VM-NR1677-01-1-a

VM-NR1677-01-1-a
NUREG/CR-1677: Volume 1, Benchmark Problem No. 1

Overview

Reference:P.Bezler, M. Hartzman & M. Reich, Dynamic Analysis of Uniform Support Motion Response Spectrum Method, (NUREG/CR-1677), Brookhaven National Laboratory, August 1980, Problem 1, Pages 24-47.
Analysis Type(s):
Modal analysis (ANTYPE = 2)
Spectral analysis (ANTYPE = 8)
Element Type(s):
Elastic straight pipe elements (PIPE16)
Elastic curved pipe elements (PIPE18)
Structural Mass element (MASS21)
Input Listing:

Test Case

This benchmark problem contains three straight sections, two bends, and two fixed anchors (refer to Figure 619: FE Model of Benchmark Problem). The total mass of the system is represented by structural mass element (MASS21) specified at individual nodes. Modal and response spectrum analysis is performed on the piping model. Frequencies obtained from modal solve and the nodal/element solution obtained from spectrum solve are compared against reference results.

Figure 619: FE Model of Benchmark Problem

FE Model of Benchmark Problem

Material PropertiesGeometric PropertiesLoading

Pipe Elements:

E = 24.00 x 106 psi
Nu = 0.3

3D mass (lb-sec2/in):

Mass @ node 2 = 0.0398
Mass @ node 3 = 0.0503
Mass @ node 4 = 0.02088
Mass @ node 5 = 0.0169
Mass @ node 6 = 0.0130
Mass @ node 7 = 0.0169
Mass @ node 8 = 0.0104
Mass @ node 9 = 0.0179
Mass @ node 10 = 0.0150

Straight Pipe:

Type 1, real 1
Outer Diameter = 7.288 in
Wall Thickness = 0.241 in

Bend Pipe:

Outer Diameter = 7.288 in
Wall Thickness = 0.241 in
Radius of Curvature = 36.30 in

Acceleration response spectrum curve defined by SV and FREQ commands.

Results Comparison

Table 34: Frequencies Obtained from Modal Solution

ModeTargetMechanical APDLRatio
128.53028.5341.00
255.77055.7711.00
381.50081.4991.00
4141.700141.7391.00
5162.800162.8161.00

Table 35: Maximum Displacements and Rotations Obtained from Spectrum Solve

Result NodeTargetMechanical APDLRatio
UX at node5 0.0078 0.00780.999
UY at node7 0.0025 0.00251.00
UZ at node4 0.0174 0.01741.00
ROTX at node3 0.0002 0.00021.00
ROTY at node7 0.0002 0.00021.00
ROTZ at node30.00010.00011.00

Table 36: Element Forces and Moments Obtained from Spectrum Solve

ResultTargetMechanical APDLRatio
Element 1
PX(I)4.9644.958 1.001
VY(I)17.87917.880 1.000
VZ(I)36.43636.430 1.000
TX(I)629.715629.600 1.000
MY(I)3226.871 3227.000 1.000
MZ(I)1393.8411394.0001.000
 
PX(J) 4.9644.958 1.001
VY(J) 17.87917.880 1.000
VZ(J) 36.43636.430 1.000
TX(J) 629.715629.600 1.000
MY(J) 1259.8791260.000 1.000
MZ(J)473.244474.2000.998
Element 10
PX(I) 24.02224.0201.000
VY(I) 7.4787.4721.001
VZ(I) 34.80034.7801.001
TX(I) 112.957113.0001.000
MY(I) 1871.6541871.0001.000
MZ(I) 650.273650.1001.000
 
PX(J) 24.02224.0201.000
VY(J) 7.4787.4721.001
VZ(J) 34.80034.7801.001
TX(J) 112.957113.0001.000
MY(J) 2477.4152477.0001.000
MZ(J)774.763774.5001.000
Element 4
PX(I)9.3049.3001.000
VY(I)10.63510.6301.000
VZ(I)9.2459.2391.001
TX(I)142.101142.1001.000
MY(I)289.991289.9001.000
MZ(I)828.470828.400 1.000
 
PX(J)12.38812.381.001
VY(J)10.63510.631.000
VZ(J)4.3104.3051.001
TX(J)423.720423.71.000
MY(J)261.350261.31.000
MZ(J)541.933541.91.000


Note:  PX (I) and PX (J) = Section axial force at node I and J.

VY (I) and VY (J) = Section shear forces along Y direction at node I and J.

VZ (I) and VZ (J) = Section shear forces along Z direction at node I and J.

TX (I) and TX (J) = Section torsional moment at node I and J.

MY (I) and MY (J) = Section bending moments along Y direction at node I and J.

MZ (I) and MZ (J) = Section bending moments along Z direction at node I and J.

The element forces and moments along Y and Z directions are flipped between Mechanical APDL and NRC results.