*HEADING ** *AQUA and *WIND loading tests for PIPE32 elements *NODE,NSET=ALL 1,10.,0., 10. 2,15.,0.,15. 3,20.,0., 20. *ELEMENT,TYPE=PIPE32,ELSET=ALL 1,1,2,3 *BEAM SECTION,SECTION=PIPE,MATERIAL=A1, ELSET=ALL 1.0,0.05 *MATERIAL,NAME=A1 *ELASTIC 30.0E9,0.3 **A negligible density is added to allow dynamic steps. *DENSITY 1E-10, *AMPLITUDE, NAME=TEN 0.,10., 1.,10. *AMPLITUDE, name= A1_5 0., 1.5, 1., 1.5 *AMPLITUDE,NAME=ZERO 0.,0., 1.,0. ********************* Node sets for loading and B.C.s *NSET,NSET=support all, *NSET,nset=twist 1, *Nset,nset=end1 1, *nset, nset=end2 3, ** **INCLUDE,INPUT=aqua3d_loads.inp ** ********************* ** Aqua loads are tested. The beam ** is kept straight and constrained, and is subject to ** buoyancy and different drag loads in different steps. For this ** simple case of a straight beam the correct total drag force ** can be calculated easily. This is measured by checking the ** reaction force at the beam nodes. ** ** i and j represent unit vectors in the 1- and z- ** directions respectively, where z represents the "upward" ** direction, i.e. the 2-direction in a 2D model and the 3-direction in ** a 3D model. Vectors vf, vp, Delta-v, ** n, and t, ap and apn have the meanings given in the ** user's manual for Aqua loads. ** ** Beam length = 10 root2 ** Beam tangent, t = (i + j)/root2 ** ** grav=32.2, rho_water=1.987. Steady-current velocity is ** 2 i + 1 j, and is scaled up in some of the steps. ** ** The reference height for wind is 10., and the exponent for ** its profile is 0.2. The velocity at the reference height is ** 36.844; this produces a velocity of 40 i at ** a height of 15. ** ** For effective section force output, pressure is evaluated at z ** coordinate of the section point. External pressure at this point is ** rho_water*grav*(40-z_section) for static case. For all steps, ** ESF1=SF1 except for Step A:1 (PB load). ** *BOUNDARY support,1,3 support,4,6 *AQUA 0.,40.0,32.2,1.987 2., 0., 1. , 0. 2., 0., 1., 2000. *WIND 1.987E-2, 10., 36.844, 0., 1., 0., 0.2 ** rho, Z0, c_x, c_y, dx, dy, alpha ** ------------------------------------------- ** Buoyancy test in Static Step ** ------------------------------------------- *STEP,NLGEOM Step A:1: Buoyancy, PB, closed end loading conditions. Total RF of -1983.8 k distributed over supporting nodes. Hoop stress S22=1.86E6 (Pipe elements only). Effective axial force: Two integration point cases: ESF1_1 = SF1_1 + 7.829e3 - 2.842e5 = SF1_1 - 2.764e5 ESF1_2 = SF1_2 + 4.891e3 - 2.829e5 = SF1_2 - 2.780e6 One integration point cases: ESF1_1 = SF1_1 + 6.360e3 - 2.836e5 = SF1_1 - 2.772e5 *STATIC 0.5,1. *DLOAD ALL,PB,,2.25,1.25,1.9,2500. *EL PRINT,ELSET=ALL LOADS, S, SF1, ESF1 *ENERGY FILE *NODE FILE RF, *EL FILE,ELSET=ALL LOADS, S, SF, ESF1 *NODE PRINT,totals=yes RF, *END STEP ** Intermediate step; clear buoyancy load ** Without this, the buoyancy load will be ** ramped down in the drag load step. *STEP,NLGEOM Intermediate step. *STATIC 0.1,0.1 *ENERGY FILE, F=0 *NODE FILE,F=0 RF, *EL FILE,ELSET=ALL,F=0 LOADS, *DLOAD,OP=NEW *END STEP ** ------------------------------------------- ** Normal Drag test in Static Step ** ------------------------------------------- *STEP,NLGEOM Step B:3: Normal drag, static, FDD. Total RF = -9000 i + 9000 j distributed over supporting nodes. *STATIC .1,.2 ** ** ** vf = Delta-v = 20 i + 10 j ** n = (i - j)/root2 ** 1/2 rho Cd D = 9.0 ** ** Testing: that a value of 2.0 is applied as a magnitude scaling factor ** to the drag force. (amag0=2.), and also that the force is ramped up ** over the step (there are two increments). ** ** Testing: that the steady-current velocity is being scaled up by a ** factor of 10. using the amplitude definition TEN. ** ** Testing: that the structural velocity prescribed here ** is not taken into account when determinining ** Delta-v, since the step is a ** static one. ** *BOUNDARY,type=velocity support,1,1,10.0 support,3,3,-20.0 *DLOAD,OP=NEW all, fdd, 2.0, 7.5491, 1.2, 0.8, ten **el,FDD, amag0, D, Cd, Alphr, Ampc, Ampw *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ------------------------------------------- ** Tangential Drag test in Static Step ** ------------------------------------------- ** *STEP,NLGEOM Step C:4: Tangential Drag, Static, FDT. Total RF = -4929.3 i - 4929.3 j distributed over supporting nodes. *STATIC .1,.1 ** ** ** vf = Delta-v = 20 i + 10 j ** t = (i + j)/root2 ** 1/2 rho pi Ct D = 0.9 ** ** Two drag loads are applied simultaneously on ** the beam for testing purposes. ** 1st load ** h (exponent) = 1.5 ** 2nd load ** h (exponent) = 2.0 (default) ** ** Testing: that the exponent on the tangential drag is ** used correctly and that it defaults to 2. ** ** *BOUNDARY,type=velocity support,1,1,10.0 support,3,3,-20.0 *DLOAD,OP=NEW all, fdt, , 7.5491, 0.0381972, 0.8, 1.5, ten all, fdt, , 7.5491, 0.0381972, 0.8, , ten **el,FDT, amag0, D, Ct, Alphr, h, Ampc, Ampw *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ------------------------------------------------------- ** Section 2 --- DYNAMIC DRAG ** ------------------------------------------------------- ** ** ** ------------------------------------------- ** Normal Drag test in Dynamic Step ** ------------------------------------------- *STEP,NLGEOM Step D:5: Normal drag, dynamic, FDD. Total RF = 9000 i - 9000 j distributed over supporting nodes. *DYNAMIC .1,.1 ** In this dynamic step the velocity of the structure ** is taken into account, and, with the alpha parameter of 0.8, ** the value of effective relative velocity is now ** ** vf = 20 i + 10 j ** vp = 12.5 i - 12.5 j, alpha=0.8 ** ... Delta-v = 10 i + 20 j ** ** ** n = (i - j)/root2 ** 1/2 rho Cd D = 9.0 ** ** Testing: that the structural velocity prescribed here ** is being taken into account when determinining ** Delta-v, and that the parameter alpha is being used correctly. ** ** ** As before: the steady-current velocity is being scaled up by a ** factor of 10. using the amplitude definition TEN. ** ** *BOUNDARY,type=velocity support,1,1,12.5 support,3,3,-12.5 *DLOAD,OP=NEW all, fdd, 2.0, 7.5491, 1.2, 0.8, ten **el,FDD, amag0, D, Cd, Alphr, Ampc, Ampw *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ** ** ------------------------------------------- ** Tangential Drag test in DYNAMIC Step ** ------------------------------------------- ** *STEP,NLGEOM Step E:6: Tangential drag, dynamic, FDT. Total RF = -162 i - 162 j distributed over supporting nodes. *DYNAMIC .1,.1 ** ** vf = 2 i + 1 j ** vp = -2 i - 1 j, (Prescribed in BC), alpha=1. ** ** ... Delta-v = 4 i + 2 j ** ... Delta-v_t = 3 i + 3 j ** ... Delta-v_n = 1 i - 1 j ** ** 1/2 rho pi Ct D = 0.9 ** ** h (exponent) = 2. (default) ** ** Testing: that in the absence of a scalimg amplitude ** the steady-current velocity is not scaled. ** ** Testing: that the alpha parameter defaults to 1.0. ** *BOUNDARY,type=velocity support,1,1,-2.0 support,3,3,-1.0 *DLOAD,OP=NEW all, fdt, , 7.5491, 0.0381972, , , **el,FDT, amag0, D, Ct, Alphr, h, Ampc, Ampw *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ** ** ** ------------------------------------------- ** Inertial Drag test. ** ------------------------------------------- ** Expected Result: total RF of 3394 i -3394 j ** distributed over nodes. ** *STEP,NLGEOM Step F:7: Inertial drag, FI. Total RF = 3394 i -3394 j distributed over supporting nodes. *DYNAMIC .1,.2 ** ** af = 0 i + 0 j (No waves) ** ap = 2 i - 6 j (Prescribed in B.C.) ** ... apt = -2 i - 2 j ** ... apn = 4 i - 4 j ** rho pi D^2/4 Ca = 60.0 ** ** ** ** *BOUNDARY,type=acceleration support,1,1, 2.0 support,3,3,-6.0 *DLOAD,OP=NEW all, fi, , 7.4111, 0.0, 0.7 **el, fi, amag0, D , Cm, Ca, Ampw *NODE PRINT,totals=yes rf, *NODE PRINT a, *NODE FILE,F=2 RF, *ENERGY FILE,F=2 *EL FILE,ELSET=ALL,F=2 LOADS, SF,ESF1 *END STEP ** ** ** Intermediate step; clear last load and ** move beam upwards so it is only half-submerged. ** *STEP,NLGEOM Intermediate step. (2) *STATIC 0.1,0.1 *DLOAD,OP=NEW *BOUNDARY support, 1, support, 3, 3, 25. *NODE FILE,f=0 RF, *ENERGY FILE,f=0 *EL FILE,ELSET=ALL,f=0 LOADS, *NODE PRINT U, *NODE PRINT,totals=yes RF, *END STEP ** ** --------------------------------------------------------- ** Partial submergence test: Normal drag in dynamic step ** ---------------------------------------------------------- *STEP, NLGEOM Step G:9: Normal drag, dynamic, partial immersion, FDD. RF = 4500 i - 4500 j distributed over supporting nodes. *DYNAMIC 1E-6,1E-6 ** ** This load is a copy of that in Step D, but now ** half of the beam is submerged so half of the ** beam will be loaded and half of the force of step D ** will be returned. ** ** NLGEOM must be active to ensure that current coordinates are used ** in the calculations. ** Testing: that partial submergence is correctly handled. ** ** *BOUNDARY,type=velocity support,1,1,12.5 support,3,3,-12.5 *EL PRINT,ELSET=ALL COORD, *DLOAD,OP=NEW all, fdd, 2.0, 7.5491, 1.2, 0.8, ten **el,FDD, amag0, D, Cd, Alphr, Ampc, Ampw *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ** Intermediate step; clear last load and ** return beam to its original configuration. ** *STEP,NLGEOM Intermediate step. (3) *STATIC 0.1,0.1 *DLOAD,OP=NEW *BOUNDARY support, 1, 3 *NODE FILE,F=0 RF, *ENERGY FILE,F=0 *EL FILE,ELSET=ALL,F=0 LOADS, *END STEP ** ------------------------------------------- ** Normal End-Drag test in Dynamic Step ** ------------------------------------------- *STEP,NLGEOM Step H:11: End-drag, dynamic, FD1,FD2. Total RF = - 5728 i - 5728 j (only load at node 1 acts) *DYNAMIC .1,.1 ** In this dynamic step he velocity of the structure ** is taken into account, and, with the alpha parameter of 0.8, ** the value of effective relative velocity is now ** ** vf = 20 i + 10 j ** vp = 12.5 i - 12.5 j, alpha=0.8 ** ... Delta-v = 10 i + 20 j ** ... Deltav_t = 15 i + 15 j ** ** Amag 1/2 rho Cd A = 18.0 ** ** P = 18.0 ( 15 root2) (15 i +15 j) = 5728 (i + j) ** ** Testing: that the structural velocity prescribed here ** is being taken into account when determinining ** Delta-v, and that the parameter alpha is being used correctly. ** ** Testing: that the load force is zero when the relative velocity ** has a positive projection on the outward normal (this happens here ** at node 2. ** ** As before: the steady-current velocity is being scaled up by a ** factor of 10. using the amplitude definition TEN. ** ** Testing: that the force magnitude scaling factor of 2.0 is being applied. ** *BOUNDARY,type=velocity support,1,1,12.5 support,3,3,-12.5 *DLOAD,OP=NEW all, fd1, 2.0, 7.5491, 1.2, 0.8, ten all, fd2, 2.0, 7.5491, 1.2, 0.8, ten **el,FDn, amag0, area, C, Alphar, Ampc, Ampw *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** --------------------------------------------------- ** Normal End-Drag test in Dynamic Step as CLOAD ** --------------------------------------------------- *STEP,NLGEOM Step I:12: End-drag, dynamic, CLOAD, TFD. Total RF = - 5728 i - 5728 j (only load at node 1 acts) *DYNAMIC .1,.1 ** ** This places drag CLOADS on the two end-nodes of the beam ** equivalent to the end-drag DLOADS of step H ** *BOUNDARY,type=velocity support,1,1,12.5 support,3,3,-12.5 *DLOAD,OP=NEW *CLOAD,OP=NEW end1, tfd, 2.0, 7.5491, 1.2, 0.8, ten -1, 0, -1 end2, tfd, 2.0, 7.5491, 1.2, 0.8, ten 1,0,1 **nod ,TFD, amag0, area, Cn, Alphar, Ampc, Ampw **tx,ty,tz *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ------------------------------------------- ** Inertial End-Drag test. ** ------------------------------------------- ** *STEP,NLGEOM Step J:13: Inertial End-drag, FI1, FI2. Total RF = - 240 i - 240 j . *DYNAMIC .1,.2 ** ** af = 0 i + 0 j (No waves) ** ap = 2 i - 6 j (Prescribed in B.C.) ** ... apt = -2 i - 2 j ** ... apn = 4 i - 4 j ** amag rho Lts F2s = 60.0 ** ** ** ** *BOUNDARY,type=acceleration support,1,1, 2.0 support,3,3,-6.0 *CLOAD,OP=NEW *DLOAD,OP=NEW all, fi1, , 1., 1.3, 60.3925, 0.5 all, fi2, , 1., 1.3, 60.3925, 0.5 **el ,FIn, amag0, Kts, F1s, Lts, F2s, Ampw *NODE PRINT,totals=yes rf, *NODE PRINT a, *NODE FILE,F=2 RF, *ENERGY FILE,F=2 *EL FILE,ELSET=ALL,F=2 LOADS, SF,ESF1 *END STEP ** ** ------------------------------------------- ** Inertial End-Drag test. -- CLOAD ** ------------------------------------------- ** *STEP,NLGEOM Step K:14: Inertial End-drag, CLOAD, TSI Total RF = -240 i - 240 j *DYNAMIC .1,.2 ** ** This places drag CLOADS on the two end-nodes of the beam ** equivalent to the end-drag DLOADS of step J ** *BOUNDARY,type=acceleration support,1,1, 2.0 support,3,3,-6.0 *DLOAD,OP=NEW *CLOAD,OP=NEW end1,TSI, , 1., 1.3, 60.3925, 0.5 -1, 0, -1 end2,TSI, , 1., 1.3, 60.3925, 0.5 1, 0, 1 **nod ,TSI, amag0, Kts, F1s, Lts, F2s, Ampw **tx,ty,tz *NODE FILE,F=2 RF, *ENERGY FILE,F=2 *EL FILE,ELSET=ALL,F=2 LOADS, SF,ESF1 *END STEP ** ** ** Intermediate step; clear last load and ** return beam to its original configuration. ** *STEP,NLGEOM Intermediate step. (4) *STATIC 0.1,0.1 *CLOAD,OP=NEW *DLOAD,OP=NEW *BOUNDARY support, 1, 3 *NODE PRINT,TOTALS=YES RF, U, *NODE FILE,F=0 RF, *ENERGY FILE,F=0 *EL FILE,ELSET=ALL,F=0 LOADS, *END STEP ** ** ** Intermediate step; rotate beam about ** one of its supporting nodes 90 degrees ** *STEP,NLGEOM Intermediate step. (5) *STATIC 0.1,.5 *BOUNDARY,op=new twist, 1,3 end2, 2 twist, 5, ,-1.570796 twist,6 *END STEP ** ----------------------------------------------------- ** End- Drag test II in DYNAMIC Step (after rotation) ** ----------------------------------------------------- ** Expected Result: ** *STEP,NLGEOM Step L:17: Transition-section buoyancy, TSB. Total RF = 9501 i - 9501 j *STATIC .1,.1 ** ** Note: The t-vector is now ( - i + j ) * (-1 for node 1, 1 for second node) /root2 ** ** Amag rho g depth area = 13.436E3 ** ** Testing: that the load force is zero when the relative velocity ** has a positive projection on the outward normal (this happens here ** at node 1). ** ** Testing: that the alpha parameter defaults to 1.0. ** ** Testing: that the t-vector is correctly rotated for the drag end-loads. ** *BOUNDARY,FIXED, OP=NEW support,1,3 support,4,6 *CLOAD,OP=NEW end1,TSB, 1., 7.0, -1, 0, -1 **nod ,TSB, amag0, area, tx,ty,tz *DLOAD,OP=NEW *NODE PRINT,TOTALS=YES U, RF, *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** *STEP,NLGEOM Step M:18: End-drag, dynamic, FDD. Total RF = -12.727 i + 12.727 j (only load at node 3 acts) *DYNAMIC .1,.1 ** ** Note: The t-vector is now ( - i + j ) * (-1 for node 1, 1 for second node) /root2 ** ** vf = 2 i + 1 j ** vp = -2 i - 1 j, (Prescribed in BC), alpha=1. ** ** ... Delta-v = 4 i + 2 j ** ... Delta-v_t = 1 i - 1 j ** ** Amag 1/2 rho Cd A = 9.0 ** ** Testing: that the load force is zero when the relative velocity ** has a positive projection on the outward normal (this happens here ** at node 1). ** ** Testing: that the alpha parameter defaults to 1.0. ** ** Testing: thata the t-vector is correctly rotated for the drag end-loads. ** *BOUNDARY,type=velocity,op=new support,1,1,-2.0 support,2 support,3,3,-1.0 support,4,6 *CLOAD,OP=NEW *DLOAD,OP=NEW all, fd1, , 7.5491, 1.2, , all, fd2, , 7.5491, 1.2, , **el,FDn, amag0, area, C, Alphar, Ampc, Ampw *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ----------------------------------------------------- ** End- Drag test II in DYNAMIC Step (after rotation) - CLOAD ** ----------------------------------------------------- ** Expected Result: RF of -12.727 i + 12.727 j for each end-node ** *STEP,NLGEOM Step N:19: End-drag, dynamic, CLOAD, TFD. Total RF = - 12.727 i + 12.727 j (only load at node 3 acts) *DYNAMIC .1,.1 ** ** ** This places drag CLOADS on the two end-nodes of the beam ** equivalent to the end-drag DLOADS of step M ** ** Testing: that the alpha parameter defaults to 1.0. ** ** Testing: that the t-vector is correctly rotated in the drag CLOADS ** *BOUNDARY,type=velocity,op=new support,1,1,-2.0 support,2 support,3,3,-1.0 support,4,6 *DLOAD,OP=NEW *CLOAD,OP=NEW end1, tfd, , 7.5491, 1.2, -1, 0, -1 end2, tfd, , 7.5491, 1.2, 1,0,1 **nod ,TFD, amag0, area, Cn, Alphar, Ampc, Ampw **tx,ty,tz *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP ** ** ** ** ** Intermediate step; clear last load and ** place node 3 end of beam 15 units above water level. ** ** Node 3 end of beam must move up from 20 j to 55 j. *STEP,NLGEOM Intermediate step. (6) *STATIC 0.1,0.1 *CLOAD,OP=NEW *DLOAD,OP=NEW *BOUNDARY,op=new twist, 3, , 35 *BOUNDARY,FIX,op=new twist,1,2 twist,4,6 *NODE FILE,F=0 RF, *ENERGY FILE,F=0 *EL FILE,ELSET=ALL,F=0 LOADS, *CONTROLS,PARAMETERS=FIELD,FIELD=DISPLACEMENT 10., *END STEP ** ** *STEP,NLGEOM Intermediate step. *STATIC 0.1,0.1 *ENERGY FILE, F=0 *NODE FILE,F=0 RF, *EL FILE,ELSET=ALL,F=0 LOADS, *DLOAD,OP=NEW *END STEP ** *STEP,NLGEOM Step O:22: Wind-drag, dynamic, WDD. Total RF = -1563. (i + j) *DYNAMIC .01,.01 ** Note: The t-vector is now ( - i + j ) * (-1 for node 1, 1 for second node) ** The current height of beam mid point is 50, 10 above still water level. ** vf = cx (10/10)^0.2 *Ampx= 36.844 * Ampx ** vf = 55.266 i + 0 j (Ampx=1.5) ** alpha vp = -2 i - 2 j, (Prescribed in BC), alpha=0.5 ** ** ... Delta-v = 57.266 i + 2 j ** ... Delta-v_t = 27.633 i - 27.633 j ** ... Delta-v_n = 29.633 i + 29.633 j ** ** Amag 1/2 rho Cd A = 9.0 E-2 ** ** Testing: that the height-dependence of the wind is being correctly done. ** ** Testing: that the alpha parameter is correctly used. ** *BOUNDARY,type=velocity,op=new support,1,1,-4.0 support,2 support,3,3,-4.0 support,4,6 *CLOAD,OP=NEW *DLOAD,OP=NEW **el,WDD, amag0, D, Cd, Alphr, Ampx, Ampy all, wdd, , 7.5491, 1.2, 0.5, A1_5, zero *NODE PRINT,NSET=ALL, TOTALS=YES U, RF, V, RF, *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *CONTROLS,RESET *END STEP ** ** ------------------------------------------- ** Wind End-Drag test in Dynamic Step ** ------------------------------------------- *STEP,NLGEOM Step P:23: Wind End-drag, dynamic, WD1, WD2. Total RF = -79.55 i + 79.55 j (only load at node 3 acts) *DYNAMIC .01,.01 ** Note: The t-vector is now ( - i + j ) * (-1 for node 1, 1 for second node)/root2 ** ** vf = 40 i + 0 j ( as in step N but now Ampx=1 (default) ** vp = -20 i - 10 j, (Prescribed in BC), alpha=1. ** ** ... Delta-v = 60 i + 10 j ** ... Delta-v_t = 25 i - 25 j ** ** Amag 1/2 rho C A = 9.0 E-2 ** ** Testing: that the load force is zero when the relative velocity ** has a positive projection on the outward normal (this happens here ** at node 1). ** ** Testing: that the alpha parameter defaults to 1.0. ** ** Testing: that the t-vector is correctly rotated for the drag end-loads. ** ** Testing: that the default value for the amplitude Ampx is 1.0 ** *BOUNDARY,type=velocity support,1,1,-20.0 support,3,3,-10.0 *CLOAD,OP=NEW *DLOAD,OP=NEW all, wd1, , 7.5491, 1.2, , all, wd2, , 7.5491, 1.2, , **el,WDn, amag0, area, C, Alphar, Ampx, Ampy *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *CONTROLS,RESET *END STEP ** ** ** ------------------------------------------- ** Wind End-Drag test in Dynamic Step - CLOADS ** ------------------------------------------- *STEP,NLGEOM Step Q:24: Wind End-drag, dynamic, CLOAD, TWD. Total RF = -79.55 i + 79.55 j (only load at node 3 acts) *DYNAMIC .01,.01 ** ** This places drag CLOADS on the two end-nodes of the beam ** equivalent to the end-drag DLOADS of step P ** ** ** Testing: that the load force is zero when the relative velocity ** has a positive projection on the outward normal (this happens here ** at node 1). ** ** Testing: that the alpha parameter defaults to 1.0. ** ** Testing: that the t-vector is correctly rotated for the drag end-loads. ** ** Testing: that the default value for the amplitude Ampx is 1.0 ** *BOUNDARY,type=velocity,op=new support,1,1,-20.0 support,2,2 support,3,3,-10.0 support,4,6 *DLOAD,OP=NEW *CLOAD,OP=NEW end1, twd, , 7.5491, 1.2, -1, 0, -1 end2, twd, , 7.5491, 1.2, 1,0,1 **nod,TWD, amag0, area, Cn, Alphar, Ampx, Ampy **tx,ty,tz *NODE FILE RF, *ENERGY FILE *EL FILE,ELSET=ALL LOADS, SF,ESF1 *END STEP **