Hutton - Fundamentals of Finite Element Analysis, страница 18

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 2004ProblemsThe system of equations to be solved for the displacements of node 4 are 5.76 1.440 U10   0 105  1.44 1.080  U11 = −5000 U120002.16and simultaneous solution yieldsU10 ⫽ 0.01736 in.U11 ⫽ ⫺0.06944 in.U12 ⫽ 0While the complete analysis is not conducted in the context of this example, the reaction forces, element strains, and element stresses would be determined by the same procedures followed in Section 3.7 for the two-dimensional case. It must be pointed out thatthe procedures required to obtain the individual element resultants are quite readilyobtained by the matrix operations described here.

Once the displacements have been calculated, the remaining (so-called) secondary variables (strain, stress, axial force) arereadily computed using the matrices and displacement interpolation functions developedin the formulation of the original displacement problem.3.9 SUMMARYThis chapter develops the complete procedure for performing a finite element analysis ofa structure and illustrates it by several examples. Although only the simple axial elementhas been used, the procedure described is common to the finite element method for allelement and analysis types, as will become clear in subsequent chapters. The direct stiffness method is by far the most straightforward technique for assembling the systemmatrices required for finite element analysis and is also very amenable to digital computerprogramming techniques.REFERENCES1. DaDeppo, D.

Introduction to Structural Mechanics and Analysis. Upper SaddleRiver, NJ: Prentice-Hall, 1999.2. Beer, F. P., and E. R. Johnston. Vector Mechanics for Engineers, Statics andDynamics, 6th ed. New York: McGraw-Hill, 1997.PROBLEMS3.1In the two-member truss shown in Figure 3.2, let ␪1 = 45 ◦ , ␪2 = 15 ◦ , andF5 = 5000 lb, F6 = 3000 lb.a. Using only static force equilibrium equations, solve for the force in eachmember as well as the reaction force components.b. Assuming each member has axial stiffness k = 52000 lb/in., compute theaxial deflection of each member.c. Using the results of part b, calculate the X and Y displacements of node 3.83Hutton: Fundamentals ofFinite Element Analysis843.

Truss Structures: TheDirect Stiffness MethodCHAPTER 33.23.33.43.5Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodCalculate the X and Y displacements of node 3 using the finite element approachand the data given in Problem 3.1. Also calculate the force in each element. Howdo your solutions compare to the results of Problem 3.1?Verify Equation 3.28 by direct multiplication of the matrices.Show that the transformed stiffness matrix for the bar element as given byEquation 3.28 is singular.Each of the bar elements depicted in Figure P3.5 has a solid circular crosssection with diameter d = 1.5 in.

The material is a low-carbon steel havingmodulus of elasticity E = 30 × 106 psi. The nodal coordinates are givenin a global (X, Y ) coordinate system (in inches). Determine the element stiffnessmatrix of each element in the global system.22(5, 30)(30, 30)2(30, 15)Y11(0, 0)X1(20, 10)(a)(0, 0)(b)(c)(⫺20, 30)2(0, 0)11(10, 10)2(d)(40, ⫺10)(e)Figure P3.53.6Repeat Problem 3.5 for the bar elements in Figure P3.6.

For these elements,d = 40 mm, E = 69 GPa, and the nodal coordinates are in meters.1(0, 0)2(0.4, 0.2)12(0.2, ⫺0.2)(0.1, 0.1)(a)Figure P3.6(b)Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 2004Problems12(0, 1.2)(3, 4)2(⫺0.3, 3)112(⫺0.5, 0)(1, 2)(0, 0)(d)(c)(e)Figure P3.6 (Continued)3.7For each of the truss structures shown in Figure P3.7, construct an elementto-global displacement correspondence table in the form of Table 3.1.63255310771413119918151117211911212448612816(a)4671086435532211(b)Figure P3.797102085Hutton: Fundamentals ofFinite Element Analysis863.

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Method3665848121151323271110947(c)69105861239211751142171338471415101116(d)232315657144(e)Figure P3.7 (Continued)3.83.9For each of the trusses of Figure P3.7, express the connectivity data for eachelement in the form of Equation 3.39.For each element shown in Figure P3.9, the global displacements have beencalculated as U1 = 0.05 in., U2 = 0.02 in., U3 = 0.075 in., U4 = 0.09 in.

Usingthe finite element equations, calculatea. Element axial displacements at each node.b. Element strain.c. Element stress.d. Element nodal forces.Do the calculated stress values agree with ␴ = F/A? Let A = 0.75 in.2,E = 10 × 106 psi, L = 40 in. for each case.Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 2004ProblemsU4U4U3U4U3110⬚U3U2U2U245⬚30⬚U1U1U1(b)(a)(c)Figure P3.93.10The plane truss shown in Figure P3.10 is subjected to a downward vertical loadat node 2. Determine via the direct stiffness method the deflection of node 2 inthe global coordinate system specified and the axial stress in each element.

Forboth elements, A = 0.5 in.2, E = 30 × 106 psi.Y(0, 0)(40, 0)X132(30, ⫺10)1500 lbFigure P3.103.11The plane truss shown in Figure P3.11 is composed of members having a square15 mm × 15 mm cross section and modulus of elasticity E = 69 GPa.a. Assemble the global stiffness matrix.b. Compute the nodal displacements in the global coordinate system for theloads shown.c. Compute the axial stress in each element.3 kN35 kN1.5 m2411.5 mFigure P3.1187Hutton: Fundamentals ofFinite Element Analysis883. Truss Structures: TheDirect Stiffness MethodCHAPTER 33.123.13Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodRepeat Problem 3.11 assuming elements 1 and 4 are removed.The cantilever truss in Figure P3.13 was constructed by a builder to support awinch and cable system (not shown) to lift and lower construction materials.

Thetruss members are nominal 2 × 4 southern yellow pine (actual dimensions1.75 in. × 3.5 in.; E = 2 × 106 psi). Using the direct stiffness method, calculatea. The global displacement components of all unconstrained nodes.b. Axial stress in each member.c. Reaction forces at constrained nodes.d. Check the equilibrium conditions.430⬚30⬚253500 lbNode XY10020 96396 96496 151.45192 96(inches)Y45⬚1XFigure P3.133.14Figure P3.14 shows a two-member plane truss supported by a linearly elasticspring.

The truss members are of a solid circular cross section having d = 20 mmand E = 80 GPa. The linear spring has stiffness constant 50 N/mm.a. Assemble the system global stiffness matrix and calculate the globaldisplacements of the unconstrained node.b. Compute the reaction forces and check the equilibrium conditions.c. Check the energy balance. Is the strain energy in balance with themechanical work of the applied force?3m15 kN50⬚4mkFigure P3.14Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 2004Problems3.153.163.17Repeat Problem 3.14 if the spring is removed.Owing to a faulty support connection, node 1 in Problem 3.13 moves 0.5 in.horizontally to the left when the load is applied. Repeat the specifiedcomputations for this condition. Does the solution change? Why or why not?Given the following system of algebraic equations   x10 −1000F 1  1  x2   F2  −10 20 −100= 0−10 20 −10  x  F  3  300−10 10x4F4and the specified conditionsx1 = 03.18x 3 = 1.5F2 = 20F4 = 35calculate x2 and x4. Do this by interchanging rows and columns such that x1 andx3 correspond to the first two rows and use the partitioned matrix approach ofEquation 3.45.Given the system   50 −5000U30 1   −50 100 −50  U2   F2 0= 0U  40 −50 75 −25   3  U44000−25 253.19and the specified condition U2 = 0.5, use the approach specified in Problem 3.17to solve for U1, U3, U4, and F2.For the truss shown in Figure P3.19, solve for the global displacementcomponents of node 3 and the stress in each element.

The elements have crosssectional area A = 1.0 in.2 and modulus of elasticity 15 × 106 psi.472 in.60⬚330⬚160⬚2Figure P3.193.20Each bar element shown in Figure P3.20 is part of a 3-D truss. The nodalcoordinates (in inches) are specified in a global (X, Y, Z) coordinate system.Given A = 2 in.2 and E = 30 × 106 psi, calculate the global stiffness matrix ofeach element.89Hutton: Fundamentals ofFinite Element Analysis903. Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Method(30, 30, ⫺20)2212(0, 80, 0)1(0, 0, 0)1(0, 0, 0)2(0, ⫺80, 0)(50, 40, 30)(0, 0, 0)1(10, 10, 0)(a)(b)(c)(d)Figure P3.203.213.22Verify Equation 3.59 via direct computation of the matrix product.Show that the axial stress in a bar element in a 3-D truss is given bydN 1␴ = Eε = Edx3.233.243.253.26dN 2dx(e)1(e)u2u1= E −L1[ R] U (e)Land note that the expression is the same as for the 2-D case.Determine the axial stress and nodal forces for each bar element shown inFigure P3.20, given that node 1 is fixed and node 2 has global displacementsU4 = U5 = U6 = 0.06 in.Use Equations 3.55 and 3.56 to express strain energy of a bar element in terms ofthe global displacements.

Apply Castigliano’s first theorem and show that theresulting global stiffness matrix is identical to that given by Equation 3.58.Repeat Problem 3.24 using the principle of minimum potential energy.Assemble the global stiffness matrix of the 3-D truss shown in Figure P3.26 andcompute the displacement components of node 4.

Also, compute the stress ineach element.Y1324ZFY ⫽ ⫺1500 lbXNode1234X004030Y000⫺20Z030025Figure P3.26 Coordinates given in inches. For eachelement E = 10 × 106 psi, A = 1.5 in.2.Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 2004C H A P T E R4Flexure Elements4.1 INTRODUCTIONThe one-dimensional, axial load-only elements discussed in Chapters 2 and 3 arequite useful in analyzing the response to load of many simple structures. However, the restriction that these elements are not capable of transmitting bendingeffects precludes their use in modeling more commonly encountered structuresthat have welded or riveted joints.