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标准椭圆形封头稳定性校核:取?n?18mm,?e?18?2.8?15.2mm,D0?1000?36?1036mm,查表4?5得系数K1?0.9,R0?K1D0?0.9?1036?932.4mmA?0.1250.125?15.2??0.002R0?e932.4B查图4-8得:B?160MPa?p???R0?e??160?15.2?2.608MPa?0.5MPa932.4
?n?18mm满足稳定性要求□2夹套壁厚设计
圆筒壁厚:??pD2?????pt?0.4?1100?1.525mm2?170?0.85?0.4pD0.4?1100?1.524mm2?170?0.85?0.5?0.4
?n???C1?C2?1.525?1?2.525mm,取?n?4mm?3mm标准椭圆形封头壁厚:??2?????0.5pt??n???C1?C2?1.524?1?2.524mm,取?n?4mm?3mm7. 有一受内压圆筒形容器,两端为椭圆形封头,内径Di=1000mm,设计(计算)压力为2.5MPa,设计温度300℃,材料为16MnR,厚度δn=14 mm,腐蚀裕量C2=2.0mm,焊接接头系数φ=0.85;在圆筒和封头焊有三个接管(方位见图),材料均为20号无缝钢管,接管a规格为φ89×6.0,接管b规格为φ219×8,接管c规格为φ159×6,试问上述开孔结构是否需要补强?
答:根据GB150规定,接管a不需要另行补强。接管b、c均需计算后确定。
椭圆形封头的计算厚度:
16MnR在300℃时许用应力,查表D1,6~16mm时,[σ]t= 144 MPa,查表4-2,C1=0.8mm;查表D21,≤10mm时,[σ]tt= 101 MPa;fr=101/144=0.701。
标准椭圆形封头壁厚:??pD2?????0.5pt?2.5?1000?10.265mm2?144?0.85?0.5?2.5接管b的计算厚度?t?pD2?????pt?2.5?203?2.557mm2?101?2.52.5?147?1.851mm2?101?2.5接管c的计算厚度?t?pD2?????pt?接管b的补强计算:
开孔直径:d?219?2?8?2??2?0.8??197.4mm所需最小补强面积接管的有效厚度:?et?8?2.8?5.2mm,封头的有效厚度:?e?14?0.8?2?11.2mmA?d??2??et?1?fr??197.4?10.265?2?10.265?5.2??1?0.701??2058.231mm2B?2d?2?197.4?394.8mm,h1?150mm,h2?0A1??B?d???e????2?et??e????1?fr??197.4?0.935?2?5.2?0.935?0.299?181.662mm2A2?2h1??et??t?fr?2h2??et?C2?fr?2?150?2.557?0.701?537.737mm2A3?36mm2Ae?A1?A2?A3?181.662?537.737?36?755.399mm2?A,需补强增加补强金属面积:A4?A?A4?2058.231?755.399?1302.832mm2接管c的补强计算:
开孔直径:d?159?2?6?2??2?0.8??152.6mm所需最小补强面积接管的有效厚度:?et?6?2.8?3.2mm,封头的有效厚度:?e?14?0.8?2?11.2mmA?d??2??et?1?fr??152.6?10.265?2?10.265?3.2??1?0.701??1635.204mm2B?2d?2?152.6?305.2mm,h1?150mm,h2?0A1??B?d???e????2?et??e????1?fr??152.6?0.935?2?3.2?0.935?0.299?140.892mm2A2?2h1??et??t?fr?2h2??et?C2?fr?2?150?1.349?0.701?283.695mm2A3?36mm2Ae?A1?A2?A3?140.892?283.695?36?460.587mm2?A,需补强
8. 具有椭圆形封头的卧式氯甲烷(可燃液化气体)储罐,内径Di=2600mm,厚度δn=20 mm,储罐总长10000mm,已知排放状态下氯甲烷的汽化热为335kJ/kg,储罐无隔热保温层和水喷淋装置,试确定该容器安全泄放量。 解:容器安全泄放量
增加补强金属面积:A4?A?A4?1635.204?460.587?-1174.617mm2Ar?容器受热面积,m2。椭圆形封头的卧式压力容器,Ar??D0?L?0.3D0????2.64??10?0.3?2.64??83.595m2F?系数,压力容器装在地面以下,用沙土覆盖时,取F?0.3;压力容器在地面上时,取F?1;当设置大于10L/m2?min的喷淋装置时,取F?0.6q?在泄放压力下液化气体的气化潜热,kJ/kg,q?335kJ/kg2.55?105FAr0.822.55?105?1?83.5950.82Ws???28686.85kg/h?7.969kg/sq335
??9. 求出例4-3中远离边缘处筒体内外壁的应力和应力强度。(例4-3:某一钢制容器,内径Di=800mm,
厚度t=36mm,工作压力pw=10MPa,设计压力p=11MPa。圆筒与一平封头连接,根据设计压力计算得到圆筒与平封头连接处的边缘力Q0=-1.102×106N/m,边缘弯矩M0=5.725×104N·m/m,如图所示。设容器材料的弹性模量E=2×105MPa,泊松比μ=0.3。若不考虑角焊缝引起的应力集中,试计算圆筒边缘处的应力及应力强度)
解:远离边缘处筒体的应力和应力强度为不考虑边缘效应时,按拉美公式计算的应力分量,按应力分类可分解成一次总体薄膜应力、沿厚度的应力梯度-二次应力,并计算
出其应力强度。
K?各应力分量沿圆筒厚度的平均值—一次总体薄膜应力Pm:
436?1.09400
??p,1?1R0?Ri?R0Ri??dr1?R0?Ri???R0Ri2?pRi2pR0Ri21?p?dr??2?22K?1R0?Ri2r2??R0?Ri11?122.222MPa1.09?1pr,1
1?R0?Ri?R0Ri1?rdr?R0?Ri?R0Ri2?pRpR0Ri21?p11?dr??????5.263MPa?22222?K?11.09?1R?RR?Rri0i?0?
2i?pz,11?R0?Rip?R0Ri1?zdr?R0?Ri?R0Ri筒体内壁各应力分量的应力梯度—内壁处的二次应力Q:
pRi2p11dr???58.48MPa2222R0?RiK?11.09?1
?p?,2?????p?,1?rp,2??rp??rp,1?zp,2?0K2?1p11?1.092?1?p2???122.222?5.737MPa2K?1K?11.09?1p??p???11?5.263??5.737MPaK?1
??筒体外壁各应力分量的应力梯度—外壁处的二次应力Q:
??p,2???p???p,1?p?rp,2??rp??rp,1?zp,2?02p2?11???122.222?-5.263MPa22K?1K?11.09?1p??5.263K?1
筒体内壁处的各应力强度:
一次总体薄膜应力强度一次加二次应力强度一次总体薄膜应力强度一次加二次应力强度
SⅠ??1??3?122.222?5.263?127.485MPaSⅣ??1??3?5.737?5.737?11.474MPaSⅠ??1??3?122.222?5.263?127.485MPaSⅣ??1??3?5.263?5.263?10.526MPa
筒体外内壁处的各应力强度: