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ME309 Homework 5
Problem 2: A Study of Material Plasticity in Pure Bending
Instructor: Prof. Sheri Sheppard
Student: Cheng-Chieh Chao
05361779
6.11.2007
mailto:[email protected]:[email protected]7/28/2019 ME 309 ANSYS Material Plasticity in Pure Bending
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Table of Contents
I. Problem Statement................................................................................................ 3
II . Objective of the analysis....................................................................................... 3III. Physical description of the part to be analyzed.................................................. 4
IV. Finite element program and computer system used for the analysis............... 4
V. Finite element model............................................................................................. 5
VI . Result plots (Stress contour plots) ....................................................................... 8
VII. Stresses and displacements for critical sections of the model ......................... 10
VIII. Theoretical solutions........................................................................................... 12
IX. Conclusions and recommendations................................................................... 13
Appendix: Log file for the solution of both models...................................................... 14
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I. Problem StatementConsider a rectangular beam loaded in pure bending between the two forces, as shown
(this is referred to as the four point bend specimen). For elastic-perfectly plastic
stress-strain behavior, show using finite element analysis, that the beam remains elastic at
Myp=ypbh2/6 and is completely plastic at Mult=1.5 Myp.
II . Objective of the analysisThe objective of this problem is to understand plastic-elastic deflection behavior of the
specimen under load. The finite element analysis simulation result will be compared to
the theoretical computation.
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III. Physical description of the part to be analyzed
The beam specimen will be loaded with the configuration of four-point bending testing
with adjustable forces F and supports at both ends. It is usually used in a test to
determine the yield strength of the material. The cross section of the specimen is a
rectangle, dimensions are shown in figure. The coordinate axes are shown in the figure,
where x axis passes through the neutral axis of the beam with the origin at the left end. y
and z axes are oriented as shown.
Dimensions:
b=1, h=2
Material properties:
Youngs Modulus E=30e6 psi
Poisson Ratio =0.3
Yield Strength yp=36000 psi
Loading conditions:
Two external forces, F, of equal magnitude and direction, are applied from the top of the
beam. Supports in the y direction are at the two ends of the specimen.
IV. Finite element program and computer system used for the analysisANSYS v7 is used for the FEM analysis in the elaine cluster at Terman.
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V. Finite element model(a) Analysis using PLANE42 element
Element used: PLANE42
Real constraint: thickness 1
Mesh density: 0.5x0.5 per element.
Boundary conditions:
UX=UY=0 at node 1 (bottom left corner)
UY=0 at node 2 (bottom right corner).
Applied loads:
The load F required for applying Myp=ypbh2/6=24000 lb-in is F=6000 lb, and the load
required for Mult=1.5 Myp=36000 lb-in is F=9000 lb.
The load F is applied incrementally as 5999, 6001, 6500, 7000, 7500, 8000, 8500, and
9000 lbs to avoid the sudden loading that may lead to the divergence of solution in
ANSYS. The F is applied at nodes 38 and 46 as in the figure above.
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Assumptions: It is assumed the material behavior of this steel beam is bilinear with zero
tangent modulus after yield point. Bilinear kinematic hardening plasticity behavior is
used for solving this problem in ANSYS.
Bilinear material behavior with zero tangent modulus, yield strength y=36000 psi
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(b) Analysis using SOLID45 element
Element used: SOLID45
Mesh density: 0.5x0.5x0.5 per element
Boundary conditions:
UX=UY=0 for all nodes at x=y=0 (bottom left corner)
UY=0 for all nodes at x=12, y=0 (bottom right corner)
Applied loads:
The load F required for applying Myp=ypbh2/6=24000 lb-in is F=6000 lb, and the load
required for Mult=1.5 Myp=36000 lb-in is F=9000 lb.
The load F is applied incrementally as 5999, 6001, 6500, 7000, 7500, 8000, 8500, and
9000 lbs to avoid the sudden loading that may lead to the divergence of solution in
ANSYS. One F is uniformly distributed along all the 3 nodes at x=4, y=2 with the
magnitude of F/3. The other F is also uniformly distributed along all the 3 nodes at x=8,
y=2 with the magnitude of F/3, as in the figure above.
Assumptions: It is assumed the material behavior of this steel beam is bilinear with zero
tangent modulus after yield point. Bilinear kinematic hardening plasticity behavior is
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used for solving this problem in ANSYS.
VI . Result plots(a) Analysis using PLANE42 element
xx stress contour plots using PLANE42 element at F=8500 lbs
Discussion: The figure shows the xx stresses when the beam is under the load of F=8500
lbs (1.42 Myp). The region between x=4 and x=8 is under the stresses over 30000 psi,
indicating the structure behavior is nonlinear. At this time, the beam is in the
elastic-plastic region.
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(b) Analysis using SOLID45 element
xx stress contour plots using SOLID45 element at F=8500 lbs
Discussion: The figure shows the xx stresses when the beam is under the load of F=8500
lbs (1.42 Myp). The region between x=4 and x=8 is under the stresses over 30000 psi,
indicating the structure behavior is nonlinear. At this time, the beam is in the
elastic-plastic region.
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VII. Stresses and displacements for critical sections of the model(a) Analysis using PLANE42 element
PLANE42
-50000
-40000
-30000
-20000
-10000
0
10000
20000
30000
40000
50000
-1 -0.5 0 0.5 1
Distance from neutral axis
S
tress
(psi)
F=5999
F=6001
F=6500
F=7000
F=7500
F=8000
F=8500
Material behavior using PLANE42 elements
node 5999 6001 6500 7000 7500 8000 8500 9000
(7, -1) -35771 -35783 -38758 -36015 -36016 -36006 -35982 N/A
(7, -0.5) -18008 -18014 -19511 -24639 -28782 -33035 -39120 N/A
(7, 0) -189.7 -189.76 -207.87 -26.236 -25.609 89.043 215.35 N/A
(7, 0.5) 17957 17963 19456 24664 28807 32948 38896 N/A
(7, 1) 36254 36266 39284 36017 36019 36003 35998 N/A
xx at nodes on x=7 vs. distance from the neutral axis in y direction under different F
Discussion:
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The stresses in the table are xx on the 5 nodes on x=7. As we can see in the figure
above, the material is elastic at F=5999 and 6001 lbs (~Myp), and xx=My/I. When the
load is increasing from 6500 lbs, the material behavior becomes nonlinear. The
solutions still converge for F=6500, 7000, 7500, 8000, 8500 lbs, indicating the material is
in the elastic-plastic region. When the load F is 9000 lbs, ANSYS fails to converge in 5
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steps, indicating that the beam is fully plastic.
(b) Analysis using SOLID45 element
SOLID45
-50000
-40000
-30000
-20000
-10000
0
10000
20000
30000
40000
50000
-1 -0.5 0 0.5 1
Distance from neutral axis
Stress
(psi)
F=5999
F=6001
F=6500
F=7000
F=7500
F=8000
F=8500
Material behavior using SOLID45 elements
node 5999 6001 6500 7000 7500 8000 8500 9000
(7, -1, 1) -35360 -35372 -38274 -36115 -36306 -36326 -36295 N/A
(7, -0.5, 1) -18856 -18862 -20452 -25071 -29238 -33715 -39524 N/A
(7, 0, 1) -82.572 -82.599 -100.61 -63.159 -102.32 -125.35 35.954 N/A
(7, 0.5, 1) 17914 17920 19415 24795 28966 33332 39163 N/A
(7, 1, 1) 36425 36437 39487 36065 36203 36348 36278 N/A
xx at nodes on x=7, z=1 vs. distance from the neutral axis in y direction under different F
Discussion:
The stresses in the table are xx on the 5 nodes on x=7. As we can see in the figure
above, the material is elastic at F=5999 and 6001 lbs (~Myp), and xx=My/I. When the
load is increasing from 6500 lbs, the material behavior becomes nonlinear. Thesolutions still converge for F=6500, 7000, 7500, 8000, 8500 lbs, indicating the material is
in the elastic-plastic region. When the load F is 9000 lbs, ANSYS fails to converge in 5
steps, indicating that the beam is fully plastic. The result agrees with PLANE42 elements.
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VIII. Theoretical solutionsReference: Case, J.; Chilver, L.; Ross, C.T.F. (1999). Strength of Materials and Structures
(4th Edition). (pp. 350-366). Elsevier
Myp=ypbh2/6
Mult=1.5 Myp
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M < Myp M > MultMyp < M < Mult
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IX. Conclusions and recommendationsThe analysis presented using PLANE42 and SOLID45 matches the theoretical material
behavior. In summary, the result is listed in the following table:
F 6000 6500 7000 7500 8000 8500 9000
Mult / Myp 1 1.083 1.167 1.25 1.333 1.417 1.5
Theory ElasticElastic-
plastic
Elastic-
plastic
Elastic-
plastic
Elastic-
plastic
Elastic-
plastic
Fully
plastic
Converge Converge Converge Converge Converge ConvergeNot
converge
ANSYS
PLANE42
&SOLID45
Elastic Elastic-plastic
Elastic-plastic
Elastic-plastic
Elastic-plastic
Elastic-plastic
Fullyplastic
Therefore, it can be concluded that ANSYS is suitable for nonlinear material plasticity
simulation. The result of finite element simulation and theoretical values matches well.
It is confirmed that a rectangular beam is pure elastic at Myp=ypbh2/6 and completely
plastic at Mult = 1.5 Myp. With proper models and settings, ANSYS can be used to
predict the material behavior for elastic and elastic-plastic.
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Appendix: Log file for the solution of both models
(a) Analysis using PLANE42 element
/SOL
solcontrol,0
neqit,5 ! MAXIMUM 5 EQUILIBRIUM ITERATIONS PER STEP
ncnv,0 ! DO NOT TERMINATE THE ANALYSIS IF THE
SOLUTION FAILS TO CONVERGE
cnvtol,u ! CONVERGENCE CRITERION BASED UPON
DISPLACEMENTS
f,46,fy,-5999
f,38,fy,-5999
solve
f,46,fy,-6001
f,38,fy,-6001
solve
*do,I,1,8
f,46,fy,-6000-(I*500)
f,38,fy,-6000-(I*500)
solve
*enddo
Finish
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(b) Analysis using SOLID45 element
/SOL
solcontrol,0neqit,5 ! MAXIMUM 5 EQUILIBRIUM ITERATIONS PER STEP
ncnv,0 ! DO NOT TERMINATE THE ANALYSIS IF THE
SOLUTION FAILS TO CONVERGE
cnvtol,u ! CONVERGENCE CRITERION BASED UPON
DISPLACEMENTS
f,163,fy,-1999.66
f,293,fy,-1999.66
f,14,fy,-1999.66
f,171,fy,-1999.66
f,285,fy,-1999.66
f,22,fy,-1999.66
solve
f,163,fy,-2000.33
f,293,fy,-2000.33
f,14,fy,-2000.33
f,171,fy,-2000.33
f,285,fy,-2000.33
f,22,fy,-2000.33
solve
*do,I,1,6
f,163,fy,-2000-(I*166.667)
f,293,fy,-2000-(I*166.667)
f,14,fy,-2000-(I*166.667)
f,171,fy,-2000-(I*166.667)
f,285,fy,-2000-(I*166.667)
f,22,fy,-2000-(I*166.667)
solve
*enddo
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FINISH