To carry out compression test on wooden cubes when load is applied, Parallel to grains, Perpendicular to grains

Job # 4

To carry out compression test on wooden cubes when load is applied

• Parallel to grains
• Perpendicular to the grains.

Objective:

• To determine compressive strength of wood
• To determine modulus of elasticity € and modulus of stiffness (K)
• To observe an-isotropic behavior of wood.

Apparatus:

1. Wooden samples
2. Vernier calipers
3. Deflection gauge
4. Shimadzu 500kN UTM

Related theory:

• Compressive strength:

The maximum stress that a material can bear is called its compressive strength.

• Strength:

The maximum value of stress that a material can bear.

• Modulus of elasticity:

It is the ratio of stress to strain and is determined by the slope of stress strain diagram from 0 to proportional limit. (i.e within elastic limit)

• Elasticity:

Ability of a body to retain its shape is called its elasticity.

• Stiffness (k):

It is the force required to produce unit deformation.

• Isotropic materials:

The materials which exhibit same properties in different directions are called isotropic. e.g. steel

• Anisotropic materials:

The materials which exhibit different properties in different directions are called an-isotropic. e.g. wood

Failure to wooden cubes when load is applied:

1. Parallel to grains:

The wooden sample will take more load to fail on applying load. This is because each fiber act as a column to parallel load.

     2. Perpendicular to grains:

When load is applied wooden sample will take comparatively less load because each fiber acts as beam and the failure of a single fiber will cause the failure of whole sample.

Procedure:

1. Measure the cross-section and length of the specimen and record the dimensions on the data sheet. The ends of the specimens should be plane and at right angles to the axis of the specimen.
2. Place the specimen in the machine.
3. Apply the load continuously until the specimen fails. Record the maximum load.
4. Draw a sketch, in perspective, indicating the grain of the wood and the manner of failure.
5. For the specimen with the grains oriented at 90^o, plot the load vs. deformation and determine the proportional limit from the curve.
6. Compute the compressive strength (for the specimen with 0^o grain orientation) or the proportional limit (for the specimen with 90^o grain orientation).

 

Observations and calculations

 

Specimen	L	W	H	A = L x W
	(mm)	(mm)	(mm)	(mm2)
Parallel to grains	52.35	51.8	48.3	2711.73
Perpendicular to grains	47.95	48.80	50.05	2339.96

 

Parallel to grains:

 

Obs.	Load ‘P’	Deflection gauge reading	Deformation ‘δ’	%age strain= $\in =(\delta \times 100)/H$	σ = $P/A$	K= $P/\delta$
	(kN)		(mm)		N/mm2	N/mm
1	0	100
2	5	106	0.1524	0.3155	1.843	32808
3	10	108	0.2032	0.4207	3.687	49212
4	15	110	0.254	0.5258	5.531	59055
5	20	113	0.3302	0.6836	7.375	60569
6	25	114	0.3556	0.7362	9.219	70303
7	30	115	0.381	0.7888	11.063	78740
8	35	117	0.4318	0.8939	12.906	81056
9	40	119	0.4826	0.9991	14.75	82901
10	45	121	0.5334	1.1043	16.594	84364
11	50	123	0.5842	1.2095	18.438	85587
12	55	124	0.6096	1.2621	20.282	90223
13	60	126	0.6604	1.3672	22.126	90854
14	65	128	0.7112	1.4724	23.969	91394
15	70	130	0.762	1.5776	25.813	91863
16	75	132	0.8128	1.6828	27.657	92273
17	80	135	0.889	1.8405	29.501	89988
18	85	138	0.9652	1.9983	31.345	88064
19	90	143	1.0922	2.2612	33.189	82402
20	95	150	1.27	2.6293	35.032	74803
21	99.5	176	1.9304	3.9966	36.692	51543

                 Stress strain graph when wood is placed parallel to grains

Hence slope = 10.017

Perpendicular to grains:

Obs.	Load ‘P’	Deflection gauge reading	Deformation ‘δ’	%age strain= $\in =(\delta \times 100)/H$	σ = $P/A$	K= $P/\delta$
	(kN)		(mm)		N/mm2	N/mm
1	0	100
2	1	103	0.0762	0.1522	0.4273	13123
3	1.5	112	0.3048	0.6089	0.6410	4921
4	2	135	0.889	1.7762	0.8547	2249
5	2.5	203	2.6162	5.2271	1.0683	955
6	3	294	4.9276	9.8453	1.2820	608
7	3.5	390	7.366	14.7172	1.4957	475
8	4	464	9.2456	18.4727	1.7094	432
9	4.5	536	11.0744	22.1266	1.9231	406
10	5	597	12.6238	25.2223	2.1367	396
11	5.5	664	14.3256	28.6225	2.3504	383

          

               Stress to strain graph when wood is placed perpendicular to grains

Hence slope = 0.3886