Compression Test On Wooden Cubes When The Load Is Applied

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7. To perform compression test on wooden cubes when the load is applied (17-12-08)  Parallel to grains  Perpendicular to grains

Objectives   

To determine the compressive strength of the wooden cubes To observe the anisotropic behavior of the wood. To determine the modulus of elasticity and modulus of stiffness.

Apparatus 500 KN SHMADZU UTM, Wooden Cubes, Vernier Caliper, Deflection Gauges.

Related Theory Compressive Strength Maximum stress that a material can bear in compression is termed as compressive strength.

Modulus of Elasticity (E) It is the ratio of unit stress to unit strain and is determined by slop of straight line from zero to proportional limit in a stress strain diagram.

𝑬=

𝝈 𝑷 𝜹 𝑷𝑳 𝟏 → 𝝈 = 𝑬𝝐 → = 𝑬 → 𝜹 = 𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑎𝑦 𝜹 ∝ 𝝐 𝑨 𝑳 𝑨𝑬 𝑬

Modulus of Stiffness (K) It is defined as force required producing unit deformation.

𝑷 ∝ 𝜹 → 𝑷 = 𝑲𝜹 𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑎𝑦 𝑲 =

𝑷 𝜹

Isotropic Materials Isotropic materials are those materials which exhibits same properties in different direction for example steel.

Anisotropic Materials Anisotropic materials are those materials which do not show same properties in different directions.

1 Compression Test on Wooden Cubes

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Failure Mechanism in Different Directions  Failure of Wooden Cube when load is applied o Parallel to Grains When the load is applied parallel to grains, the wooden sample will take more load to fail, the ability of wood to take more load parallel to grains before failure is because each fiber act as column to the applied load and even after the failure of the single fiber the rest of the fibers will keep on taking the load.

 Failure of Wooden Cube when load is applied o Perpendicular to Grains When the load is applied perpendicular to the grains, the wooden sample takes comparatively less load. This is because the failure of the single fiber will lead to the failure of the whole sample. The strength of the wooden sample when the load is applied parallel to the grains is about ten times more as compare to when the load is applied perpendicular to grains.

Procedure    

First of all determine the dimension of all three sides of the wooden cube by the Vernier caliper. Then fix the cube in the machine as shown in figure. We increased the load in increments. We noted the load and deflection readings on machine and deflection gauges respectively.

Calculations and Observations Cube Dimensions Specimen Parallel To Grains Perpendiular To Grains

Length (mm)

Width (mm)

Height (mm)

49.1

49.65

54.5

49.5

50.5

50.2

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 Readings of Experiment when load is applied o Parallel to Grains L = 49.1 mm Sr. No

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Deflection Deformation Gauge (mm) Reading

Load KN 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115

A = 2437.82 mm2

H = 54.5 mm

N 0 5000 10000 15000 20000 25000 30000 35000 40000 45000 50000 55000 60000 65000 70000 75000 80000 85000 90000 95000 100000 105000 110000 115000

δ 100 110 117 121 125 128 131 134 137 140 143 146 148 152 155 158 162 166 171 176 181 188 196 219

0 0.254 0.4318 0.5334 0.635 0.7112 0.7874 0.8636 0.9398 1.016 1.0922 1.1684 1.2192 1.3208 1.397 1.4732 1.5748 1.6764 1.8034 1.9304 2.0574 2.2352 2.4384 3.0226

%strain (δ/H)*100 0.0000 0.4661 0.7923 0.9787 1.1651 1.3050 1.4448 1.5846 1.7244 1.8642 2.0040 2.1439 2.2371 2.4235 2.5633 2.7031 2.8895 3.0760 3.3090 3.5420 3.7750 4.1013 4.4741 5.5461

Stress (Mpa) σ= load/area 0.0000 2.0510 4.1020 6.1530 8.2041 10.2551 12.3061 14.3571 16.4081 18.4591 20.5101 22.5611 24.6122 26.6632 28.7142 30.7652 32.8162 34.8672 36.9182 38.9692 41.0203 43.0713 45.1223 47.1733

E (Mpa)

K (N/mm)

σ/ε

P/δ

#DIV/0! 440.079516 517.740607 628.6850228 704.1272256 785.8562786 851.7668051 906.0460623 951.5232778 990.178911 1023.440735 1052.36406 1100.19879 1100.19879 1120.202404 1138.136679 1135.689074 1133.538147 1115.694548 1100.19879 1086.616089 1050.189754 1008.515557 850.5738544

#DIV/0! 19685.03937 23158.86985 28121.48481 31496.06299 35151.85602 38100.0762 40528.02223 42562.24729 44291.33858 45779.16133 47072.92023 49212.59843 49212.59843 50107.37294 50909.58458 50800.1016 50703.88929 49905.73361 49212.59843 48605.03548 46975.66213 45111.54856 38046.71475

50.0000 40.0000

30.0000 20.0000 10.0000

0.0000 0.0000

1.0000

2.0000

3.0000

3 Compression Test on Wooden Cubes

4.0000

5.0000

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 Readings of Experiment when load is applied o Perpendicular to Grains H = 50.5 mm A = 2499.75 mm2

L = 49.5 mm Sr. No

Deflection Deformation Gauge (mm) Reading

Load KN

1 0 2 2 3 4 4 6 5 8 6 10 7 12 8 14 9 16 10 18 11 20 12 22 13 24 14 26 15 28 16 30 17 32 18 34 19 36 20 38 21 40 22 42 23 43.9

N 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 22000 24000 26000 28000 30000 32000 34000 36000 38000 40000 42000 43940

%strain

Stress (Mpa)

E (Mpa)

K (N/mm)

δ

(δ/H)*100

σ= load/area

σ/ε

P/δ

0 0.4064 0.508 0.5842 0.635 0.6731 0.7112 0.7366 0.762 0.8128 0.8382 0.8636 0.889 0.9144 0.9652 1.016 1.0668 1.1176 1.1684 1.2446 1.2954 1.4224 1.6002

0 0.8210101 1.0262626 1.180202 1.2828283 1.359798 1.4367677 1.4880808 1.5393939 1.6420202 1.6933333 1.7446465 1.7959596 1.8472727 1.949899 2.0525253 2.1551515 2.2577778 2.360404 2.5143434 2.6169697 2.8735354 3.2327273

100 116 120 123 125 126.5 128 129 130 132 133 134 135 136 138 140 142 144 146 149 151 156 163

0 0.8001 1.6002 2.4002 3.2003 4.0004 4.8005 5.6006 6.4006 7.2007 8.0008 8.8009 9.6010 10.4010 11.2011 12.0012 12.8013 13.6014 14.4014 15.2015 16.0016 16.8017 17.5778

#DIV/0! 97.45068995 155.9211039 203.3753529 249.4737663 294.1907621 334.1166513 376.3612853 415.7896105 438.5281048 472.4881937 504.4506303 534.586642 563.0484308 574.4461723 584.7041397 593.9851578 602.422447 610.1260588 604.5920356 611.4553095 584.7041397 543.7439132

#DIV/0! 4921.259843 7874.015748 10270.45532 12598.4252 14856.63349 16872.89089 19006.24491 20997.37533 22145.66929 23860.65378 25474.75683 26996.62542 28433.94576 29009.5317 29527.55906 29996.25047 30422.33357 30811.36597 30531.8978 30878.49313 29527.55906 27459.06762

20.0000

15.0000 10.0000 5.0000 0.0000

0

0.5

1

1.5

4 Compression Test on Wooden Cubes

2

2.5

3

3.5

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Comparison between Two Graphs %age strain on x-axis and Stress (MPa) on y axis 50.0000 45.0000 40.0000 35.0000 30.0000 25.0000

Parallel

20.0000

Perpendicular

15.0000 10.0000 5.0000 0.0000 0.0000

1.0000

2.0000

3.0000

4.0000

5.0000

6.0000

Cube after Fracture

Figure 2 when the load is applied parallel to grains

Figure 1 when the load is applied perpendicular to grains

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Comments When the load is applied parallel to grains, the failure of the sample is due to shear failure, which results in cracks at edges, and during load applied perpendicular to grains, the failure is also shear failure where fiber have slide over one another. If we are designing any wood structure, we should keep in mind that load should be applied on wood to the parallel to its grains because doing so strength is approximately ten times more than when the load is applied perpendicular to grains.

When the Load is Applied Parallel to Grains



In this type of loading shear failure cracks start developing near the edges at about 45 degrees.



But due to Platen Effect and Eccentricity, failure is not purely shear failure, so that’s way we see a in between crack in the sketch of wooden cube after fracture.

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When the Load is Applied Perpendicular to Grains



In this type of loading, failure is also due to shear failure because in this layers of wood have slides over each other.



We can easily see bulging effect.

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