*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
**