3.1: 1, 3, 5, 9, 10, 12, 14, 18

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1 3.:, 3, 5, 9,,, 4, 8 ) We want to solve d d c() d = f() with c() = c = constant and f() = for different boundary conditions to get w() and u(). dw d = dw d d = ( )d w() w() = w() = w() ( ) c d d = u() u() = w() c 6c (a) Boundary conditions: u() = and w() =. u() u() = w() c 6c u() = 6c w() ( ) ] d ( ) 3 + ] ( ) 3 + ] w() = w() ( ) ( ) 3 ] w() = ( ) (b) Boundary conditions: u() = and u() =. u() u() = w() ( ) 3 + ] c 6c u() = = w() c 6c w() = 6 w() = w() ( ) u() = 6c ( ) 3 ( ) ] w() = 6 ( ) 3) Looking at the case of a bar free at both ends, dw d = f() with w() = w() =, we want to know under which conditions a solution eists. Integrating both sides of the differential equation from to gives dw d d = w() + w() = f()d f()d So, for there to be a solution to this problem, f() must satisfy the condition f()d =. Physically, this means that the net force on the bar must be zero. 5) Again we have dw d = f() and c() d = w, with boundary conditions u() = and w() =. This time, f() = f = constant and c() jumps from c = for to c = for >. So, c() = + S( ). dw d d = fd w() + w() = ( )f w() = ( )f /7/ Page of 9

2 d = w() c() = f d = + S( ) f ( )d for ( )d for < f ( )d + f f u() u() = ] for 3f 8 + f 3 8 for < u() = f ] for f ] for < 9) The solution to cu +u = is u() = d +d ep( c )+. With the boundary conditions u() = u() =, d and d can be found as follows. u() = d + d = d = d u() = d + d ep( c ) = d ( ep( c ) ) = d = d = ep( c ) Now, we can find the limit of d, d, and u() as c from the positive side. lim d = lim d = lim c + c + c + ep( c ) = lim ep( c ) c + ep( c ) = = ep( ( ) c ) c + lim u() = + lim c + d + lim d ep( c + c + c ) = + lim ep( c ) = for = lim u() = c + for + for = + for When is away from = the limit of u() as c + clearly goes to. However, near = the limit rapidly diverges from and heads to. And if much faster than c, then the limit of u() will go to. The boundary condition u() = is always satisfied but the end = has a boundary layer. ) We want to solve u = with u() = u() = using finite elements with three hat functions (φ i ()) and h = 4. So, c() = and f() =, and we need to calculate the components of the matri K and vector F in the equation KU = F where U = (U, U, U 3 ) and K ij = c() dφ j d dv i d F i = f()φ i ()d U() = U φ () + U φ () + U 3 φ 3 () The test functions v i () are taken to be the trial functions φ i (), or v i () = φ i (). This makes the K matri symmetric. The components of K and F are given below. F = F = F 3 = h = ( ) K = K = K 33 = h h = 8 K 3 = K 3 = K = K = h ( h ) = 4 K 3 = K 3 = h ( h ) = 4 Now, we can solve the matri equation KU = F to get the vector U and the finite element solution U(). 8 4 U KU = F U = U U = U 3 U 3 3 /7/ Page of 9

3 Finally, we can compare to the finite element solution to the eact solution u() =. Clearly U() cannot equal u() since u() is quadratic and U() is piecewise linear. However, we can compare the value at the mesh points: = 4, =, and = 3 4. It is shown below that the values of U() and u() are identical at the meshpoints. ( ) u(h) = u = 3 ( ) 4 6 = U u(h) = u = 4 ( ) 6 = U 3 u(3h) = u = = U 3 ) The mass matri M is defined as M ij = V i V j d. We would like to know the mass matri for the three hat functions. So V i () would be a hat function centered about = hi, where h is the spacing between mesh points. It is clear that M is symmetric, so M ij = M ji. Also, the integral of the proct of two hat functions is non-zero only if the two hat functions are the same or are adjacent; in other words, M ij = if i j >. So, M is a tri-diagonal, symmetric matri when hat functions are used. Furthermore, since the integral of the square of a hat function is identical regardless of the which hat function is used, the values along the main diagonal are all equal to the same value (calculated to be h 3 ). Also, since the integral of the proct of two adjacent hat functions is identical regardless of which two adjacent pairs of hat functions is selected, the values along the diagonal adjacent to the main diagonal are all equal to the same value (calculated to be h 6 ). For the specific case of three hat functions dividing the interval to with h = 4, we get the following 3 3 matri for M: M = ) We can use Simpson s 6, 4 6, 6 rule to integrate and find the area underneath the bubble function φ 5(). h h φ 5 ()d = h 6 φ 5(h) + 4h 6 φ 5( 3h ) + h 6 φ 5(h) = h 6 () + 4h 6 () + h 6 () = h 3 8) To find the finite element solution to a fied-free hanging bar with c() = (boundary conditions u() = and u () = ), we must add an additional half-hat function, φ N+ () at the end in addition to the N interior hat functions φ i (). The spacing between meshpoints is h = N+ and the test functions are taken to be equal to the trial functions (V i () = φ i ()). (a) The stiffness matri K would now be (N + ) (N + ) instead of N N as in the fied-fied case. The etra row and column mostly have zero entries ecept near the lower right corner. The entry in row N, column (N + ) (which is identical to the entry in row (N + ), column N because K is symmetric) would be equal to h. The entry in row (N + ), column (N + ) would be equal to h. So multiplying the column vector U with the (N + ) row of K gives the epression U N+ U N h which is a first difference of U; specifically, it is a backward difference of U at the point =, which is analogous to u (). This last row of the stiffness matri is representing the boundary condition u () =. (b) With a constant load f() = f, the new last component of the column vector F would be F N+ = hf. The 3 3 matri K and 3 column vector F from problem can be generalized for the N N case. But in addition to that, we must add the etra row and column to get the larger (N + ) (N + ) stiffness matri K and the (N +) column vector F for this fied-free case. The form for K and F (for a fied-free hanging bar with c() = and f() = f ) are given by K = h... F = hf. /7/ Page 3 of 9

4 Now, we can solve KU = F to get the column vector U and the finite element solution U() = Σ (N+) i= U i φ i (), and then compare it with the true solution of u() = f ( ). Particularly, we would like to compare the values evaluated at the mesh points: for the finite element solution this is simply given by U i ; for the true solution it is given by u(hi) = h f ( h i)(i) = h f (N + ) i](i). U = K F U.. U N U N = h f N N... N (N + ) Through some careful counting we can see that for i =,..., (N + ): U i = h f (N + i)i ] i + Σi j=j = h f (N + )i i ] (i)(i + ) i + = h f (N + )i i i + i + i ] U i = h f (N + ) i] (i) This equation is identical to the equation for u(hi), so u(hi) = U i. 3.:, 3, 5, 6. ) We want to solve the equation for a cantilevered beam with a unit force at the midpoint, u () = δ( ) with boundary conditions u() = u () = and M() = M () =. The differential equation can be split into M () = δ( ) and u () = M(), and then solved as follows. M ()d = δ( )d M () M () = S( ) M ()d = (S( ) d M() M() = ( ) (R( ) ) u ()d = R( ) + + ] d u () u () = Q( ) + + u ()d = Q( )d + ] ( + ) d u() u() = C( ) + ( ) u() = 6 (3 + 3 ) for 6 (3 + 3 ) + ( )3 for < /7/ Page 4 of 9

5 3) We want to solve u () = δ() with boundary conditions u( ) = u ( ) = and u() = u () =. The general solution to u () = δ() is given by A + B + C + D 3 u() = 6 3 for A + B + C + D 3 for We can use the boundary conditions to get four equations that can be used to solve for the four unknowns A, B, C, and D. u( ) = A B + C D + 6 = u() = A + B + C + D = u ( ) = B C + 3D = u () = B + C + 3D = A A B 3 C = 6 B C = 3 D D u() = for for 5) Which of the 8 cubic finite elements, φ d (), φ s (),..., φ d 3(), φ s 3() based at the meshpoints =, 3, 3,, are dropped because of the essential boundary conditions below for u = f? (a) Fied-fied: u() = u() =. φ d () and φ d 3() would be dropped. (b) Fied-free: u() = u () =. Only u() = is an essential boundary condition. Thus, φ d () would be dropped. 6) Which of the 8 cubic finite elements, φ d (), φ s (),..., φ d 3(), φ s 3() based at the meshpoints =, 3, 3,, are dropped because of the essential boundary conditions below for u = f? (a) Built-in beam: u = u = at both ends. φ d (), φ s (), φ d 3(), and φ s 3() would be dropped. (b) Simply supported beam: u = u = at both ends. Only u() = u() = are the essential boundary conditions. Thus, φ d () and φ d 3() would be dropped. (c) Cantilevered beam: u() = u () = u () = u () =. Only u() = u () = are the essential boundary conditions. φ d () and φ s () would be dropped. 4 8 /7/ Page 5 of 9

6 MATLAB Assignment We want to convert between a geocentric rectangular coordinate system and a geodetic coordinate system. This conversion is useful in GPS applications. The conversion from geodetic coordinates (latitude φ, longitude λ, altitude h) to geocentric rectangular coordinates (X, Y, Z) is straightforward and is given by X = (N + h) cos φ cos λ () Y = (N + h) cos φ sin λ () Z = ( f) N + h ] sin φ (3) where N = a f( f) sin φ ] (4) and a = m is the length of the semi-major ais of the Earth and f = 97 is the polar flattening of the Earth. The conversion from geocentric rectangular coordinates to geodetic coordinates is more difficult because it requires solving a system of non-linear equations. The number of equations to solve can be reced from three to two by solving for λ analytically. By substituting equations and into the quotient Y X we get tan λ = Y X (5) By squaring both sides of equations and, adding the two resulting equations, and finally taking the square root of both sides of the summed equation, we get X + Y = (N + h) cos φ cos λ + sin λ = (N + h) cos φ The above equation is used to define the function g (φ, h), and equation 3 is used to define the function g (φ, h): g (φ, h) (N + h) cos φ X + Y = (6) g (φ, h) ( f) N + h ] sin φ Z = (7) where N is given in equation 4. Now, we need to calculate the Jacobian, g J = φ g φ g h g h The partial derivatives with respect to altitude h are easy to calculate. They are simply g h = cos φ and N = sin φ. To simplify calculating the partial derivatives with respect to φ, we will first calculate g h N φ = a f( f) sin φ ] 3 ( f( f) sin φ cos φ) = N 3 f( f) sin φ cos φ a φ. g φ = N 3 N cos φ (N + h) sin φ = φ a f( f) sin φ cos φ N sin φ h sin φ = N 3 ] f( a sin φ f) cos φ a N h sin φ = N 3 a = N 3 a sin φ f( f) cos φ + f( f) sin φ ] h sin φ sin φ + f f ] N 3 h sin φ = a ( f) + h sin φ /7/ Page 6 of 9

7 g φ N = ( f) φ sin φ + ( f) N + h ] cos φ N = ( f) 3 a f( f) sin φ cos φ + ( f) N cos φ + h cos φ = ( f) N 3 ] f( cos φ f) sin φ + a + h cos φ a = ( f) N 3 a cos φ f( f) sin φ + f( f) sin φ ] + h cos φ = ( f) N 3 N 3 cos φ + h cos φ = a a ( f) + h cos φ If we define D N 3 a ( f) + h, then the Jacobian can be simply written as D sin φ cos φ J = D cos φ sin φ and the inverse of the Jacobian is J sin φ cos φ sin φ cos φ = = det(j) D cos φ D sin φ D sin φ D cos φ D cos φ D sin φ = sin φ cos φ = D sin φ D cos φ ] D D cos φ D sin φ cos φ sin φ Now, we can find the solution vector u = (φ, h) using Newton s Method. We start with an initial guess vector u init and set the current vector u cur equal to it. For each iteration in the loop, we compute a new vector u new through the computation u new = u cur J g or φnew h new ] = φcur h cur N ] D sin φ cur cos φ cur D cos φ ] cur g (φ cur, h cur ) sin φ cur g (φ cur, h cur ) (8) where D = N 3 a ( f) + h (9) and the functions g and g are defined in equations 6 and 7 respectively, ecept the functions would of course not equal zero when evaluated at the current u cur unless u cur happened to eactly be the true solution. Every time the loop is repeated, u new of the prior loop becomes u cur of the current loop. The new calculated vector u new should be closer to the actual solution. Eventually, u new should be close enough to the actual solution, within some tolerance, that we can break out of the loop and treat u new as the solution. The MATLAB function that does this conversion essentially uses equations to 4 to convert the current estimate for the geodetic coordinates into geocentric rectangular coordinates (X cur, Y cur, Z cur ), and then it calculates the Euclidean distance between this point and the point (X, Y, Z) specified by the user. If this distance is smaller than some threshold distance, which can be specified by the user through an optional argument in the MATLAB function, it will break out of the loop. Newton s method requires an initial guess to start from; however, testing showed that regardless of where the initial guess was, the unique correct solution was quickly obtained in a few iterations. Therefore, the MATLAB function makes specifying the initial guess of the latitude and altitude optional, and by default it sets both to zero (at the equator with zero altitude). Finally, the longitude is eplicitly calculated from X and Y alone, using equation 5. The code for the MATLAB function, gps, that converts from geocentric rectangular to geodetic coordinates is displayed in Listing. /7/ Page 7 of 9

8 We can look at an eample where we wish to convert the geocentric rectangular coordinates (X, Y, Z) to geodetic coordinates latitude, longitude, and altitude. X, Y, and Z are specified as follows: X = Y = Z = m m m Then the following MATLAB code can be used to call the function and get the converted coordinates: phi, lambda, h] = gps( , , ) The variable h has the altitude (h) in meters, the variable phi has the latitude (φ) in degrees, and the variable lambda has the longitude (λ) in degrees. The values for these three variables, for this eample, are φ 57.3 λ 9.95 h m We can verify the results of the function by plugging these values into equations to 4 and getting the original specified values for X, Y, and Z. Listing : MATLAB function that converts from geocentral rectangular to geodetic coordinates. function l a t, long, a l t ] = gps ( X, Y, Z, v a r a r g i n ) %GPS Converts from g e o c e n t r i c r e c t a n g u l a r to g e o d e t i c c o o r d i n a t e s. 3 % l a t, long, a l t ] = gps (X, Y, Z, i n i t l a t =, i n i t a l t =,... 4 % t o l = e 6, ma iter = ) 5 % c o n v e r t s from g e o c e n t r i c r e c t a n g u l a r c o o r d i n a t e s (X, Y, Z) in meters to 6 % g e o d e t i c c o o r d i n a t e s : l a t i t u d e ( l a t ) and l o n g i t u d e ( long ) in d e g r e e s 7 % and a l t i t u d e ( a l t ) in meters. The conversion i s done by s o l v i n g a 8 % system o f non l i n e a r e q u a t i o n s using Newton s method, so an i n i t i a l 9 % guess f o r t he l a t i t u d e ( i n i t l a t ) in d e g r e e s and the a l t i t u d e % ( i n i t a l t ) in meters i s necessary. By d e f a u l t the i n i t i a l l a t i t u d e i s % s e t to d e g r e e s ( equator ) and the i n i t i a l a l t i t u d e i s s e t to m. % 3 % The t o l e r a n c e ( t o l ), which s p e c i f i e s the maimum a c c e p t i b l e d i s t a n c e in 4 % meters between the s p e c i f i e d (X, Y, Z) p o i n t and the e s t i m a t e d p o i n t 5 % g iven by the c a l c u l a t e d l a t i t u d e, l o n g i t u d e, a l t i t u d e, i s s e t to e 6 m 6 % by d e f a u l t. The maimum number o f i t e r a t i o n s ( ma iter ) b e f o r e a b o r t i n g 7 % Newton s method and p r i n t i n g an e r r o r message i s s e t to by d e f a u l t. 8 9 % Constants f o r Earth a = ; % m] l e n g t h o f semi major a i s o f Earth f = /97; % p o l a r f l a t t e n i n g 3 % Set d e f a u l t v a l u e s f o r t o l e r a n c e and ma iter i f not s p e c i f i e d. 4 numvarargs = length ( v a r a r g i n ) ; 5 i f ( numvarargs > 4) 6 error ( Requires at most 4 o p t i o n a l inputs ) ; 7 end 8 optargs =,, e 6, }; % Set d e f a u l t s 9 newvals = c e l l f u n (@( ) isempty ( ), v a r a r g i n ) ; 3 optargs ( newvals ) = v a r a r g i n ( newvals ) ; 3 i n i t l a t, i n i t a l t, t o l, ma iter ] = optargs : } ; 3 /7/ Page 8 of 9

9 33 % Newton s Method 34 u = degtorad ( i n i t l a t ) ; i n i t a l t ] ; 35 g r i g h t = sqrt (X X + Y Y) ; Z ] ; 36 g right norm = norm( g r i g h t ) ; % g r i g h t n o r m = s q r t (X X + Y Y + Z Z ) ; 37 for i =: ma iter ] 38 phi = u ( ) ; h = u ( ) ; 39 N = a ( f ( f ) ( sin ( phi )ˆ))ˆ( /); 4 D = h + ( f )ˆ Nˆ3 / a ˆ ; 4 g l e f t = (N+h ) cos ( phi ) ; (N ( f )ˆ + h ) sin ( phi ) ] ; 4 g = g l e f t g r i g h t ; 43 u = u sin ( phi )/D, cos ( phi )/D ; cos ( phi ), sin ( phi ) ] g ; % C a l c u l a t e c u r r e n t d i s t a n c e from d e s i r e d p o i n t ( e r r o r ) and check i f i t 46 % i s w i t h i n t he s p e c i f i e d t o l e r a n c e. 47 i f ( abs (norm( g l e f t ) g right norm ) < t o l ) 48 break ; % Tolerance s a t i s f i e d, break out o f loop. 49 end 5 end 5 5 % Check i f loop aborted because o f maimum i t e r a t i o n s 53 i f ( i >= ma iter ) 54 error ( gps : maiteration, Eceeded maimum number o f i t e r a t i o n s (%d ) without f i n d i n g, a s o l u t i o n within the t o l e r a n c e ], ma iter ) ; 57 end % Set output v a r i a b l e s 6 long = radtodeg ( atan (Y, X) ) ; 6 l a t = radtodeg ( phi ) ; 6 a l t = h ; 63 end /7/ Page 9 of 9