2016年中国石油大学(北京《工程力学(双语)》2011-2012学年第二学期期末考试试卷A卷

合集下载

中国石油大学现代远程教育《工程力学》期末考试答案

中国石油大学现代远程教育《工程力学》期末考试答案

中国石油大学(北京)远程教育学院期末考试《工程力学》学习中心:____姓名:___学号:____关于课程考试违规作弊的说明1、提交文件中涉嫌抄袭内容(包括抄袭网上、书籍、报刊杂志及其她已有论文),带有明显外校标记,不符合学院要求或学生本人情况,或存在查明出处的内容或其她可疑字样者,判为抄袭,成绩为“0”。

2、两人或两人以上答题内容或用语有50%以上相同者判为雷同,成绩为“0”。

3、所提交试卷或材料没有对老师题目进行作答或提交内容与该课程要求完全不相干者,认定为“白卷”或“错卷”,成绩为“0”。

一、题型简答题,8题,每题5分,共40;计算题,4题,每题15分,共60分)二、题目学号尾数为奇数的同学做题目序号后有“A”字样的题,学号尾数为偶数的同学做题目序号后有“B”字样的题简答题:1A 在铸铁压缩试验中,破坏后的铸铁试样断口平滑呈韧性,与轴线近似成45°。

破坏前,该断口所在斜截面的应力有何特点?答:剪应力最大1B 在铸铁扭转试验中,铸铁断口的形态就是什么样的?答:断口呈螺旋面、与轴线近似成45°。

2A 根据铸铁试件扭转破坏断口可以推断,铸铁的扭转破坏与什么因素有很大的关系? 答:最大拉应力2B 电阻应变片(简称电阻片或应变片)应用广泛,它就是利用什么原理实现电测的? 答:金属丝的电阻随机械变形而发生变化 3A 冲击韧性的物理意义就是什么? 答:试样断裂过程中断面单位面积吸收的能量 3B 矩形截面梁在截面B 处沿铅垂对称轴与水平对称轴方向上分别作用有P F ,如图所示。

请问最大拉应力与最大压应力发生在危险截面A 的哪些点上?答:4A 构件中危险点的应力状态如图所示。

构件为钢制:x σ=45MPa,y σ=135MPa,z σ=0,xy τ=0,许用应力[]σ=160MPa 。

请用第三强度理论校核该构件的强度。

答:选用第三强度理论,构件满足强度要求4B 构件中危险点的应力状态如图所示。

《工程力学》期中试卷(静力学)

《工程力学》期中试卷(静力学)


。(选择填写:一定平衡;一定不平衡;平衡与否不能确定;)
5. 对于均质物体,其重心和
心总是在同一点上。
6. 作用于刚体上主动力的合力,无论其大小,只要其作用线与接触面法线的夹角

擦角(选择填写:<;>;≤;≥)物体便处于平衡,这种现象称为

7. 按照接触物体之间可能会相对滑动或相对滚动,摩擦可分为滑动摩擦和
F3
A、 力系可合成为一个力偶;B:力系可合成为一个力; C:力系可简化为一个力和一个力偶;D:力系的合力为零,力系平衡;
G F2
图 2.5
6.空间平行力系有( )个独立的平衡方程;
A.6 个; B.5 个; C.4 个;
D.3 个
7.空间平行力系简化的最简结果最可能的结果是( ) A.平衡力系;B:合力;C:力偶;D:力螺旋;
=
−12.75kN
…..(2 分)
(2). 以 AC 杆为研究对象:
ΣFx = 0 : FAx − FCx = 0
FAx = −10.39kN ………………(1 分)
MA
ΣFy = 0 : FAy − FCy − qL = 0
Fcy Fcx
qL
FAx
FAy = FCy + qL = −8.75kN ….(1 分)
(1 分)
FEx = −400(N )
(1 分)
FBy FBx
FDy FDx
Ff
FDx FDy
FEx
M
FEy
相,
相同,则两
,不改变该力偶对物体的转动效果。
3. 空间力偶对刚体的作用效果取决与三个因素


GG
GG
4. 作用在刚体上仅有两个力,且 FA 、 FB ,满足 FA + FB = 0 ,则此刚体

中国石油大学(北京)《工程力学》期末考试答案37567教学资料

中国石油大学(北京)《工程力学》期末考试答案37567教学资料

中国石油大学(北京)《工程力学》期末考试答案37567中国石油大学(北京)远程教育学院期末考试《工程力学》学习中心:____姓名:___学号:____ 关于课程考试违规作弊的说明1、提交文件中涉嫌抄袭内容(包括抄袭网上、书籍、报刊杂志及其他已有论文),带有明显外校标记,不符合学院要求或学生本人情况,或存在查明出处的内容或其他可疑字样者,判为抄袭,成绩为“0”。

2、两人或两人以上答题内容或用语有50%以上相同者判为雷同,成绩为“0”。

3、所提交试卷或材料没有对老师题目进行作答或提交内容与该课程要求完全不相干者,认定为“白卷”或“错卷”,成绩为“0”。

一、题型简答题,8题,每题5分,共40;计算题,4题,每题15分,共60分)二、题目学号尾数为奇数的同学做题目序号后有“A”字样的题,学号尾数为偶数的同学做题目序号后有“B”字样的题简答题:1A 在铸铁压缩试验中,破坏后的铸铁试样断口平滑呈韧性,与轴线近似成45°。

破坏前,该断口所在斜截面的应力有何特点?答:剪应力最大1B 在铸铁扭转试验中,铸铁断口的形态是什么样的?答:断口呈螺旋面、与轴线近似成45°。

2A 根据铸铁试件扭转破坏断口可以推断,铸铁的扭转破坏和什么因素有很大的关系?答:最大拉应力2B 电阻应变片(简称电阻片或应变片)应用广泛,它是利用什么原理实现电测的?答:金属丝的电阻随机械变形而发生变化 3A 冲击韧性的物理意义是什么?答:试样断裂过程中断面单位面积吸收的能量3B 矩形截面梁在截面B处沿铅垂对称轴和水平对称轴方向上分别作用有,如图所示。

请问最大拉应力和最大压应力发生在危险截面A的哪些点上?答:4A 构件中危险点的应力状态如图所示。

构件为钢制:=45MPa,=135MPa,=0,=0,许用应力=160MPa。

请用第三强度理论校核该构件的强度。

答:选用第三强度理论,构件满足强度要求4B 构件中危险点的应力状态如图所示。

中国石油大学工程力学第二次作业答案

中国石油大学工程力学第二次作业答案

各位石油大学机械制造及自动化专业的朋友们此答案为1309级工程力学第二次作业答案以下没有显示试题目,答案全是对的,可以抄写过去即可。

有意可单选题 (共40道题)收起1.(分)A、.B、.C、.D、.我的答案:D 此题得分:分2.(分)A、.B、.C、.D、.我的答案:C 此题得分:分3.(分)A、.B、.C、.D、.我的答案:A 此题得分:分4.(分)A、.B、.C、.D、.我的答案:A 此题得分:分5.(分)A、.B、.C、.D、.我的答案:A 此题得分:分6.(分)A、.B、.C、.D、.我的答案:A 此题得分:分7.(分)A、.B、.C、.D、.我的答案:A 此题得分:分8.(分)A、.B、.C、.D、.我的答案:C 此题得分:分9.(分)A、.B、.C、.D、.我的答案:B 此题得分:分10.(分)A、.B、.C、.D、.我的答案:C 此题得分:分11.(分)A、.B、.C、.D、.我的答案:C 此题得分:分12.(分)A、.B、.C、.D、.我的答案:D 此题得分:分13.(分)A、.B、.C、.D、.我的答案:D 此题得分:分14.(分)A、.B、.C、.D、.我的答案:B 此题得分:分15.(分)A、.B、.C、.D、.我的答案:C 此题得分:分16.(分)A、.B、.C、.D、.我的答案:A 此题得分:分17.(分)A、.B、.C、.D、.我的答案:B 此题得分:分18.(分)A、.B、.C、.D、.我的答案:D 此题得分:分19.(分)A、.B、.C、.D、.我的答案:B 此题得分:分20.(分)A、.B、.C、.D、.我的答案:D 此题得分:分21.(分)A、.B、.C、.D、.我的答案:B 此题得分:分22.(分)A、.B、.C、.D、.我的答案:D 此题得分:分23.(分)A、.B、.C、.D、.我的答案:B 此题得分:分24.(分)A、.B、.C、.D、.我的答案:B 此题得分:分25.(分)A、.B、.C、.D、.我的答案:D 此题得分:分26.(分)A、.B、.C、.D、.我的答案:C 此题得分:分27.(分)A、.B、.C、.D、.我的答案:A 此题得分:分28.(分)A、.B、.C、.D、.我的答案:C 此题得分:分29.(分)A、.B、.C、.D、.我的答案:B 此题得分:分30.(分)A、.B、.C、.D、.我的答案:C 此题得分:分31.(分)A、.B、.C、.D、.我的答案:A 此题得分:分32.(分)A、.B、.C、.D、.我的答案:B 此题得分:分33.(分)A、.B、.C、.D、.我的答案:C 此题得分:分34.(分)A、.B、.C、.D、.我的答案:A 此题得分:分35.(分)A、.B、.C、.D、.我的答案:D 此题得分:分36.(分)A、.B、.C、.D、.我的答案:C 此题得分:分37.(分)A、.B、.C、.D、.我的答案:A 此题得分:分38.(分)A、.B、.C、.D、.我的答案:C 此题得分:分39.(分)A、.B、.C、.D、.我的答案:B 此题得分:分40.(分)A、.B、.C、.D、.我的答案:A 此题得分:分。

中国石油大学(北京)2012—2013学年第二学期《工程力学(Ⅰ)(双语)》期中考试试卷标准答案A

中国石油大学(北京)2012—2013学年第二学期《工程力学(Ⅰ)(双语)》期中考试试卷标准答案A

中国石油大学(北京)2012--2013学年第二学期《工程力学(Ⅰ)(双语)》期中考试试卷标准答案A(闭卷考试)Ⅰ (Each 6 points, total 30 points)1 a =b -c2 ︒453 ④, ⑥, ⑦4 The instant center for velocities of bar BC is point O.5()()[]222122221kR R R R R k W s --=⎥⎦⎤⎢⎣⎡---=Ⅱ (Total 10 points) 提到下划线加粗部分就给满分1(4 points)轮胎与地面间的摩擦力是赛车前进的动力,赛车前进时,空气的流动使扰流板产生向下的压力,赛车速度越快,该压力越大,从而产生更大的地面压力,根据库仑摩擦定律,这就增大了轮胎与地面间的最大静摩擦力,从而在不增加自身重量的条件下,提高赛车的最大极限速度。

2(3 points)轮子只滚不滑,与地面的接触点是速度瞬心,轮子的下部分离速度瞬心近,线速度慢,上部分离速度瞬心远,线速度快,相机爆光需要时间,导致了轮子照片下部分较清晰, 而上部分较模糊。

3(3 points)如果没有尾旋翼或尾旋翼没有工作,单螺旋桨直升飞机在空中时,在重力和升力(空气产生)的作用下,对过重心的垂直轴的角动量守恒,螺旋桨转动时,直升飞机其它部分会朝相反的方向旋转,螺旋桨转速改变时,直升飞机其它部分的转速也会发生改变,增加尾旋翼并调整尾旋翼的转速,产生一个对过重心的垂直轴的力矩,来控制单螺旋桨直升飞机在空中的姿态。

Ⅲ (20 points)Fig. III-2 Fig. III-1 Studying beam BC, draw FBD (Fig. III-1) (2 points)∑=0BM 02122=-⨯-⨯ql l ql l R C(4 points) ∑=0y F 0=-+ql R R CB(4 points) Studying beam AB, draw FBD(Fig. III-2) (2 points)∑=0yF 0=--ql R R B A(4 points) ∑=0AM 0232=+⨯-⨯-A B M l ql l R 22ql M A = (4 points) Ⅳ (20 points)moving particle: A; moving reference frame: OB; transport motion: rotation Velocity: (6 points)r e a v v v +=θωcos b v e =θωθ2c o s c o s b v v e a == θθωθc o s t a n t a n b v v e r ==aa τea nea ra ca x 'av rv ev2B qBCAccelerate: (8 points)Taking the projection on x' axis yieldr a v a ωθ2cos =Forces: (6 points)Studying collar and pin A, draw FBDθsin rc sp x F F ma -= 0=x aθcos rc y F mg ma +-= θθω22c o s t a n 2b a a a y == θθωθtan sec 2sec 32mb mg F rc +=θθωθ222tan sec 2tan mb mg F sp +=Ⅴ (20 points)(1) the angular velocity and angular acceleration of the bar: FBD:mg , A N , B NKinematics analysis:The instant center for velocities of bar AB is point O Assuming the bar in the release (θ = 0) is position (1) and in an arbitrary position (2) defined by the angle θ21-W mg=()θcos 12-L 1T 0= 2T 221ωO J =223121ω⎪⎭⎫ ⎝⎛=mL (6 points) 1221T T W -=- ()03121cos 1222-⎪⎭⎫ ⎝⎛=-ωθmL L mg(2 points) ()θωcos 132-=L g(1 point) ()ωθωαsin 32Lg =(1 point) aa =++n ea τea ra magnitude direction√√√√? 2ω⋅OA ? +c a √rv ω20y(2) the reactions at A and B: Draw FBD:mg , A N , B NAnalysis the motion: (5 points) 下面两种方法皆可,选其一 Absolute Motion Method:⎪⎭⎪⎬⎫==θθcos 2sin 2Ly L x C C ⎪⎭⎪⎬⎫-==θθθθsin 2cos 2L yL xC C ⎪⎭⎪⎬⎫--=+-=θθθθθθθθsin 2cos 2cos 2sin 222 L L yL L x C C Relative Motion Method:θθθθcos 2sin 22L L x a C Cx +-==θθθθsin 2cos 22L L y a C Cy --==Writing the differential equations of plane motion:⎭⎬⎫==∑∑Y ym X xm C C⎪⎪⎭⎪⎪⎬⎫-=⎪⎭⎫⎝⎛--=⎪⎭⎫ ⎝⎛+-mg N L L m N L L m B Aθθθθθθθθsin 2cos 2cos 2sin 222 (4 points)(1 point) B a =Ca +τCB a / +nC B a / Aa =Ca +τCA a / +nC A a/。

中石油北京20春石大远程在线考试——《工程力学》答案

中石油北京20春石大远程在线考试——《工程力学》答案

中国石油大学(北京)远程教育学院期末考试《工程力学》学习中心:石家庄桥西奥鹏学习中心姓名:姚军_学号:959048关于课程考试违规作弊的说明1、提交文件中涉嫌抄袭内容(包括抄袭网上、书籍、报刊杂志及其他已有论文),带有明显外校标记,不符合学院要求或学生本人情况,或存在查明出处的内容或其他可疑字样者,判为抄袭,成绩为“0”。

2、两人或两人以上答题内容或用语有50%以上相同者判为雷同,成绩为“0”。

3、所提交试卷或材料没有对老师题目进行作答或提交内容与该课程要求完全不相干者,认定为“白卷”或“错卷”,成绩为“0”。

一、题型简答题,8题,每题5分,共40;计算题,4题,每题15分,共60分)二、题目学号尾数为奇数的同学做题目序号后有“A”字样的题,学号尾数为偶数的同学做题目序号后有“B”字样的题简答题:1A 在铸铁压缩试验中,破坏后的铸铁试样断口平滑呈韧性,与轴线近似成45°。

破坏前,该断口所在斜截面的应力有何特点?1B 在铸铁扭转试验中,铸铁断口的形态是什么样的?答:断口呈现螺旋面,与轴线近似45°斜截面破坏,2A 根据铸铁试件扭转破坏断口可以推断,铸铁的扭转破坏和什么因素有很大的关系?2B 电阻应变片(简称电阻片或应变片)应用广泛,它是利用什么原理实现电测的?答:金属丝的电阻随机械变形而发生变化。

3A 冲击韧性的物理意义是什么?3B 矩形截面梁在截面B处沿铅垂对称轴和水平F,如图所示。

请对称轴方向上分别作用有P问最大拉应力和最大压应力发生在危险截面A 的哪些点上?答:δ+max 发生在d 点。

δ¯max 发生在b 点。

4A 构件中危险点的应力状态如图所示。

构件为钢制:x σ=45MPa ,yσ=135MPa ,z σ=0,xy τ=0,许用应力[]σ=160MPa 。

请用第三强度理论校核该构件的强度。

4B 构件中危险点的应力状态如图所示。

构件材料为铸铁:x σ=20MPa ,y σ=-25MPa ,z σ=40MPa ,xy τ=0,许用应力[]σ=30MPa 。

中国石油大学(北京)2012—2013学年第二学期《工程力学(Ⅰ)(双语)》期中考试试卷A

中国石油大学(北京)2012—2013学年第二学期《工程力学(Ⅰ)(双语)》期中考试试卷A

A卷中国石油大学(北京)2012—2013学年第二学期《工程力学(Ⅰ)(双语)》期中考试试卷考试方式:闭卷考试班级:姓名:学号:(试卷不得拆开,所有答案均写在题后相应位置)Ⅰ (Each 6 points, total 30 points) Answer the following questions briefly.1A force system is shown in Fig. I-1 that consists of three forces. When(a =b -c , b =c -a , c =a -b ), it can be reduced to a force only.Fig. I-1 Fig. I-22A homogeneous slender bar of weight W and length L is shown in Fig. I-2. The coefficients ofstatic friction at contact point A is 0.5. Neglect friction between the bar and wall at B. The smallest positive angle for which the bar can remain at rest is . 3A truss is shown in Fig. I-3, Find the zero members (the bars in which the internal force is zero).Fig. I-3 Fig. I-44 For the linkage shown in Fig. I-4, determine the instant center for velocities of bar BC. 5The collar of weight W slides on a smooth circular arc ofradius R. The ideal spring attached to the collar has the free length L 0 = R and stiffness k. When the slider moves from A to B, the work done by the spring is .Fig. I-5Ⅱ(Total 10 points) Answer the following questions using the knowledge you learned in the engineering mechanics course.1 (4 points) What role does the spoiler(扰流板) of a racing car play?Fig. II-12 (3 points) The photograph of a wheel, which is rolling without slipping on the ground, is shownin Fig. II-2. Its lower part is clear, while the upper is relatively blurry, why?Fig. II-23 (3 points) Explain the function of the tail rotor(尾旋翼) of helicopter.Fig. Ⅱ-3Ⅲ(20 points) For a structure shown in Fig. III that consists of two beams AB and BC which are connected by a smooth pin B. Determine the reactions of supports A, C as well as pin B.2Fig. IIIⅣ (20 points) The collar A of mass m slides on the weightless rod OB , which rotates with a constantangular velocity ωθ= . A pin attached to the collar engages the fixed vertical slot. Neglecting friction, determine (a) the absolute acceleration of the collar A ; (b) the force exerted on the pin by the slot; (c) the force exerted on the collar by the rod. Express you answers in terms of θ, ω, m, b, and g .Fig. IVⅤ(20 points) Fig. V shows a homogeneous slender bar AB of mass m and length L. When the bar was in the position θ=0, it was displaced slightly and released from rest. Neglect friction and assume that endA does not lose contact with the vertical surface.(1) Determine the angular velocity and angular acceleration of the bar as functions of the angle θusingthe theorem of the kinetic energy;(2) Find the reactions at A and B as functions of the angle θusing the differential equations of planemotion of rigid body.Fig. V。

中国石油大学工程力学课件tm406ma

中国石油大学工程力学课件tm406ma
method 6.3 Analysis of plane stress state - graphical
method 6.4 Triaxial stress 6.5 Theory of strength
2020/11/3
Kylinsoft
MOM-6-2
C
6.1 Stress state at a point
s1
P Lt
pri L Lt
pri t
2P pr
t
2
2
0
pLri
cosd
2 pri L
Hoop stress. Where: t – thickness of the vessel
2020/11/3
Kylinsoft
MOM-6-24
From figure f ,
s2
p(ri2 ) (ro2 ri2 )
(ro
pri2 ri )(r0 ri )
pri2 2ri t
pri 2t
pr 2t
Longitudinal stress. here, for thin-walled vessels, rorir
s3 0
2020/11/3
Kylinsoft
MOM-6-25
Sample 6.3(a): 圆球形容器(Spherical pressure vessels)的应力状态
Kylinsoft
MOM-6-27
6.2 Analysis of plane stress state -
C
analytic method
❖问题的提法
已知sx、sy和xy,求给定斜面 (用来表示)上的应力s和
由x正向逆时针转到n正向为 正;反之为负。

中国石油大学工程力学考试Exam(2009-2010-1-fin)(A)取消

中国石油大学工程力学考试Exam(2009-2010-1-fin)(A)取消

中国石油大学(北京)2009—-2010学年第1学期 《 Engineering Mechanics 》期末考试试卷/AClass: Name : Registration No .: Score : No. I II III IV V VI VII TotalScoreDate: Jan 22, 2010I. Choose the correct answer for the following questions1. (4 Points) A overhanging beam with its bending moment diagram is shown in the figure. Its allowable tensile and compressive stresses are [σT ] =39.3MPa and [σC ] =58.8MPa respectively. The most reasonable section is .(a) (b) (c) (d)2. (4 Points) If a circular bar in tension fails along its cross section, it is made of _______ material. If a circular bar in tension fails along its section inclined at an angle 45︒, it is made of _______ material.a. Ductileb. Brittlec. Cannot be judged.3. (4 Points) The deflectionexpression of a prismatic beam with length l is)3(6)(2x l EIPx x v -=Then, the supports at two ends are:a. At x = 0 roller and at x =l fixed end;b. At x = 0 fixed end and at x =l pin;c. At x = 0 roller and at x =l pin;d. At x = 0 fixed end and at x =l free.4. (4 points) All the bars with equal section in two trusses are slender and made of same material. Assume the P 1 and P 2 are the critical loads for two trusses respectively. The correct relation following is(a) P 1 > P 2; (b) P 1 < P 2; (c) P 1 = P 2; (d) The relation can’t be determined.5. (4 points) Weight of the body is W, and pushing force is P. W=100N, P=500N. μS =0.3. The static and kinetic sliding friction coefficients between the weight and the wall are μS =0.3, and μk =0.25 respectively. Then the friction between the weight and the wall is (a) 150N; (b) 125N; (c) 100N; (d) 500N.II Answer following questions simply. 1. (3 Points) In pure bending beam test,8 strain gages are pasted up to the beam as shown. Please explain the function of the 8th gage pasted up to the bottom at right angle with axis.2. (4 points) For a real member undergoing repeated loading, the mean stress and amplitude ofthe stress at its critical point are σm and σa respectively as point P shown in following figure of polygonal line for calculating fatigue failure for asymmetric circle. Please write down theaaa2a1σmin , σmax , and character of circle r . For this circle, is it necessary to check member ’s staticstrength?III. (10 points) Determine the force Q required for the equilibrium of the compound lever if P =240N.IV. (12 points) For the beam as shown, draw bending moment and shearforce diagrams.V. (18 points) A circular shaft (d=85 mm ) with tensile load P =280 kN and torque T =10 kN-m isshown in the figure. Let σallow = 160 MPa. (a) Draw the stress element at A and label the stress values on it ;(b )Exam the shaft with 3th strength theory.A TxP PTA20kN30kN/m30kN/mA B C1mD E 1m1m1mVI. (15 points) Three vertical bars of the same material and same diameter support a rigid horizontal beam at points A, B and C.Determine the axial forces of these three bars.VII. (18 points) A thin curved bar AB has a centerline in the form of quarter circle of radius R, as shown in the figure. The bar is fixed atsupport A and free at B. A horizontal load P acts at free end. Determine the horizontal deflection δh , the vertical deflectionδv .中国石油大学(北京)2009—-2010学年第1学期 《 Engineering Mechanics 》期末考试试卷/BClass: Name : Registration No .: Score :No. I II III IV V VI VII TotalScoreI. Choose the correct answer for the following questions1. (4 points) Weight of the body is W, and pushing force is P. W=100N, P=500N. μS =0.3. The static and kinetic sliding friction coefficients between the weight and the wall are μS =0.3, and μk =0.25 respectively. Then the friction between the weight and the wall is (a) 150N; (b) 125N; (c) 100N; (d) 500N.2. (4 Points) The deflection expression of a prismaticbeam with length l is)3(6)(2x l EIPx x v -=Then, the supports at two ends are:a. At x = 0 roller and at x=l fixed end;b. At x = 0 fixed end and at x=l pin;c. At x = 0 roller and at x=l pin;d. At x = 0 fixed end and at x=l free.3. (4 points) All the bars with equal section in two trusses are slender and made of same material. Assume the P1and P2 are the critical loads for two trusses respectively. The correct relation following is(a) P1 > P2; (b) P1 < P2; (c) P1 = P2; (d) The r elation can’t be determined.4. (4 Points) A overhanging beam with its bending moment diagram is shown in the figure. Itsallowable tensile and compressive stresses are [σT] =39.3MPa and [σC] =58.8MPa respectively.The most reasonable section is.(a) (b) (c) (d)5. (4 Points) If a circular bar in tension fails along its cross section, it is made of _______ material. If a circular bar in tension fails along its section inclined at an angle 45︒, it is made of _______ material.a. Ductileb. Brittlec. Cannot be judged.aaa2a1II. (18 points) A circular shaft (d=85 mm ) with tensile load P =280 kN and torque T =10 kN-m isshown in the figure. Let σallow = 160 MPa. (a) Draw the stress element at A and label the stress values on it ;(b )Exam the shaft with 4th strength theory.A TxP PTIII. (12 points) For the beam as shown, draw bending moment and shear force diagrams.IV . (15 points) Three vertical bars of the same material and same diameter support a rigid horizontal beam at points A, B and C.Determine the axial forces of these three bars.V. (18 points) A thin curved bar AB has a centerline in the form of quarter circle of radius R, as shown in the figure. The bar is fixedat support A and free at B. A horizontal load P acts at free end. Determine the horizontal deflection δh , the vertical deflectionδv . VI. (10 points) Determine the force Q required for the equilibrium of the compound lever if P =240N.VII. Answer following questions simply. 1. (3 Points) In pure bending beam test,A20kN30kN/m30kN/mA B C1mD E 1m1m1m8 strain gages are pasted up to the beam as shown. Please explain the function of the 8th gagepasted up to the bottom at right angle with axis.2.(4 points) For a real member undergoing repeated loading, the mean stress and amplitude ofthe stress at its critical point are σm and σa respectively as point P shown in following figure of polygonal line for calculating fatigue failure for asymmetric circle. Please write down the σmin, σmax, and character of circle r. For this circle, is it necessary to check member’s static strength?中国石油大学(北京)2009—-2010学年第1学期《Engineering Mechanics》期末考试试卷(留学生) Class: Name:Registration No.: Score:No. I II III IV V VI VII TotalScoreI. Choose the correct answer for the following questions1. (5 Points) A overhanging beam with its bending moment diagram is shown in the figure. Itsallowable tensile and compressive stresses are [σT] =39.3MPa and [σC] =58.8MPa respectively.The most reasonable section is .(a) (b) (c) (d)2. (5 Points) For a real member undergoing repeated stress, its character of cycle r=. When r = , its endurance limit is the smallest.a.maxminσσb.minmaxσσc. +1d. –1e. 0f. –∞g. 0.53. (5 Points) Weight of the body is W, and pushing force is P. W=100N, P=500N. μS=0.3. The static and kinetic sliding friction coefficients between the weight and the wall are μS=0.3, and μk=0.25 respectively. Then the friction between the weight and the wall is(a) 150N; (b) 125N;(c) 100N; (d) 500N.4. (5 Points) For two slender compressing bars with same length, same support conditions and made of same material but different cross sectional areas as following, P Cr2=______P Cr1.a. 4b. 8c. 12d. 165. (5 Points) The following figure is a typical low-carbon2aa2a2asteel σ-ε diagram in tension. Points a, b, c, d, e, f and g represent different aspects of mechanical behavior for the material. Please connect the label with its meaning. a yield stress b ultimate stress c elastic limit e percent elongation f proportional limitII. (10 points) Determine the force P required for the equilibrium of the compound lever if Q =4200N. IV . (12 points) For the beam as shown, draw bending moment and shear force diagrams.VI. (20 points) A circular shaft (d=85 mm ) with tensile load P =280 kN and torque T =10 kN-m isshown in the figure. Let σallow = 160 MPa. (a) Draw the stress element at A and label the stress values on it ;(b )Exam the shaft with 4th strength theory.A TxP PTVI. (15 points) Three vertical bars of the same material and same diameter support a rigid horizontal beam at points A, B and C.Determine the axial forces of these three bars.VII.(18 points) A thin curved bar AB has a centerline in the form of quarter circle of radius R, as shown in theA20kN30kN/m30kN/mA B C1mD E 1m1m1mfigure. The bar is fixed at support A and free at B. A horizontal load P acts at free end.Determine the horizontal deflection δh, the vertical deflection δv.中国石油大学(北京)2009—-2010学年第1学期《Engineering Mechanics》期末考试试卷/C Class: Name:Registration No.: Score:No. I II III IV V TotalScoreI. Choose the correct answer for the following questions1. (5 Points) A overhanging beam with its bending moment diagram is shown in the figure. Itsallowable tensile and compressive stresses are [σT] =39.3MPa and [σC] =58.8MPa respectively.The most reasonable section and beam’s position is .(a) (b) (c) (d)2. (5 Points) For a real member undergoing repeated stress, its character of cycle r=. When r = , its endurance limit is the smallest.a. max minσσ b. minmax σσ c. +1 d. –1 e. 0 f. – ∞ g. 0.53. (5 Points) The block weights 200N and T=100N, F=80N. Neglecting the friction at pulley, the friction between the block and the earth is .4. (5 Points) For two slender compressing bars with same length, same support conditions and made of same material but different cross sectional areas as following, P Cr2=______P Cr1. a. 4 b. 8 c. 12 d. 165. (5 Points) The following figure is a typical low-carbon steel σ-ε diagram in tension. Points a, b, c, d, e, f and g represent different aspects of mechanical behavior for the material. Please connect the label with its meaning.a yield stressb ultimate stressc elastic limit e percent elongation f proportional limitII. (12 points) For the beam as shown, draw bending moment and shear force diagrams.2aa2a2aμS =0.2III .(18 points) A circular shaft (d=85 mm ) with tensile load P =280 kN and torque T =10 kN-m isshown in the figure. Let σallow = 160 MPa. (a) Draw the stress element at A and label the stress values on it ;(b )Exam the shaft with 3th strength theory.A TxP PTIV . (15 points) A rigid bar AB of length L is hinged to a wall at A and supported by two verticalbars labeled 1 and 2 attached at points C and D. The bars 1 and 2 have the same cross-sectional area A and are made of the same material (modulus E), but the bar 2 is twice as long as the bar 1. (a) Find the tensile forces N 1 and N 2 in the bars due to the vertical load P acting at end B of the bar. (b) Find the downward displacement B at end B of the bar.V . (15 points) For a closed cylindrical pressure vessel, if r i ≈ r o ≈ r=1000mm, t=10mm, p=0.8MPa. Determine: (a) the hoop and longitudinal stresses in the cylinder and (b) the change in diameter of the cylinder. Let elastic modulus E=200GPa, and Poission ’s ratio ν=0.25.APB CDAh2ha b L1 2V.(15 points) A horizontal bracket ABC is fixed at support A and free at C. The bracket is constructed from a circular cross-sectional bar with constant diameter d. A vertical load P acts at C. Find the vertical displacement δv at C with energy method. E and G are constant.(With other method only 15 60%=9 points are possible).[文档可能无法思考全面,请浏览后下载,另外祝您生活愉快,工作顺利,万事如意!]。

工程力学(中国石油大学(华东))知到章节答案智慧树2023年

工程力学(中国石油大学(华东))知到章节答案智慧树2023年

工程力学(中国石油大学(华东))知到章节测试答案智慧树2023年最新绪论单元测试1.工程力学的任务就是到现场去从事工程技术研究。

参考答案:错2.工程力学的英文翻译为mechanical engineering。

参考答案:错3.古希腊最伟大的力学家是:参考答案:阿基米德4.德国的应用力学学派的领袖是:参考答案:普朗特5.哪一位不是中国力学事业的奠基人?参考答案:林家翘6.达芬奇是一位著名的应用力学家。

参考答案:对7.力学在医学领域几乎没什么影响,因为医生根本不学习力学知识。

参考答案:错8.哥廷根学派的弟子有:参考答案:冯卡门9.郭永怀的主要研究领域为原子弹。

参考答案:错10.生物力学之父是:参考答案:冯元桢第一章测试1.二力平衡条件的适用范围是()。

参考答案:刚体2.图示三铰刚架ACB,受水平力P作用而平衡,有以下四种说法,其中错的是()。

参考答案:R A的方向指向C,R B的方向不确定3.运用哪个公理或推论可将力的三要素中的作用点改为作用线()。

参考答案:力的可传性原理4.只限物体任何方向移动,不限制物体转动的支座称为()。

参考答案:固定铰支座5.力的作用线都相互平行的平面力系称为()。

参考答案:平面平行力系6.根据二力平衡原理中的两个力应该满足的条件,指出下面说法错误的一种是( )。

参考答案:这两个力是作用力与反作用力7.二力杆对它所约束的物体的约束反力作用在物体上( )。

参考答案:必定沿着两作用点的连线8.“二力平衡公理”和“力的可传性原理”只适用于( ) 。

参考答案:刚体9.如图所示,用一绳将重物吊在天花板上,其中FT为绳对重物的拉力,FTˊ为重物对绳的拉力,其中属于作用力与反作用力是()。

参考答案:F T与F Tˊ10.力在某轴上的投影和沿某轴方向上的分力,正确的是( )。

参考答案:分力为矢量, 投影为矢量第二章测试1.图示(a)、(b)两图代表平面汇交力系的两个力多边形,以下四种说法,正确的是().参考答案:图(a)是平衡力系,图(b)的F4是合力2.平面汇交力系平衡的必要和充分条件是该力系的()为零。

中国石油大学(北京)现代远程教育《工程力学》期末考试问题详解

中国石油大学(北京)现代远程教育《工程力学》期末考试问题详解

中国石油大学(北京)远程教育学院期末考试《工程力学》学习中心:____姓名:___学号:____关于课程考试违规作弊的说明1、提交文件中涉嫌抄袭内容(包括抄袭网上、书籍、报刊杂志及其他已有论文),带有明显外校标记,不符合学院要求或学生本人情况,或存在查明出处的内容或其他可疑字样者,判为抄袭,成绩为“0”。

2、两人或两人以上答题内容或用语有50%以上相同者判为雷同,成绩为“0”。

3、所提交试卷或材料没有对老师题目进行作答或提交内容与该课程要求完全不相干者,认定为“白卷”或“错卷”,成绩为“0”。

一、题型简答题,8题,每题5分,共40;计算题,4题,每题15分,共60分)二、题目学号尾数为奇数的同学做题目序号后有“A”字样的题,学号尾数为偶数的同学做题目序号后有“B”字样的题简答题:1A 在铸铁压缩试验中,破坏后的铸铁试样断口平滑呈韧性,与轴线近似成45°。

破坏前,该断口所在斜截面的应力有何特点?答:剪应力最大1B 在铸铁扭转试验中,铸铁断口的形态是什么样的?答:断口呈螺旋面、与轴线近似成45°。

2A 根据铸铁试件扭转破坏断口可以推断,铸铁的扭转破坏和什么因素有很大的关系?答:最大拉应力2B 电阻应变片(简称电阻片或应变片)应用广泛,它是利用什么原理实现电测的?答:金属丝的电阻随机械变形而发生变化3A 冲击韧性的物理意义是什么?答:试样断裂过程中断面单位面积吸收的能量3B 矩形截面梁在截面B 处沿铅垂对称轴和水平对称轴方向上分别作用有P F ,如图所示。

请问最大拉应力和最大压应力发生在危险截面A 的哪些点上? 答:4A 构件中危险点的应力状态如图所示。

构件为钢制:x σ=45MPa ,yσ=135MPa ,z σ=0,xy τ=0,许用应力[]σ=160MPa 。

请用第三强度理论校核该构件的强度。

答:选用第三强度理论,构件满足强度要求4B 构件中危险点的应力状态如图所示。

构件材料为铸铁:x σ=20MPa ,y σ=-25MPa ,z σ=40MPa ,xy τ=0,许用应力[]σ=30MPa 。

中国石油大学(华东)力学2012-2013学年第二学期地球物理学《力学》A卷参考答案

中国石油大学(华东)力学2012-2013学年第二学期地球物理学《力学》A卷参考答案

A卷2012—2013学年第二学期地球物理学专业《力学》期末试卷A卷参考答案与评分标准专业班级姓名学号开课系室物理与光电工程系考试日期 2013年6月1日08:00-10:00二总分题号一1 2 3 4 5 6 7 8得分阅卷人1.请在试卷正面答题,反面及附页可作草稿纸;2.答题时请注意书写清楚,保持卷面整洁;3.本试卷分填空选择题和计算题两大类,满分100分;试卷本请勿撕开,否则作废;4. 本试卷正文共4页。

一、填空选择题(共10小题,每小题3分,共30分,请将正确答案填在题目中相应位置!!) 1、D 2、D 3、16/3 4、3g/4; mg/4 5、34π; 4.5 6、D 7、B 8、D 9、C 10、C二、计算题(共8小题,共70分)1、(本题8分)解(1)j t R i t R dt r d V ωωωωcos sin +-== 2分 j t R i t R dtV d aωωωωsin cos 22--== 2分(2)⎪⎩⎪⎨⎧=+=t R y t R R x ωωsin cos 2 则有:2222R y R x =+⎪⎪⎭⎫ ⎝⎛- 可知质点在圆心为(R/2 , 0)半径为R 的圆周上运动。

2分j t R i t R aωωωωsin cos 22--=')2(22r i R rωω-=--= 2分2、(本题10分)解:选升降机为参考系,设m1、m2对升降机的加速度为a受力分析如图所示: 2分()⎪⎪⎪⎩⎪⎪⎪⎨⎧==-=-+=--ββR a mR R T T a m g m a m T a m a m T g m 221220********* 3分 解之得:()()0212112/2/2a g m m m m m m T -+++= 2分()()021212/a g m m m m ma -++-=1分m1运动加速度(对地)为:()())2/(2/22102211111m m m a m m g m m m T g m a ++++-=-=2分解:(1)由角动量守恒定律有:222312202lv m ml lvm+=ω 2分 则 lv830=ω 2分 (2)xdx k kvdx df ω==dx x k xdf dM f 2ω==30231l k dx x k M lf ωω==⎰ 3分 (3)323131l k dt d ml dt d IM f ωωω===- ⎰⎰-=2/000ωωωωd klmdt t2分 2ln klmt =1分4、(本题10分)解: 2分2分2分2分2分s222rad/4//===ttRv k ω,42t=∴ωtR 24=ωR v =时,s 1=t tv R t24=s2m/4=dt dv at/=Rt 8=s 2m/8=R v a n /2=s2/m 16=)(222/1a a t n a +=s2m/58=解: 设运动表达式 由图可见: A = 2m , 当t = 0 时有: 2分解得:40πϕ-= 3分当t = 1 s 时有:43πω=3分 振动表达式: 2分 6、(本题10分)解:1)由题意知:ω=2πν=500π波的传播方向向左。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
160 120
.
(MPa)
30
80 40
zC
400 60
0 -40
t
M
M
b Dimensions in mm Fig. I-2
-80
Fig. I-3 , the amplitude of
3.
(6 points) A stress cycle is shown in Fig. I-3, the mean stress m = stress a = , the character of cycle r = .
P
P
(a) Fig. I-1 2.
(b)
(6 points) A T-beam made of cast iron is in pure bending. Knowing the ratio of tensile allowable stress to compressive allowable stress is [t]/[c]=1/4, the reasonable width b of the flange is
4.
(6 points) Four different cases of supports for a column are shown in Fig. I-4, case highest critical load.
has the
A卷 第 2 页 共 5 页
y B
Mz
A
h/2
z
h/2
My
b
x C
Pl 3 for the case shown in Fig. IV(c). 3EI
(5) If the free end C is connected to a vertical bar (elastic modulus E2, cross section area A, length h) before vertical load P is applied, as shown in Fig. IV(b), determine the stress in the bar.
A卷
中国石油大学(北京)2011—2012 学年第二学期
《工程力学(双语) 》期末考试试卷
考试方式(闭卷考试)
班级: 姓名: 学号:
No.




Total
Score
(试卷不得拆开,所有答案均写在题后相应位置)
A卷 第 1 页 共 5 页
I 1.
(Total 30 points) Fill up the blanks with correct answer. (6 points) Stress-strain curves for three materials are shown in Fig. I-1, along with a loading device which can make specimens made of each material and with same dimensions to have equal end displacements. Material increasing, Material (1, 2, 3) has the highest ultimate stress. With the tensile load P (1, 2, 3) in the device will fracture first.
=0.25. Assume that ri ro r .
Fig. III
A卷 第 4 页 共 5 页
IV (40 points) A horizontal bracket ABC is fixed at support A and free at C, as shown in Fig. IV(a). The bracket is constructed from a circular cross-sectional bar with constant diameter d. A vertical load P acts at the free end C. E1 is known. (1) Determine the places of the critical points. (2) Draw stress elements to show the stress state of the critical points. (3) Determine the equivalent stresses of 3th strength condition at critical points. (4) Find the vertical displacement v at C. Note: the deflection at the free end of a cantilever beam is given by vB
E2
h
v
C A a (a) B
E1
C P A a (b) Fig. IV B
E1
a
a
P
EI
P
B
l
(c)
A卷 第 5 页 共 5 页
II
(15 points) For a beam shown in Fig. II, draw its shear force and bending moment diagrams.
30kN/m 20kN 30kN/m
A 1m
C 1m
D 1m
E 1m
B
Fig. II
A卷 第 3 页 共 5 页
III (15 points) Consider a closed cylindrical steel pressure vessel, as shown in Fig. III. The radius of the cylinder is 1000mm and its wall thickness is 10mm. (a) Determine the hoop and the longitudinal stresses in the cylindrical wall caused by an internal pressure of 0.80MPa. (b) Calculate the change in diameter of the cylinder caused by pressurization. Let E =200GPa, and
DHale Waihona Puke AB Fig. I-4
C
D Fig. I-5
5.
(6 points) The internal forces on a cross section of a doubly symmetric beam made of ductile material are shown in Fig. I-5, are critical points.
相关文档
最新文档