Are there any examples of an MSUP transient dynamic analysis? I cannot
get my acceleration load to have any effect in the results.


Yes, please see the ANSYS 9.0 spring-mass example below (and attached).
It is important to note how the loads are applied. In this model, the gravity
(acceleration) load is specified in the modal analysis and then applied via a
load vector with the LVSCALE command. An additional force is applied in the
second transient load step, but this force should also be initialized as zero in
the preceding static load step. If the force were not applied, then the mass
would stay at rest in the second load step after reaching static equilibrium in
the first load step. This allows static analyses to be run as MSUP transients.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

fini
/clear

/TITLE, MSUP Transient Example

/plopts,info,1
/view,,1,2,3

/PREP7
ET,1,MASS21,,,2 ! 3D mass with no rotary inertia
R,1,10.0 ! mass (10.0 Kg)
ET,2,COMBIN14,,3 ! 1D longitudinal spring-damper (UZ DOF)
R,2,100.0 ! spring constant (100.0 N / meter)
N,1,0,0,0
N,2,0,0,0 ! coincident node ...
TYPE,1
REAL,1
E,1 ! mass element at node 1
TYPE,2
REAL,2
E,2,1 ! spring element from node 2 to node 1
eplot
fini

/SOLU
ANTYPE,MODAL
MODOPT,LANB,3
MXPAND,3,,,yes
OUTRES,ALL,ALL
ACEL,0,0,9.8 ! include base acceleration in load vector
D,2,ALL,0.0 ! fix one end
SOLVE
save
fini

/solu
ANTYPE,TRANS
TRNOPT,MSUP,3 ! MSUP transient
TIME,0.0 ! initial static solution defaults to TIME=0
DELTIM,0.001 ! DELTIM may not be changed in subsequent load steps
OUTRES,ALL,all

acel,0,0,0 ! apply gravity via load vector only ...
f,1,fz,0.0 ! introduce nodal force in first load step ...

!!! LVSCALE,0! do not include gravity (no load vector)
LVSCALE,1 ! initial load step for frontal solver static solution

! Resulting initial static displacement should be:
!
! X = F/K = ma/K = (10.0 Kg)(-9.8 m/sec^2)/(100 N/m) = -0.98 meters

kbc,1 ! step-apply loads
solve
save

time,1.0 ! DELTIM cannot be changed ... already set ...

!!! LVSCALE,0 ! do not include gravity (no load vector)
LVSCALE,1 ! gravity via load vector = -9.8 m/sec^2

f,1,fz,-150.0 ! a = f/m = -150 N / 10 kg = -15.0 m/sec^2 (initially)

kbc,1 ! step-apply loads
solve
save
fini

/POST26
numvar,200
file,file,rdsp
NSOL,2,1,U,Z,uz2 ! displacement
DERIV,3,2,1,,vz3 ! velocity
DERIV,4,3,1,,az4 ! acceleration
prvar,2,3,4
PLVAR,4 ! acceleration from force load only ...

/eof


! Note: Acceleration from gravity is used to establish the initial
! static displacement (-0.98 m). The step-applied force in
! the second load step causes an initial acceleration per F=ma,
! but then the spring stiffness kicks in and alters the motion
! of this spring-mass system. If the force load is not applied
! (in the second load step), then the mass, which is in static
! equilibrium at the end of the first load step, will not move
! in the second load step (i.e., UZ = VZ = AZ = 0). For the
! model conditions shown above, the results are shown below.


***** ANSYS POST26 VARIABLE LISTING *****

TIME 1 UZ 3 DERI 4 DERI
uz2 vz3 az4
0.0000 -0.980000 0.371230E-02 -12.2155
0.10000E-02 -0.980004 -0.943125E-02 -13.1436
0.20000E-02 -0.980019 -0.225748E-01 -14.0716
0.30000E-02 -0.980049 -0.375744E-01 -14.9994
0.40000E-02 -0.980094 -0.525737E-01 -14.9990
0.50000E-02 -0.980154 -0.675724E-01 -14.9984
0.60000E-02 -0.980229 -0.825704E-01 -14.9976
0.70000E-02 -0.980319 -0.975677E-01 -14.9967
0.80000E-02 -0.980424 -0.112564 -14.9957
0.90000E-02 -0.980544 -0.127559 -14.9945
0.10000E-01 -0.980679 -0.142553 -14.9931
0.11000E-01 -0.980829 -0.157545 -14.9916
. . . .
. . . .
. . . .
0.99600 -3.97993 0.306791E-01 14.9992
0.99700 -3.97989 0.456781E-01 14.9988
0.99800 -3.97984 0.606767E-01 14.9983
0.99900 -3.97977 0.756747E-01 14.9980
1.0000 -3.97969 0.906727E-01 14.9977



.





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