浙江大学宁波理工学院结构力学2011-2012(C) 答案
结构力学试题与答案汇总(完整版)
院（系） 学号 姓名 .密封线内不要答题 密封……………………………………………………………………………………………………………………………………………………结构力学试题答案汇总结构力学课程试题 ( B )卷考 试 成 绩题号 一二三四成绩得分一、选择题(每小题3分,共18分)1. 图 示 体 系 的 几 何 组 成 为 ： （ ） A. 几 何 不 变 ， 无 多 余 联 系 ； B. 几 何 不 变 ， 有 多 余 联 系 ； C. 瞬变 ； D. 常 变 。
2. 静 定 结 构 在 支 座 移 动 时 ， 会 产 生 ： （ ）A. 内 力 ；B. 应 力 ；C. 刚 体 位 移 ；D. 变 形 。
3. 在 径 向 均 布 荷 载 作 用 下 ， 三 铰 拱 的 合 理 轴 线 为： （ ）A ．圆 弧 线 ；B ．抛 物 线 ；C ．悬 链 线 ；D ．正 弦 曲 线 。
4. 图 示 桁 架 的 零 杆 数 目 为 ： （ ）A. 6；B. 7；C. 8；D. 9。
5. 图 a 结构的最后弯矩图为：（）A．图 b； B．图 c ； C．图 d ； D．都不对。
6. 力法方程是沿基本未知量方向的：（）A．力的平衡方程；B．位移为零方程；C．位移协调方程； D．力的平衡及位移为零方程。
二、填空题（每题3分,共9分）1.从几何组成上讲，静定和超静定结构都是_________体系，前者_________多余约束而后者_____________多余约束。
2. 图 b 是图 a 结构 ________ 截面的 _______ 影响线。
3. 图示结构 AB 杆 B 端的转动刚度为 ________, 分配系数为________, 传递系数为 _____。
三、简答题（每题5分，共10分）1.静定结构内力分析情况与杆件截面的几何性质、材料物理性质是否相关？为什么？2.影响线横坐标和纵坐标的物理意义是什么？四、计算分析题，写出主要解题步骤(4小题,共63分)1.作图示体系的几何组成分析（说明理由），并求指定杆1和2的轴力。
结构力学-参考答案
模块1参考答案1.结构有哪几种分类?答：结构主要有：杆件结构，薄壁结构和实体结构三类。
2.结构力学的研究对象和研究任务是什么?答：结构力学的研究对象：结构力学的研究对象是杆件结构，薄壁结构和实体结构的受力分析将在弹性力学中进行研究。
严格地说，一般的杆件结构是空间结构，但它们中的大多数均可简化为平面结构。
所以，本门课程主要研究平面杆件结构，即组成结构的所有杆件及结构所承受的外荷载都在同一平面内的结构。
结构力学是研究结构的合理形式以及结构在受力状态下内力、变形、动力反应和稳定性等方面的规律性的科学。
研究的目的是使结构满足安全性、适用性和经济方面的要求。
建筑物、构筑物、结构物在各类工程中大量存在：（1）住宅、厂房等工业民用建筑物；（2）涵洞、隧道、堤坝、挡水墙等构造物；（3）桥梁、轮船、潜水艇、飞行器等结构物。
结构力学的任务：结构力学与材料力学、弹性力学有着密切的联系，他们的任务都是讨论变形体系的强度、刚度和稳定性，但在研究对象上有所区别。
材料力学基本上是研究单个杆件的计算，结构力学主要是研究杆件的结构，而弹性力学则研究各种薄壁结构和实体结构，同时对杆件也作更精确的分析。
结构力学研究杆件结构的强度、刚度和稳定性问题，其具体任务包括以下几个方面：（1）杆件结构的组成规律和合理的组成方式。
（2）杆件结构内力和变形的计算方法，以便进行结构强度和刚度的验算。
（3）杆件结构的稳定性以及在动力荷载作用下的反应。
结构力学是土木工程专业的一门重要的专业基础课，在各门课程的学习中起着承上启下的作用。
结构力学的计算方法很多，但所有方法都必须满足以下几个三个基本条件：（1）力系的平衡条件。
在一组力系作用下，结构的整体及其中任何一部分都应满足力系的平衡条件。
（2）变形的连续条件，即几何条件。
连续的结构发生变形后，仍是连续的，材料没有重叠和缝隙；同使结构的变形和位移应该满足支座和结点的约束条件。
（3）物理条件。
把结构的应力和变形联系起来的条件，即物理方程或本构方程。
《结构力学(2)》试题-2012-A-参考答案
M A = 350 × 3+300 × 1 = 1350kN.m （4）综上， M A,max = 1475kN.m ，下侧受拉。
《结构力学（2） 》试卷 A 卷
第 2 页 共 6 页
题 4 解：
EI
1） 自由度 （如图） ： （5 分） 2）δ11 计算（5 分）
δ 11 = 1 1 a3 5a 3 2 × × (0.5a ) × a + = EI 3 48 EI 48 EI
M2
1
1
第二振型
题 7 解： 1）2 个自由度如图： （5 分） m1 = 2m 2）建立运动方程： （5 分）
δ 11 = 1 1 1 a 2 3 (a ) 2a + (a ) = 3 EI 3 EI a3 4 EI
3
y1
m2 = m
m
P(t ) = P0 sinθt
A
m
C
y2
3m
专业
3m
M A 影响线
tan α1 = -1 ， tan α 2 = 1 ， tan α 3 = -1 （2）将各个集中荷载依次移动到 B 点，求解最大负弯矩 ----------（5 分） P4 作用在 B 点时，P P2 不在结构上： 左移， 右移， ∑ Pi tan α i <0 ； ∑ Pi tan α i = -50<0 ， 1、 不是临界位置。 P3 作 用 在 B 点 时 ， P 1 不 在 结 构 上 ： 左 移，
( 32 − λ ) A3 +8 A4 = 0 16 A3 + ( 5 − λ ) A4 = 0
特征方程为 32 − λ 8 =0 16 5−λ 解得： λ1 = 36.11 ， λ2 = 0.886 ，即 ω1 = 24 EI EI 24 EI EI = 0.815 ， ω2 = = 5.20 。代入振型方 3 3 3 λ1ma ma λ2 ma ma3
结构力学课后习题答案
结构⼒学课后习题答案附录B 部分习题答案2 平⾯体系的⼏何组成分析2-1 （1）× （2）× （3）√ （4）× （5）× （6）×。
2-2 （1）⽆多余约束⼏何不变体系；（2）⽆多余约束⼏何不变体系；（3）6个；（4）9个；（5）⼏何不变体系，0个；（6）⼏何不变体系，2个。
2-3 ⼏何不变，有1个多余约束。
2-4 ⼏何不变，⽆多余约束。
2-5 ⼏何可变。
2-6 ⼏何瞬变。
2-7 ⼏何可变。
2-8 ⼏何不变，⽆多余约束。
2-9⼏何瞬变。
2-10⼏何不变，⽆多余约束。
2-11⼏何不变，有2个多余约束。
2-12⼏何不变，⽆多余约束。
2-13⼏何不变，⽆多余约束。
2-14⼏何不变，⽆多余约束。
5-15⼏何不变，⽆多余约束。
2-16⼏何不变，⽆多余约束。
2-17⼏何不变，有1个多余约束。
2-18⼏何不变，⽆多余约束。
2-19⼏何瞬变。
2-20⼏何不变，⽆多余约束。
2-21⼏何不变，⽆多余约束。
2-22⼏何不变，有2个多余约束。
2-23⼏何不变，有12个多余约束。
2-24⼏何不变，有2个多余约束。
2-25⼏何不变，⽆多余约束。
2-26⼏何瞬变。
3 静定梁和静定刚架3-1 (1) √；(2) ×；(3) ×；(4) √；(5) ×；(6) √；(7) √；(8) √。
3-2 (1) 2，下；(2) CDE ，CDE ，CDEF ；(3) 15，上，45，上；(4) 53，-67，105，下； (5) 16，右，128，右；(6) 27，下，93，左。
3-3 (a) 298AC M ql =-，Q 32AC F ql =；(b) M C = 50kN·m ，F Q C = 25kN ，M D = 35kN·m ，F Q D = -35kN ；(c) M CA = 8kN·m ，M CB = 18kN·m ，M B = -4kN·m ，F Q BC = -20kN ，F Q BD = 13kN ； (d) M A = 2F P a ，M C = F P a ，M B = -F P a ，F Q A = -F P ，F Q B 左 = -2F P ，F Q C 左 = -F P 。
结构力学单选题及答案
《结构力学》单选题1.弯矩图肯定发生突变的截面是（）。
A.有集中力作用的截面；B.剪力为零的截面；C.荷载为零的截面；D.有集中力偶作用的截面。
2.图示梁中C截面的弯矩是（）。
4m2m4mA.12kN.m(下拉)；B.3kN.m(上拉)；C.8kN.m(下拉)；D.11kN.m(下拉)。
3.静定结构有变温时，（）。
A.无变形，无位移，无内力；B.有变形，有位移，有内力；C.有变形，有位移，无内力；D.无变形，有位移，无内力。
4.图示桁架a杆的内力是（）。
A.2P；B.－2P；C.3P；D.－3P。
5.图示桁架，各杆EA为常数，除支座链杆外，零杆数为（）。
A.四根；B.二根；C.一根；D.零根。
l= a66.图示梁A点的竖向位移为（向下为正）（）。
A.)24/(3EIPl； B.)16/(3EIPl； C.)96/(53EIPl； D.)48/(53EIPl。
P7. 静定结构的内力计算与（ ）。
A.EI 无关；B.EI 相对值有关；C.EI 绝对值有关；D.E 无关，I 有关。
8. 图示桁架，零杆的数目为：（）。
A.5；B.10；C.15；D.20。
9. 图示结构的零杆数目为（ ）。
A.5；B.6；C.7；D.8。
10. 图示两结构及其受力状态，它们的内力符合（ ）。
A.弯矩相同，剪力不同；B.弯矩相同，轴力不同；C.弯矩不同，剪力相同；D.弯矩不同，轴力不同。
P P P P2l l11. 刚结点在结构发生变形时的主要特征是（ ）。
A.各杆可以绕结点结心自由转动；B.不变形；C.各杆之间的夹角可任意改变；D.各杆之间的夹角保持不变。
12. 若荷载作用在静定多跨梁的基本部分上，附属部分上无荷载作用，则（）。
A.基本部分和附属部分均有内力；B.基本部分有内力，附属部分没有内力；C.基本部分无内力，附属部分有内力；D.不经过计算，无法判断。
13.图示桁架C 杆的内力是（）。
A.P；B.－P/2；C.P/2；D.0。
结构力学答案（Structuralmechanicsanswer）
结构力学答案（Structural mechanics answer）"The test of structural mechanics" chapter 01 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)The object of study, 1 structural mechanics BThe structure of A, B, single pole rodC, shell D, entity structure2, the structure strength calculation purpose is to ensure that the structure of AA, B, economical and safe without excessive deformationC D, beautiful and practical and non rigid motion3, the structure stiffness calculation, is to ensure that the structure of CA, no rigid motion B, beautiful and practicalC, without excessive deformation of D, economical and safe4, there are several fixed hinge support constrained force component? BA, a B, twoC, three D, four5, there are several movable hinge support constrained force component AA, a B, twoC, three D, fourSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, the stability of structure is DEA, structure of resistance failureThe ability of B and non rigid motionC, the structure of the ability to resist deformationD, the resistance ability of instabilityThe structure of E, the ability to keep the original balance formWhat kind of situation is not 2, the following BCDE plane structureA, all of the axis of the rod are positioned in the same plane, the load also function in the planeB, all of the axis of the rod are positioned in the same plane, and the plane vertical loadC, all of the axis of the rod are positioned in the same plane, and the plane parallel loadD, all of the axis of the rod are not located in the same planeEffect of E and load is not in the structure planeWhich of the following conditions should be 3, according to the spatial structure of ABDEA, all of the axis of the rod are positioned in the same plane, and the plane vertical loadB, all of the axis of the rod are not located in the same planeC, all of the axis of the rod are positioned in the same plane, the load also function in the planeD, all of the axis of the rod are positioned in the same plane, and the plane parallel loadEffect of E and load is not in the structure plane4, in order to ensure that the structure is economical and safe, to calculate the structure of BA, strengthB, stiffnessC, stabilityD, forceE, displacement5, the constraints of rigid node is ABA, the rod end cannot move relative to constraintB, binding the rod end can not rotateThe rod end C, constraints can be relatively moved along one directionD, the relative rotation rod end constraintsE, the relatively movable rod end constraintsThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, shell thickness is far less than the other two dimensions.Correct2, the entity structure and thickness of the other two dimensions are of the same order of magnitude.Correct3, in order to ensure that the structure is economical and safe, to carry out stiffness calculation of structure.error4, structural mechanics is a subject of bar structure strength, stiffness and stability.Correct5, the hinge node constraint cannot move relative to the rod end, but the relative rotation.Correct"The test of structural mechanics" chapter 02 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, the three rigid geometric composition without redundant constraint invariant system, the number of constraints is necessary for DA, 3 B, 4C, 5 D, 62, what is the hinge to connect four rigid plates node? CA, B, single hinge node incomplete hinge nodeC, D, combined hinge node node3, connecting two rigid plates hinge has several constraints? AA, 2 B, 3C, 4 D, 54, there are several fixed hinge support constraints? BA, a B, twoC, three D, four5, a link system can reduce several degrees of freedom? AA, 1 B, 2C, 3 D, 4Second questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)Calculate the degrees of freedom ABC 1 geometry systemA, may be greater than zeroB, may be equal to zeroC, may be less than zeroD, must be greater than zeroE, must be equal to zero2, if the system is not a geometric system, then the BCD A, calculate the degrees of freedom may be greater than zero B, calculate the degrees of freedom may be equal to zero C, calculate the degrees of freedom may be less than zero D, degree of freedom must be equal to zeroE, calculate the degrees of freedom must be equal to zero3, from two yuan to remove a non redundant constraint geometric invariant system after the new system ACEA, is free of unnecessary restrictions of geometric invariant systemB, geometry variable systemC, the same degree of freedomD, is a redundant constraint geometric invariant systemE, is a geometric transient system4, the plane system in a link BCEA, can reduce the system of two degrees of freedomB, can reduce the system of one degree of freedomC, there is a constraintD, which has two degrees of freedomE, which has three degrees of freedom5, the following discussion is correctA, transient system in small load will produce great force ABCEB, the only geometric invariant system can be used as a building structureC, building structure degree of freedom is equal to zeroD, building structure calculation of degrees of freedom is equal to zeroE, degree of freedom equal to zero system that can be used as building structureThird questions to determine the problem (1 points for eachquestion 5 questions, a total of 5 points)1, the connection of two restriction effect two link rigid body is equivalent to a single hinge.Correct2 degrees of freedom, equal to zero system that can be used as building structure.Correct3, a link is equivalent to a constraint.Correct4, a single hinge can reduce the system of two degrees of freedom.Correct5, a single hinge is equivalent to two constraints.Correct"The test of structural mechanics" chapter 03 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, if a beam axis is parallel to the shear diagram and bendingmoment diagram for CA, B, parallel to the axis of zeroC, D, two parabolic oblique line2, if the shear diagram of a beam is zero line, then the moment diagram for BA, B, zero line parallel axisC, D, two parabolic oblique line3, if a beam axis is parallel to the shear diagram, the load on the beam is AA, B, no load uniform loadC, D, a couple of concentrated forceIn 4, the mutation at the moment of beam is what force? CA, axial force B, transverse concentrated forceC, D, a couple of forcesThrust 5, three hinge arch has nothing to do with the following factors? DA, high B, load vectorC, D, the span of arch axis shapeSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, change the following factors that will not change the internal force of statically determinate beam? BCEA, loadB, section sizeC, material propertiesD, the span of the beamE, the stiffness of the beam2, which of the following factors on static beam internal force does not produce BCDEA, loadB, temperature changeC, mobile supportD, manufacturing errorE, material shrinkage3, if the shear diagram of a simply supported beam is a parallel axis, then the load on the beam may be ABCA, a left support action of a coupleB, have the right support action of a coupleC, a cross between a concentrated momentD, a cross between the uniform load.E, a cross between a concentrated forceThe characteristics of internal force diagram 4, beam without loading section is BA, the shear diagram parallel axisB, linear oblique shear diagramC, two parabolic shear diagramD, the bending moment diagram of parallel axisE, the bending moment diagram of oblique line5, static beam section size change, which of the following factors do not change? The displacement of ABCDA, axial forceB, shearC, momentD, bearing forceE, displacementThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, the external force in the basic part, internal force, deformation and displacement of accessory parts are zero.error2, static structure only by the equilibrium conditions we can find all the internal force and reacting force.Correct3, static structure in the movement of the bearing under the action of internal force, do not produce.Correct4, static structure only by the equilibrium conditions we can find all the internal force and reacting force.Correct5, zero rod force, so it is not needed in the truss rod can be removed.error"The test of structural mechanics" chapter 04 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, the parallel chord truss beam (upper and lower internodes, when the upper bearing alignment) and lower load influence line is different is that? DInfluence of A, B, axial force of upper chord line oblique axial force lineThe influence line of effect of C and the bottom chord axial force, axial force vertical line D2, overhanging beam a line shape feature is AA B, a straight line, two lines lineC, the two parallel lines D, parabola3, support the extended beam shear influence line shape feature is CA B, a straight line, two lines lineC, the two parallel lines D, parabolaSection 4, overhanging beam bending moment influence line between supports is BA B, a straight line, two lines lineC, the two parallel lines D, parabola5, by the master-slave structure stress characteristics show that the internal force influence line affiliated part in the basic part of AA, B, all is zeroC andD are all negative, positive or negativeSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)Shear effect of 1 section C, beam line, vertical scale C is left a, right C vertical scale is B, the following discussion is correctABA, a P = C 1 C in the left section shear generatedB, B P = C 1 C shear in the right when theC, a P = C C 1 in the left section shear point generatedD, B for P = 1 at C C during right shearE, a P = C 1 C shear in the right when the2, the bearing section overhanging beam bending moment influence line shape is ABCDIn A, the bearing vertical scale is zeroExcept B, the bearing is a straight line.C, the bearing is in the zero lineD, the other at the bearing vertical scale is zeroE, a straight line3, with the static part of the statically indeterminate beam, statically indeterminate part of the internal force influence line is characteristic of ABA, in the static part is on the lineB, in the indeterminate part is curveC, in the static part is curveD, in the indeterminate part is linearIn E, the entire structure is curve4, drawing method of influence line ABA, static methodB, mobile methodC, force methodD, displacement methodE, moment distribution methodEffect of line 5, which of the following is the dimensionless value? ABEA, bearing forceB, shearC, momentD, the restriction momentE, axial forceThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)K effect of line 1, C beam section bending moment in Shubiao said the moment P = K section 1 in the K point of the.error2, if there is a concentrated force on the influence line of vertex, will load the critical position.errorThe maximum vertical bending moment influence line 3 and the fixed end of the cantilever beam section of standard section at the free end.Correct4, beam bending moment influence line is broken.Correct5, the static truss influence line in a straight line between adjacent nodes will.Correct"The test of structural mechanics" chapter 05 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, static frame displacement in the movement of the bearing under the action by what is produced? DA B, the bending deformation and axial deformationC, D, shear deformation and rigid body motion2, using the principle of virtual work, the state should meet what conditions? DA, B, continuous constraintsC, D, physical balance3, the plane frame displacement under load deformation is mainly caused by what? BA B, the bending deformation and axial deformationC D, shear deformation and torsion deformation4, support for mobile indeterminate structure will produce CA B, the internal force and support reaction forceC, D displacement and deformationIn 5, the reciprocal theorem in delta = delta 12, 21, AA, 12 B, 11 sigma deltaC, 22 D, 31 sigma deltaSecond questions, multiple choice questions (2 points for eachquestion 5 questions, a total of 10 points)1, when the force acting on the part ofA, a subsidiary of the internal force is not zero.B, a subsidiary of the displacement is not zeroC, a subsidiary of the deformation is not zero.D, part of the basic force is zeroE, the basic part of the deformation is zero2, the static properties will change the structure of the material, which of the following factors do not change?A, displacementB, deformationC, axial force and shear forceD, momentE, bearing force3, two static frame geometry and size, load, the same material but different section size, the two factors are the sameA, forceB, forceC, stressD, displacementE, deformation4, two static two static frame geometry and the same size, load and section size but with different materials, the two factors are the sameA, forceB, forceC, stressD, displacementE, deformation5, the master-slave structure, only part of temperature rise, the following discussion is correct ACDEA, the internal force of structureB, the whole structure deformationC, part of the deformationD, part of the zero displacementE, the basic part of the deformation is zeroThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, the structure deformation, will cause displacement, in turn, will have deformed structure displacement.error2, static structure temperature changes caused by the displacement calculation formula is only applicable to the static set structure, is not suitable for statically indeterminate structure.Correct3, the virtual work is a fictional work, is actually not possible.error4, static structure in non load under the action of external factors, but does not produce force, displacement.Correct5, can not use the displacement diagram multiplication forthree hinge arch.Correct"The test of structural mechanics" chapter 06 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, the typical force equations, pay coefficient DA, B, constant constant greater than zero is less than zeroC, D, equal to zero can be positive or negative to zero2, get rid of a movable hinge support, equivalent to remove several constraints? AA, 1 B, 2C, 3 D, 43, get rid of a directional support, equivalent to remove several constraints? BA, 1 B, 2C, 3 D, 44, get rid of a fixed hinge bearing, equivalent to remove several constraints? AA, 2 B, 3C, 4 D, 55, the essence of force equation is DA, B, physical condition of equilibrium conditionsC, theorem D and displacement conditionsSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, single span symmetric structure in antisymmetric loads, section ACD on the symmetry axisA, displacement along a symmetry axis direction is zeroThe displacement of B, vertical symmetry axis direction is zeroC, the bending moment is zeroD, the axial force is zeroE, the angular displacement is zero2, two span symmetric structure under symmetrical load, the following discussion is correctRigid node line displacement A, the symmetric axis is zeroB, on the axis of symmetry of the rigid joint angular displacement is zeroC, in the column no momentD, no shear columnE, in the column without axial force3, in the typical force equations in coefficient is ABCDA, can be positiveB, may be negativeC, may be zeroD, there is reciprocal relationE, and the external cause4, the typical force equations can be positive or negative zero isA, 11.B, 12.C, 21.D, 22.E, Delta 1PThe basic system of computing 5, force method can be ABDA, static structureB, no extra bound geometric invariant systemC, geometric transient systemD, a statically indeterminate structureE, geometric constant variable systemThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, open a single hinge, equivalent to removing two constraints.Correct2, get rid of a fixed hinge bearing, equivalent to removing two constraints.CorrectThe basic unknown quantity 3, force method is redundant force.Correct4, the main factor in the typical force equations of constant greater than zero.Correct5, the internal force calculation of statically indeterminate structure should consider the equilibrium condition and deformation continuity condition.Correct"The test of structural mechanics" chapter 07 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, fixed at both ends with statically indeterminate beam angle, the known A terminal of the 0 A bar is internal force error is CA MB, A B, MA 2I = theta theta A = 4IC QAB, 3I A/l D, theta = - QBA = - 6I A/l.In 2, the typical displacement method equation R11 * Z1+r12 * Z2+R1P * Z1+r22 * R21 = 0, Z2+R2P = 0, the following formula is correct BA, R11 B, R12 = R21 = R22C, R12 * Z2 = R21 * Z1 D, R1P = R2P3, one end is fixed with one end hinged with the statically indeterminate beam angle, known fixed end A theta is A, shear force of bar terminal for AA, 3I A/l B, 6I - theta theta A/lC, 12I A/l D, 0.4, A end fixed B end hinged with statically indeterminate beam, fixed end known angle theta A, the following conclusion is wrong CA, MB = 0, MA = B 3I A.C, QBA = 0, QAB = D 3I 9 A/l5, the beam stiffness of single span single infinite non hinged rectangular frame, the following discussion about the basic unknown displacement method, the right is CA, three B, two basic unknown basic unknown quantityC, just two node angle is zero, only one unknown quantity D, just two node angle is not zero, only an unknown quantity.Second questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, if the additional constraint in the basic system ofdisplacement force (torque) is equal to zero, then AEA, the basic system is consistent with the original structural stress, deformation.B, the basic system and the original structure of the force is not consistent, inconsistent deformation.C, the basic system is consistent with the original structure of the force, but the deformation is not consistent.D, the basic system and the original structure of the force is not the same, but the same deformation.E, the basic system of the original structure and equivalent.2, single span symmetric structure in antisymmetric loads, section ADE on the symmetry axisA, displacement along a symmetry axis direction is zeroThe displacement of B, vertical symmetry axis direction is zeroC, the angular displacement is zeroD, the bending moment is zeroE, the axial force is zero3, two cross symmetric structure under antisymmetric loads, the following discussion is correct AEThe vertical displacement of rigid node A, the symmetric axis is zeroB, on the axis of symmetry of the rigid joint angular displacement is zeroC, in the column no momentD, no shear columnE, in the column without axial force4, using the displacement method to calculate the loads of statically indeterminate structure, the rod relative stiffness is calculated by ABThe real displacement A, node displacement is not structureB, the internal force is correctC, node displacement is the real displacement of structureD, the internal force is not correctE, node displacement and internal force are correct5, the following discussion about the basic system of displacement method is correct ACThe basic system of A, the displacement method is a group ofsingle span statically indeterminate beamThe basic system of B, the displacement method is static structureThe basic system of displacement method C, a structure is uniqueD, a method of structure displacement basic system has many choicesThe basic system of E, the displacement method is geometry variable systemThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, one end is fixed with one end hinged with the statically indeterminate beam, fixed end angle theta A is known as B, angle hinged = - 0 A (CCW).error2, the displacement method can calculate the statically indeterminate structure can also calculate the static structure.Correct3, hyperstatic structure calculation of total principle is the desire of statically indeterminate structure to take a basic system, and then make the basic system in terms of stress anddeformation of the original structure and the same.Correct4, one end is fixed with one end hinged with the statically indeterminate beam, fixed end angle theta A is known as MA, bending moment of the fixed end theta = 3I A.Correct5, one end is fixed with one end hinged with the statically indeterminate beam, fixed end angle theta A is known as the QAB, bar shear -3i theta = A/l.Correct"The test of structural mechanics" chapter 08 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, what kind of structure can not be calculated by the method of moment distribution? DThe structure of A, B, non continuous beam node linear displacementC, frame D, node displacement of structure2, member AB A end is a fixed end, no load span, known B end moment is m, then A = AB rod end bending momentC, m/23, the multi node structure, obtained the moment distribution method is CA B, the approximate solution and exact solutionC, D, analytic solution asymptotic solution4, A rod end is fixed and the B end of the sliding line stiffness I, rotational stiffness of C A terminalA, 4I, B, 3IC, I, D, 05, when the distal end is a fixed end, the transfer coefficient is equal to BA, 1 B, 0.5C, -1, D, 0Second questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, only what moment to the distal transfer? BDA, fixed end momentB, the distribution of bending momentFinally, the bending moment CThe proximal bending moment D, the rotation of the rod endThe distal end of the rotating bending moment E, the2, A rod end is fixed and the B end of the sliding line stiffness is I, the following formula is correctA, SBA = IB, SAB = 3IC, SAB = ID, SBA = 3IE, CAB = 13, the moment distribution method can be used to calculate the structure of what kind of? ABCA, continuous beamB, a non sway frameThe structure of C, no node displacementThe structure of D, no node displacementE, the beam stiffness is infinite structure4, continuous beam in a bearing of two adjacent span live load, to both sides of every span live load is the most unfavorable load on what the layoutA, the support section of negative momentB, the force of supportC, the positive moment spanD, the seat on the right side of the maximum shear forceE, the left side of the smallest section shear bearing5, the following discussion about the moment distribution method is correct?A, moment distribution method to obtain exact solutionsB, get the asymptotic solution of moment distribution methodC, from the first node unbalanced torque larger absolute value of node to startD, you want to change the unbalanced joint torque distributionAt the same time, E can relax the adjacent nodesThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, AB A said the moment distribution coefficient of node function unit couple, AB lever A is sharing the rod end moment.Correct2, the moment distribution method are derived based on the displacement method, so the structure can be calculated using displacement method can be calculated by the method of moment distribution.error3, in the moment distribution method, non adjacent nodes can also relax.CorrectScope 4, moment distribution method is continuous beam and non sway frame.CorrectThe end of 5, at a bar of the nodal moment distribution coefficient equals 1.Correct"The test of structural mechanics" chapter 09 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, when the harmonic loads for the single degree of freedom system with the particle damping, if the load frequency is far greater than the natural frequency of the system, and then the dynamic load balance is mainlyD, inertial force2, when the harmonic loads for the single degree of freedom system without damping on the particle, if the load frequency is far less than the natural frequency of the system, and then the dynamic load balance is mainly AA, the elastic restoring force3, a single degree of freedom vibration system, the damping ratio zeta, dynamic coefficient, resonance when the following results are correctA, f = 0.05, P = 104, a single degree of freedom system, the initial displacement of 0.685cm, the initial velocity is zero free vibration, vibration of a cycle after the maximum displacement is 0.50cm, the damping ratio for the systemA, 0.05In 5, low damping system can not be ignored the influence of damping on what?C, the free vibration amplitudeSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, a simple point motor, in order to improve the vibration frequency of the beam, the following measures are correctA, shorten the spanB, increasing sectionC, the end of the beam to the fixed endD, reduced qualityE, the motor speed increases2, the free vibration of multi degree of freedom system of the main calculation ABCA, frequencyB, cycleC, modeD, dynamic force。
浙江大学考研结构力学2012年真题答案
浙江大学2012年攻读硕士学位研究生入学考试结构力学试题参考答案一、选择题（共10小题，每题3分，共计30分） 【答案】1.A; 2.C; 3.A; 4.D; 5.B; 6.D; 7.C; 8.A;【提示】1.先去掉支座，判断主体结构，将水平的三根链杆作为I 、II 、III 刚片，然后进行判断可知是无多余约束的几何不变体系。
2.瞬变体系必定存在多余约束，不一定存在瞬铰且总体约束的数目不定，不可以作为结构使用。
3.将结构分成正对称和反对称两个结构来反对，在Δ/2正对称位移作用下，弯矩图为0.在Δ/2反对称位移作用下，弯矩图为反对称，所以可以排除B ，又因无荷载作用，可排除C 、D 。
4.取受P 作用的节点分析，将力往斜杆方向和与斜杆垂直的方向分解进行求解。
5.弯矩无影响，无需画弯矩图。
先作出轴力图，根据轴力图求得0001t Nds t a t a ααα∆==⨯⨯=⎰，故选B 。
6.改变支座B 的方向只影响轴力，弯矩和剪力都不变。
7.力法4个，将钢架中间处断开，去除3个约束，再将结构右侧支座处链杆断开，去除1个约束，共计4个。
位移法5个。
三个钢节点处刚结，并在左侧加一水平的约束，最右侧水平支座处加一个竖向约束，共计5个。
8．可根据各自的弯矩图判断，知极限荷载值最大的是A 。
9．可根据各自的弯矩图判断，知极限荷载值最小的是C 。
10.极限荷载的计算只需要考虑结构最终的平衡状态，不需要考虑其变形状态。
二、填空题（共7小题，每空2分，共计26分）【答案】11A.-4q; 11B.12A. 2596Pl EI ; 12B. 532P-；13A.34EI l ; 13B.34EIl; 14A.0; 14B.0; 15A.224ql -; 15B.22ql -;16A.212236112EI EI ql l l θ-∆-; 16B. 212226112EI EI ql l l θ-∆+; 17A. 25-.【提示】11.先算出支座反力，再根据曲线方程求出曲线在D 点的斜率，取D 得有半部分受力分析，易得出D M 和QD F 。
结构力学课后习题答案[1]
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浙江大学宁波理工学院2012–2013学年一学期
《结构力学（2）》结业生返校考试标准答案
开课分院：土建分院，考试形式：闭卷，允许带计算器入场
考试日期：__ 2012 ___年____月____日，考试所需时间： 90 分钟
考生姓名学号考生所在分院：土建分院专业班级： .
一、单项选择题（每小题3分，共15分）
1、单元刚度矩阵是表示一个单元的下列两组量值之间的相互关系（ A ）。
A：杆端力与杆端位移 B：杆端力与结点外力
C：杆端位移与结点位移 D：结点外力与结点位移
2、关于自振周期与结构质量的关系，以下叙述正确的是（ C ）
A：自振周期与质量成正比 B：自振周期与质量成反比
C：自振周期与质量的平方根成正比 D：自振周期与质量的平方根成反比
3、关于可破坏荷载和可接受荷载，以下叙述正确的是（ B ）
A：可破坏荷载恒小于可接受荷载 B：可破坏荷载恒不小于可接受荷载
C：可破坏荷载恒等于可接受荷载 D：都可能
4、能量法的势能驻值条件等价于用位移表示的平衡方程，当荷载等于临界荷载时，该平衡方程是（ D ）
A：位移有零解的齐次方程 B：位移有零解的非齐次方程
C：位移有非零解的非齐次方程 D：位移有非零解的齐次方程
5、在动力计算中，集中质量的数目与体系自由度数（ B ）
A 总是相等
B 不一定相等
C 质点数目总是多于自由度数目
D 质点数目总是少于自由度数目
二、填空题（每空格3分，共15分）
1、结构失稳有两种基本形式：极值点失稳和（分支点失稳）
2、上限定理可叙述为：可破坏荷载是极限荷载的（上限）
3、结构在动力荷载作用下的振动称为（强迫振动）。
4、用矩阵位移法对下图所示桁架进行分析计算，整体刚度矩阵的阶数为（ 5 ）
5、下图所示体系有2个质点，如忽略轴向变形，只考虑弯曲变形，对该体系进行结构动力计算分析，其自由度数为（ 2 ）
三、试求图示连续梁的整体刚度方程。
（20分）
解：各单元的刚度矩阵为：
(1)单元(两端结点1,2)：(1)42[]24i i K i i ⎡⎤=⎢⎥⎣⎦ (2)单元(两端结点2,3)：
(2)
]8[844i i i i K ⎡⎤=⎢⎥
⎣⎦
将上述2个单元刚度矩阵组装成如下原始整体刚度矩阵[K ]I
由于结点3为固端，去掉第3行与列得整体刚度矩阵[K ]
原结构结点3为固端，故不需考虑M 3。
等效结点荷载M 1，M 2为：
212122************F
F F p ql M M M M M ql F l ⎧⎫
⎪⎪⎧⎫⎧⎫⎪⎪⎪⎪=-=⎨⎬⎨⎬⎨⎬+⎪⎪⎩⎭⎪⎪⎩⎭-+⎪⎪⎩⎭
整体刚度方程为：22231421211212128p
ql i i i i ql F l θθ⎧⎫
⎪⎪⎧⎫⎡⎤⎪⎪=⎨⎬⎨⎬⎢⎥
⎣⎦⎩⎭⎪⎪
-+⎪⎪⎩⎭
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A
B
l
2212
12
F
F AB
BA
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M
=-
=8
8
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AB BA F l M M =-
=
四、试求图示单自由度体系的自振圆频率ω 和自振周期T 。
（杆件质量忽略不计，EI =常数） (15分)
解：（1）作M 图利用图乘法求得柔度系数
自振圆频率
ω=
=
自振周期
2T π
ω==
4
12
3
12*22243448l l l M l ds EI EI EI δ⋅⋅⋅===
⎰
五、试求下图所示单自由度体系的临界荷载F pcr (20分)
解：设B 点的位移为y 能量法：
体系应变能：21
2
εV ky =
荷载势能：222
32*224p p p p
y y y V F F F l l l λ⎛⎫=-⋅=-+=- ⎪⎝⎭
总势能：2
21324p p p y E V V ky F l
ε=+=-
由总势能驻值条件
0p E y
∂=∂得
302p y
ky F l
-=
y 有非0解条件的荷载为临界荷载：
23
cr p F kl =
静力平衡法：
弹簧反力：()RB F ky =← A 、 C 支座反力：2
1
(),
()3
3
RA RC F ky F ky =
→=→
取结点B 的上部作为隔离体，由0B M =∑得
2
20
03
RC p p F l F y kyl F y ⋅-=⇒
-= 上式平衡方程中y 有非0解条件的荷载为临界荷载：
23
cr p F kl =
=∞
=∞
六、试求图示连续梁的极限荷载q u 。
（15分）
解：破坏机构有以下2种
破坏机构1
对于破坏机构1，虚功方程为
1
()2
u q l M θθ⋅∆=+
将 1
2l θ∆= 代入得 128u M q l
=
破坏机构2
对于破坏机构
22u u q l M M θθ⋅∆=+⋅
将 1
2l θ∆= 代入得 2210u M q l
=
q 1<q 2
所以极限荷载为：28u u M
q l
=
命题（组）老师签名：____________________ 年 月 日
研究所（教研室）教学负责人签名：_______________
年 月 日。