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06级函授A卷答案

06级函授A卷答案
06级函授A卷答案

2008~2009学年第二学期期末考试 《铁路轨道》试题A 答案 (适用班级:函授铁道工程07级)

1、钢轨是铁路轨道的组成部件 ,要求钢轨具有足够的 强度 、 韧性

和 耐磨性 。

2、锰、硅可提高钢轨的质量。磷、硫都是 有害成分 。

3、锁定轨温:是指无缝线路的长钢轨无温度力状态时的轨温。通常将铺设长

钢轨的两端落槽时的平均轨温作为锁定轨温。

4、三角坑:在不足18米的距离内,先是左股较右股钢轨高,两个最大水平误

差超过4㎜。

5、应力放散:无缝线路在运用过程中,原锁定轨温不明或发生变化时,在适当

轨温时,将中间扣件、防爬设备松开,采取措施使钢轨伸缩,释放内部应力,

再重新锁定的过程。

二、填空题(每空1分,共25分)

1、 无缝线路的锁定轨温,是长钢轨无 温度应力或温度力 时的轨温。通常将铺设

长钢轨的两端正常就位时的 平均轨温 作为锁定轨温。

2、无缝线路钢轨内温度力和温度应力的大小,与轨温 变化量的大小 有直接关系,

适当控制轨温的 变化量 ,可降低钢轨内部温度力。

3、在长钢轨两端接头处,防止钢轨纵向位移的阻力,叫 接头阻力 。它是由钢轨

与夹板间的 摩擦 阻力所提供。

4、道床阻力通常取轨枕位移量为 2㎜ 时的相应阻力值作为 计算阻力值 。

5、无缝线路道床横向阻力是防止 涨轨跑道 、 轨距变化 ,保证轨道稳定的主要因素。

6、木枕失效的原因是 腐朽 、 机械磨损 、 裂缝 。

7、每千米最多铺设混凝土枕为 1840根 ,木枕为 1920根 。

8、连续弹性基础梁理论将钢轨视为一根支撑在 弹性支座 上的无限长梁。

9、曲线轨道的主要病害是: 方向不良 、 磨耗超限 。

10、无缝线路依据处理钢轨内部温度应力方式不同,分为 温度应力

式 和 放散温度应力式 两种类型。

11、无缝线路轨道除承受列车荷载外,还要承受巨大的 温度力 ,它仅与 温度变化幅度 成正比关系,而与 线路的长度 无关。

12、温度应力式无缝线路的两端钢轨联接有两种型式:一种是 普通夹板接头 ;另一种为 尖轨接头 。

三、选择题(每小题2分,共12分)

1、辙叉由(A 、B 、C )及联结零件组成。

A 、心轨

B 、翼轨

C 、护轨

D 、基本轨

2、导曲线支距计算中,坐标原点为(A )。

A 、 基本轨作用边正对尖轨根端的O 点

B 、基本轨作用边正对尖轨尖端的O 点

C 、基本轨作用边正对圆曲线起点

3、铺碴由集中开采的碴场以列车运碴铺设时,可根据情况采取(A 、B )。

A 、自碴场铺设

B 、向碴场铺设

C 、背碴场铺设

D 、离碴场铺设

4、轨道不平顺的情况有(A 、B 、C 、D ) A 、垂直不平顺 B 、横向不平顺 C 、复合不平顺 D 、曲线头尾几何偏差

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6、应力放散的方法有(A 、B 、C )和列车碾压法。 A 、滚筒放散法 B 、机械拉伸法 C 、撞轨法 D 、局部调整法 四、判断题(每小题2分,在括号中打上对错号,共计12分) 1、日常轨缝的不均匀,是指日常某一个轨温条件下检查的实际轨缝值与正确的检算值之差,如单个实际轨缝与检算值之差在±2㎜以上,视为“不均匀”;如连续20个轨缝中有一半及以上是在±2㎜以上,视为“严重不均匀”。 ( √ ) 2、拨道时,一次拨道量不超过20㎜,由于线路的弹性作用,实际拨道量要比计算值多拨一些,混凝土枕线路要多拨30%,木枕线路要多拨20%。 ( × ) 3、无缝线路的应力放散与应力调整的根本区别在于,应力放散的范围大一些,而应力调整的范围小。 ( × ) 4、上弯夹板是利用专用设备把已经存在下弯的夹板矫正, 并使它产生2㎜以内的反向矢度,即上弯量为2㎜以内。( √ ) 5、线路大中修纵断面设计是在既有建筑物的限制条件下,寻求最经济、合理的线路改善方案。 ( √ ) 6、轨排组装作业过程:锚固 →配轨散轨→匀枕、散配件→ 上扣件 → 紧螺帽 → 检查 → 装车。 ( × ) 班级 姓名 学号

五、简答题(每小题5分,共计25分)

1、曲线轨道结构与直线轨道的不同点有哪些?

①曲线轨道需要设置外轨超高;

②小半径曲线轨道需要进行轨距加宽;

③曲线轨道需要配置缩短轨;

④曲线轨道需要设置缓和曲线;

⑤曲线轨道需要进行方向整正;

⑥曲线轨道需要进行轨道加强。

2、影响线路爬行的因素有哪些?

①在长大坡道、进站地段,列车减速、限速、制动;

②运量大,爬行量也大,爬行方向与列车运行方向一致;

③列车轴重大、速度高,则沿着运行方向爬行也大;

④线路状态不良,扣件松弛,道床松散,爬行加大。

3、引起胀轨跑道的主要原因是什么?

①锁定轨温偏低;

②冬季不正常收缩及严重不均匀爬行;

③线路维修作业不良,高温情况下阻力未恢复;

④线路设备状态不良,尤其是道床不符合标准,阻力下降;

⑤扣件扣压力不足或道钉浮起,造成轨道框架刚度降低;

⑥线路方向不良,钢轨死弯多,从而增加了轨道原始不平顺。

4、提高过岔速度的措施有哪些?

①加大道岔的导曲线半径,减小车轮对道岔各部位的冲击;

②采用对称道岔;

③以曲线尖轨取代直线尖轨,采用曲线辙叉;

④采用变曲率的导曲线,减小冲击角降低动能损失。

5、道床的作用是什么?

①传递由钢轨、轨枕传来的机车车辆动荷载,使其均匀地分布在地基基床面上,防止路基基床压陷;

②提供抵抗轨道框架纵、横向位移的阻力,保持轨道稳定和正确的几何位置,保证行车安全;

③提供排水能力,使基床面干燥,有足够强度,防止翻浆冒泥及轨道下沉;

④提供轨道弹性,起到缓冲、减振作用;

⑤调节轨道框架的水平与方向,保持良好的线路和纵断面。

六、计算题(每小题8分,共计16分)

1、某一曲线位于一般地段,曲线半径为800m,通过该曲线的平均行车速度为90km/h,

最高行车速度为140km/h,试确定该曲线的外轨超高和缓和曲线长度。

已知:R=1200m,ν

平=90km/h,ν

max

=140km/h

求:h、l0

解:

①h=11.8×ν

2÷R=11.8×902÷1200=79.65 (㎜)

根据取值原则,取为5㎜的倍数,此曲线超高为80㎜。②l0≥9 hν

max

≥9×80×140≥117600(㎜) ≈100.8 m

i= h ÷l0 =80 ÷100.8= 0.79‰﹤2‰

根据取值原则,进整为10m的倍数,此缓和曲线为110m。

答:曲线的外轨超高和缓和曲线长度分别为80㎜和110m。

2、某地区的无缝线路,其缓冲区两侧的长钢轨长度分别

为l1=1000m,l2=800m,长钢轨中部固定,分向两侧放散,

由此得放散长度l

放1

=500m,l放2=400m;放散后计划锁定轨

温t1=24℃,原锁定轨温t2=10℃,缓冲区由4根25 m长的缓

冲轨组成,并已知Δ1=Δ2=8㎜,缓冲区原有轨缝之和为20

㎜。计算其锯轨量(线路无爬行) 。

已知:α=0.0118,t1=24℃,t2=10℃,

L=l

放1

+ l放2=500+400=900 m,

△t= t1__ t2 =24—10=14℃

求:λ

解:

①放散量计算

Δl=αL△t=0.0118×900×14=149 (㎜)

②缓冲区预留轨缝计算

缓冲区预留轨缝的总和:

∑a=8×5=40 (㎜)

缓冲区预留轨缝的总和:

∑b=20(㎜)

③锯轨量计算

λ=Δl+(∑a-∑b)=149+(40-20)=169 (㎜)

因线路无爬行,故锯轨量为169 ㎜。

班级

姓名

学号

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