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Table-l: Deflection at various points along the length of the bar using 2, 4 and 8 elements and compare results with exact so

Solve example Problem # 1.1 of your book using Direct Method for 2, 4 and 8 elements and compare your results with exact solu

USE ANSYS (no hand calculations)

engineering   Mechanical-Engineering   
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Answer #1

Material Property Applied

rial M Linear Isotropic Properties for Material Number 1 X Linear Isotropic Material Properties for Material Number 1 T1 Temp

Ansys result for 2 element design

PRINT U NODAL SOLUTION PER NODE

***** POST1 NODAL DEGREE OF FREEDOM LISTING *****

LOAD STEP= 1 SUBSTEP= 1   
TIME= 1.0000 LOAD CASE= 0   

THE FOLLOWING DEGREE OF FREEDOM RESULTS ARE IN THE GLOBAL COORDINATE SYSTEM

NODE UY
1 0.0000
2 0.0000
3 -0.12821E-02

MAXIMUM ABSOLUTE VALUES
NODE 3
VALUE -0.12821E-02

1 DISPLACEMENT ANSYS R16.0 STEP=1 SUB = 1 TIME=1 DMX = .001282 AUG 1 2020 03:28:55

Deformed + undeformed

Now for 4 element system

Global Element Sizes [ESIZE] Global element sizes and divisions (applies only to unsized lines) SIZE Element edge length ND

4 element meshed

Center line is the keyframe line

LINES ANSYS R16.0 TYPE NUM Y AUG 1 2020 03:33:55

Result

1 DISPLACEMENT ANSYS R16.0 STEP=1 SUB =1 TIME=1 DMX = .005317 AUG 1 2020 03:38:27

PRINT U NODAL SOLUTION PER NODE

***** POST1 NODAL DEGREE OF FREEDOM LISTING *****

LOAD STEP= 1 SUBSTEP= 1   
TIME= 1.0000 LOAD CASE= 0   

THE FOLLOWING DEGREE OF FREEDOM RESULTS ARE IN THE GLOBAL COORDINATE SYSTEM

NODE UY
1 0.0000
2 -0.53171E-02
3 -0.10256E-02
4 -0.22091E-02
5 -0.36077E-02

MAXIMUM ABSOLUTE VALUES
NODE 2
VALUE -0.53171E-02

8 element solution

A Global Element Sizes [ESIZE] Global element sizes and divisions (applies only to unsized lines) SIZE Element edge length

Meshed to 8 elements

1 LINES ANSYS R16.0 TYPE NUM AUG 1 2020 03:42:08

Deformed shape

1 DISPLACEMENT ANSYS R16.0 Y STEP=1 SUB =1 TIME=1 DMX = .005328 AUG 1 2020 03:45:10


PRINT U NODAL SOLUTION PER NODE

***** POST1 NODAL DEGREE OF FREEDOM LISTING *****

LOAD STEP= 1 SUBSTEP= 1   
TIME= 1.0000 LOAD CASE= 0   

THE FOLLOWING DEGREE OF FREEDOM RESULTS ARE IN THE GLOBAL COORDINATE SYSTEM

NODE UY
1 0.0000
2 -0.53282E-02
3 -0.49628E-03
4 -0.10268E-02
5 -0.15966E-02
6 -0.22120E-02
7 -0.28809E-02
8 -0.36135E-02
9 -0.44232E-02

MAXIMUM ABSOLUTE VALUES
NODE 2
VALUE -0.53282E-02

Theoretical calculation

y width Thickness Area P E dL = P* / A*E
1 1.9 0.125 0.2375 1000 10400000 2.07665E-05
2 1.8 0.125 0.225 1000 10400000 9.00517E-05
3 1.7 0.125 0.2125 1000 10400000 0.000220614
4 1.6 0.125 0.2 1000 10400000 0.000429122
5 1.5 0.125 0.1875 1000 10400000 0.000737646
6 1.4 0.125 0.175 1000 10400000 0.001175851
7 1.3 0.125 0.1625 1000 10400000 0.001784308
8 1.2 0.125 0.15 1000 10400000 0.002619619
9 1.1 0.125 0.1375 1000 10400000 0.00376261
10 1 0.125 0.125 1000 10400000 0.005331901
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