英汉双语弹性力学6课件.pptVIP

其中常数A，B由边界条件确定。 在平面应变的情况下，只需在以上各式中将 例2： 设有一厚壁圆筒，内半径为a，外半径为b。从一均匀温度加热，内表面增温Ta ，外表面增温Tb，如图所示。试求筒内无热源，热流稳定后的热应力。 得无热源，热流稳定后的热传导微分方程为 解：首先求温度场。由热传导微分方程 a b Ta Tb From the boundary conditions we figure out A and B and substitute them back, then we obtain the temperature field Integrating two times yields or For axisymmetric temperature field 对于轴对称温度场有 积分两次得： 或 由边界条件： 求出A，B后回代，得温度场： Integrating yields Substitute T into the stress expressions of the problems of plane strain 积分后得 将T代入平面应变问题应力表达式 Solution Let We can arrived at Therefore x y o a a b b Exercise 6.1 Show that the rectangular thin plate undergoes temperature change Please evaluate the temperature stress (assume ab). 练习6.1 图示矩形薄板中发生变温 试求温度应力（假定a远大于b） 解： 取 可解得 所以 x y o a a b b Hence,we can arrived at Let Then 由此得 取 则 Therefore The boundary conditions are satisfied apparently. From i.e. 所以 边界条件 显然满足 由 即 So We can arrived at While the boundary condition is satisfied permanently. 得 而边界条件 恒成立。 故 Exercise 6.2 It is known that a homogeneous disk with a radius b is placed in a isothermal rigid hoop. The disk and the hoop are made of same material. Suppose the disk is heated with the following rule: The temperature of the hoop is kept normal temperature T0. And the strain generated due to this temperature can be ignored. Evaluate the value of the compression stress with a distance r from the center of the disk. Solution: The expression of displacement of the plane stress problems of the axisymmetric plate is Because u is a finite value at the center, i.e. r=0, so c2=0. 练习6.2 已知半径为b的均质圆盘，置于等温刚性套箍内，圆盘和套箍由相同的材料制成，设圆盘按如下规律加热 套箍温度则保持为常温T0，而由此温度所引起的应变可以忽略，试求距圆盘中心为r处的压应力值。 解： 轴对称平板的平面应力问题的位移的表达式为 由于在中心处，即r=0处，u为有限值，因此c2=0。 When r=b，u=0. Substitute it into（a）, i.e. Substituting the value of c1 It can be obtained from expression (b) that 当r=b时，u=0，代入（a）即 由（b）式得 将c1的值代入 yields 得 将应力分量用形变分量和变温T表示的物理方程为： 几何方程仍然为： 将几何方程代入物理方程，得用位移分量和变温T 表示的应力分量 Introducing the above equations into the differen