Received () Revised ()
GLASSY STATES IN A SHAKEN SANDBOX
PETER F.STADLER
Institut f¨u r Theoretische Chemie und Molekulare Strukturbiologie,
Universit¨a t Wien,W¨a hringerstra?e17,A-1090Wien,Austria;and
The Santa Fe Institute,1399Hyde Park Rd.,Santa Fe NM87501,USA
ANITA MEHTA
S N Bose National Centre for Basic Sciences,Block JD Sector3,Salt Lake,Calcutta700098,
India;and
ICTP,Strada Costiera11,I-34100Trieste,Italy
JEAN-MARC LUCK
Service de Physique Th′e orique(URA2306of CNRS),CEA Saclay,
91191Gif-sur-Yvette cedex,France
Received()
Revised()
Our model of shaken sand,presented in earlier work,has been extended to include a more realistic‘glassy’state,i.e.,when the sandbox is shaken at very low intensities of vibration.We revisit some of our earlier results,and compare them with our new results on the revised model.Our analysis of the glassy dynamics in our model shows that a variety of ground states is obtained;these fall in two categories,which we argue are representative of regular and irregular packings.
1.Introduction
The test of a good lattice model of a complex system is whether it succeeds in capturing the essential physics of a real system in its bid to reduce its technical complexity.Areas as diverse as tra?c?ow[10],plate tectonics[12,13]and granular ?ow[28]are examples where lattice models have been used rather successfully despite their apparent simplicity,to describe at least a few of the salient features of some genuinely complex systems.
In this spirit,we present two versions of a model of shaken sand in the following; both models exhibit behaviour that is representative of shaken sand between the ?uidised and the glassy regimes.While the second model includes rather complex interactions between the‘grains’in the frozen state,in contrast to the?rst,the latter is nevertheless surprisingly successful in replicating at least some of the qualitative features associated with the glassy regime.One of the aims of this contribution is then to identify some of the essential features that are needed for a(discrete)
1
2Peter F.Stadler,Anita Mehta,Jean-Marc Luck
minimal model of the system in question.
The earlier model(‘old model’)[32]is the generalisation of a cellular-automaton (CA)model[4,28]of an avalanching sandpile.This model shows both fast and slow dynamics in the appropriate regimes:in particular,it reduces to an exactly solvable model of noninteracting grains in the frozen(‘glassy’)regime,and provides one with a toy model for ageing in vibrated sand[9,24].
Next,we present a more realistic version of the model(‘new model’)[26],which is the topic of ongoing research.While identical in the?uidised regime,the latter model is less of a toy model for the glassy state,in that the grains are no longer non-interacting,but are coupled to each other based on their orientations.Our analysis shows that a rich variety of ground states is obtained,which we analyse in terms of a particular parameter that has an interpretation in terms of the irregularity of the grains.
2.The Model
We consider a rectangular lattice of height H and width W with N≤HW grains located at the lattice points,shaken with vibration intensityΓ.Each grain is a rectangle with sides1and a≤1,respectively.Consider a grain(i,j)in row i and column j whose height at any given time is given by h ij=n ij?+an ij+,where n ij?is the number of vertical grains and n ij+is the number of horizontal grains below(i,j).
The dynamics of the model are described by the following rules:
(i)If lattice sites(i+1,j?1),(i+1,j),or(i+1,j+1)are empty,grain
(i,j)moves there with a probability exp(?1/Γ),in units such that the
acceleration due to gravity,the mass of a grain,and the height of a lattice
cell all equal unity.
(ii)If the lattice site(i?1,j)below the grain is empty,it will fall down.
(iii)If lattice sites(i?1,j±1)are empty,the grain at height h ij will fall to either lower neighbour,provided the height di?erence h ij?h i?1,j±1≥2.
(iv)The grain?ips from horizontal to vertical with probability exp(?m ij(?H+?h)/Γ),where m ij is the mass of the pile(consisting of grains of unit
mass)above grain(i,j).For a rectangular grain,?H=1?a is the height
di?erence between the initial horizontal and the?nal vertical state of the
grain.Similarly,the activation energy for a?ip reads?h=b?1,where
√
b=
Glassy states in a shaken sandbox3
Fig.1.Snapshots of a sandbox with W=30,H=100,n=2500grains with a=0.7,?h=0.05, and shaking intensitiesΓ=0.1(top row:glassy regime)and0.8(bottom row:?uidised regime), for times t=1(before tapping),t=2(each grain on average touched once by the MC simulation), t=5,10,30,and100.
are the mechanism for each particle to?nd local stability.In related work[7,8], this was associated with a threshold,called the single particle relaxation threshold (SPRT).
In line with recent investigations of compaction[1,2,3,11,16,19,20,27,30,31],we have examined the behaviour of the packing fraction of our model,as a function of the vibration intensityΓ.Let N?and N+be the numbers of vertical and horizontal grains in the box.The packing fractionφis:
N+?aN?
φ=
4Peter F.Stadler,Anita Mehta,Jean-Marc Luck
3.A Spin-Model for the Ordered Regime
The frozen regime is characterised by an absence of holes within the sandbox,and negligible surface roughness.Here,the earlier model[32]reduces to an exactly solvable model of W independent columns of H noninteracting grain orientations σn(t)=±1,withσ=+1denoting a horizontal grain,andσ=?1denoting a vertical grain.The orientation of the grain at depth n,measured from the top of the system,evolves according to a continuous-time Markov dynamics,with rates
w(?→+)=exp(?n?h/Γ),
(3.1)
w(+→?)=exp(?n(?H+?h)/Γ),
as m ij=n=H+1?i.The parameters?H and?h correspond to two characteristic lengths of the model,
ξeq=Γ/?H,ξdyn=Γ/?h.(3.2) The equilibrium lengthξeq is the typical depth below which all grains are frozen into their horizontal ground-state orientation,while the dynamical lengthξdyn is characteristic of the divergence of the relaxation time with depth,τ～exp(n/ξdyn). As a consequence,any perturbation propagates only logarithmically slowly down the system,over an ordering lengthΛ(t)≈ξdyn ln t.
The detailed analysis of this model in earlier work,despite its‘toy model’nature, was remarkably successful in reproducing certain features of the glassy state.In this work,we present an improved version where the spins are no longer noninteracting, in the frozen regime.They are in fact coupled in a rather complex way;these more realistic interactions lead to a rich variety of ground states depending on the analogue of the aspect ratio a,which we argue is representative of the shape of the grains.We present this model in the next section.
4.A Generalisation
The generalisation of our earlier model involves the insertion of eq.(2.1)into the transition rates of the system.We thus require that,for a given value of a,the transitions are such that the packing fraction of the system is locally minimised. We now allow a to take arbitrary values(while always remaining positive):1?a can then be visualised as the size of the‘void’associated with the‘wrong’orientation of the grain.For convenience,we write these rules in a more general form below.
Thus,the dynamical rules(iv)and(v)of the rectangular grain model may be regarded as a special case of a more general model with transition rates
w(?→+)=exp(?(λij+ηij)/Γ),
w(+→?)=exp(?(λij?ηij)/Γ),(4.1) where
ηij=Am ij++Bm ij?,λij=Cm ij++Dm ij?,(4.2)
Glassy states in a shaken sandbox5 are,respectively,the ordering?eld and the activation energy felt by grain(ij).In these expressions,m ij±is the number of horizontal and vertical grains above the grain(i,j),respectively.Note that grains are only counted from(i,j)to the?rst void above level j.Thus m ij=m ij++m ij?.
The earlier model is recovered by setting
A=B=?H/2,C=D=?H/2+?h.(4.3) In the general situation where the equalities A=B,C=D are not obeyed, the rates(4.1)depend on the orientations of all the grains above the grain under consideration.Our new model is therefore a fully directed model of interacting grains,where causality acts both in time and in space,as the orientation of a given grain only in?uences the grains below it and at later times.
The key parameter which governs the statics and dynamics of the model at low shaking intensity turns out to be the dimensionless ratioε=A/B.Consider the ordered regime,where there are no holes,in the zero-temperature limit(Γ→0). In this regime,the steady-state values of the grain orientations are given by the deterministic,recursive equation
σn=sign(εn+(n)+n?(n)),(4.4) where n±(n)is the number of horizontal and vertical grains at depths0,...,n?1, so that n+(n)+n?(n)=n=1,2,...We assumeσ0=+1.
As long asε≥0,(4.4)leads to the trivial ground state where all the grains are horizontal(σn=+1),generalising thus the case of the earlier model(ε=1).
In the frustrated regime(ε<0),ground states have a richer structure.They contain non-trivial fractions of horizontal and vertical grains,
f+=1/(1?ε),f?=?ε/(1?ε).(4.5) The grain orientations are distributed in a way which depends,rather unexpectedly, on whetherεis rational or not.
?Rational case:ifε=?p/q is a negative rational number(in irreducible
form),(4.4)determines the grain orientationσn whenever the depth n is
not a multiple of r=p+q.The orientations of the latter grains are left free.
We thus obtain an extensively degenerate set of ground states,each of them
being a random sequence of two types of r-mers,i.e.,clusters of r grains.
The associated con?gurational entropy per grain readsΣ=(ln2)/r.The
simplest example isε=?1,hence r=2and f?=1/2,where the clusters
are+?and?+,so that the ground states are all the dimerised grain
con?gurations.Two examples consist of trimers(r=3),namelyε=?1/2,
with f?=1/3and clusters+?+and?++,andε=?2,with f?=2/3
and clusters+??and?+?.
?Irrational case:ifεis a negative irrational number,(4.4)determines all
the grain orientations,so that the model admits a unique,non-degenerate
6Peter F.Stadler,Anita Mehta,Jean-Marc Luck
ground state,where the grain orientations are distributed in a quasiperi-
odic fashion.The rule(4.4)is indeed equivalent to the cut-and-project algo-
rithm used to build quasiperiodic binary chains,which are one-dimensional
analogues of perfect quasicrystals[15,17,18,21,22,23,25].For instance,for
?ε=(√
0.7in each case.For the old model(left),A=B=0.35,C=D=0.4,while for the new model (right)A=?0.2,B=C=D=0.4.The shaking intensities are,for bottom to top,Γ=0.3(old model only),0.5,0.8,1,1.5,2,5,and10.
jamming limitφjam,identi?ed with a dynamical phase transition in related work[6].
While we have presented in earlier work[32]a full analysis of two-time correla-tion functions for the old model,work is currently in progress to investigate this in the rather more complex new model.
We?nally investigate the analogue of‘annealed cooling’,whereΓis increased and decreased cyclically,and the response of the packing fraction observed[30,31]. The results obtained here are similar to those[5]seen using more realistic models of shaken spheres,but the simplicity of the present lattice-based models allows for a greater transparency.
Starting with the sand in a?uidised state,as in the experiment[30,31],we submit the sandbox to taps at a given intensityΓfor a time t tap and increase the intensity in steps ofδΓ;at a certain point,the cycle is reversed,to go from higher to lower intensities.The entire process is then iterated twice.Fig.3shows the resulting behaviour of the volume fractionφas a function ofΓ,where an‘irreversible’branch and a‘reversible’branch of the compaction curve are seen,which meet at the ‘irreversibility point’Γ?[30,31].The left-and right-hand side of Fig.3correspond respectively to high and low values of the‘ramp rate’δΓ/t tap[30,31],while the upper and lower panels correspond respectively to the old and new versions of our model.As the ramp rate is lowered,we note that:
?the width of the hysteresis loop in the so-called reversible branch decreases,
in both cases.The‘reversible’branch is thus not reversible at all;more
realistic simulations of shaken spheres[2,3,27]con?rm the?rst-order,irre-
versible nature of the transition,which allows the density to attain values
pointΓ?to the‘shoulder’Γjam,as the ramp rateδΓ/t tap is lowered.The packing fraction tends to its close-packing limit in the limit of low intensities for the old model,while it asymptotes towards the jamming limit for the new model(see text).
that are substantially higher than random close packing,and quite close to
the crystalline limit[29].Precisely such a transition has also recently been
observed experimentally in the compaction of rods[34].
?In both panels,the‘irreversibility point’Γ?approachesΓjam(the shaking
intensity at which the jamming limitφjam is approached),in agreement
with results on other discrete models[14].However,in the upper panel,
the packing fraction at low intensities tends towards close-packing(the so-
called dynamical transition referred to in[8]);this is at odds with the results
of real experiment,which models with a greater degree of complexity[7]
have been able to replicate.In the lower panel,which corresponds to our
new model,we see tentative indications of an improvement in this respect
vis-a-vis the old model;the packing fraction here asymptotes towards the
jamming limit,rather than rising inde?nitely towards the close-packing
limit[7,30].
6.Discussion
We have presented in the above two models of shaken sandboxes;while their design was such that they would show identical behaviour in regimes where there were a ?nite density of voids,the modelling of the densely packed regime was completely
Glassy states in a shaken sandbox9 distinct in each case.In the?rst case,the model reduced to a model of noninteract-ing spins,while in the second,the insistence that all allowed transitions minimised a suitably de?ned local packing fraction led in fact to an intricate coupling between the grains.This physically motivated interaction was extremely nonlocal as well as directional.In this way we were able to generate a model that,despite being one-dimensional,has an extremely complex ground-state structure,depending on the regularity of the grain shape.Our present investigations,to be published else-where[26],concern the e?ect of zero-temperature and?nite-temperature tapping of this system;our preliminary studies indicate that for regularly shaped grains, strong metastability in the achievable ground states is observed.For irregularly shaped grains,as in reality,a far better packing is achievable,since orientations of irregularly shaped grains are much better able to?ll space[33].
It is however a rather salutary exercise to see that despite the relative sophis-tication of the new model in its inclusion of non-trivial interactions,most of the qualitative behaviour of the packing fraction as a function of steady as well as annealed tapping,remains unchanged.We expect that quantitative features such as two-time correlation functions will be far more non-trivial in the second model than the?rst,although we expect their overall features to be rather similar.It is tempting to speculate that the directionality due to gravity(which leads to strongly non-Hamiltonian behaviour,since grain couplings are propagating down the pile) which unites both?rst and second models might well be the most important ingre-dient that is needed to describe such lattice-based models of shaken sand. References
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一榀框架结构荷载计算书
毕业设计 题目一榀框架计算书 班级土木工程2006级高本学生姓名孟凡龙 指导老师
2011.5 摘要 本工程为济南某综合教学楼楼，主体三层，钢筋混凝土框架结构。梁板柱均为现浇，建筑面积约为3000m2，宽35米，长为60米，建筑方案确定。建筑分类为乙类公共类建筑，二类场地，抗震等级三级。 .
目录 第一章框架结构设计任务书 (1) 1.1工程概况 (1) 1.2设计资料 (2) 1.3设计内容 (2) 第二章框架结构布置及结构计算图确定 (2)
2.1梁柱界面确定 (2) 2.2结构计算简图 (2) 第三章荷载计算 (5) 3.1恒荷载计算： (5) 3.1.1屋面框架梁线荷载标准值 (5) 3.1.2楼面框架梁线荷载标准值 (5) 3.1.3屋面框架节点集中荷载标准值 (6) 3.1.4楼面框架节点集中荷载标准值 (7) 3.1.5恒荷载作用下结构计算简图 (8) 3.2活荷载标准值计算 (9) 3.2.1屋面框架梁线荷载标准值 (9) 3.2.2楼面框架梁线荷载标准值 (9) 3.2.3屋面框架节点集中荷载标准值 (9) 3.2.4楼面框架节点集中荷载标准值 (10) 3.2.5活荷载作用下的结构计算简图 (10) 3.3风荷载计算 (11) 第四章结构内力计算 (15) 4.1恒荷载作用下的内力计算 (15) 4.2活荷载作用下的内力计算 (25) 4.3风荷载作用下内力计算 (33) 第五章内力组合 (34) 5.1框架横梁内力组合 (38) 5.2柱内力组合 (46) 第六章配筋计算 (60) 6.1梁配筋计算 (60) 6.2 柱配筋计算 (75) 6.3楼梯配筋计算 (80) 6.4基础配筋计算 (84) 第七章电算结果 (80) 7.1结构电算步骤 (86) 7.2结构电算结果 (87) 参考文献 (112)
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Don't over do it！别做过头了！ Don't sit there daydreaming！别闲着做白日梦！Don't stand on ceremony！别太拘束！ Drop me a line！要写信给我！ Easy come easy go！来得容易去得也快！ First come first served！先到先得！ Get a move on！快点吧！ Get off my back！不要嘲笑我！ Give him the works！给他点教训！ Give me a break！饶了我吧！ Give me a hand！帮我一个忙！ Great minds think alike！英雄所见略同！ I'll treat you to lunch.午餐我请你！ In one ear，out the other ear.一耳进，一耳出！I'm spaced-out！我开小差了！ I beg your pardon！请你再说一遍！ I can't afford that！我付不起！ I can't follow you！我不懂你说的！ I can't help it！我情不自禁！ I couldn't reach him！我联络不上他！ I cross my heart！我发誓是真的！ I don't mean it！我不是故意的！
一榀框架计算内力计算
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图8-19 恒载作用下计算简图（括号内数值为梁柱相对线刚度） 1111 5.32 0.33615.836 A B A B A S S μ= ==∑ 1212 5.796 0.36615.836 A A A A A S S μ= ==∑ 节点B 1:11112 1.12 2.24B D B D S i ==?= 18.076B S =∑
老外喜欢的简短的英语口语(很实用)
God works. 上帝的安排。 Not so bad. 不错。 No way! 不可能! Don't flatter me. 过奖了。 Hope so. 希望如此。 Go down to business. 言归正传。 I'm not going. 我不去了。 Does it serve your purpose? 对你有用吗? I don't care. 我不在乎。 None of my business. 不关我事。 It doesn't work. 不管用。 Your are welcome. 你太客气了。 It is a long story. 一言难尽。 Between us. 你知，我知。 Sure thin! 當然! Talk truly. 有话直说。 I'm going to go. 我這就去。 Never mind. 不要緊。 Why are you so sure? 怎么这样肯定? Is that so? 是这样吗? Come on, be reasonable. 嗨，你怎么不讲道理。 When are you leaving? 你什么时候走?
You don't say so. 未必吧，不至于这样吧。Don't get me wrong. 别误会我。 You bet! 一定，当然! It's up to you. 由你决定。 The line is engaged. 占线。 My hands are full right now. 我现在很忙。Can you dig it? 你搞明白了吗? I'm afraid I can't. 我恐怕不能。 How big of you! 你真棒! Poor thing! 真可怜! How about eating out? 外面吃饭怎样? Don't over do it. 别太过分了。 You want a bet? 你想打赌吗? What if I go for you? 我替你去怎么样? Who wants? 谁稀罕? Follow my nose. 凭直觉做某事。 Cheap skate! 小气鬼! Come seat here. 来这边坐。 Dinner is on me. 晚饭我请。 You ask for it! 活该! You don't say! 真想不到! Get out of here! 滚出去!
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一框架结构设计任务书 1.1 工程概况： 本工程为成都万达购物广场----成仁店，钢筋混凝土框架结构。梁板柱均为现浇，建筑面积约为5750m2，宽27米，长为45米，建筑方案确定。建筑分类为乙类公共类建筑，二类场地，抗震等级三级。 图1-1 计算平面简图 1.2 设计资料 1）气象条件： 基本风压3155KN/m2 2）抗震设防： 设防烈度7度 3）屋面做法： 20厚水泥砂浆面层 一层油毡隔离层 40厚挤塑聚苯板保温层 15厚高分子防水卷材 20厚1:3水泥砂浆找平 1:6水泥焦渣1%找坡层，最薄处30厚 120厚现浇钢筋混凝土板 粉底 4）楼面做法： 8~13厚铺地砖面层
100厚钢筋砼楼板 吊顶 1.3设计内容 1)确定梁柱截面尺寸及框架计算简图 2)荷载计算 3)框架纵横向侧移计算； 4)框架在水平及竖向力作用下的内力分析； 5)内力组合及截面设计； 6)节点验算。 二框架结构布置及结构计算简图确定 2.1 梁柱截面的确定 通过查阅规范，知抗震等级为3级，允许轴压比为[μ]=0.85
由经验知n=12~14kn/m2 取n=13kn/m2 拟定轴向压力设计值N=n?A=13kn/m2×81m2×5=5265KN 拟定柱的混凝土等级为C30，f c=14.3N/mm2柱子尺寸拟定700mm× 700mm μ===0.75<[μ]=0.85 满足 初步确定截面尺寸如下： 柱：b×h=700mm×700mm 梁（BC跨、CE、EF跨）=L/12=9000/12=750mm 取h=800mm，b=400mm 纵梁=L/12=9000/15=600mm 取h=600mm，b=300mm 现浇板厚取h=120mm
《工程力学》第4次作业解答(杆件的内力计算与内力图).
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超经典英语日常高频常用口语
老友记中经典高频口语2008-07-2419:33 won’’t let her go without a fight!我不会轻易放过她的 1、I won 2、It could happen to anyone./It happens to anybody./That ha ppens.谁都可能会遇到这种情况 3、I’m a laundry virgin.（注意virgin的用法，体会老美说话之鲜活） 4、I hear you.我知道你要说什么。/我懂你的意思了 5、Nothing to see here!这里没什么好看的/看什么看！ 6、Hello?Were we at the same table?有没有搞错？（注意hell hello o 的用法，用疑问语气表示“有没有搞错？”） 7、You are so sweet/that that’’s so sweet.你真好。 8、I think it works for me.（work为口语中极其重要的小词） 9、Rachel,you are out of my league（等级，范畴）.你跟我不是
同一类人 10、You are so cute.你真好/真可爱 11、Given your situation,the options with the greatest chances f or success would be surrogacy.（given 表示考虑到的意思；非常简洁好用） 12、Let Let’’s get the exam rolling.现在开始考试了(get get………………rolling rolling 的用法) 13、Why don don’ ’t we give this a try?我们为何不试一下呢14、Bravo on the hot nanny!为那个性感的保姆喝彩！/赞一下那个性感的保姆！（重点是brave on sth/sb 这个句型，表示为……喝彩/赞叹的意思） 15、My way or the highway.不听我的就滚蛋！（很漂亮的习语，压后韵）
一榀框架结构设计手算+电算
一榀框架结构设计手算+电算
前言 毕业设计是学生在毕业之前在专业知识上面的一次检验，是学生从学校学习到工作岗位的过渡，在毕业设计阶段，要求要学会综合应用以前大学四年学到的专业课程，还有必要的设计规范和施工图集。通过学习、研究与实践，使理论深化、知识拓宽、专业技能延伸。通过毕业设计的实践，使学生能够深刻理解框架结构体系的布置特点、结构传力途径以及计算简图的确定方法，掌握风荷载及地震作用的计算方法、框架结构内力与位移计算的实用方法；掌握现浇多层框架结构的抗震概念设计，框架的截面设计原理、抗震构造要求及地基基础的设计方法；熟练阅读工程地质报告，熟悉施工图的内容、工作步骤及表达方法，培养学生综合运用所学专业知识来分析和解决实际工程问题的能力。 本次设计要求布图合理，图线清晰，尺寸齐全，注文工整，能最大程度的表达设计意图，符合国家制图标准及有关设计规范的规定。结构设计计算书要求方法合理，计算正确，排版工整，逻辑通顺。 由于时间和水平有限，不足之处，请各位专家、老师给予批评指正。
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必背英文面试口语第1期:请做自我介绍Tell me about yourself Tell me about yourself 请做自我介绍 I currently work as a secretary to the General Manager in the sales office of a Japanesetelecommunications company. 目前我在一家日本电信公司的营业厅担任总经理秘书一职。 Since he doesn't speak Japanese, my main function is to act as liaison betw een him and eightJapanese staff members. 因为总经理不会说日语，所以我的主要任务就是协助总经理与八位日本同事进行沟通。In addition to general office duties, I also perform accounting, including pay rolls. 除了办公室的日常事务之外，我也做一些与会计相关的业务，包括做工资表。 My boss frequently travels and I effectively manage the office during his abs ence. 我的上司经常出差，在他出差期间，我就负责打理办公室的事物。 I'm a self-starter and he fully trusts my ability to work without his supervisio n and he valuesmy judgment in a variety of contexts. 我工作很主动，所以没有上司的监督，我也能做好一切，由此深得上司的信赖。此外，遇到各种情况，上司也很尊重我的判断。 必背英文面试口语第2期:你认为自己有哪些优点? What are some of your strengths? What are some of your strengths? 你认为自己有哪些优点？ My biggest strength is tenacity. 我最大的优点是坚韧。 All my past employers and business associates have said I an tenacious.
水平地震作用下的框架侧移验算和内力计算
水平地震作用下的框架侧移验算和力计算 5.1 水平地震作用下框架结构的侧移验算 5.1.1抗震计算单元 计算单元：选取6号轴线横向三跨的一榀框架作为计算单元。 5.1.2横向框架侧移刚度计算 1、梁的线刚度： b /l I E i b c b = (5-1) 式中：E c —混凝土弹性模量s I b —梁截面惯性矩 l b —梁的计算跨度 I 0—梁矩形部分的截面惯性矩 根据《多层及高层钢筋混凝土结构设计释疑》，在框架结构中有现浇层的楼面可以作为梁的有效翼缘，增大梁的有效侧移刚度，减少框架侧移，为考虑这一有利因素，梁截面惯性矩按下列规定取，对于现浇楼面，中框架梁Ib=2.0Io,，边框架梁Ib=1.5Io ，具体规定是：现浇楼板每侧翼缘的有效宽度取板厚的6倍。 2、柱的线刚度： c c c c h I E i /= (5-2) 式中：Ic —柱截面惯性矩 hc —柱计算高度 一品框架计算简图： 3、横向框架柱侧移刚度D 值计算： 212c c c h i D α= (5-3) 式中：c α—柱抗侧移刚度修正系数
K K c +=2α（一般层）；K K c ++=25.0α（底层） K —梁柱线刚度比，c b K K K 2∑= （一般层）；c b K K K ∑=（底层） ① 底层柱的侧移刚度： 边柱侧移刚度： A 、E 轴柱：68.010 5.61045.41010=??==∑c b i i K 中柱侧移刚度： C 、 D 轴柱：18.1105.6102.345.410 10=??+== ∑）（c b i i K ② 标准层的侧移刚度 边柱的侧移刚度： A 、E 轴柱：51.010 72.821045.4221010=????==∑c b i i K 中柱侧移刚度： C 、 D 轴柱：88.01072.82102.345.42210 10 =???+?== ∑）（c b i i K
超全面的实用简洁英语口语大全
超全面的实用简洁英语口语大全 After you. 您先。 Allow me. 让我来。 Any day will do. 哪一天都行夕 Are you kidding 你在开玩笑吧! Are you sure 你肯定吗 Any messages for me 有我的留言吗 Any thing else 还要别的吗 Any urgent thing 有急事吗 As soon as possible! 越快越好! Absence makes the heart grow fonder. 小别胜新婚。Be careful! 注意! Be quiet! 安静点! Believe it or not! 信不信由你! Between us. 你知，我知。 Big mouth! 多嘴驴! Blast! 该死，讨厌! Bless you! 祝福你! Bottle it! 閉嘴! Bottoms up! 干杯(见底)! Boy! (表示惊奇，兴奋等)哇!好家伙!
Break the rules. 违反规则。 Can I have this. 可以给我这个吗 Can I help you 我能帮你吗 Can you dig it 你搞明白了吗 Can-do. 能人。 Cheap skate! 小气鬼! Cheer up! 振作起来! Chin up. 不气，振作些。 Close-up. 特寫鏡頭。 Clothes make the man. 人要衣装。 e on! 快点，振作起来! e on, be reasonable. 嗨，你怎么不讲道理。 e on. 来吧(赶快) e seat here. 来这边坐。 Congratulations! 祝贺你! Control yourself! 克制一下! Count me on 算上我。 Daring! 亲爱的! Dead end. 死胡同。 December heartbeat. 黄昏恋。 Did you miss the bus 你错过公共汽车了Dinner is on me. 晚饭我请。
毕业设计新规范框架内力计算
6.3.8 基础顶面恒载计算 由于本工程为五层框架结构，建筑高度较低，跨度基本相等，刚度比较均匀，风荷载影响较小。因此，为了简化计算，本设计的风荷载仅按一榀框架单独承担其受荷面积，忽略空间整体作用。 6.3.8.1 设计资料 基本风压：ω0＝0.30KN/m 2，地面粗糙度类别为B 类。KJ6承受风荷载的计算宽度B ＝（6+6）÷2＝6m 6.3.8.2 荷载计算 风荷载近似按阶梯形分布，首先应将其简化为作用在框架节点上的节点荷载。 作用在屋面梁和楼面梁节点处的集中风荷载标准值： ()/2k z s z o i j W h h B 式中 基本风压ω0＝0.30KN/m 2 βZ —风振系数，因为建筑物高度H ＝21m<30m ，因此βZ ＝1.0； s μ—风荷载体型系数，根据建筑物体型查得 1.3s ； z μ—风压高度变化系数，建设地点位于城市郊区，所以地面粗糙度为B 类； h i —下层柱高； h j —上层柱高，对顶层取女儿墙高度的2倍,即1.24m ； B —迎风面宽度，B ＝（6+6）÷2＝6m 。 计算过程如表6-1所示：
风荷载受荷简图见图6-26所示。 图6-26 框架风载受荷简图 6.3.8.3 框架柱D值计算 梁、柱的相对线刚度见表6-2 所示，侧移刚度D值计算如表6-2 、表6-3所示： 表6-2 KJ-3 2～5层柱D值计算 D 2 b c k K k2K K2 12 ** c D i h （KN/m） 边柱（A轴柱）2.08 2.08 1.68 2 1.24 1.68 0.457 2 1.68 4 2 12 0.457 1.24103855 4.2 中柱（C轴柱）2.082 5.582 6.18 2 1.24 6.18 0.756 2 6.18 4 2 12 0.756 1.24106377 4.2 中柱（D轴柱）2.082 5.582 6.18 2 1.24 6.18 0.756 2 6.18 4 2 12 0.756 1.24106377 4.2
弯曲的内力与强度计算 习题
弯曲的内力与强度计算 一、判断题 1.如图1示截面上，弯矩M和剪力Q的符号是：M为正，Q为负。（） 图1 2.取不同的坐标系时，弯曲内力的符号情况是M不同，Q相同。（） 3、在集中力作用的截面处，Q图有突变，M连续但不光滑。（） 4、梁在集中力偶作用截面处，M图有突变，Q图无变化。（） 5.梁在某截面处，若剪力Q=0，则该截面的M值一定为零值。（） 6.在梁的某一段上，若无荷载作用，则该梁段上的剪力为常数。（） 7.梁的内力图通常与横截面面积有关。（） 8.应用理论力学中的外力定理，将梁的横向集中力左右平移时，梁的Q图，M图都不变。（） 9.将梁上集中力偶左右平移时，梁的Q图不变，M图变化。（） 10.图2所示简支梁跨中截面上的内力为M≠0，Q=0。（） 图 2 图 3 11.梁的剪力图如图3所示，则梁的BC段有均布荷载，AB段没有。（） 12.上题中，作用于B处的集中力大小为6KN，方向向上。（） 13.右端固定的悬臂梁，长为4m，M图如图示，则在x=2m处，既有集中力又有集中力偶。（）
图 4 图 5 14.上题中，作用在x=2m处的集中力偶大小为6KN·m，转向为顺时针。（） 15.图5所示梁中，AB跨间剪力为零。（） 16.中性轴是中性层与横截面的交线。（） 17.梁任意截面上的剪力，在数值上等于截面一侧所有外力的代数和。（） 18.弯矩图表示梁的各横截面上弯矩沿轴线变化的情况，是分析梁的危险截面的依据之一。（） 19.梁上某段无荷载作用，即q=0，此段剪力图为平行x的直线；弯矩图也为平行x轴的直线。 （） 20.梁上某段有均布荷载作用，即q=常数，故剪力图为斜直线；弯矩图为二次抛物线。（） 21.极值弯矩一定是梁上最大的弯矩。（） 22.最大弯矩Mmax只可能发生在集中力F作用处，因此只需校核此截面强度是否满足梁的强度条件。（） 23.截面积相等，抗弯截面模量必相等，截面积不等，抗弯截面模量必不相等。（） 24.大多数梁都只进行弯曲正应力强度核算，而不作弯曲剪应力核算，这是因为它们横截面上只有正应力存在。（） 25.对弯曲变形梁，最大挠度发生处必定是最大转角发生处。（） 26.两根不同材料制成的梁，若截面尺寸和形状完全相同，长度及受力情况也相同，那么对此两根梁弯曲变形有关量值，有如下判断： （1）最大正应力相同；（） （2）最大挠度值相同；（） （3）最大转角值不同；（） （4）最大剪应力值不同；（） （5）强度相同。（） 27.两根材料、截面形状及尺寸均不同的等跨简支梁，受相同的荷载作用，则两梁的反力与内力相同。（）
必背英文面试口语(超级实用!!!)
必背英文面试口语第 1 期:请做自我介绍Tell me about yourself Tell me about yourself 请做自我介绍 I currently work as a secretary to the General Manager in the sales office of a Japanesetelecommunications company. 目前我在一家日本电信公司的营业厅担任总经理秘书一职。 Since he doesn't speak Japanese, my main function is to act as liaison betw een him and eightJapanese staff members. 因为总经理不会说日语，所以我的主要任务就是协助总经理与八位日本同事进行沟通。 In addition to general office duties, I also perform accounting, including pay rolls. 除了办公室的日常事务之外，我也做一些与会计相关的业务，包括做工资表。 My boss frequently travels and I effectively manage the office during his abs ence. 我的上司经常出差，在他出差期间，我就负责打理办公室的事物。 I'm a self-starter and he fully trusts my ability to work without his supervisio n and he valuesmy judgment in a variety of contexts. 我工作很主动，所以没有上司的监督，我也能做好一切，由此深得上司的信赖。此外，遇到 各种情况，上司也很尊重我的判断。 必背英文面试口语第2 期:你认为自己有哪些优点? What are some of your strengths? What are some of your strengths? 你认为自己有哪些优点？ My biggest strength is tenacity . 我最大的优点是坚韧。 All my past employers and business associates have said I an tenacious.
第五章.竖向荷载作用下的框架内力计算
5.1 计算单元的确定 取6号轴线一榀框架进行计算，计算宽度为(6.6+6.6)/2=6.6m 。如图下图所示 横向框架荷载传递图 5.2 荷载计算 5.2.1 恒荷载的计算 1、五层、 （1）q 、q 0、q 0′、q 0″分别为女儿墙、边跨横梁(走道纵梁)、走道横梁、次梁自重（扣除板自重），为均布荷载形式；β为考虑梁粉刷自重时的放大系数，取β=1.05。 女儿墙：q=3.47×0.9=3.12 kN/m 边跨横梁（走道纵梁）：q 0=1.05×0.3×(0.6-0.1)×25=3.94kN/m 走道横梁：q 0′=1.05×0.3×(0.4-0.1)×25=2.36kN/m 次梁：q 0″=1.05×0.2×(0.5-0.1)×25=2.1kN/m （2）q 1、q 1′分别为屋面板自重传给横梁的梯形和三角形荷载等效为均布荷载值 q 1=[1-2×(3.3/6.6×2) 2+(3.3/6.6×2) 3]×4.38×3.3/2=6.44kN/m q 1′=8 5 ×4.38×3.0/2=4.11kN/m （3）q 2、q 2′分别为屋面板自重传给纵梁上的梯形和三角形荷载等效为均布荷载值 梯形：q 2=[1-2×(3.0/6.6×2) 2+(3.0/6.6×2) 3]×4.38×3.0/2=5.96kN/m 三角形：q 2′=8 5 ×4.38×3.3/2=4.52kN/m P 1为由板传给次梁及次梁自重传给纵梁的集中力 P 1= q 1×6.6+ q 0″×6.6/2=49.43kN P 2为由板传给外纵梁及外纵梁、女儿墙自重传给柱子的集中力 P 2=（ q 2′+ q 0+q ）×3.3×2=76.42 kN P 3为由板传给内纵梁及内纵梁自重传给柱子的集中力。