# Simply Supported Plate Problem # # References: # 1. MSC/DYNA verification manual # 2. T. Belytschko, J. Lin and C. S. Tsay, "Explicit Algorithms # for the Nonlinear Dynamics of Shells", Computational # Methods in Applied Mechanics Engineering. 42, pp 225-251, 1984 # 3. S. Timoshenko and J.N. Goodier, "Theory of Elasticity", 2nd ed., # McGraw-Hill, New York, 1951. # # Problem Description # # A square plate is simply supported on all four sides and subjected to a suddenly # applied uniform load. # # Reference Results # # The analytical result for the deflection in the centre of the plate is shown # by reference 3. # # # Elastic material law # -------------------- # Maximum deflection Period # Analytical 0.2129 in 1.070 ms # LS-Dyna 0.220 in 1.103 ms # Impact 0.2209 in 1.095 ms # # ElastioPlastic material law # -------------------- # Maximum deflection # Analytical N/A in # LS-Dyna 3 pts 0.3143 in # Impact 3 pts 0.3118 in # LS-Dyna 5 pts 0.2963 in # Impact 5 pts 0.2971 in # # # Engineering data # # Length L = 10 in # Thickness t = 0.5 in # Young modulus E = 10E7 psi # Density rho = 2.588E-4 lb-sec^2/in^4 # Poissons ratio nu = 0.3 # Yield stress yield = 30000 psi # Plastic modulus Ep = 0.107 psi # Pressure load P = 300 psi # Isotropic hardening beta = 1.0 # ANALYSIS RUN TIME CONTROL CONTROLS run from 0.00000 to 0.00120 print every 0.00010 step print tracker every 0.000010 step #------------------------------------------------------------------- # ANALYSIS MODEL MATERIAL CONFIGURATION MATERIALS OF TYPE ELASTOPLASTIC mat_1 E = 10000000.00000 RHO = 0.0002588000 NU = 0.300 YIELD_STRESS = 30000.00000 EP = 0.107 #MATERIALS OF TYPE ELASTIC #mat_1 E = 10000000.00000 RHO = 0.0002588000 NU = 0.300 #------------------------------------------------------------------- # ANALYSIS MODEL NODAL POINT CONFIGURATION NODES 1 x = 0.00000000 y = 0.00000000 z = 0.00000000 constraint = Constr_10 2 x = 1.25000000 y = 0.00000000 z = 0.00000000 constraint = Constr_13 3 x = 0.00000000 y = 1.25000000 z = 0.00000000 constraint = Constr_12 4 x = 1.25000000 y = 1.25000000 z = 0.00000000 5 x = 2.50000000 y = 0.00000000 z = 0.00000000 constraint = Constr_13 6 x = 0.00000000 y = 2.50000000 z = 0.00000000 constraint = Constr_12 7 x = 2.50000000 y = 1.25000000 z = 0.00000000 8 x = 1.25000000 y = 2.50000000 z = 0.00000000 9 x = 2.50000000 y = 2.50000000 z = 0.00000000 10 x = 3.75000000 y = 0.00000000 z = 0.00000000 constraint = Constr_13 11 x = 0.00000000 y = 3.75000000 z = 0.00000000 constraint = Constr_12 12 x = 3.75000000 y = 1.25000000 z = 0.00000000 13 x = 1.25000000 y = 3.75000000 z = 0.00000000 14 x = 3.75000000 y = 2.50000000 z = 0.00000000 15 x = 2.50000000 y = 3.75000000 z = 0.00000000 16 x = 5.00000000 y = 0.00000000 z = 0.00000000 constraint = Constr_15 17 x = 0.00000000 y = 5.00000000 z = 0.00000000 constraint = Constr_14 18 x = 5.00000000 y = 1.25000000 z = 0.00000000 constraint = Constr_11 19 x = 1.25000000 y = 5.00000000 z = 0.00000000 constraint = Constr_11 20 x = 3.75000000 y = 3.75000000 z = 0.00000000 21 x = 5.00000000 y = 2.50000000 z = 0.00000000 constraint = Constr_11 22 x = 2.50000000 y = 5.00000000 z = 0.00000000 constraint = Constr_11 23 x = 5.00000000 y = 3.75000000 z = 0.00000000 constraint = Constr_11 24 x = 3.75000000 y = 5.00000000 z = 0.00000000 constraint = Constr_11 25 x = 5.00000000 y = 5.00000000 z = 0.00000000 constraint = Constr_11 #------------------------------------------------------------------- # ANALYSIS MODEL TRIANGULAR ELEMENT CONFIGURATION #------------------------------------------------------------------- # ANALYSIS MODEL QUADRILATERAL ELEMENT CONFIGURATION ELEMENTS OF TYPE shell_BT_4 1 nodes = [2,4,3,1] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 2 nodes = [5,7,4,2] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 3 nodes = [10,12,7,5] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 4 nodes = [16,18,12,10] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 5 nodes = [4,8,6,3] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 6 nodes = [7,9,8,4] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 7 nodes = [12,14,9,7] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 8 nodes = [18,21,14,12] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 9 nodes = [8,13,11,6] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 10 nodes = [9,15,13,8] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 11 nodes = [14,20,15,9] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 12 nodes = [21,23,20,14] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 13 nodes = [13,19,17,11] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 14 nodes = [15,22,19,13] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 15 nodes = [20,24,22,15] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 16 nodes = [23,25,24,20] nip = 5 t = 0.500 material = mat_1 shear_factor = 1.000 mhc = 0.100 oophc = 0.100 rhc = 0.100 load = Load_41 #------------------------------------------------------------------- # ANALYSIS MODEL HEXAHEDRONAL ELEMENT CONFIGURATION #------------------------------------------------------------------- # ANALYSIS MODEL BOUNDARY CONDITIONS CONSTRAINTS OF TYPE boundary_condition Constr_10 ax = 0 vx = 0 ay = 0 vy = 0 arx = 0 vrx = 0 ary = 0 vry = 0 Constr_11 az = 0 vz = 0 Constr_12 ax = 0 vx = 0 ary = 0 arz = 0 vry = 0 vrz = 0 Constr_13 ay = 0 vy = 0 arx = 0 arz = 0 vrx = 0 vrz = 0 Constr_14 ax = 0 vx = 0 az = 0 vz = 0 ary = 0 arz = 0 vry = 0 vrz = 0 Constr_15 ay = 0 vy = 0 az = 0 vz = 0 arx = 0 arz = 0 vrx = 0 vrz = 0 CONSTRAINTS OF TYPE Rigid_Body #------------------------------------------------------------------- # ANALYSIS MODEL LOADS LOADS Load_41 p = 300.0 #------------------------------------------------------------------- # TRACKERS TRACKERS OF TYPE NODEDISPLACEMENT 1 node = [1] direction = z filename = Ver_04.trk target = [0.00075,0.000001,0.2971,0.0001]