December 10, PROBLEM NO points max.

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1 PROBLEM NO points max.

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5 PROBLEM NO points max. B 3A A C D A H k P L 2L Given: Consider the structure above that is made up of rod segments BC and DH, a spring of stiffness k and rigid connectors C and D. The two rod segments are made up of the same material having a Young s modulus of E. Segment BC has a tapered cross section, with the cross-sectional area varying linearly from 3A at B to A at C. Segment DH has a constant cross-sectional area of A over its length. The spring joining connectors C and D has a stiffness of k = 4EA / L. A load of P acts to the right on connector C. It is desired to set up and solve a finite element model of the system, with segments BC and DH each being represented by a single element. This model will have four nodes: B, C, D and H. Find: For this problem, a) Draw a free body diagram of the entire system (BC, DH and the spring). Include all forces acting on the system, including both applied and reaction forces. b) Write down the stiffnesses of segments BC and DH in terms of E, A and L. c) Write down the (4x4) stiffness matrix [K] for the system corresponding to the axial displacements of nodes B, C, D and H. d) Write down the forcing vector {F}, where {F} has a length of 4. e) Enforce the boundary conditions on [K] and {F}. f) Solve for the displacements of nodes C and D. g) Determine the stress in element DH. F B P F H

6 k 1 = 1 2 E ( L 3A + A ) = 2 EA L k 2 = k = 4 EA L k 3 = 1 EA 2 L Therefore, K = EA L / 2 1/ / 2 1/ 2 F B P { F} = 0 F H Enforcing the boundary conditions ( u C = u D = 0 ) gives: or: K { F} = = EA L P / 2 6u C 4u D = PL EA 4u C u D = 0 u C = 9 8 u D Therefore: u D 4u D = PL EA u D = 4 PL 11 EA u C = 9 PL 22 EA Load carried by DH: F DH = k 3 u D = 1 EA 4 PL 2 L 11 EA = 2 11 P ( compression ) Stress in element DH: σ DH = F DH A = 2 P ( 11 A compression )

7 PROBLEM NO points max. Given: The member shown below is subject to a distributed load with a magnitude P. It is known that the member has a uniform cross section area A, second area moment I, and a shear shape factor f!. The member is composed of a material whose Young s modulus is E and shear modulus G. Find: Using Castigliano s theorem, determine the slope and vertical deflection at the free end O of the member. Please include the shear, normal and bending shear strain energies in your solution. Leave your answer in terms of, at most, the known parameters of w, L, H, E, G, A, I, and f!. NOTE: For full credit, you need to draw all necessary free body diagrams and internal cuts, and write down all relevant equilibrium equations.

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9 restart; N d 0; dndmd d diff N, Md ; dndfd d diff N, Fd ; N d 0 dndmd d 0 dndfd d 0 V d P$x CFd; dvdmd d diff V, Md ; dvdfd d diff V, Fd ; V d P x CFd dvdmd d 0 dvdfd d 1 M d MdK P$x2 2 KFd$x; dmdmd d diff M, Md ; dmdfd d diff M, Fd ; M d MdK 1 2 P x2 KFd x dmdmd d 1 dmdfd dkx N1 d P$LCFd; dn1dmd d diff N1, Md ; dn1dfd d diff N1, Fd ; N1 d P LCFd dn1dmd d 0 dn1dfd d 1 V1 d 0; dv1dmd d diff V1, Md ; dv1dfd d diff V1, Fd ; V1 d 0 dv1dmd d 0 dv1dfd d 0 M1 d MdK P$L2 2 Md d 0; Fd d 0; KFd$L; dm1dmd d diff M1, Md ; dm1dfd d diff M1, Fd ; M1 d MdK 1 2 P L2 KFd L dm1dmd d 1 dm1dfd dkl Md d 0 Fd d 0 deflmdx d 1 fs $int M$dMdMd, x = 0..L C EI AG $dndmd, x = 0..L ; deflmdx dk P L3 6 EI deflmdy d 1 fs $int M1$dM1dMd, y = 0..H C EI AG $dn1dmd, y = 0..H ; deflmdy dk P L2 H 2 EI $int V$dVdMd, x = 0..L C 1 AE $int N 1 $int V1$dV1dMd, y = 0..H C $int N1 AE (1) (2) (3) (4) (5) (6) (7) (8) (9)

10 delfmd d deflmdx CdeflMdy; delfmd dk P L3 6 EI K P L2 H 2 EI (10) deflfdx d 1 EI = 0..L ; fs 1 $int M$dMdFd, x = 0..L C $int V$dVdFd, x = 0..L C $int N$dNdFd, x AG AE deflfdx d P L4 8 EI deflfdy d 1 fs $int M1$dM1dFd, y = 0..H C EI AG $dn1dfd, y = 0..H ; deflfdy d P L3 H 2 EI delffd d deflfdx CdeflFdy; delffd d P L4 8 EI C fs P L2 2 AG 1 $int V1$dV1dFd, y = 0..H C $int N1 AE C P L H AE fs P L2 C 2 AG C P L3 H 2 EI C P L H AE (11) (12) (13)

11 PROBLEM NO. 4 - PART A: 10 points max. A slab is brought to failure subjected to the loading configuration depicted in the following figure. P P L L L Circle the correct answer in the following statements: (a) The internal resultant for cross section between the two applied loads correspond to: (i) An unloaded configuration (ii) Pure bending (iii) Axial loading (iv) Pure shear (v) A combined loading condition comprised of shear force and bending moment (b) For any cross section between the two loads, the state of stress of a point on the top face of the slab is represented by the following Mohr s circle: (i) (ii) (iii) (iv) (c) For any cross section between the two loads, the state of stress of a point on the bottom face of the specimen is represented by the following Mohr s circle: (i) (ii) (iii) (iv)

12 PROBLEM NO. 4 - PART A (continued) (d) If the slab is made of steel, a ductile material, then failure will occur first between the two loads and on: (i) the top face of the specimen at a cross section between the two loads (ii) the bottom face of the specimen at a cross section between the two loads (iii) the top and bottom faces of the specimen simultaneously (iv) the neutral surface of the specimen (e) If the specimen is made of concrete, a brittle material with an ultimate compressive strength larger than the ultimate tensile strength, then failure will occur first between the two loads and on: (i) the top face of the specimen at a cross section between the two loads (ii) the bottom face of the specimen at a cross section between the two loads (iii) the top and bottom faces of the specimen simultaneously (iv) the neutral surface of the specimen

13 PROBLEM NO. 4 - PART B: 9 points max. P P Side view Front view cross section y x 25t t 4t A column of height 25t and rectangular cross section of dimensions 4t by t is subject to a compressive load P. The column is made of a material with Young s modulus E and yield strength σ! = E/100. Determine the critical buckling load following the steps below. (a) The lowest buckling load corresponds to a [circle the correct answer]: (i) Pinned-pinned column (ii) Pinned-fixed column (iii) Fixed-fixed column (iv) Pinned-free column

14 PROBLEM NO. 4 - PART B (continued) (b) Determine the cross-sectional area, the second area moments with respect to the x- and y-axes of the column. A = 4t 2 I yy = 4t I xx = t ( )( t) 3 12 ( )( 4t) 3 12 = t4 3 = 16t4 3 Notice that buckling about the y-axis will require a lower critical load than the buckling about the x-axis. (c) The column buckles following: (i) Euler s theory (ii) Johnson s theory.[circle the correct statement] S r = L eff r S r = ( ) c = L eff r L eff I yy / A = 0.7L t 4 / 3 c ( ) / 4t 2 = = π 2E σ Y = 10 2π = 44.4 Since S r ( S r ) Euler theory to be used c (d) Based on the above assessment, determine the critical buckling load of the column, P cr. P cr = π 2 EI yy 2 L eff = Et 2

15 PROBLEM NO. 4 - PART C: 6 points max. Fill in the blanks with words. Please do not use any symbols or numbers. (a) One can define only independent elastic properties to characterize a material. For example, if the and the are given, then the Young's modulus Poisson'sratio shearmodules two is determined as! =!/2(1 +!). (b) The mechanical design of any mechanical members must the structure from buckling and from plastic deformations or sudden fracture, depending on whether the material is or, respectively. and failure modes occur at a critical point in the structure, and they are determined from the at such critical point. In contrast, buckling is greatly affected by the overall of the structure. dimensions (c) The internal loads at a certain cross section of a three-dimensional deformable solid are conveniently decomposed into internal resultants, namely and. prevent ductile brittle Ductile brittle state ofstress geometry six axialforce twoshear forces two bending moments torque (d) For pure, Euler-Bernoulli beam theory assumes that cross sections remain plane bending and to the deflection curve of the deformed beam. In addition, it assumes that the remains free of as the beam deforms. perpendicular neutralsurface strain