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1、<p>  7 Rigid-Frame Structures</p><p>  A rigid-frame high-rise structure typically comprises parallel or orthogonally arranged bents consisting of columns and girders with moment resistant joints. Resi

2、stance to horizontal loading is provided by the bending resistance of the columns, girders, and joints. The continuity of the frame also contributes to resisting gravity loading, by reducing the moments in the girders.&l

3、t;/p><p>  The advantages of a rigid frame are the simplicity and convenience of its rectangular form.Its unobstructed arrangement, clear of bracing members and structural walls, allows freedom internally for t

4、he layout and externally for the fenestration. Rigid frames are considered economical for buildings of up to' about 25 stories, above which their drift resistance is costly to control. If, however, a rigid frame is c

5、ombined with shear walls or cores, the resulting structure is very much stiffer so th</p><p>  As highly redundant structures, rigid frames are designed initially on the basis of approximate analyses, after

6、which more rigorous analyses and checks can be made. The procedure may typically include the following stages:</p><p>  1. Estimation of gravity load forces in girders and columns by approximate method.<

7、/p><p>  2. Preliminary estimate of member sizes based on gravity load forces with arbitrary increase in sizes to allow for horizontal loading.</p><p>  3. Approximate allocation of horizontal load

8、ing to bents and preliminary analysis of member forces in bents.</p><p>  4. Check on drift and adjustment of member sizes if necessary.</p><p>  5. Check on strength of members for worst combin

9、ation of gravity and horizontal loading, and adjustment of member sizes if necessary.</p><p>  6. Computer analysis of total structure for more accurate check on member strengths and drift, with further adju

10、stment of sizes where required. This stage may include the second-order P-Delta effects of gravity loading on the member forces and drift..</p><p>  7. Detailed design of members and connections.</p>

11、<p>  This chapter considers methods of analysis for the deflections and forces for both gravity and horizontal loading. The methods are included in roughly the order of the design procedure, with approximate method

12、s initially and computer techniques later. Stability analyses of rigid frames are discussed in Chapter 16.</p><p>  7.1 RIGID FRAME BEHAVIOR</p><p>  The horizontal stiffness of a rigid frame is

13、 governed mainly by the bending resistance of the girders, the columns, and their connections, and, in a tall frame, by the axial rigidity of the columns. The accumulated horizontal shear above any story of a rigid frame

14、 is resisted by shear in the columns of that story (Fig. 7.1). The shear causes the story-height columns to bend in double curvature with points of contraflexure at approximately mid-story-height levels. The moments appl

15、ied to a joint f</p><p>  The overall moment of the external horizontal load is resisted in each story level by the couple resulting from the axial tensile and compressive forces in the columns on opposite s

16、ides of the structure (Fig. 7.2). The extension and shortening of the columns cause overall bending and associated horizontal displacements of the structure. Because of the cumulative rotation up the height, the story dr

17、ift due to overall bending increases with height, while that due to racking tends to decrease. Cons</p><p>  The response of a rigid frame to gravity loading differs from a simply connected frame in the cont

18、inuous behavior of the girders. Negative moments are induced adjacent to the columns, and positive moments of usually lesser magnitude occur in the mid-span regions. The continuity also causes the maximum girder moments

19、to be sensitive to the pattern of live loading. This must be considered when estimating the worst moment conditions. For example, the gravity load maximum hogging moment adjacent to </p><p>  The dependence

20、of a rigid frame on the moment capacity of the columns for resisting horizontal loading usually causes the columns of a rigid frame to be larger than those of the corresponding fully braced simply connected frame. On the

21、 other hand, while girders in braced frames are designed for their mid-span sagging moment, girders in rigid frames are designed for the end-of-span resultant hogging moments, which may be of lesser value. Consequently,

22、girders in a rigid frame may be smaller than </p><p>  7.2 APPROXIMATE DETERMINATION OF MEMBER FORCES CAUSED BY GRAVITY LOADSIMG</p><p>  A rigid frame is a highly redundant structure; consequen

23、tly, an accurate analysis can be made only after the member sizes are assigned. Initially, therefore, member sizes are decided on the basis of approximate forces estimated either by conservative formulas or by simplified

24、 methods of analysis that are independent of member properties. Two approaches for estimating girder forces due to gravity loading are given here.</p><p>  7.2.1 Girder Forces—Code Recommended Values</p&g

25、t;<p>  In rigid frames with two or more spans in which the longer of any two adjacent spans does not exceed the shorter by more than 20 %, and where the uniformly distributed design live load does not exceed thre

26、e times the dead load, the girder moment and shears may be estimated from Table 7.1. This summarizes the recommendations given in the Uniform Building Code [7.1]. In other cases a conventional moment distribution or two-

27、cycle moment distribution analysis should be made for a line of girders at </p><p>  7.2.2 Two-Cycle Moment Distribution [7.2].</p><p>  This is a concise form of moment distribution for estimat

28、ing girder moments in a continuous multibay span. It is more accurate than the formulas in Table 7.1, especially for cases of unequal spans and unequal loading in different spans.</p><p>  The following is a

29、ssumed for the analysis:</p><p>  1. A counterclockwise restraining moment on the end of a girder is positive and a clockwise moment is negative.</p><p>  2. The ends of the columns at the floor

30、s above and below the considered girder are fixed.</p><p>  3. In the absence of known member sizes, distribution factors at each joint are taken equal to 1 /n, where n is the number of members framing into

31、the joint in the plane of the frame.</p><p>  Two-Cycle Moment Distribution—Worked Example. The method is demonstrated by a worked example. In Fig, 7.4, a four-span girder AE from a rigid-frame bent is show

32、n with its loading. The fixed-end moments in each span are calculated for dead loading and total loading using the formulas given in Fig, 7.5. The moments are summarized in Table 7.2.</p><p>  The purpose of

33、 the moment distribution is to estimate for each support the maximum girder moments that can occur as a result of dead loading and pattern live loading. A different load combination must be considered for the maximum mom

34、ent at each support, and a distribution made for each combination. </p><p>  The five distributions are presented separately in Table 7.3, and in a combined form in Table 7.4. Distributions a in Table 7.3 ar

35、e for the exterior supports A and E. For the maximum hogging moment at A, total loading is applied to span AB with dead loading only on BC. The fixed-end moments are written in rows 1 and 2. In this distribution only .th

36、e resulting moment at A is of interest. For the first cycle, joint B is balanced with a correcting moment of - (-867 + 315)/4 = - U/4 assigned to MBA </p><p>  The second cycle involves the release and balan

37、ce of joint A. The unbalanced moment of 936 is balanced by adding -U/3 = -936/3 = -312 to MBA (row 5), implicitly adding the same moment to the two column ends at A. This completes the second cycle of the distribution. T

38、he resulting maximum moment at A is then given by the addition of rows 4 and 5, 936 - 312 = 624. The distribution for the maximum moment at E follows a similar procedure.</p><p>  Distribution b in Table 7.3

39、 is for the maximum moment at B. The most severe loading pattern for this is with total loading on spans AB and BC and dead load only on CD. The operations are similar to those in Distribution a, except that the T first

40、cycle involves balancing the two adjacent joints A and C while recording only their carryover moments to B. In the second cycle, B is balanced by adding - (-1012 + 782)/4 = 58 to each side of B. The addition of rows 4 an

41、d 5 then gives the maximum hoggin</p><p>  The complete set of operations can be combined as in Table 7.4 by initially recording at each joint the fixed-end moments for both dead and total loading. Then the

42、joint, or joints, adjacent to the one under consideration are balanced for the appropriate combination of loading, and carryover moments assigned .to the considered joint and recorded. The joint is then balanced to compl

43、ete the distribution for that support.</p><p>  Maximum Mid-Span Moments. The most severe loading condition for a maximum mid-span sagging moment is when the considered span and alternate other spans and to

44、tal loading. A concise method of obtaining these values may be included in the combined two-cycle distribution, as shown in Table 7.5. Adopting the convention that sagging moments at mid-span are positive, a mid-span tot

45、al; loading moment is calculated for the fixed-end condition of each span and entered in the mid-span column of row 2. Th</p><p>  7.2.3 Column Forces</p><p>  The gravity load axial force in a

46、column is estimated from the accumulated tributary dead and live floor loading above that level, with reductions in live loading as permitted by the local Code of Practice. The gravity load maximum column moment is estim

47、ated by taking the maximum difference of the end moments in the connected girders and allocating it equally between the column ends just above and below the joint. To this should be added any unbalanced moment due to ecc

48、entricity of the girder co</p><p><b>  第七章框架結(jié)構(gòu)</b></p><p>  高層框架結(jié)構(gòu)一般由平行或正交布置的梁柱結(jié)構(gòu)組成,梁柱結(jié)構(gòu)是由帶有能承擔(dān)彎矩作用節(jié)點(diǎn)的梁、柱組成。具有抗彎能力的梁、柱和節(jié)點(diǎn)共同作用抵抗水平荷載。連續(xù)框架可降低梁的跨中彎矩而有利于抵抗重力荷載。</p><p>  框架結(jié)

49、構(gòu)有簡(jiǎn)捷和便于采用矩形體系的優(yōu)點(diǎn)。由于這種布置形式?jīng)]有斜支撐和結(jié)構(gòu)墻體,因此,沒(méi)有不便利之處,內(nèi)部可以自由布置,外部可以自由設(shè)計(jì)門、窗??蚣芙Y(jié)構(gòu)對(duì)于25層以內(nèi)的建筑是經(jīng)濟(jì)的,超過(guò)25層由于要限制其位移而花費(fèi)的代價(jià)高,顯得很不經(jīng)濟(jì)。如果框架與剪力墻及芯筒相結(jié)合,剛度能夠大幅度提高,可以建造50層以上的建筑。板柱結(jié)構(gòu)與框架結(jié)構(gòu)非常相似,不同之處僅是用板代替了梁。和框架結(jié)構(gòu)一樣,板柱結(jié)構(gòu)是通過(guò)其水平和豎向構(gòu)件之間的連續(xù)抗彎作用來(lái)抵抗水平和豎向

50、荷載。</p><p>  對(duì)于高次超靜定框架結(jié)構(gòu),應(yīng)根據(jù)近似分析進(jìn)行初步設(shè)計(jì),隨后進(jìn)行精確分析和校核。分析過(guò)程一般包括以下幾步:</p><p>  1.按近似方法確定梁和柱所受重力荷載;</p><p>  2.初步確定在重力荷載作用下構(gòu)件的截面尺寸,考慮水平荷載的作用進(jìn)行構(gòu)件截面尺寸的任意調(diào)整;</p><p>  3.將水平荷載分配到

51、各梁柱結(jié)構(gòu)上,對(duì)這些結(jié)構(gòu)構(gòu)件的內(nèi)力進(jìn)行初步分析;</p><p>  4.檢驗(yàn)位移并對(duì)構(gòu)件截面尺寸做必要的調(diào)整;</p><p>  5.按最不利的重力荷載和水平荷載組合檢驗(yàn)構(gòu)件強(qiáng)度,做必要的構(gòu)件截面尺寸調(diào)整;</p><p>  6.為了更精確地驗(yàn)算構(gòu)件強(qiáng)度和位移,利用計(jì)算機(jī)對(duì)結(jié)構(gòu)進(jìn)行整體分析,需要時(shí)則近一步調(diào)整構(gòu)件截面尺寸。這一階段中應(yīng)包括考慮重力荷載對(duì)構(gòu)件內(nèi)力

52、和位移產(chǎn)生的Ρ一△二階效應(yīng);</p><p>  7.構(gòu)件和節(jié)點(diǎn)的詳細(xì)設(shè)計(jì)。</p><p>  本章討論在重力和水平荷載作用下結(jié)構(gòu)的變形和內(nèi)力分析方法。這些方法基本上按照設(shè)計(jì)過(guò)程中的次序介紹,首先是近似法,然后介紹計(jì)算機(jī)分析技術(shù)??蚣芙Y(jié)構(gòu)的穩(wěn)定性分析將在第十六章中討論。</p><p>  7.1框架結(jié)構(gòu)的性能</p><p>  框架結(jié)構(gòu)

53、的側(cè)向剛度主要取決于梁、柱及節(jié)點(diǎn)的抗彎能力,在較高的框架中主要取決于柱子的軸向剛度。作用于框架任一層間的水平集中剪力由該層柱子的抗剪能力抵抗(圖7. 1)。剪力使框架結(jié)構(gòu)每層的柱產(chǎn)生雙曲率彎曲,其反彎點(diǎn)大約在層高的中間部位。上、下柱引起的作用于節(jié)點(diǎn)處的彎矩由相鄰梁承擔(dān),該梁、柱的變形引起框架的整體變形,使各層間產(chǎn)生水平位移。在水平推力作用下結(jié)構(gòu)的整體變形和剪力圖如圖7. 1所示,其凹面朝向風(fēng)荷載作用方向,最大傾角在基底附近,最小傾角在頂

54、端。</p><p>  外部水平荷載產(chǎn)生的總彎矩由各層間兩個(gè)邊柱中的軸向拉、壓力組成的力矩抵抗(圖7.2 )。柱子的伸、縮引起結(jié)構(gòu)的整體彎曲變形,并產(chǎn)生相應(yīng)的水平位移。因?yàn)檗D(zhuǎn)角沿建筑高度累加,所以整體彎曲變形引起的層間位移隨高度增加而增加,而剪切變形引起的層間位移隨高度的增加而減小。其結(jié)果在建筑的最頂部整體彎曲對(duì)層間位移的貢獻(xiàn)會(huì)大大超過(guò)剪切變形對(duì)層間位移的貢獻(xiàn)。但是,整體彎曲變形對(duì)總位移的貢獻(xiàn)與剪切變形對(duì)總位移

55、的貢獻(xiàn)之比不會(huì)超過(guò)10%,除非在極高或細(xì)長(zhǎng)的框架中。因此,高層框架結(jié)構(gòu)變形型式為剪切型。</p><p>  從梁的連接受力性能來(lái)看,高層建筑采用的剛性節(jié)點(diǎn)連續(xù)的框架不同于一般簡(jiǎn)單連接的普通框架。梁在柱邊附近產(chǎn)生負(fù)彎矩,跨中正彎矩值常常很小。這種連續(xù)性能使梁中最大彎矩對(duì)活荷載的作用方式非常敏感。如果能夠估計(jì)出產(chǎn)生最不利彎矩的因素,則必須加以認(rèn)真的考慮。例如,重力荷載作用下梁在邊柱附近產(chǎn)生的最大負(fù)彎矩只會(huì)在活荷載作

56、用于邊跨和相間跨時(shí)才能發(fā)生,如圖7.3a中的A點(diǎn)。而梁在內(nèi)柱附近產(chǎn)生的最大負(fù)彎矩只會(huì)在活荷載作用于相鄰跨時(shí)才能發(fā)生,如圖7.3a中的B點(diǎn)。當(dāng)活荷載作用于本跨和相間跨時(shí),梁的跨中正彎矩最大,如圖7. 3a中的AB和CD跨。</p><p>  框架的尺寸取決于柱子在水平荷載作用·下的抗彎強(qiáng)度,這往往會(huì)使框架柱的截面尺寸大于相應(yīng)全對(duì)角支撐簡(jiǎn)單連接框架的柱截面尺寸。另外,框架支撐結(jié)構(gòu)中的梁被設(shè)計(jì)為只具有跨中正

57、彎矩,而框架結(jié)構(gòu)的梁則被設(shè)計(jì)為端部為負(fù)彎矩和跨中為正彎矩,跨中彎矩值較小。因此,框架結(jié)構(gòu)中梁的截面尺寸會(huì)小于相應(yīng)的框架支撐結(jié)構(gòu)中梁的截面尺寸。梁截面的減小將會(huì)降低其造價(jià),有時(shí)可以降低層高,經(jīng)濟(jì)效益明顯。但是,由于剛性節(jié)點(diǎn)的處理相當(dāng)復(fù)雜,代價(jià)較高,使上述經(jīng)濟(jì)優(yōu)勢(shì)被削弱。</p><p>  7.2重力荷載作用下構(gòu)件內(nèi)力的近似計(jì)算</p><p>  框架結(jié)構(gòu)是多次超靜定結(jié)構(gòu),因此,只有在確定

58、了構(gòu)件截面尺寸后才能進(jìn)行精確分析。所以,在初步設(shè)計(jì)階段,可根據(jù)傳統(tǒng)的公式和不考慮構(gòu)件特征值的簡(jiǎn)化分析法近似確定構(gòu)件中的內(nèi)力,以此為基礎(chǔ)確定構(gòu)件的截面尺寸。下面將討論在重力荷載作用下構(gòu)件內(nèi)力計(jì)算的兩種方法。</p><p>  7.2.1梁的內(nèi)力—規(guī)范推薦值</p><p>  對(duì)于兩跨以上的框架結(jié)構(gòu),當(dāng)任何相鄰兩跨中的長(zhǎng)跨不超過(guò)短跨的20%跨度,同時(shí)設(shè)計(jì)均布活荷載不超過(guò)3倍的恒載時(shí),梁的彎

59、矩和剪力可以按表7.1確定。表中各數(shù)值是根據(jù)統(tǒng)一建筑規(guī)范【7.1】中的推薦值給出。對(duì)于其它情況,可按照樓面連續(xù)梁采月傳統(tǒng)彎矩分配法或兩次循環(huán)彎矩分配法進(jìn)行分析確定。</p><p>  7.2.2彎矩分配法【7.2】</p><p>  彎矩分配法用于計(jì)算多跨連續(xù)梁的彎矩是非常便利的形式。該方法的計(jì)算結(jié)果比表7.1中的推薦公式計(jì)算結(jié)果更精確,特別是對(duì)于不等跨和荷載變化較大的情況。</

60、p><p>  彎矩分配法的分析假定如下:</p><p>  1.梁端約束彎矩以反時(shí)針?lè)较驗(yàn)檎?,順時(shí)針?lè)较驗(yàn)樨?fù);</p><p>  2.被分析的梁與上、卞柱的連接為固接;</p><p>  3.當(dāng)構(gòu)件尺寸尚未確定時(shí),每個(gè)節(jié)點(diǎn)的分配系數(shù)取1/n,n是框架平面內(nèi)連接在各節(jié)點(diǎn)上的構(gòu)件總數(shù)。</p><p>  兩次循環(huán)彎矩

61、分配實(shí)例. </p><p>  現(xiàn)以一個(gè)實(shí)例具體說(shuō)明兩次循環(huán)彎矩分配的過(guò)程。在圖7.4中表示出一個(gè)取自框架單榻結(jié)構(gòu)的中跨連續(xù)梁AE和作于其上的荷載。恒載和全部荷載作用下每跨的固端彎矩采用圖7.5中的公式計(jì)算。這些彎矩值匯總于表7.2 中。</p><p>  彎矩分配的目的是為了確定每個(gè)支座處梁在恒載和標(biāo)準(zhǔn)活荷載作用下可能產(chǎn)生的最大彎矩。計(jì)算梁在支座處最大彎矩時(shí)必須考慮不同的荷載組合,而

62、每種荷載組合作用都應(yīng)進(jìn)行一次計(jì)算。</p><p>  表7.3分別給出了5組彎矩分配,表7.4給出的是混合形式。表7. 3中a組分配為邊支座A和E。為了求出A點(diǎn)的最大負(fù)彎矩。在AB跨施加全部荷載,在BC跨只布置恒載。固端彎矩列于1、2行。這組分配主要是為了求出A點(diǎn)的結(jié)果。第一循環(huán)分配,節(jié)點(diǎn)B由修正彎矩-(-867+315 )/ 4=-U4平衡指定為MBA,U是不平衡彎矩。這些計(jì)算可不作記錄,而將修正彎矩的一半即

63、(-U/4)/2傳于MAB。將此值記錄在第3行,同時(shí)與固端彎矩相力盯,結(jié)果寫在第4行。</p><p>  第二循環(huán)包括節(jié)點(diǎn)A的釋放和平衡。通過(guò)加上-U/3=-936/3 =-312(MBA第5行)來(lái)平衡原有的不平衡彎矩936,這也隱含地說(shuō)明對(duì)節(jié)點(diǎn)A處的兩個(gè)柱端增加了同樣大小的彎矩。彎矩分配的第二循環(huán)到此完成,節(jié)點(diǎn)A處最終的最大彎矩為第4、5行數(shù)值之和,即936-31=624。節(jié)點(diǎn)E處的最大彎矩分配應(yīng)遵循同樣的方

64、式。</p><p>  表7.3中b組分配是針對(duì)B節(jié)點(diǎn)的最大彎矩。最不利的荷載作用方式是BC作用全部荷載,CD跨只有恒載。計(jì)算過(guò)程類似于a組分配,但首次循環(huán)應(yīng)平衡兩相間節(jié)點(diǎn)A和C,同時(shí)僅將它們的傳遞彎矩記錄在節(jié)點(diǎn)B。第二次循環(huán),將-(-1012+782)/4=58與節(jié)點(diǎn)B兩邊值分別相加,以此平衡節(jié)點(diǎn)B。將第4、5行的彎矩相加后就得出節(jié)點(diǎn)B的兩端最大負(fù)彎矩。節(jié)點(diǎn)C和D的彎矩分配即c組和d組分配按照b組分配的同樣的

65、方式進(jìn)行。</p><p>  全部運(yùn)算過(guò)程可以一同列于表7,4中,表中首先記錄每個(gè)節(jié)點(diǎn)處相應(yīng)于恒載和全部荷載的固端彎矩。選擇合理的荷載組合,將所關(guān)心的節(jié)點(diǎn)兩側(cè)相鄰節(jié)點(diǎn)加以平衡,然后將傳遞彎矩分配于所關(guān)心的節(jié)點(diǎn)并記錄。平衡該節(jié)點(diǎn),完成這個(gè)支座的分配。</p><p><b>  最大跨中彎矩</b></p><p>  與跨中最大正彎矩出現(xiàn)的相

66、應(yīng)最不利荷載條件即為該跨和相間各跨均作用全部荷載。該值計(jì)算的最簡(jiǎn)便方法是將兩次循環(huán)彎矩分配組合起來(lái),如表7.5所示。采用跨中正彎矩為正號(hào)的規(guī)則就可以計(jì)算各跨為固端條件時(shí)全茍祠苛載在跨中產(chǎn)生的彎矩,并記錄在第2行的跨中一列。實(shí)際結(jié)構(gòu)的節(jié)點(diǎn)處應(yīng)該能夠轉(zhuǎn)動(dòng),所以需要對(duì)跨中彎矩加以修正。修正的方法是:將該跨左端第3行的傳遞彎矩乘以(1+0.5D.F.)/2,右端的傳遞彎矩乘以-(1+0. 5D.F. )/2,其中D.F.為合理性分配系數(shù),這些結(jié)

67、果記錄在跨中一列上。例如,從A傳遞給AB跨中的彎矩為〔(1+0.5/3)/2〕×69=40,從B傳給AB跨中的彎矩為-〔(1+0.5/4)/2〕×(-145)=85。這些修正彎矩與固端情況下的跨中彎矩相加就得到了最大跨中正彎矩,即733+40+82=855。</p><p><b>  7.2.3柱的內(nèi)力</b></p><p>  重力荷載使柱子

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