Lab Report 2.docx

MIDDLE EAST TECHNICAL UNIVERSITY DEPARTMENT OF CIVIL ENGINEERING CE 241 : MATERIAL SCIENCE

LABORATORY REPORT #2 : ISOTROPY AND ANISTROPY

LAB GROUP 3

Buse ONAY 2026326 Section 4 Muammer AKCA 2025401 Section 5 Tuğberk GUNER 2026029 Section 5

Submission Date : 14/12/2015

Table of Contents 1) Object and Scope...............................................................................................................................1 2) Preliminary Remarks..........................................................................................................................1 3) Test Specimen....................................................................................................................................1 4) Apparatus...........................................................................................................................................1 5) Test Procedure....................................................................................................................................1 

For parallel loading case to fibers...............................................................................................1



For perpendicular loading case to fibers.....................................................................................1

6) Calculations.......................................................................................................................................2 

For parallel loading case to fibers...............................................................................................2



For perpendicular loading case to fibers.....................................................................................5

7) Results................................................................................................................................................8 

For parallel loading case to fibers...............................................................................................8



For perpendicular loading case to fibers.....................................................................................8

8) Discussion of Results.........................................................................................................................8 9) References..........................................................................................................................................8

1) Object and Scope 2) Preliminary Remarks 3) Test Specimen 4) Apparatus 5) Test Procedure  For parallel loading case to fibers  For perpendicular loading case to fibers

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6) Calculations  For parallel loading case to fibers From given data the force-time graphs look like as below :

Axial Force (N)



Time ( s ) *For next steps we will neglect the data from first loading-unloading part. 

Loading Rate We calculate loading rate using the slope of the force-time graph Loading Rate = △F/△t

 For 2nd Loading:

△F = F2 –F1 = 30905.995N - 49.727 N = 30856.267N △t = t2 – t1 = 89.937 s - 59.958 s

= 29.97925s

Loading Rate = △F/△t = 30856.267N/ 29.979 s 4

= 1029.254 N/s

 For 3rd Loading:

△F = F2 –F1 = 30925.726N - 66.635 N = 30859.091N △t = t2 – t1 = 149.946 s - 119.967 s

= 29.979s

Loading Rate = △F/△t = 30859.091N/ 29.979 s

= 1029.348 N/s

 Axial Stress Axial Stress equal to force over area of surface which force applied. = F/A For example t= 120.017 s for steel axial stress is : = F/A= 106.288 N / (0.08)2 = 166075 Pa = 0.166075 MPa

 Axial Strain Axial Strain equal to change in axial length over default length. For example t= 120.017 s for steel axial strain : From Lab manual ”L” is 80mm.  = △L/L = 0.0696 mm /80mm = 0.00087  Elastic Modulus (Young’s Modulus) Elastic Modulus equal to slope of stress-strain graph. E= Stress/Strain =

/

5

Axial Stress ( MPa)

Axial Strain

 For 2nd Loading

△

=

2

–

1

= 4.829 MPa - 0.0078 MPa

= 4.821 MPa

△ =  2 –  1 = 0.00401 - 0.000793 = 0.00322 Elastic Modulus =>

E = △ /△ = 4.821 MPa / 0.00322 = 1497 MPa

 For 3rd Loading

△

=

2

–

1

= 4.832 MPa - 0.0104 MPa

= 4.822 MPa

△ =  2 –  1 = 0.00403 - 0.000865 = 0.00317 Elastic Modulus =>

E = △ /△ = 4.822 MPa / 0.00317 = 1521 MPa

6

 For perpendicular loading case to fibers From given data the force-time graphs look like as below :

Axial Force (N)



Time ( s ) *For next steps we will neglect the data from first loading-unloading part. 

Loading Rate We calculate loading rate using the slope of the force-time graph Loading Rate = △F/△t

 For 2nd Loading:

△F = F2 –F1 = 10085.025N - 121.587 N = 9963.437N △t = t2 – t1 = 29.979 s - 20.069 s

= 9.909s

Loading Rate = △F/△t = 9963.437 N/ 9.909 s 7

= 1005.426 N/s

 For 3rd Loading:

△F = F2 –F1 = 10074.669N - 117.611 N = 9957.058 N △t = t2 – t1 = 49.948 s - 40.039 s

= 9.909s

Loading Rate = △F/△t = 9957.058 N/ 9.909 s

= 1004.782 N/s

 Axial Stress Axial Stress equal to force over area of surface which force applied. = F/A For example t= 12.31 s for steel axial stress is : = F/A= 7771.795 N / (0.08)2 = 1214343 Pa = 1.214 MPa

 Axial Strain Axial Strain equal to change in axial length over default length. For example t= 12.31 s for steel axial strain : From Lab manual ”L” is 80mm.  = △L/L = 0.596 mm /80mm = 0.00745  Elastic Modulus (Young’s Modulus) Elastic Modulus equal to slope of stress-strain graph. E= Stress/Strain =

/

8

Axial Stress ( MPa )

Axial Strain

 For 2nd Loading

△

=

2

–

1

= 1.576 MPa - 0.0189 MPa

= 1.557 MPa

△ =  2 –  1 = 0.00869 - 0.00132 = 0.00737 Elastic Modulus =>

E = △ /△ = 1.557 MPa / 0.00737 = 211 MPa

 For 3rd Loading

△

=

2

–

1

= 1.575 MPa - 0.0189 MPa

= 1.556 MPa

△ =  2 –  1 = 0.00876 - 0.00146 = 0.00731 Elastic Modulus =>

E = △ /△ = 1.556 MPa / 0.00731 = 213 MPa 9

7) Results  For parallel loading case to fibers Modulus of Elasticity (Young's Modulus) The modulus of elasticity of the wood specimen was found to be 1521 MPa.

 For perpendicular loading case to fibers Modulus of Elasticity ( Young' Modulus) The modulus of elasticity of the wood specimen was found to be 213 MPa.

8) Discussion of Results  Wood specimen was loaded in two different directions in order to observe that whether it has different behaviors in different directions and see it is isotropic or anisotropic material. The specimen is stronger in perpendicular direction to the its fibers as it is seen in the axial stress- axial strain graphs in perpendicular case stress values are greater.

 Anisotropy means having different properties in different directions for a material. Anisotropic materials exhibits different behaviors in different directions. Unlike isotropic materials anisotropic materials are directionally dependent materials.

 It is not possible for an anisotropic material to calculate the Bulk and Shear modulus by using the formulas which are used for isotropic materials since anisotropic materials have different mechanical properties in different directions. Thus it is needed to carry out different calculations which are taken into consider the dependency of direction of material to calculate Bulk and Shear modulus for an anisotropic material. 

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9) References

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