674705.9Combined243.8426.1234041.73939240.4in3 Bending stress ,σ = BM max/SM =17.91N/mm2, or 2.6ksi OK Shear stress ,τ = RF max/A web =9.15N/mm2, or 1.3ksi OK4.3 Check supporting APV' end box mounting structure on TF-12 transverse bulkheadBending stress ,σ = BM max/SM =72.79N/mm2, or10.6ksi OK Shear stress ,τ = RF max/A web =17.26N/mm2, or 2.5ksi OKthk(cm)width(cm)sectionarea(cm2)ctr.dist. toplt top(cm)d(cm)I0 (cm4)mom. ofinert.(cm4)SM(cm3)top flg 1.310130.65 1.8871.8web 1.3 6.28.06 4.425.8184.0btm flg 1.957.5109.258.4532.948.7web 1.2581013.453.3262.1top flg 1.25121518.025 2.01270.1Combined155.318.8302636.726916.4in3Bending stress ,σ = BM max/SM =74.44N/mm2, or10.8ksi OK Shear stress ,τ = RF max/A web =17.26N/mm2, or 2.5ksi OK。

型材的剖面模数

1360
1380 1400 1450
1500
1550 1600
1650
1700 1750 1800 1850 1900
1950
2000 2050
2100
2150 2200 2250 2300 2350 2400 2450
具有附连翼板的剖面，（mm）
- 370×14 - 370×15
- 360×28 - 380×26 - 400×24
- 280×12
580
- 280×22
肘板尺寸
折
无折边
有折边边
55
mm - 280×9.5 - 280×8.0
- 290×9.5 - 290×8.0
折边 60
- 310×10.5 - 310×8.5
mm
- 315×10.5 - 315×8.5 - 320×10.5 - 320×8.5
- 330×11.0 - 330×9.0
- 200×17
240 - 150×100×12
- 220×15
250 - 180×90×10 -200×11.5 - 200×18
260 - 160×80×14
- 220×16 - 240×14
肘板尺寸
无折边
有折边
- 180×7.0 - 180×6.5
折边 - 190×7.0 - 190×6.5
- 370×13
肘板尺寸
无折边
有折边
- 370×12.0 -370×9.5
折边 - 380×12.0 -380×10.0
70 mm
- 390×12.5 -390×10.0
折 - 400×13.0 - 400×10.5 边

第十四讲型材剖面设计

M W1
及
NS It
则首先要建立型材剖面要素与剖面几何尺寸之间的关系式。
船体教研室吴春芳编 Email:qiuyuan_cat@
船体强度与结构设计
1.以焊接于船体钢板上的T型材为例
1）参考轴的选取：以小翼板厚度中点的轴线О'—О'为参考轴，则剖面中和轴至参考轴的距离为：
h f2h f 2 h1 f1 f 2 f
…… （1）移轴定律
船体教研室吴春芳编 Email:qiuyuan_cat@
船体强度与结构设计
2)剖面对中和轴的惯性矩为：
2 h fh I f 2h2 f ( f1 f 2 f )h12 ……（2） 12 2 2
在剖面积F和高度h相同的情况下,剖面利用系数的大小表明了材料在剖面中分布的合理程度.
比面积: 1.剖面模数比面积 2.剖面惯性矩比面积意义:产生单位剖面模数所需的剖面积.
船体教研室吴春芳编 Email:qiuyuan_cat@
船体强度与结构设计
二、型材的强度要求及剖面要素计算
为保证型材有足够强度，必须使翼板的最大正应力和腹板上的最大剪应力小于许用应力，即
对最小板厚的要求；
４）剖面内材料分布合理，使得结构重量最轻，这是船体结构工程师的重要目标之一。
船体教研室吴春芳编 Email:qiuyuan_cat@
船体强度与结构设计
2、衡量型材剖面积内材料分布合理程度的指标有：剖面利用系数、比面积梁材受横荷作用时其抵抗弯矩的能力由剖面的最小剖面模数保证，最小剖面模数W1
—— 剖面上最远纤维至中和轴的距离 Z max
因而，在给定剖面积F时，若
Z Z max

第五型材剖面设计

大家知道，梁材受横荷重作用时，其抵抗弯矩
的能力由剖面的最小剖面模数保证。剖面的最
小剖面模数等于：
W1
I Z
z 2 dF
F
Z
(
zLeabharlann 2) dFZ
max F Z
max
max
max
式中 I F Z 2dF——型材剖面积对其中和轴的惯性矩
Z——微面积dF至中和轴的距离
Z m—ax —剖面上最远纤维至中和轴的距离
第5页/共64页
剖面模数比面积：
Cw
F W 2/3
F-型材剖面积；W-包括带板在内的型材的剖面模数
Cw小，表明利用较少的材料，获得了较大的剖面模数，所以剖面材料的利用率越高，显然Cw越小越有利。
第6页/共64页
惯性矩比面积：
Ci
F I 1/ 2
I包括带板在内的剖面惯性矩，Ci小，表明利用较少的材料，获得了较大的惯性矩，所以剖面
, N 0.85 f N [ ] 0.85 f
满足腹板剪切强度要求腹板的面积应该为：
f
N
0.85
第17页/共64页
3.抗弯强度与抗剪强度关系
要求梁具有相同的剪切强度和弯曲强度，则由于材料的许用剪切应力为
[ ] [ ]
3
剪切和弯曲强度相同的条件为：
max [ ] 0.57 max [ ]
型材侧向失稳如下图所示。
第25页/共64页
图构件侧向失稳
第26页/共64页
2）.型材总稳定性临界压力集度的确定
型材总稳定性的分析模型为：仓壁板对扶强材提供抗转约束，扶强材端部的约束取为自由支持。
图承受水压的扶强材
第27页/共64页