Question

X=0 x = 1/2 x= L u U2 Uz (a) Trial solution for a 1-D quadratic elastic bar element can be written as follows: ū(x) = [N]{u} where, [N] = [N1 N2 N3] and {u} u2 13 1 and Ni L2 L2 [N] and {u} are known as interpolation function matrix and nodal displacement, respectively. (272 – 3L + L´), N= = (22- La), Ns = 12 (2=– LE) Derive the expression for element stiffness matrix, (Kelem) and element force vector {F} using following expression and exact integration procedure. dN NT dr Here, (B) is known as strain-displacement matrix. (b) Use finite element formulation derived in (a) to generate numerical values of element stiffness matrix and force vector for the following bar with uniformly distributed load. The bar is fixed at r = 0 and a point load, P = 10 kN/m is applied. First, solve for displacement for the problem using 1-element model. 9(x) = 5 kN/m E = 120 Gpa PA = 1x10-3 m² L = 1 m (c) Now repeat the problem using uniformly meshed 2-element model as shown below. Generate element stiffness matrix and force vector for both elements, assemble them and find the nodal displacement. Compare displacement solution from (b) and (c) by plotting against the length (you can use MATLAB or MS Excel to generate the plot). ez u uz us

Answer 1

. AN dn dN2 dN INI dre 4e3L -4 (2-2) + (an-L) L2 [ Kalimi 3 J ETE AB de BTB dac CEA 0 da 43L 40-3 L 는 는 (-) - EA D (n-1) 눈(m EA 13. L I do I 여에 의에에에 You olmalm

Boundary conditions: e=0 16 So REA o 1046 3 L uz 3 3 3 un - 0 - 8/3 16½-½ ²-8/7/ 546 10416 54/12+ P us o Solving, Uz 0.0308 0.0590 uz = 10PX 0.0846 'un 0.1076 ug uz -- [143[MM] va is en {N; N. Ng} for e { N / N 3 Ng} for ez

x = linspace(0,1,11)';
L = 1;
N1 = [(2*x.^2-3*L*x+L^2)/L^2 -4*(x.^2-L*x)/L^2 (2*x.^2-L*x)/L^2];
u1 = 1.0e-03 *[ 0 ; 0.0573; 0.1042];
u = N1*u1;
plot(x,u)
hold on

x = linspace(0,1/2,6)';
L = 1/2;
N1 = [(2*x.^2-3*L*x+L^2)/L^2 -4*(x.^2-L*x)/L^2 (2*x.^2-L*x)/L^2];
u1 = 1.0e-03 *[0; 0.0308; 0.0590];
u1 = N1*u1;
x = linspace(0,1/2,6)';
L = 1/2;
N1 = [(2*x.^2-3*L*x+L^2)/L^2 -4*(x.^2-L*x)/L^2 (2*x.^2-L*x)/L^2];
u2 = 1.0e-03 *[ 0.0590; 0.0846; 0.1076];
u2 = N1*u2;
u = [u1(1:end-1);u2];
plot( linspace(0,1,11),u)
legend('1 element','2 elements')

*10-4 1.2 1 element 2 elements 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1