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复合材料结构力学作业

复合材料结构力学设计作业

1 一. 对材料AS4/3501-6进行设计

已知61.1,134.0,3.0,86.6,65.9,2.147ρυmmtGPaGMPaEMPaETL

MPaSMPaYMPaYMPaXMPaXCTcT105,186,4.49,1468,2356

最大正应力准则为pipiTpiTpiCpiTSYYXXR1222221111,,min

1 2

50 mm 100 mm

STEP I Special Stacking Sequence (SSS)

(一) 在Task I载荷作用下

已知Longitudinal Load =100 kN,Transverse Load =-5 kN, Shear Load =30 kN

外加载荷可等效为mkNNNNNTT/600502000122211

对0nS度铺设层合板,

MPaT447837314925}{,带入最大正应力准则得

N=max{6.3349,2.0054,42.6476}=42.6476,所以0nS所需的最小层数为42.6层,且12σ先破坏

对90nS度铺设层合板

MPaT447814925373

N=max{0.2541,302.1255,42.6476}=302.1255,所以90nS所需的最小层数为302.1255层,且22σ先破坏

对(45)nS度铺设层合板

45度 MPaT3.19125.1801.5496, N=max{2.3328,3.6538,18.2124}=18.2124

-45度 MPaT3.19127.3808.1218, N=max{0.51732,7.7065 ,18.2124}=18.2124 复合材料结构力学设计作业

2 所以对(45)nS度铺设层合板,共需要18.21*4=72.84层,且12σ先破坏

对(0/60)nS度铺设层合板

0度 MPaT7.2272.65.6366, N=max{2.7022,0.0333,2.1686}=2.7022

+60度MPaT8.4507.1852267, N=max{0.9622,3.7591,4.2933}=4.2933

-60度MPa2237.3815.1918T, N=max{1.3069,7.7267,2.1248}=7.37267

所以对(0/60)nS度铺设层合板,共需要7.37*6=44.22层,且22σ先破坏

绘制在表格中,如下所示:

层合板铺层 层合板层数 层合板厚度(mm) 层合板质量(g) 取整数层 取整后每层厚度(mm) 最先破坏层

0ns 42.6 5.7084 47.4628 44 0.1297 12σ

90ns 302.13 40.485 325.9043 302 0.1341 22σ

45ns 72.84 9.7606 78.5728 74 0.1319 12σ

0/60ns 44.22 5.9255 47.7003 48 0.1234 22σ

从上表中可以看到,0ns所需的层数最少,即质量最轻

对])45/45/(90/0[zyx铺层,设+45度和-45度的层数相同

(1) 当0度铺层占10%,90度铺层占0%时, 则45度和-45度各占45%时

0度 MPaT43987030809}{, N=13.08

+45度 MPaT6.24899.1484.9050}{, N=23.71

-45度 MPaT6.24891.5855.268}{, N=23.71

代入最大正应力准则进行校核,经比较得,N=23.71,所以共需要23.71*2=47.42层

(2) 当0度铺层占10%,90度铺层占10%时, 则45度和-45度各占40%时

0度,MPaT48243327954}{, N=max{11.865,2.328,4.5905}=11.865

90度, MPaT482155514524}{, N=max{9.8937,31.48,4.5905}=31.48

45度, MPaT200232211831}{, N=max{5.0216,6.5182,19.07}=19.07

-45度, MPaT6.20017.8005.1598}{, N=max{0.6783,16.2085,19.0629}=19.0629 复合材料结构力学设计作业

3 代入最大正应力准则进行校核,经比较得,N=31.48,所以共需要31.48*2=62.96层

(3) 当0度铺层占10%,90度铺层占20%时, 则45度和-45度各占35%时

0度 MPaT53518327248}{, N=11.57

90度 MPaT534158310498}{, N=32.04

+45度 MPaT177943414048}{, N=16.94

-45度 MPaT17785.9652.2702}{, N=19.53

经比较得,N=32.04,所以共需32.04*2=64.08层

(4) 当0度铺层占10%,90度铺层占25%时, 45度和-45度各占32.5%

0度,MPaT5659027327}{, N= 11.6

90度, MPaT3.5655.16149092}{, N= 32.6822

45度, MPaT171648115117}{, N= 16.3429

-45度,MPaT.1.17169.10427.3118}{, N= 21.1113

经比较得,N=32.68,所以共需要32.68*2=65.36层

(5)当0度铺层占10%,90度铺层占30%时, 则45度和-45度各占30%时

0度 MPaT606927618}{, N=11.72

90度 MPaT8.59916557933}{, N=15.95

+45度 MPaT167552516207}{, N=22.67

-45度 MPaT16756.11202.3477}{, N=33.49

经比较得,N=33.49,所以共需要33.49*2=66.98层

(6)当0度铺层占10%,90度铺层占40%时, 则45度和-45度各占25%时

0度 MPaT68313028732}{, N=12.2

90度 MPaT6839.17608.6107}{, N=15.64

+45度 MPaT164260618561}{, N=26.0

-45度 MPaT164212854062}{, N=35.65

经比较得,N=35.65,所以共需要35.65*2=71.3层

(7)当0度铺层占10%,90度铺层占50%时, 45度和-45度各占20%

0度,MPaT79325530507}{, N= 12.95 复合材料结构力学设计作业

4 90度,MPaT5.7936.19028.4691}{, N= 38.5

45度,MPaT165968521327}{, N= 15.8

-45度,MPaT6.16589.14724888}{, N=14.3

经比较得,N=38.5,所以共需要38.5*2=77层

(8)当0度铺层占10%,90度铺层占60%时, 则45度和-45度各占15%时

0度 MPaT94637832990}{, N=14

90度 MPaT2.9463.20863607}{, N=42.2

+45度 MPaT172076224781}{, N=16.3

-45度 MPaT172017024701}{, N=34.46

经比较得,N=42.2,所以共需要42.2*2=84.4层

(9)当0度铺层占10%,90度铺层占75%时, 则45度和-45度各占6.25%时

0度 MPaT145062239942}{, N=16.95

90度 MPaT8.144925721729.6-}{, N=52.07

+45度 MPaT196487734490}{, N=18.7

-45度 MPaT5.196323173722}{, N=46.9

经比较得,N=52.07,所以共需要52.07*2=104.14层

(10)当0度铺层占25%,90度铺层占0%时, 则45度和-45度各占37.5%时

0度 MPaT50756820231}{, N=8.5883

+45度 MPaT16329.1-8284.7}{, N=15.55

-45度 MPaT9.16326.4949.2475}{, N=10.01

经比较得,N=15.55,所以共需要15.55*2=31.10层

(11)当0度铺层占25%,90度铺层占10%时, 则45度和-45度各占32.5%时

0度 MPaT56526519118}{, N=8.11

90度 MPaT5.5635.10725.9458}{, N=21.71

+45度 MPaT134712610829}{, N=13.85

-45度 MPaT5.13465.6844.1169}{, N=12.83 复合材料结构力学设计作业

5 经比较得,N=21.71,所以共需要21.71*2=43.42层

(12) 当0度铺层占25%,90度铺层占25%时, 则45度和-45度各占25%时

0度,MPaT6831819099}{, N= 8.1

90度,MPaT2.68314117.5669}{, N=23.09

45度,MPaT116722213964}{, N=max{5.93,4.5,11.1}=11.1

-45度,MPaT1.11676.9006.534}{, N=max{0.36,18.23,11.1}=18.23

经比较得,N=23.09,所以共需要23.09*2=46.18层

(13)当0度铺层占25%,90度铺层占50%时, 则45度和-45度各占12.5%时

0度 MPaT104722721050}{, N=9.97

90度 MPaT10473.13271.2452}{, N=26.8

+45度 MPaT110725720408}{, N=10.54

-45度 MPaT110712978.1809}{, N=26.21

经比较得,N=26.8,所以共需要26.8*2=53.6层

(14)当0度铺层占50%,90度铺层占0%时, 则45度和-45度各占25%时

0度 MPaT68335712870}{, N=6.5

90度 MPaT2.6837.4718.9106}{, N=9.55

+45度 MPaT6.10351.1823.9131}{, N=9.86

-45度 MPaT6.10356.4967.5367}{, N=11.05

经比较得,N=11.05,所以共需要11.05*2=22.1层

(15)当0度铺层占50%,90度铺层占10%时, 则45度和-45度各占20%时

0度 MPaT79312712568}{, N=7.55

90度 MPaT793.5-4.7185492.7-}{, N=14.54

+45度 MPaT8519811957}{, N=8.1

-45度 MPaT8518.6894882}{, N=13.96

经比较得,N=14.54,所以共需要14.54*2=29.08层

(16)当0度铺层占50%,90度铺层占25%时, 则45度和-45度各占12.5%时

0度 MPaT10475212812}{, N=9.97

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