结构动力学习题答案

2.1解：a刚质产单设为）度即是使点生位位移所需施加的力，k则当质产时，：点生位移Δ，有：KΔ＝K1Δ+K2刚Δ；所以，等效度K＝K1+K2()()12sIpWkuukkuuWmuuWptuδδδδδδδ−==+−==ii虚由功原理得：()()12kkumupt++=iib）112212kkk∆=∆=∆⎧⎨∆+∆=∆⎩联立解得：1212kkkkk=+与a）同理，得：()1212kkumuptkk+=+iiic）()3312113kkkkk∆=∆⎧⎪∆=+∆⎨⎪∆+∆=∆⎩联立解得：312123()kkkkkkk+=++与a）同理，得：()312123()kkkumuptkkk++=++iii2.2解：a刚单载）先求框架度，用位荷法求δ1112h12h12h12h所以，31111112222236cchhhhEIEIδ⎛⎞⎛⎞=×××××=⎜⎟⎜⎟⎝⎠⎝⎠（其中EIb＝∝，故1/EIb＝0）∴31161cEIkhδ==()36ckIpEIWkuuuuhWmuuWptuδδδδδδδ−==−==ii虚由功原理：0kIpδδδ++=得：()36cEIumupth+=iib）33131241684119314ccEIEkhh×+=×=×+I运动∴方程：()316819cEIumupth+=ii2.3解：()222224()339111339149243342424115222sDIpWkuukuuWCuuCuullrlluullWrllWptuδδδδδδπuulruuδδπδπδδ−==⎛⎞−==⎜⎟⎝⎠⎛⎞⎡⎤⎜⎟⎛⎞⎛⎞−=++=⎢⎥⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎢⎥⎜⎟⎣⎦⎝⎠=iiiiiiiiδ椭圆纸注意：①平行于面放置题须虑椭圆为时②此不考板的重力做功。因列方程，u时静为以平衡位置零点，重力一弹参见教直被簧力所抵消。（材P32响释重力影的解）虚由功原理：0sIDpδδδδ+++=得：241149991152kuCulrupπ++=iii义质广量21491152Mlrπ=义刚广度K:4/9K义广阻尼C：1/9C义载广荷：P2.4解：()2222211122222221221121123212212111121cossinsin2sIpWkqbqWmLqqmqqqmqqmqqqmqqqqWmgqqmgqqqmgLqqδδ1δδδδδδδδδ−=−−=−+++=−−iiiiiiiiiiδ由0sIpδδδ++=得：22212111222121222222111020sinsin32cos00mqqqqmLmqmgqqmgLqqqmgqkbmkmq⎡⎤⎡⎤11⎧⎫⎧⎫+−−⎧⎫⎪⎪⎪⎪⎢⎥⎢⎥+=⎨⎬⎨⎬⎨⎬⎢⎥⎢⎥⎩⎭⎪⎪⎪⎪+−⎢⎥⎢⎥⎩⎭⎩⎭⎣⎦⎣⎦iiiiiiii2.5解：()122122222sinsDIpWkuuWCuuWmuumuLuLmuumuumLumLumLWmgLδδδδδδθδθδδθδδθθδθδθ−=−=⎛⎞−=+++⎜⎟⎝⎠=++++=−iiiiiiiiiiiiiiiiiiδθ虚功原理：0sIDpδδδδ+++=得：1222222000sin0000mmmLuCukumLmLmgLθθθθ⎧⎫⎧⎫+⎡⎤⎧⎡⎤⎡⎤⎧⎫⎪⎪⎪⎪++=⎨⎬⎨⎬⎨⎬⎨⎬⎢⎥⎢⎥⎢⎥−⎣⎦⎣⎦⎩⎭⎣⎦⎩⎪⎪⎪⎪⎩⎭⎩⎭iiiiiii⎫⎭2.6解：()121212122122222111224sinsDIpWkuukuuWCuuCuuCuuCuuWmuumuLuLmuumuumLumLumLWmgLδδδδδδδδδθδθδδθδδθθδθδθδθ−=+⎛⎞⎛⎞−=+×=+⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞−=+++⎜⎟⎝⎠=++++=−iiiiiiiiiiiiiiiiiiiiδi虚功原理：0sIDpδδδδ+++=得：1221212222210004sin0000mmmLuukkuCCmLmLmgLθθθθ⎡⎤⎧⎫⎧⎫+++⎡⎤⎧⎡⎤⎧⎫⎪⎪⎪⎪⎢⎥++=⎨⎬⎨⎬⎨⎬⎨⎬⎢⎥⎢⎥⎢⎥−⎣⎦⎩⎭⎣⎦⎩⎪⎪⎪⎪⎩⎭⎩⎭⎣⎦iiiiiii⎫⎭3.1解：由题可知：sTD64.02=，即sTD28.1=cmu1.3)0(=cmTuD2.2)2(−=由⎥⎦⎤⎢⎣⎡++=−)sin)0()0((cos)0()(twwuwutwuetuDDnDtwnξξ得：)2(u)0(DTu=2.21.3−=()0ue)0(2uDnTwξ−−=πξξ21−−e所以：%9.10109.03429.012==⇒=−ξξξπsradTwwDDn/938.4109.0128.12121222−×=−=−=πξπξmKNmNmwKn/5.2194/105.2194938.410903232=×=××==阻尼系数：msKNmwCn/9.96938.41090%9.10223•=××××==ξ3.2解：sradmkwn/21087510350033=××==设()twBtwAtunnsincos+=将及代入解得：()6.40=u()6.42.1=u6.4=A8.11=B()tttu2sin8.112cos6.4+=⇒()cmu4.114.2−=⇒振幅：cmBAu7.12220=+=3.3解：总刚度mKNK/5602.01120==(1)sradmkwn/36.2211256000==(2)%5.16841212==+njinIuuIjππξ(3)sradwwnD/06.22165.0136.22122=−×=−=ξ3.4解：以体系静平衡位置作为原点km−1则,共同作用的静平衡位置1m2mkgmust2=碰撞之前的速度2m222vmgh=碰撞之后：动量守恒122()(0)2mmumgh+=i即212(0)2mugmm=+ih动力方程：()()()120ststmmuuKuu′′+−+−=()122mmuKumg⇒++=ii解得：⎟⎟⎠⎞⎜⎜⎝⎛+=++=212sincosmmKwKgmtwBtwAunnn将及(0)0u=212(0)2mugmm=+ih代入解得：22122,()mgghABmkkm=−=+m于是：22212122cossin()nnnmgmgghKuwtmwtwkkmmKm⎛⎞=−++=⎜⎟⎜⎟++⎝⎠m3.5解：()[]2220211nnstdξ+⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−==当1=nww时，发生共振有：ξ210.51==stduR(1)当101=nww时，()()22211.021.0115.0×+−==ξstduR(2)由式(1),(2)可以解得%95.4=ξ3.6解：()[]()[][]22222121nnnξξ+−+=0=ξ()[]2225211nwTR×−=πnww,解得：sradwn/36.47=mKNmwKn/6.203636.4790822=×==3.7解：对于任意简支梁跨中δ−F关系：EIFl483=δ3148lEIK=⇒所以体系的刚度：mKNlEIKK/1.59514.21016.41006.29696236831=××××===−mKPust50105.41.5951267.0−×===sradmkwn/5.104545101.59513=×==sradw/10602300ππ=×=301.05.10410===πβnww01.0=ξ()[]()()10.1301.001.02301.0112112222220=××+−=+⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−==nnstd−−×=××=•=设()φ−=wtuusin0则()φ−−=′′wtwuusin20所以：加速度振幅()2225200/109.410100.5smwua−−×=××==π3.8解：(1)从图中可以看出：cmu38.00=NKu17300=mKNK/3.4551038.017302=×=⇒−(2)()()msNKuwwEnDeq/2.6347%0.71038.0103.455129.2222320•==×××××==−ππξ(3)%0.14%0.722=×==ξη3.9．解：当ωπ/0≤≤t时)sin2sin2(2]})cos[(]){cos[(2)](sin[sin1)(222200000wt−−−=+−−−+=−=∫∫τττωτωωωτωτω当ωπ/t时]sin)[sin()()sin2)sin(2(20)](sin[sin1)(2202222000t−−=−+−−=+−=∫∫ππτωτωπτπτ3.10．解：当dTt≤≤0时)cos1()](sin[1)(2000twmpdtpmtunntnn−=−=∫ωτωωτ当时dTt}cos)]({cos[0)](sin[1)(2000twTtwmpdtpmtundnnTnnd−−=+−=∫ωτωωτ3.11．解：当时，令dTtα=ndTT/（0.2,5.1,0.1,.5.0,1.0=α）)2sin(sin2)]2/1/(sin[2sin2)(200αππαπωω−=−=nndnddnnTtmpTtwTTwmptu4.1解：（1）列出动平衡方程为：⎪⎪⎭⎪⎪⎬⎫=+−=−++00)(2212..2222121..11qkqkqmqkqkkqm故为静力耦联（2）列出动平衡方程为：⎪⎪⎭⎪⎪⎬⎫=++=+++00)22..22..1211..22..121(qkqmqmqkqmqmm故为动力力耦联（3）列出动平衡方程为：⎪⎪⎭⎪⎪⎬⎫=++=+++00)22..22..1211..22..121(qkqmqmqkqmqmm故为动力力耦联（4）列出动平衡方程为：⎪⎪⎭⎪⎪⎬⎫=++++=+++++++0)2()0)()222111..22221..1212111..221..121()42(2(qkhhkqhkqhmhmqhmhmqhkqkqhmhmqmm故为动力力耦联4.2解：利用虚功原理：ukuukuWs22112δδδ+=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−+⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛+=llmmuuuuluuuuWI12..1..2221..2..11222δδδ由Wsδ＋WIδ＝0可得：0)632()63(2..1..221..2..11=+++++uuuuuuuummKmmKδδ；由、u1δu2δ为任意数方程成立，故：⎪⎪⎩⎪⎪⎨⎧=++=++0632063..1..22..2..11uuuuuummKmmK写成矩阵形式为：⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡00200366321..2..1uuuuKKmmmm运动特征方程为：（[K]-ω2[M]）{φ}=0；可得：0[M]-[K]2=ω即：03266322=−−−−ωωmkmmmk⇒结构得自振频率为：.1=ω1.5925mk；.2=ω3.0764mk。设：；其中n＝1,21)(2=φn代入特征方程可得：7342.2)1(1=φ；7321.0)2(1−=φ故结构的一阶振型为：{}⎭⎬⎫⎩⎨⎧=17342.21φ故结构的二阶振型为：{}⎭⎬⎫⎩⎨⎧−=17321.02φ4.4解：可知刚度：k1＝k2＝hEI324列出动平衡方程为：()()⎪⎭⎪⎬⎫=+−=−++ttpukukumpukukkum22212..22122121..11)(代入刚度得：()()⎪⎪⎭⎪⎪⎬⎫=+−=−+tEIEItEIEIpuhuhumpuhuhum22313..2212313..1124242448写成矩阵形式为：()()⎪⎪⎭⎪⎪⎬⎫⎪⎪⎩⎪⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡ttEIEIEIEIppuuhhhhuumm21213333..2.