内容发布更新时间 : 2024/5/16 12:03:25星期一 下面是文章的全部内容请认真阅读。

6.10万方油罐，己定半径R=40m，高H=21.5m，选用16MnR材料，板高2.0m，该材料强度极限?b=490Mpa，材料屈服极限为?s=325MPa，每层圈板高度为2.0m，焊缝系数取?=0.9,若己知储液密度?液=900Kg/m3，试水密度为?水=1000Kg/m3，试用变点法确定试水条件下1至2层板壁厚。(20分) 第九次课ppt中

?b/2.5=196Mpa ?b/2.33=210Mpa ?s/1.5=216.7Mpa ?s/1.33=244Mpa

Sd=min(?b/2.5,?s/1.5)=196Mpa St=min(?b/2.33,?s/1.33)=210Mpa (1)第一圈板壁厚计算:

?1油定点=?1水定点=4.9?(H?0.3)D4.9?0.9?(21.5?0.3)?80??42.4mm t[?]?196?0.94.9(H?0.3)D4.9?(21.5?0.3)?80??47.1mm

[?]t?196?0.9H?油4.9?油HD）?41.89mm Sd?Sd?H4.9HD）?43.37mm St?St?0.0696D?1油变点=（1.06-H?1水变点=（1.06-0.0696DH=47.1mm ?1定点=max（?1油定点,?1水定点）=43.37mm ?1变点=max（?1油变点,?1水变点）=43.37mm ?1=min（?1定点,?1变点）(2)第二圈壁厚计算：

?a02=4.9?油(H2?0.3)D4.9?0.9?(19.5?0.3)?80??38.4

Sd?196?0.9K??i?143.37??1.13 0?ai38.4K(K?1)?0.06278

1?KKi?1C?Hi?H??hi?21.5?2?19.5

n?10??0.61R?ai?0.32CHi?1.14775???min?CHi?1.22421?=1.14775

???1.22R?0?1.22?40?0.0384=1.512?ai??xminj?ai=4.9?油(Hi?xmin)D4.9?0.9?(19.5?1.14775)?80??36.7045mm

Sd?196?0.90j?=?ai??ai?38.4?36.7045?1.6955?0.01

第二次试算

0令?ai=36.7045

K??i?143.37??1.1816 0?ai36.7045K(K?1)?0.08641

1?KKi?1C?Hi?H??hi?21.5?2?19.5

n?10??0.61R?ai?0.32CHi?1.2783???min?CHi?1.684995?=1.2783

???1.22R?0?1.22?40?0.0367045=1.478?ai??xminj?ai=4.9?油(Hi?xmin)D4.9?0.9?(19.5?1.2783)?80??36.4434mm

Sd?196?0.90j?=?ai??ai?36.7045-36.4434?0.2611?0.01

第三次试算

0令?ai=36.4434

K??i?143.37??1.19 0?ai36.4434K(K?1)?0.09019

1?KKi?1C?Hi?H??hi?21.5?2?19.5

n?10??0.61R?ai?0.32CHi?1.2993???min?CHi?1.758705?=1.2993

???1.22R?0?1.22?40?0.0364434=1.473?ai??xminj?ai=4.9?油(Hi?xmin)D4.9?0.9?(19.5?1.2993)?80??36.4014mm

Sd?196?0.90j?=?ai??ai?36.4434-36.4014?0.042

其值已经很小了，我们就不再往下迭代了。 确定第二层壁板变点计算厚度

?=hi?12000??1.518 R?i?140000?43.37因为1.375???2.625

jj?2=?ai?(?i?1??ai)(2.1??/1.25)?36.4014?(43.37?36.4014)(2.1?1.518/1.25)?42.573mm