最新哈工大计算力学考试题及答案

（1）For a plane stress state, if the y-coordinate is regarded as a symmetric axis, try to make a sketch and write down the displacement boundary conditions at the symmetric axis in finite element modeling. (6 points) Solution:As shown in the figure, for a symmetric problem, we may define 0==A A v u at point A ; and at point B , 0=B u（2）Try to use the Castigliano’s first theorem to obtain the matrix equilibrium equations for the system of springs shown in the following Figure.(10 points)Solution ：For the spring element, the strain energy is given by ()221u k U e ∆=In which, k – stiffness of the spring, u ∆ - deflection of the spring.The total strain energy of the system of four springs is expressed by means of the nodal displacements and spring constants as2454234322322121)(21)(21)(21)(21U U k U U k U U k U U k U e -+-+-+-=By application of the Castigliano’s first theorem for each element12111211)()1)((F U U k U U k U U e=-=--=∂∂23221212)()(F U U k U U k U U e=-+-=∂∂ 34332323)()(F U U k U U k U U e=-+-=∂∂ 45443434)()(F U U k U U k U U e=-+-=∂∂ 54545)(F U U k U U e=-=∂∂ The system stiffness matrix can be written as⎪⎪⎪⎭⎪⎪⎪⎬⎫⎪⎪⎪⎩⎪⎪⎪⎨⎧=⎪⎪⎪⎭⎪⎪⎪⎬⎫⎪⎪⎪⎩⎪⎪⎪⎨⎧⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡--+--+--+--54321432444433332222111100000000000000F F F F F U U U k k k k k k k k k k k k k k k k（3）The interpolation functions for a beam element of length L arewrite down a polynomial representation of the displacement field )(x v in terms of the above interpolation functions and show that 22/dx v d = constant for the beam element subject to pure bending. (12 points) Solution ：The displacement field for a beam element is332220)(x a x a x a a x v +++=It can also be expressed in terms of interpolation functions and nodal variables as24231211)(θθN v N N v N x v +++=Substitute the interpolation functions into the above equation and after a few manipulations, we have24231211θθdxdN v dx dN dx dN v dx dN dx dv +++=22223221221322326634166θθ⎪⎪⎭⎫ ⎝⎛+-+⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛+-+⎪⎪⎭⎫ ⎝⎛+-=L x L x v L x L x L x L x v L x L x 2223212132226212664126θθ⎪⎭⎫ ⎝⎛+-+⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛+-+⎪⎭⎫ ⎝⎛+-=L x L v L x L L x L v L x L dx v d for the beam element subject to pure bending, we have21v v =, 21θθ-=So, constant 22411122=-=+-=θθθLL L dx v d （4）For a 2D problem, if the mid-points of each side of a triangular element are alsodefined as nodal points, try to write down an appropriate polynomial representationof the displacement field variable, and discuss its convergence conditions. (14 points ）Solution ：The polynomial representation of the displacement field variable can be written as26524321y a xy a x a y a x a a u +++++=21211210987y a xy a x a y a x a a v +++++=The convergence conditions include: (1) the compatibility conditions.Since the above equations are continuous within the element, so the displacement field iscontinuous in the element.On the common boundary, the side line is a quadratic function that has three independent constants. And since there are three nodes, the boundary curve can be uniquely determined by the quadratic function, so on the common boundary, there is no void, no material overlap either.(2) the completeness condition.The rigid body motion can be determined by the constants 1a and 7a The rigid body rotation can be realized by3a y u =∂∂, and 8a xv =∂∂ The constant strain condition can be satisfied by2a xux =∂∂=ε, 9a y v y =∂∂=ε, and 83a a x v y u xy +=∂∂+∂∂=γIn summary, convergence conditions are satisfied for the element.（5）Considering a beam element, Denoting the element length by L and the moment ofinertia of the cross-sectional area by z I , write down an appropriate function to express the displacement field, and finally, derive the finite element equation and nodal forces of the element by using the Galerkin ’s method. (18 points)Solution ：The governing equation for the problem of beam flexure is)(d d d d 2222x q x v EI x z =⎪⎪⎭⎫ ⎝⎛ The displacement solution can be written as∑==+++=4124231211)()()()()()(i i i x N x N v x N x N v x N x v δθθTherefore, the element residual equations are4,10d )(d d d d )(212222==⎥⎦⎤⎢⎣⎡-⎪⎪⎭⎫⎝⎛⎰i x x q x v EI x x N x x z iIntegrating the derivative term by parts and assuming a constant z EI , we obtain4,10d )(d d d d d d d )(2121213333==--⎰⎰i x x q N x x vx N EI xvEI x N xx i x x i z x x z iand since3322d d d d d d d d x vEI x v EI x x M V z z -=⎪⎪⎭⎫ ⎝⎛-=-= Integrating again by parts and rearranging gives4,1d d d d d d d )(d d d d d 2121212122332222=+-=⎰⎰i xvEI x N x v EI N x x q N x x v x N EI x x z i x x z i xx i x x i zThe shear forces and bending moments at element nodes now explicitly appear in the element equations.The above equation can be written in the matrix form }{}]{[F k =δ where the terms of thestiffness matrix are defined by4,1,d d d d d 222221==⎰j i x x N xN EI k jx x i z ijThe terms of the element force vector are defined by4,1d d d d d d d )(2121212233=+-=⎰i xvEI x N x vEI N x x q N F x x z i x x x x z i i ior,4,1)(d d )(d )(212121=++=⎰i x M xN x V N x x q N F xx i x x xx i i iwhere the integral term represents the equivalent nodal forces and moments produced by the distributed load.（6）Consider the three-node line element with interpolation functionsUse the element as the parent element in the isoparametric mapping332211)()()(x r N x r N x r N x ++=with 321x x x << but otherwise arbitrary nodal coordinates. a.b. How does the x coordinate vary between nodes of the isoparametric element?c. Has the basic element geometry changed from that of the parent element?d.e.Determine the Jacobian matrix for the transformation ，and calculate the Jacobianmatrix for the basic element with nodal coordinates 03and ,0,0.2321.x x x ==-=.f. Find the inverse of the Jacobian matrix, and calculate its value for the above basic element too.g. Calculate the value of determinate J at a point with 25.0=r .(20 points)Solution ：a. 332211)()()(x r N x r N x r N x ++=32212)2()44()132(x r r x r r x r r -+-++-=It can be seen that the x coordinate vary as a quadratic function between nodes of the isoparametric elementb. the basic element geometry may change from that of the parent element. The basic element is still a straight line element, however, its length may change. The length of the parent element is 1, and the length of the basic element is 13x x -.c. the Jocabian matrix can be written as[]⎥⎦⎤⎢⎣⎡∂∂=⎥⎦⎤⎢⎣⎡∂∂=∑=31i i i x r N r x JThat is 321)14()21(4)34(][x r x r x r J -+-+-=for the basic element with nodal coordinates 03and ,0,0.2321.x x x ==-=we have 3431268][+=-++-=r r r Jd. the inverse of the Jocabian matrix is 3211)14()21(4)34(1][x r x r x r J -+-+-=-for the basic element with nodal coordinates 03and ,0,0.2321.x x x ==-=we have 341][1+=-r Je. T he value of determinate J at a point with 25.0=r is calculated by434=+=r J。