答案 a1 《结构力学Ⅱ》

《结构力学Ⅱ》

试题类型：A√、B 制作人：

一、填空题（每空1分，共15分）

1.结构在大小、方向和作用点随时间变化的荷载作用下，质量运动加速度所引起的惯性力和荷载相比大到不可忽视时，则把这种荷载称为动荷载。

2.在振动过程的任一时刻，确定体系全部质量位置或变形状态所需的独立参数个数，称为体系的自由度。自由度为1的体系称为单自由度体系；自由度为有限值的体系称为有限自由度体系；质量连续分布的体系为无限自由度体系

3. 影响线作法：其一是静力法，另一为机动法（虚功法）。

4.用广义坐标法或有限单元法将无限自由度体系简化为有限自由度体系时，体系的自由度数等于广义坐标数或独立结点位移数。

5. 频率谱中最小的频率称为结构的基本频率，简称为基频。其余依次称为第二频率、第三频率等。

二、选择题（每题3分，共15分）

1.单自由度及多自由度体系中的刚度系数和柔度系数关系（A）

A．单自由度中互为倒数 B. 多自由度中互为倒数

C．均互为倒数 D.都不互为倒数

2.图1属于结构动力学研究内容的哪一类（ C ）

图 1

A．荷载识别 B. 结构的参数识别

C．结构的响应分析 D. 结构的振动控制

3.静荷载只与作用位置有关，而动荷载是（C）的函数。

A．时间 B.坐标

C．时间和坐标 D.位置

4.对多数工程结构来说，自振频率的个数与那一项相等（C）

A．结构的质量个数 B. 结构的位移个数

C ．结构的动力自由度数目 D. 结构的质点个数

5. 下列选项中那一项不属于将无限自由度体系简化为有限自由度体系的常用方法：（ D ）

A ．集中质量法 B. 广义坐标法 C ．有限单元法 D. 刚度法与柔度法

三、判断体系的动力自由度数目，并标出位移方向（每题5分，共10分） 1.

标出自由度3分，位移方向2分

2.

标出自由度3分，位移方向2分

四、计算题（共60分）

1. 利用影响线作Y A , M A , M K , Q K 的图形。（10分）

解：0=∑A m x M A -=

（2.5分）

0=∑y

F

1=A Y

（2.5分）

xl/2：MK= -(x - l/2 )，QK=1

（2.5分）

（2.5分）

2. 用力矩分配法计算图示梁,作弯矩图（15分）

解：EI EI S BA 103103=?

= 5EI S BC =6.0)2.03.0(3.0=+=EI EI BA μ 4.0)2.03.0(2.0=+=EI

EI BC μ（3分）

（9分）

（3分）

3.试用柔度法建立图 a 所示受均布动荷载 作用、无重弹性梁中点有一个集中质量 m

的简支梁运动方程。（15分）

解：因为对梁和刚架等忽略轴向变形，因此本题为单自由度体系。设质量 m 的位移为 y （向下为正）。加惯性力和阻尼力后的受力情形如图 b 所示。

（1分）

列位移方程视体系位移 y 是由惯性力 FI、阻尼力 FD 和动荷载 FP 共同作用引起的（在时刻 t 看成静力）。则由结构位移计算可知y＝DPd×[FI(t)＋FD(t)] （5分）其中 DP 是由动荷载引起的质量沿自由度方向的位移，d 为自由度方向加单位广义力所引起的自身方向位移（如图 d、e）。经结构位移计算求得：DP=5q(t)l4/384EI （2分）d=l3/48EI

可得y ＝5q(t)l4/384EI＋l3/48EI×[－m?－cy]（3分）

经整理后可表示为

或

式中 k=1/d=l3/48EI ，FE(t)= kDP=5q(t)l/8 （5分）4.如图所示两层刚架，已知横梁为刚性，各立柱的抗弯刚度EI = 6.0×106 Nm2 ，立柱的质量忽略不计，横梁的质量 m1 = m2 = 5000 kg，每层的高度 l = 5 m。试求其自振频率和振型。（20分）

(a)

解：由自由度分析可知，结构是两自由度体系，设m1的位移为y1，m2的位移为y2如图 a所示。因为是剪切型结构，所以用刚度法求解比较方便。根据结构静力分析如图（b、c）可求得

(b) (c) （5分）

，(5分)

（5分）

代入已知条件的有关数值后，刚架的自由振动频率为

，

将所得频率代入式（8-11）可得振型为

（5分）

7

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