Hutton - Fundamentals of Finite Element Analysis (523155), страница 70
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Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.9 TorsionSince nodes 2, 3, and 4 are on the outside surface, we set 2 = 3 = 4 = 0 to obtain thesolution4G A31 =We still have not addressed the problem that the angle of twist per unit length is unknownand continue to ignore that problem temporarily, since for this very simple two-elementmodel, we can continue with the hand solution. The torque is given by Equation 9.149 asT =2 dAAbut the integration is over the entire cross section.
Hence, we must sum the contributionfrom each element and, in this case, since we applied the symmetry conditions to reducethe model to one-fourth size, multiply the result by 4. For each element, the torquecontribution isT (e) = 2 N (e) d A (e)Aand for the linear triangular element this simply becomesT (e) =2 A (e)(e)(e) 1 + 2 + 33Accounting for the use of symmetry, the total torque indicated by our two-element solution isT =4642 A (1)(2) 1 + 2 =G A 239orT64=G A29Now we address the problem of unknown angle of twist per unit length.
Noting, in the lastequation, the ratio is constant for specified shear modulus and cross-sectional area, wecould simply specify an arbitrary value of , follow the solution procedure to computethe corresponding torque, compute the ratio, and scale the result as needed. If, for example, we had assumed = 10 −6 rad/mm, our result would beT =64(80)(10 3 )(10 −6 )(312.5) 2 = 55555.6 N·mm ⇒ 55.6 N·m9Thus, to answer the original question, we compute the angle of twist per unit length corresponding to the specified torque as=100(10 −6 ) ≈ 1.8(10 −6 ) rad/mm55.6381Hutton: Fundamentals ofFinite Element Analysis3829.
Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsand the total angle of twist would be = L = 1.8(10 −6 )(1000 ) = 1.8(10 −3 ) rador about 0.1 degree. The exact solution [5] for this problem shows the angle of twist perunit length to be = 1.42(10 −6 ) rad/mm . Hence, our very simple model is in error byabout 27 percent.9.10 SUMMARYIn this chapter, we present the development of the most basic finite elements usedin stress analysis in solid mechanics.
As the finite element method was originallydeveloped for stress analysis, the range of element and problem types that can beanalyzed by the method are very large. Our description of the basic concepts isintended to give the reader insight on the general procedures used to developelement equations and understand the ramifications on element, hence, model,formulation for various states of stress. As mentioned in the context of platebending in Section 9.8, finite element analysis involves many advanced topics inengineering not generally covered in an undergraduate program.
The interestedreader is referred to the many advanced-level texts on the finite element methodfor further study. The intent here is to introduce the basic concepts and generateinterest in learning more of the subject.REFERENCES1. Kreyszig, E. Advanced Engineering Mathematics, 5th ed. New York: John Wileyand Sons, 1983.2. Shigley, J. E. Mechanical Engineering Design, 6th ed. New York: McGraw-Hill,1998.3. Cowper, G.
R. “Gaussian Quadrature Formulas for Triangles.” InternationalJournal for Numerical Methods in Engineering 7 (1973).4. Huebner, K. H., and E. A. Thornton. The Finite Element Method for Engineers,2nd ed. New York: John Wiley and Sons, 1982.5. Timoshenko, S. P., and J. N. Goodier. Theory of Elasticity, New York: McGrawHill, 1970.PROBLEMS9.19.2Use the general stress-strain relations from Appendix B and the assumptions ofplane stress to derive Equations 9.2.Let {z} be the N × 1 column matrix [z 1 z 2 z 3 · · · z N ]T and let [ A] be an N ×N real-valued matrix. Show that the matrix product {z} T [ A]{z} always results ina scalar, quadratic function of the components zi.Hutton: Fundamentals ofFinite Element Analysis9.
Applications in SolidMechanicsText© The McGraw−HillCompanies, 2004Problems9.39.49.59.69.7Beginning with the general elastic stress-strain relations, derive Equation 9.50for the conditions of plane strain.Determine the strain-displacement matrix [B] for a three-node triangular elementin plane strain.Determine the strain-displacement matrix [B] for a four-node rectangular elementin plane stress.Using the interpolation functions given in Equation 9.28, determine theexplicit expression for the strain energy in a three-node triangular element inplane stress.The constant strain triangular element shown in Figure P9.7 is subjected toa uniformly distributed pressure as shown. Determine the equivalent nodalforces.3 (0.5, 2)p02 (1, 1)1 (0, 0)Figure P9.79.8The constant strain triangular element shown in Figure P9.8 is subjectedto the linearly varying pressure as shown and a body force from gravity(g = 386.4 in./s2) in the negative y direction.
Determine the equivalentnodal forces.2(0.5, 1)3(⫺0.1, 0.8)p01 (0, 0)Figure P9.89.9The element of Figure P9.7 is of a material for which the modulus of elasticity isE = 15 × 106 psi and Poisson’s ratio is = 0.3 . Determine the element stiffnessmatrix if the element is subjected to plane stress.383Hutton: Fundamentals ofFinite Element Analysis3849. Applications in SolidMechanicsCHAPTER 99.109.119.129.139.14Text© The McGraw−HillCompanies, 2004Applications in Solid MechanicsRepeat Problem 9.9 for plane strain.Repeat Problem 9.9 for an axisymmetric element.The overall loading of the element in Problem 9.7 is such that the nodaldisplacements are u1 = 0.003 in., v1 = 0, u2 = 0.001 in., v2 = 0.0005 in.,u3 = 0.0015 in., v3 = 0.
Calculate the element strain, stress, and strain energyassuming plane stress conditions.Repeat Problem 9.13 for plane strain conditions.A thin plate of unit thickness is supported and loaded as shown in Figure P9.14.The material is steel, for which E = 30 × 106 and = 0.3 . Using the fourconstant strain triangular elements shown by the dashed lines, compute thedeflection of point A. Compare the total strain energy with the work of theexternal force system.p0 ⫽ 300 psi20 in.30 in.Figure P9.149.159.16Integrate Equation 9.65 by the Gaussian numerical procedure to verifyEquation 9.66.A three-node triangular element having nodal coordinates, shown inFigure P9.16, is to be used as an axisymmetric element.
The material propertiesare E = 82 GPa and = 0.3 . The dimensions are in millimeters. Calculate theelement stiffness matrix using both the exact definition of Equation 9.99 andthe centroidal approximation of Equation 9.100. Are the results significantlydifferent?3(15, 10)1(10, 0)2(20, ⫺10)Figure P9.16Hutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 2004Problems9.179.189.19The axisymmetric element in Figure P9.16 is subjected to a uniform, normalpressure p0 acting on the surface defined by nodes 1 and 3.
Compute theequivalent nodal forces.The axisymmetric element in Figure P9.16 is part of a body rotating aboutthe z axis at a constant rate of 3600 revolutions per minute. Determine thecorresponding nodal forces.Consider the higher-order three-dimensional element shown in Figure P9.19,which is assumed to be subjected to a general state of stress.
The element has20 nodes, but all nodes are not shown for clarity.a. What is the order of the polynomial used for interpolation functions?b. How will the strains (therefore, stresses) vary with position in the element?c. What is the size of the stiffness matrix?d. What advantages and disadvantages are apparent in using this element incomparison to an eight-node brick element?Figure P9.199.209.219.229.239.249.25Show that in a uniaxial tension test the distortion energy at yielding is given byEquation 9.128.A finite element analysis of a certain component yields the maximum principalstresses 1 = 200 MPa, 2 = 0, 3 = −90 MPa.
If the tensile strength of thematerial is 270 MPa, is yielding indicated according to the distortion energytheory? If not, what is the “safety factor” (ratio of yield strength to equivalentstress)?Repeat Problem 9.21 if the applicable failure theory is the maximum shear stresstheory.The torsion problem as developed in Section 9.9 has a governing equationanalogous to that of two-dimensional heat conduction.
The stress function isanalogous to temperature, and the angle of twist per unit length term (2G ) isanalogous to internal heat generation.a. What heat transfer quantities are analogous to the shear stress componentsin the torsion problem?b. If one solved a torsion problem using finite element software for twodimensional heat transfer, how would the torque be computed?The torsion problem as developed in Section 9.9 is two-dimensional when posedin terms of the Prandtl stress function. Could three-dimensional elastic solidelements (such as the eight-node brick element) be used to model the torsionproblem? If yes, how would a pure torsional loading be applied?Figure P9.25 shows the cross section of a hexagonal shaft used in a quick-changepower transmission coupling.
The shear modulus of the material is 12 × 106 psi385Hutton: Fundamentals ofFinite Element Analysis3869. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicszy0.875 in.Figure P9.25and the shaft length is 12 in. Determine the total angle of twist when the shaftis subjected to a net torque of 250 ft-lb. Use linear triangular elements and takeadvantage of all appropriate symmetry conditions.Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 2004C H A P T E R10Structural Dynamics10.1 INTRODUCTIONIn addition to static analyses, the finite element method is a powerful tool foranalyzing the dynamic response of structures. As illustrated in Chapter 7, thefinite element method in combination with the finite difference method can beused to examine the transient response of heat transfer situations.
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