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first steps
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  前10阶
     Through analyzing the free vibration frequencies of the first 10 steps with a designed hook load of 1568 kN and their main vibration modes, it is concluded that the derrick vibration is mainly the integral bending vibration; the four columns of derrick are the main load-bearing components; and the load-bearing capacity of the front two is most unsubstantial.
     文章通过分析在设计钩载为1568 kN时的前10阶自振频率和各阶主振型,得出了以下结论:井架以整体弯曲振动为主,四根立柱为主要承载构件,且前两立柱承载能力最为薄弱。
短句来源
     This shows that the difference between the out-surface stiffness of the arch rib and the vertical stiffness is great.The out-surface stiffness of the bridge deck is weak,the distorted vibration appears in the first 10 steps.
     桥面系面外刚度相对较弱,桥梁前10阶振型中出现了扭转振动形式。
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The first step in caprolactam polymerization is an addition polymerization. The authors have studied the initiative influence of various substances on the polymerization and obtained the following results. The initiators of the free radical type, such as benzoyl peroxide and hydrogen peroxide together with their redox systems, all exerted no effect on caprolactam. While carboxylic acids gave only weak initiative effect in absence of water and amines gave no action, the mixtures of acids and amines and...

The first step in caprolactam polymerization is an addition polymerization. The authors have studied the initiative influence of various substances on the polymerization and obtained the following results. The initiators of the free radical type, such as benzoyl peroxide and hydrogen peroxide together with their redox systems, all exerted no effect on caprolactam. While carboxylic acids gave only weak initiative effect in absence of water and amines gave no action, the mixtures of acids and amines and of acids and water initiated the polymerization rapidly. Amino-acids of various types also gave rapid initiation. 6-Benzoyl-amino-caproic acid was found to be far inferior in the initiative activity than benzoic acid. Trimethylphenyl ammonium iodide did not initiate the polymerization, and attempt to polymerize N-methyl caprolactam gave negative results.From these facts, the polymerization of caprolactam is clearly neither of free radical nor of step-wise addition mechanism. The authors suggest that the initiation is exerted by both the cation and anion of the initiator. These ions activate the amide grouping by a proton transfer mechanism, which causes caprolactam to polarize in the type of a zwitter-ion and thereafter polymerization proceeds.

己內醯胺聚合的最初階段為加成聚合。作者研究若干種化合物,試驗它們對己內醯胺聚合的引發作用。結果,自由基型的引發劑如過氧化苯甲醯,過氧化氫和它們的氧化還原體系,均不能引發己內醯胺的聚合。羧酸在無水存在時其引發作用甚弱。胺類則缺乏引發效應。但羧酸与胺同時存在時能迅速的引起己內醯胺的聚合。氨基酸不論何種類型均有迅速的引發作用。苯甲醯胺基己酸的引發作用遠較苯甲酸為差。碘化三甲基苯基銨(季銨鹽)不能引發己內醯胺的聚合,而N-甲基己內醯胺也不能用這些引發劑引起聚合。根據以上結果,己內醯胺聚合顯非自由基型的聚合,亦非引發劑与單體的逐步加成。作者建議引發劑的正負兩種離子都有影響。引發機構是由於氫原子的轉移,使己內醯胺發生雙離子式的極化,因而發生聚合。

This article presents a method for calculating the gravity dam with various types of opennings.The procedure takes as follows: First step:Assuming the dam as a solid body,calculating it's stresses. Second step:Along the boundary of the opennings,applying forces which equal but opposite to the primary,and assuming the openning is placed at an infinite region,we obtain the stresses weakened by openning. Third step:Combining these stresses,the final stresses in dam are given. If the openning is...

This article presents a method for calculating the gravity dam with various types of opennings.The procedure takes as follows: First step:Assuming the dam as a solid body,calculating it's stresses. Second step:Along the boundary of the opennings,applying forces which equal but opposite to the primary,and assuming the openning is placed at an infinite region,we obtain the stresses weakened by openning. Third step:Combining these stresses,the final stresses in dam are given. If the openning is too big in size,outside boundary condition should be considered. Take the penstock,which buried in dam,as an illustrated example.

本文提出一个计算重力壩体内部孔口应力的方法,其步骤如下:第一步,假设壩为实体,求出其应力。第二步,沿孔口周边作用以和初始应力相等相反之力,并假定孔口位于无限域内,求出壩体的削弱应力。第三步,将两次应力相加,即求得壩体的实在应力。如果孔口的尺寸过大,尚需考虑外边界条件的适合。最后用输水管应力分析来说明这个方法的使用。

The anodic oxidation of antimony in sulphuric acid, in hydrochloric acid, and in potassium hydroxide solutions has been studied by measuring the polarization curves and by using the A. C. bridge to determine the electrode capacity and ohmic resistance for the investigation of the change of surface conditions in the course of oxidation. The empirical relations of polarization potential φ with current density i and pH at 25° are as follows:In 1—13.4 N potassium hydroxide solutions,φ=a + 0.116 log i — 0.115 pHin...

The anodic oxidation of antimony in sulphuric acid, in hydrochloric acid, and in potassium hydroxide solutions has been studied by measuring the polarization curves and by using the A. C. bridge to determine the electrode capacity and ohmic resistance for the investigation of the change of surface conditions in the course of oxidation. The empirical relations of polarization potential φ with current density i and pH at 25° are as follows:In 1—13.4 N potassium hydroxide solutions,φ=a + 0.116 log i — 0.115 pHin 1—12.4 N sulphuric acid solutions,φ=a′+ 0.024 log i — 0.048 pHin 1—12 N hydrochloric acid solutions,φ=a″ + 0.024 logi + 0.051 pHThe rate of dissolution in KOH and in HCl at constant potential increased with the increase of OH- and Cl- concentration respectively; while in H_2SO_4, it decreased with increasing acid concentration. The pH dependence of the dissolution rate, λ=((?)log i/(?)PH)_φ, was estimated in each case.Theoretical data analysed in terms of electrochemical kinetics suggest the following mechanism for the anodic oxidation.In KOH, OH~- → OH_(ads.)+e~-, OH_(ads.)+OH~-→H_2O+O~-, Sb + O~-→SbO_(ads.) +e~-,2 SbO_(ads.) + O~-→Sb_2O_3 + e~-, Sb_2O_3 + O~- → Sb_2O_4 + e~-, 2 Sh_2O_4 → Sb_2O_3 + Sb_2O_5. The first step is regarded as rate-controlling.In H_2SO_4, H_2O~-→O~- + 2H~+ + e~-, Sb + O~- → SbO_(ads.) + e~-, SbO_(ads.) → SbO~+ + e~-,SbO~++ O~-→SbO_2~+ + e~-; and in HCl, the first and second steps are the same, then ShO_(ads.) + 2HCl → SbCl_2 + H_2O, SbCl_2 + Cl~- → SbCl_3 + e~-, and SbCl_3 + Cl~- → SbCl_4~-. The rate of the process is determined by the third step in each case.Accumulation of the oxidation products, Sb_2O_3, Sb_2O_5 and O_(ads.,) on the electrode surface to a thickness of 3—5 molecular layers leads to passivation in both acid and alkaline solutions. The evolution of oxygen is possibly the consequence of discharge of O- on the passive surface.

本文用经典极化曲线的方法和交流电测定电极表面状态的方法,对锑在氢氧化钾、硫酸和盐酸中的阳极过程进行研究,获得如下的结果: 1.锑的阳极电位φ与极化电流密度i及pH之间的关系可由公式表示:φ=a+ blogi—b′pHo在1—13.4N氢氧化钾溶液中,b=116毫伏,b′=115毫伏;在1—14N硫酸中,b=24毫伏,b′=48毫伏;在1—12N盐酸中,b=24毫伏,b′=—48至—54毫伏。a为常数。 2.溶液pH对锑的阳极溶解速度的关系,λ=((?)logi/(?)pH)_φ,在氢氧化钾溶液中,λ=1;在硫酸中,λ=2;在盐酸中λ′=((?)logi/(?)pH)_φ=+2:即溶解速度在氢氧化钾或盐酸中随浓度增长而加速;在硫酸中则相反。 3.假设锑在这些介质中的阳极反应具共同的基本步骤:Sb+O~-→SbO_吸+e~-,在氢氧化钾溶液中,过程的进行依靠类似的步骤,使SbO继续氧化而逐步形成Sb_2O_3,Sb_2O_4;后者在溶液中因不稳定而歧化为Sb_O_3及Sb_2O_5。O~-假定通过如下步骤形成:OH~-→OH_吸+e~-,OH_吸+OH~-→O~-+H_2O。前者如为控制步骤则与实验数据相符合。在...

本文用经典极化曲线的方法和交流电测定电极表面状态的方法,对锑在氢氧化钾、硫酸和盐酸中的阳极过程进行研究,获得如下的结果: 1.锑的阳极电位φ与极化电流密度i及pH之间的关系可由公式表示:φ=a+ blogi—b′pHo在1—13.4N氢氧化钾溶液中,b=116毫伏,b′=115毫伏;在1—14N硫酸中,b=24毫伏,b′=48毫伏;在1—12N盐酸中,b=24毫伏,b′=—48至—54毫伏。a为常数。 2.溶液pH对锑的阳极溶解速度的关系,λ=((?)logi/(?)pH)_φ,在氢氧化钾溶液中,λ=1;在硫酸中,λ=2;在盐酸中λ′=((?)logi/(?)pH)_φ=+2:即溶解速度在氢氧化钾或盐酸中随浓度增长而加速;在硫酸中则相反。 3.假设锑在这些介质中的阳极反应具共同的基本步骤:Sb+O~-→SbO_吸+e~-,在氢氧化钾溶液中,过程的进行依靠类似的步骤,使SbO继续氧化而逐步形成Sb_2O_3,Sb_2O_4;后者在溶液中因不稳定而歧化为Sb_O_3及Sb_2O_5。O~-假定通过如下步骤形成:OH~-→OH_吸+e~-,OH_吸+OH~-→O~-+H_2O。前者如为控制步骤则与实验数据相符合。在硫酸中则以共同步骤形成的中间物SbO失去电子而成SbO~+为控制步骤。在盐酸中因氧化膜SbO被氯离子破坏而生成络离子SbCl_4~-为控制步骤。在酸中,O~-假定是由水分子的单电子放电产生:H_2O→O~-+2H~++e~-。 4.由于氧化产物(Sb_2O_3,Sb_2O_5及吸附氧原子)在电极表面的积累,不论在酸或碱中,形成3—5个分子层,才导致表面钝化。氧的逸出发生在稳定钝化区的电位,这可能是O~-在钝化表面继续放电的结果。

 
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