《建筑工程计量与计价》二参考答案 (2)

《建筑工程计量与计价》试题二

一．填空题（共20分，每空格1分）

1.定额按编制程序和用途分为施工定额、预算定额、概算定额与概算指标、投资估算指标

2.工程建设中使用的材料有 .一次性使用材料和周转性使用材料两种类型。

3.多层建筑坡屋顶内和场馆看台下，当设计加以利用时净高超过2.10m的部位应计算全面积；净高在1.20m至2.10m的部位应计算1/2面积；当设计不利用或室内净高不足1.20m时不应计算面积。

4.整体楼梯包括休息平台、平台梁、斜梁及楼梯的连接梁，按水平投影面积计算，不扣除宽度小于500mm 的楼梯井，伸入墙内部分不另增加。当整体楼梯与现浇楼层无梯梁连接时，按楼层的最后一个踏步外边缘加30cm为界。

5.普通窗按洞口面积以平方米计算。

6.墙裙以高度在以内为准，超过时按墙面计算，高度低于以内时，按踢脚板计算。

7.分部分项工程量清单应包括、、和。

8.应计入建筑安装工程造价的税金包括、及等。

二．选择题（共10分，每小题1分）

1．某单层混凝土结构工业厂房高15m，其一端有6层砖混车间办公楼与其相连，构成一单位工程，两部分外墙外边距离350mm，各部分首层勒脚以上外墙外边所围面积分别为2000m2和，300m2，缝长20m，则该单位工程建筑面积为( )m2。

A．2300 B．3800 C．3807 D．3842

2．整体地面面层工程量均按主墙间净面积以“平方米”计算，应扣除( )。

A．凸出地面构筑物、室内地沟所占面积B．柱、垛所占面积

C．间壁墙所占面积D．附墙烟囱所占面积

3．某建筑物采用现浇整体楼梯，楼梯共四层自然层，楼梯间净长6m，净宽3m，楼梯井宽500mm，长3m，则该现浇楼梯的混凝土工程量为( )m2。

A．18 B．72 C．70.5 D．66

4．除另有说明外，所有清单项目的工程量应以实体工程量为准，并以完成后的( )计算。

A．净值B．净值加损耗C．净值加需要增加的工程量D．实际量

5．构筑物混凝土工程量除另有规定者外，均按图示尺寸以体积计算，应扣除门窗洞口及( )孔洞所占的体积。

A．0.1m2以外B．0.1m2以内C．0.3m2以外D．0.3m2以内

6．如果是打孔后，先埋入预制混凝土桩尖，再灌注混凝土的打孔灌注方式，其桩尖的工程量按( )计算体积。

A．个B．设计规定C．图示尺寸D．钢筋混凝土章节规定

7．（）是计算单项工程和单位工程的成本和造价的基础。

A、施工定额

B、预算定额

C、概算指标

D、概算定额

8．高低联跨的建筑物，应以（）为界分别计算建筑面积；其高低跨内部连通时，其变形缝应计算在低跨面积内。A、高跨结构外边线 B. 高跨结构内边线

C. 高跨结构中心线

D.变形缝中心线

9．墙体工程量计算时不应扣（）面积。A、每个面积大于0.3m2的孔洞 B、过人洞面积

C、嵌入且平行于墙体的混凝土构件

D、混凝土梁头体积

10．下列说法不正确的是（）。

A、无墙裙的，其高度按室内地面或楼面至天棚底面之间的距离计算；

B、有墙裙的，其高度按墙裙顶至天棚底面之间的距离计算；

C、钉板条天棚的内墙面，其高度按室内地面或楼面至天棚底面另加100mm计算。

D、钉板条天棚的内墙面，其高度按室内地面或楼面至天棚底面另加200mm计算。

三、是非题（共10分，每小题1分）

1、当设计要求与定额项目的内容不完全一致时，可直接套用定额的预算基价及工料机消耗量，计算该

分项工程的直接工程费以及工料机需用量。（）

2、有永久性顶盖的室外楼梯(建筑物无室内楼梯)，其建筑面积按建筑物自然层的水平投影面积的计算。

3、干湿土以地质资料提供的地下常水位为分界线，地下常水位以上为干土，以下为湿土，如用人工降

低地下水位时，干湿土划分仍以地下常水位为准。（）

4、砖垛、三皮砖以上的腰线和挑檐等体积，并入墙身体积内计算。（）

5、设计图纸中未注明的钢筋接头和施工损耗没有综合在定额项目内，套用定额时采用钢筋的理论质量

加上损耗量套取定额。（）

6、措施项目清单为可调整清单，投标人对招标文件中所列项目，可根据企业自身特点作适当的变更增

减。（）

7．定额编制方法有技术测定法、经验估工法、统计计算法、比较类推法。（）

8．按《计价规范》规定：工程量清单计价应包括按招标文件规定，完成工程量清单所列项目的全部费用，包括分部分项工程费、措施项目费、其他项目费和规费、税金、风险费。

英语泛读教程2参考答案(1-10)

英语泛读教程（第2册）参考答案 Unit One Comprehension Points 1.Every year at the same time the stars were in the same place in the sky. 2.They are the planets. 3.He thought that … (Page 2, parag. 12) 4.The idea was that the stars and the sun did not turn around the earth. Instead, it was the earth that was turning. The sun and the stars were not really moving. 5.Because they would upset the whole science of astronomy and cause cruel persecution by the Church. 6.Content/details. https://www.360docs.net/doc/e516209178.html,mon. 8.Because radio telescopes find out new stars by picking up radio waves from them. 9.It means the group of stars that our earth is a member of. 10.The implication is that probably by then the earth has already been melted. 11.Every three days it gets very dim. It stays dim for a few hours, then it gets bright again. 12.Because a comet isn’t solid. It is thin like a cloud. Its tail is nothing but glowing gas. And its head is made of small pieces that could not hurt the earth. Exercises I. True or false 1. F 2. F 3. T 4. F 5. T 6. F 7. T 8. T 9. T 10. F II. 1.略。 2.Four stages. Childhood, manhood, old age, and the last stage. 3.The very hottest ones are blue. Those not quite so hot are white, and those less hot than the white ones are yellow and red. 4.略。 5.They help sailors to find their way and they help us to tell time. Unit Two Comprehension Points 1.It means to function properly. 2.电气史上最伟大的人物之一迈克尔·法拉第写了一篇报告，介绍了一种叫做gutta percha 的类似橡胶的物质，这种物质是从生在在马来西亚的一种树上提取出来的。 3.It means to bring out of the water on to the land. 4.To lay a cable under the Atlantic. 5.To become weaker. 6.第一个it指的是the telegraph failed，第二个it指的是it ever worked，第三个it指的是 the cable. 7.Because a big ship can’t carry the end of the cable on to the shore, it will get stranded.

机器学习第二章答案2

1.1. Give three computer applications for which machine learning approaches seem appropriate and three for which they seem inappropriate. Pick applications that are not already mentioned in this chapter, and include a one-sentence justification for each. Ans. Machine learning: Face recognition, handwritten recognition, credit card approval. Not machine learning: calculate payroll, execute a query to database, use WORD. 2.1. Explain why the size of the hypothesis space in the EnjoySport learning task is 97 3. How would the number of possible instances and possible hypotheses increase with the addition of the attribute WaterCurrent, which can take on the values Light, Moderate, or Strong? More generally, how does the number of possible instances and hypotheses grow with the addition of a new attribute A that takes on k possible values? Ans. Since all occurrence of “φ” for an attribute of the hypothesis results in a hypothesis which does not accept any instance, all these hypotheses are equal to that one where attribute is “φ”. So the number of hypothesis is 4*3*3*3*3*3 +1 = 973. With the addition attribute Watercurrent, the number of instances = 3*2*2*2*2*2*3 = 288, the number of hypothesis = 4*3*3*3*3*3*4 +1 = 3889. Generally, the number of hypothesis = 4*3*3*3*3*3*(k+1)+1. 2.3. Consider again the EnjoySport learning task and the hypothesis space H described in Section 2.2. Let us define a new hypothesis space H' that consists of all pairwise disjunctions of the hypotheses in H. For example, a typical hypothesis in H' is (?, Cold, High, ?, ?, ?) v (Sunny, ?, High, ?, ?, Same) Trace the CANDIDATE-ELIMINATATION algorithm for the hypothesis space H' given the sequence of training examples from Table 2.1 (i.e., show the sequence of S and G boundary sets.) Ans. S0= (φ,φ,φ,φ,φ,φ) v (φ,φ,φ,φ,φ,φ) G0 = (?, ?, ?, ?, ?, ?) v (?, ?, ?, ?, ?, ?) Example 1: S1=(Sunny, Warm, Normal, Strong, Warm, Same)v (φ,φ,φ,φ,φ,φ) G1 = (?, ?, ?, ?, ?, ?) v (?, ?, ?, ?, ?, ?) Example 2: S2= {(Sunny, Warm, Normal, Strong, Warm, Same)v (Sunny, Warm, High, Strong, Warm, Same)，(Sunny, Warm, ？, Strong, Warm, Same) v (φ,φ,φ,φ,φ,φ)} G2 = (?, ?, ?, ?, ?, ?) v (?, ?, ?, ?, ?, ?) Example 3: S3={(Sunny, Warm, Normal, Strong, Warm, Same)v (Sunny, Warm, High, Strong, Warm, Same)，(Sunny, Warm, ？, Strong, Warm, Same) v (φ,φ,φ,φ,φ,φ)} G3 = {(Sunny, ?, ?, ?, ?, ?) v (?, Warm, ?, ?, ?, ?), (Sunny, ?, ?, ?, ?, ?) v (?, ?, ?, ?, ?, Same), (?, Warm, ?, ?, ?, ?) v (?, ?, ?, ?, ?, Same)}

泛读教程第二册答案(全)

Keys to Reading Course 2 Unit 1 Reading Section A Word Pretest 1.B 2.A 3.B 4.A 5.B 6.C 7.B 8.C Reading Comprehension 1.B 2.A 3.B 4.B 5.C 6.C Vocabulary Building Word Search 1. assignment 2. irony 3. reverse 4. accomplish 5. assemble 6. squeeze 7. sensual 8. fragment 9. narcotic 10. adolescence Use of English 1. Bob agreed to take on the leadership of the expedition. 2. The world was taken in by his fantastic story of having got to the Pole alone. 3. He took up his story after a pause for questions and refreshments. 4. That takes me back to the time I climbed to the top of Mount Fuji. 5. The members of the party took it in turns to steer the boat. 6. They took it for granted that someone would pick up their signals and come to their aid. Stems 1. proclaim: to announce officially and publicly; to declare 2. percentage: a proportion or share in relation to a whole; a part 3. confirm: to support or establish the certainty or validity of; to verify 4. affirm: to declare positively or firmly; to maintain to be true 5. centigram: a metric unit of mass equal to one hundredth of a gram 6. exclaim: to express or utter(something) suddenly or vehemently Synonyms 1. adaptability 2. purpose 3.strained 4.hold 5.defeat Cloze important second France student bilingual monolingual serious means use difficult Section B 1.F 2.T 3.T 4.C 5.A 6.B 7.B 8.B 9.B 10.T 11.T 12.F 13.F 14.T 15.T Section C 1.F 2.T 3.T 4.F 5.T 6.F 7.F 8.F 9.F 10.F

第二章参考答案

第2章参考答案 2写出下列十进制数的原码、反码、补码和移码表示（用8位二进制数）。如果是小数，则用定点小数表示；若为整数，则用定点整数表示。其中MSB 是最高位（符号位），LSB 是最低位。 (1)－1 (2) －38/64 解：（1）-1=(-0000001)2 原码: 10000001 反码: 11111110 吧补码: 11111111 移码: 01111111 （2）-38/64=-0.59375=(-0.1001100)2 或-38/64=-（32+4+2）*2-6=-（100110）*2-6=(-0.1001100)2 原码: 1 .1001100 反码: 1 .0110011 补码: 1 .0110100 移码: 0.0110100 注：-1如果看成小数，那么只有补码和移码能表示得到，定点小数-1的补码为：1.0000000 此例类似于8位定点整数的最小值-128补码为10000000 3 有一字长为32位的浮点数，符号位1位；阶码8位，用移码表示；尾数23位，用补码表示；基数为2.请写出：（1）最大数的二进制表示，（2）最小数的二进制表示，（3）规格化数所能表示的数的范围。解：（题目没有指定格式的情况下，用一般表示法做）（1）最大数的二进制表示：0 11111111 11111111111111111111111 （2）最小数的二进制表示：1 11111111 00000000000000000000000 (1) ）（231221*27--- (2) ）（1*2127-- （3）规格化最大正数：0 11111111 11111111111111111111111 ）（231221*27---

泛读教程2、课后题答案

A :c B 1-5 baccd 6-8 cdd D.1-5 daaba 6-10 dadca Fast Reading:1 dbcca 2 cbbcd 3 dcdcd Home Reading:abcccbac Unit 2 A:c B 1-5 cdada 6-10 bdacb D 1-5 dcbad 6-10 cbbad 11-15 cabdc Fast Reading:1 bbcad 2 bbbdd 3 cddda Home Reading: abadcadbcd Unit 3 A :c B 1-5 dccad 6-10 ddcaa D 1-5 badcd 6-12 bbcdacb Fast Reading:1 dbacc 2 abdcbd 3 dabc Home Reading:abdccccbab Unit 4 A :c B bdcaab D 1-5 dbacb 6-12 ccdcdcb Fast Reading:1 cdabd 2 cdcdd 3 dccbb Home Reading:bcddcdccdd Unit 5 A: c B 1-5 cacbd 6-10 bdabc D 1-5 abacd 6-10 ababa Fast Reading:1 ddbdb 2 dbddc 3 cdbcd Home Reading:cbdcbaacbd Uint 6 A:a B 1-5 bbbad 6-10 bdddc D 1-5 ccbaa 6-12 dabada Fast Reading:1 cabdd 2 addaa 3 bccdb Home Reading:bcacbbbb Unit 7 A:b B 1-5 cadbd 6-11 abcbad D 1-5 badca 6-14 bcbbadcab Fast Reading:1 cabdb 2 dbbdd 3 dccbd Home Reading:dcdbaccadb Unit 8 A:b B 1-5 badda 6-8 bac D:dacdcabbdab Fast Reading:1 cdbbd 2 dcacb 3 adcab Home Reading:accbdbacdb Unit 9 A:c B 1-5 bdcac 6-8 bcd D:aaabccdacbb Fast Reading:1 bcdda 2 adccd 3 dcbdc Home Reading:bcbadcab Unit 10 A:a B 1-5 acdcb 6-10 dadcc D:aaadbbcdabb Fast Reading: 1 ddccc 2 bbbca 3 cdacc Home Reading:bddcddcba Unit 11 A:c B 1-5 bdbba 6-7 bc D:aabbadab Fast Reading:1 bcbcb 2 ccbdd 3 abccd Home Reading:cdddddbcba

第二章习题答案

第二章 1.在立方点阵中画出下面的点阵面和结点方向。（010）（01- 1） (113) [110] [201] [101] 2.将下面几个干涉面（属立方晶系）按面间距的大小排列。（123），（100），（2-00），（311-），（12-1），（210），（110），（2- 21），（030），（130）。解：立方晶系的面间距公式为2 2 2 d h k l = ++间距大小排列为（100），（110），（2- 00），（210），（12- 1），（2- 21）=（030），（130），（123）

3.在六方晶系中h+k=-i 。证明1：如图，任意截面交1a 和2a 于C,D;过3a 做反向延长线，并交线段CD 于B 由正弦公式可得： 11sin sin h i ABC ACB -= ∠∠ 11 sin sin k i ABD ADB - =∠∠

由三角形中几何关系可得： ABD ABC π∠=-∠ 23ACB ABC π∠= -∠ 3 ADB ABC π∠=∠- 联立上述5式可解得：2sin()cos() 3sin ABC h k i i ABC π ∠+=-=-∠ 证明2：（比较复杂）

) (0 ) ( 2 2 2 2 2 2 1 1 1 2 2 2 2 2 2 2 ) 60 cos( 2 ) 60 cos( 2 ) 120 cos( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 2 2 2 1 2 2 2 2 1 k h i i k h hk hi ki k h i ab ae be b a e abe be a e ab e b e a b a be e b ae e a e be ae ab ab b a be e b ae e a be e b ae e a d d d d c be e b d ae e a d ab b a c d d c + - == + + = + + + + + = + - - + + = + - - + + - + - + = + + + + + = - + - + + - + + - + = + + = - + = - + = - + = + = 截距的倒数等于面指数两边开方，得 4.（11-0）、（12-1）、（3-12）属于[111]晶带。解：由晶带定律知0 hu kv lw ++= 带入得1*1+(-1)*1+1*0=0 1*1+(-2)*1+1*1=0 (-3)*1+1*1+2*1=0 所以属于[111]晶带。 5.（110）、（311）、（132）是否属于同一晶带晶带轴是什么再指出属于

英语泛读教程2-王守仁-答案

4That takes me back to the time I climbed to the top of Mount Fuji. 5The members of the party took it in turns to steer the boat. 6They took it for granted that someone would pick up their signals and come to their aid. Stems 7proclaim: to announce officially and publicly; to declare 8percentage: a proportion or share in relation to a whole; a part 9confirm: to support or establish the certainty or validity of; to verify 10affirm: to declare positively or firmly; to maintain to be true 11centigram: a metric unit of mass equal to one hundredth of a gram 12exclaim: to express or utter(something) suddenly or vehemently Synonyms

第二章习题及答案

化工原理练习题五.计算题 1. 密度为1200kg.m的盐水，以25m3.h-1的流量流过内径为75mm的无缝钢管。两液面间的垂直距离为25m，钢管总长为120m，管件、阀门等的局部阻力为钢管阻力的25％。试求泵的轴功率。假设：（1）摩擦系数λ＝0.03；（2）泵的效率η＝0.6 1.答案***** Z1＋ｕ2/2ｇ＋P1/ρｇ＋He＝Z2＋ｕ2/2ｇ＋P2/ρg+∑H f Z＝0，Z＝25m，ｕ≈0，ｕ≈0，P ＝P ∴H＝Z＋∑H＝25＋∑H ∑H＝（λ×ｌ/d×ｕ/2ｇ）×1.25 ｕ＝Ｖ/A＝25/（3600×0.785×（0.07 5））＝1.573m.s ∑H＝（0.03×120/0.075×1.573/（2×9.81）×1.25 ＝7.567m盐水柱 H＝25＋7.567＝32.567m N＝Q Hρ/102＝25×32.567×120 0/

（3600×102）＝2.66kw N轴＝N/η＝2.66/0.6＝4.43kw 2.(16分) 如图的输水系统。已知管内径为d=50mm, 在阀门全开时输送系统的Σ(l+le ) =50m,摩擦系数可取λ=0.03,泵的性能曲线,在流量为6 m3.h-1至15 m3.h-1范围内可用下式描述: H=18.92-0.82Q2.，此处H为泵的扬程m,Q为泵的流量m3.h-1,问: (1)如要求流量为10 m3.h-1，单位质量的水所需外加功为多少? 单位重量的水所需外加功为多少?此泵能否完成任务? (2)如要求输送量减至8 m3.h-1 (通过关小阀门来达到)，泵的轴功率减少百分之多少?(设泵的效率变化忽略不计) 答案***** ⑴u=10/(3600×0.785×0.05)=1.415[m.s-1] Σhf =λ[Σ(l+le )/d](u2/2)

泛读教程第二版第一册unit13

Unit 13 Physical Fitness 1-5 BACBA ACB 1-5 BBAAC BCA Word Match subtract to take (a number, amount, ect.) from something larger dissolve to cause something to end or disappear persevere to continue doing something in spite of difficulties alleviate to make less hard to bear; relieve overtax to demand too much arthritis a disease that causes the joints to become swollen and painful obesity having too much fat in the body stroke an illness caused by a braking or blocked blood vessel in the brain fatigue physical or mental tiredness; exhaustion ; weariness tissue the substance of an organic body or organ diet a limited list of food or drink that one is allowed meditation focusing attention on only one thing so as to be calm and relaxed clearance official permission for someone to do something leisure the time when you are not working ratio the relationship between two things expressed in numbers Prefixes precede: to come before in time subway: an underground railway physiological: relating to physiology (function of a living organism ) preschool: of early childhood (before elementary school ) subzero: below zero physique: physical build (body size and shape ) subconscious : below the level of conscious perception posthumous : after one's death Cloze 2. joggers 4 .shoes 5. protect 7. far 8. short 9. distance 10. run SECTION B 1-5 ACBBC ABC SECTION C 1-5 FTTFT FFTTT

材料力学答案第二章

第二章拉伸、压缩与剪切第二章答案 2.1 求图示各杆指定截面的轴力，并作轴力图。 40kN 50kN 25kN （a ） 4 4F R F N 4 40kN 3 F N 3 25kN 2F N 2 20kN 11 F N 1 解： F R =5kN F N 4 =F R =5 kN F N 3 =F R +40=45 kN F N 2 =-25+20=-5 kN F N 1 =20kN 45kN 5kN 20kN 5kN

（b） 1 10kN 6kN F N 1 =10 kN F N 2 =10-10=0 F N 3 =6 kN 1—1截面： 2—2截面： 3—3截面：10kN F N 1 1 1 10kN 10kN 2 2 F N 2 6kN 3 3 F N 3 2.2 图示一面积为100mm 200mm的矩形截面杆，受拉力F = 20kN的作用，试求：（1）

6 π = θ的斜截面m-m 上的应力；（2）最大正应力max σ和最大剪应力max τ的大小及其作用面的方位角。解： 320101MPa 0.10.2 P A σ?===?2 303cos 14 σσα==?=3013sin600.433MPa 2 22 σ τ= = ?=max 1MPa σσ==max 0.5MPa 2 σ τ= =F 2.3 图示一正方形截面的阶梯形混凝土柱。设重力加速度g = 9.8m/s 2, 混凝土的密度为 33m /kg 1004.2?=ρ，F = 100kN ，许用应力[]MPa 2=σ。试根据强度条件选择截面宽度a 和b 。

b a 解： 2 4, a ρ?3 42 2.0410ρ=??11 [] a σσ=0.228m a ≥ = =22 342424431001021040.2282104a b b ρρ=?+?=??+???+???2[], b σσ≥0.398m 398mm b ≥ == 2.4 在图示杆系中，AC 和BC 两杆的材料相同，且抗拉和抗压许用应力相等，同为[]σ。BC 杆保持水平，长度为l ，AC 杆的长度可随θ角的大小而变。为使杆系使用的材料最省，试求夹角θ的值。

第二章课后习题答案

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