CHAPTER V

WOOD QUALITY TEST

A. Depreciation Wood

1. Objectives

a. To determine the percentage of wood angle.

b. To find out what happens to wood shrinkage from wet to dry. In this case

the depreciation based on the shrinkage of the wood grain.

2. Basic Theory

Determine the moisture content in wood timber volume, less water in

the wood, it also decreases the volume of the timber. Because wood is a

shrinkage of volume shrinkage, the shrinkage is happening in three directions

dimensions:

a. Shrinkage direction toward the center (radial)

b. Directions shrinkage in the direction of the tangent line type (tangential)

c. Direction shrinkage in the direction of the length of the rod (axial)

3. Materials used

a. Wood samples of I (wood camphor)

b. Wood samples II (wood meranti)

4. Apparatus

a. Calipers

b. Scale

c. Oven with a max temperature of 150 ° C made in Germany

5. Steps Work

a. Each sample was coded teams, parallel fiber = a, b = perpendicular fibers,

fiber direction = c.

b. Considering each sample timber.

c. Measure the length, width, and thickness of wood (tangential, radial, axial)

d. Incorporate wood into the oven for ± 24 hours.

e. The next day put into oxilator for ± 15 minutes, and then weighed.

f. Measuring wood shrinkage.

95

6. Workflow

Figure V.1 Workflow of Depreciation Wood

Preparing Apparatus and Materials :a. Wood samples of I (wood camphor)b. Wood samples of II (wood meranti)c. Calipersd. Scalee. Oven

Depreciation Wood

Start

Steps Work:a. Each sample was coded teams, parallel fiber = a, b =

perpendicular fibers, fiber direction = c.b. Considering each sample timber.c. Measure the length, width, and thickness of wood

(tangential, radial, axial)d. Incorporate wood into the oven for ± 24 hours.e. The next day put into oxilator for ± 15 minutes, and then

weighed.f. Measuring wood shrinkage.

Observing Test Results

Data Analysis

Finish

Conclusion

7. Observation Results.

Figure V.1 Steps Work Depreciation Wood

Informations Before oven After oven

Camphor Meranti Camphor Meranti

Length (cm)Wide (cm)High (cm)

10,86,84,9

10,26,94,7

10,76,44,6

106,74,5

Table V.2 Tables Wood Sample Observations

Samples

Weight (gram) Before oven After oven

Before oven

After oven

atangential

(cm)

Bradial(cm)

caxial(cm)

a’tangential

(cm)

b’radial(cm)

c’axial(cm)

Camphor (I)

285 265 5,2 4,2 10,8 5,1 3,9 10,7

Meranti (II)

230 210 5,8 3,6 10,2 5,7 3,5 10

8. Data Analysis

a. Sample I ( Wood Camphor )

Sringkage length woods

1) Direction axial = c – c’= 10,8 – 10,7= 0,1 cm

Large depresiation =

0,110,8 x

= 0.926 %2) Direction radial = b – b’

= 4,2 – 3,9 = 0,3 cm

Large depresiation =

0,34,2 x

= 7,143 %3) Direction tangential = a – a’ = 5,2 – 5,1 = 0,1 cm

Large depresiation =

0,15,2 x

= 1,923 % Water content of wood camphor

1) Weight of water = weight before oven – weight after oven = 285 – 265 = 20 gram

2) The water content of the dry weight =

weight waterweight after oven x 100%

=

20265 x

= 7,5471 %

3) The water content of the wet weight =

weight waterweight before oven x 100

=

20285 x

= 7,0175 %

The level of moisture

Formula : X =

(1,15 . 6 x )−6 kv6kv x 100 %

By : 6x = weight of the object before the oven

6kv = weight of the object after the oven

X = moisture content of wood

XKa =

(1,15 .285 )−265265 x

= 23,679 %

b. Sample II (Meranti wood)

The length of wood shrinkage

1) Direction axial = c – c’ = 10,2 – 10 = 0,2 cm

Large Depreciation =

0,210,2 x 100 %

= 1,96 % 2) Direction radial = b – b’ = 3,6 – 3,5 = 0,1 cm

Large Depreciation =

0,13,6 x 100 %

= 2,78 % 3) Direction tangential = a – a’ = 5,8 – 5,7 = 0,1 cm

Large Depreciation =

0,15,8 x 100 %

= 1,72 %

The water content of the wood1) Weight water = weight before oven – weight after oven = 230 – 210 = 20 gram

2) The water content of the dry weight =

weight waterweight after oven x100 %

=

20210 x 100 %

= 9,524 %

3) The water content of the wet weight =

weight waterweight before oven x100 %

=

20230 x 100 %

= 8,69 % Meranti wood moisture levels

Formula : X =

(1,15 . 6 x )−6 kv6kv

x100%

By : 6x = weight of the object before the oven

6kv = weight of the body after oven

X = moisture content of wood

XJa =

(1,15 . 230 )−210210 x 100%

= 25,9524 %

c. Average price of wood shrinkage

a = shrinkage direction c ( axial ) =

0 ., 926 %+1 ,96 %2 = 1,443 %

b = shrinkage direction b (radial) =

7 ,143 %+2 ,78 %2 = 4,9615%

c = shrinkage direction a (tangential) =

1, 923 %+1 ,72 , %2 =1,8215 %

9. Conclusion

a. From the above experiments obtained:

a) Sample I (Camphor wood)

1) Losses timber directions axial (c) = 0.926%

2) Losses timber directions radial (b) = 7.143 %

3) Losses timber directions tangensial (a) = 1.923 %

b). Sampel II (Meranti wood)

1) Losses timber directions axial (c) = 1.96 %

2) Losses timber directions radial (b) = 2.78 %

3) Losses timber directions tangensial (a) = 1.72 %

b. From the calculation of wood camphor the moisture content above is

equal to 23.679 % is about 20%, it can be said the wood was wet.

c. From the calculation of meranti wood the moisture content above is

equal to 25.9524 % is about 20%, it can be said the wood was wet.

d. From the samples I and II, the average wood shrinkage axial direction

(c) < wood shrinkage radial direction (b) > wood shrinkage tangential

(a).

e. Camphor wood moisture content > Meranti wood moisture content

10. Suggestions

a. Pieces of wood should be flat and angled corners should, because it will

affect the measurement of the length of wood.

b. In the measurement, caution should be considered properly.

c. Timber should be examined first, whether cracked or not to affect wood

shrinkage.

B. Pressure Strong Test Wood

1. Objectives

To find the powerful urge to know the strength of the wood so that

the wood properly.

2. Basic Theory

Wood for building or construction should be a good and healthy.

Provided that all the attributes and shortcomings of existing, will not

damage or diminish the value of construction (building). In Indonesia

Timber Construction Regulations of 1961 stated that for some types of

wood, which allowed voltage taken rather low. But this rule may also be

given to the evidence of the investigation can be accounted for.

Table V.3 Wood Class Data Classification

Class DensityAbsolute bending strength (kg/cm³)

Compressive strength of absolute

(kg/cm³)IIIIIIIVV

> 11001100 – 750750 – 500500 – 360

< 360

> 11001100 – 750750 – 500500 – 360

< 360

> 600600 – 425425 – 300300 – 215

< 215

3. Materials Used

Table V.4 Material Test Data Urges Strong Wood

No Data Wood Wood camphor Wood Meranti

1.2.3.4.5.6.7.

Wood defectsLength ( p )Wide ( l )High (t )

Weight ( s )Volume ( v )Density ( Bj )

-30 cm6,6 cm4,9 cm

700 gram970,2 cm3

0,72 gr/cm3

-30 cm6,6 cm5 cm

510 gram990 cm3

0,52 gr/cm3

4. Apparatus

a. Scale

b. Ruler bracket

c. Sandpaper

d. Calipers

e. Saw

f. Test equipment demanded

5. Steps Work

a. Cutting wood with a length of ±30 cm, ± 4 cm wide and ±5 cm tall,

then measure the length, width and height are actually using calipers.

b. The weighing timber.

c. Wood insert into test equipment and records pressed wood shortening

occurs, starting from the moment the needle indicates the load 2.5 kN,

5 kN; 7.5 kN, and so on (multiplier 2.5 kN)..

d. Draw a sketch of timber damage caused by the load acting on the

timber.

6. Workflow

7.Start

Pressure Strong Test Wood

Steps Work:a. Measure the width of the test surface in the central part of the

sample length, in millimeters.b. Placing the sample on recording devices, so that the test surface

facing up. Placing the press (punch) at odds with the test sample. Place the test sample and recording devices in the test machine. Perform loading sample rate continuously with constant pressure or constant rate of movement of the loading head, simultaneously recording devices will record the load and deformation. The rate of increase in load should be such that the test length not less than 2 minutes.

c. Testing continued until the deformation reaches 2.5 mm. It can be seen from the diagram the pressure readings on the machine or deformation on the gauge. Expenses associated with the deformation fmax should be noted.

Observing Test Results

Preparing Apparatus and Materials :a. Scaleb. Ruler bracketc. Sandpaperd. Caliperse. Sawf. Test equipment demanded

Data Analysis

Conclusion

Finish

Figure V.2 Workflow of Pressure Strong Wood

7. Observation Results

Table V.5 Observation data pressure strong test camphor wood

No.Weight P

(kN)Depreciations

( ∆L ) cm

1 0.02 0.13 0.24 0.35 0.4

Table V.6 Stress and strain calculations camphor wood

No.Weight (P) Depreciations σ=P/A

ε=ΔL/L0

Correction

kN Kg ΔL (cm) (Kg/cm²)Analytic

( ε' )

Correction :

x+0 .333x + 1 . 000 =

376 . 81475 . 36

475.26 X + 159.42 = 185.51 X + 185.51292.75 X = 26.09

X = 0.089

σp = 465.00 kg/cm² σmax = 492.260 kg/cm² σ0,05 = 475.00 kg/cm² εp = 1.05 εmax = 1.18 ε0,05 = 1.1

a). Modulus of elasticity =

σ p

ε p =

465 . 001 .05 = 442.857 kg/cm²

b). Modulus of elasticity = ½ . σp . εp

= ½ . 465.00 . 1.05

= 244.125 kg/cm²

c). Modulus secant =

σ0 , 05

ε0 , 05 =

475 . 0001 .1 = 431.818 kg/cm²

d). Unknown : P = 16500 kg A = 345 cm² σk = 16500 = 47,826 kg/cm² = 4,7826 MPa

345Absolute power σk = 4,7826 MPa

Table V.7 Observation data pressure strong test meranti wood

No.Weight P

(kN)Depreciation( ∆L (cm))

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Table V.8 Stress and strain calculations meranti wood

No.Weight (P) Depreciation σ=P/A

ε=ΔL/L0

Correction

kN Kg ΔL (Kg/cm²)Analytic

(ε')

Correction : X + 0. 333 = 57.971 X + 4.33 411.594 411.59 X + 137.20 = 57.971 X + 251.21 353.62 X = 114.01

X = 0.322

σp = 349.700 kg/cm² σmax = 421.59 kg/cm² σ0,05 = 391.800 kg/cm² εp = 3.667 εmax = 4.650 ε0,05 = 4.100

a). Modulus of elasticity =

σ p

ε p =

349 .7003 .667 = 95.364 kg/cm²

b). Modulus of elastic = ½ . σp . εp

= ½ x 349.700 x 3.667 = 2564.699 kg/cm²

c). Modulus secant =

σ0 , 05

ε0 , 05 =

391 .8004 .100 = 95.561 kg/cm²

d). Unknown : P = 14200 kg A = 345 cm²

σk =

14200345 = 41,159 kg/cm² = 4,1159 MPa

Absolute power σk = 4,1159 MPa

8. Conclusion

a. From the experimental results strongly urged camphor wood by using a

hydraulic insisted, is obtained:

σp = 465.00 kg/cm²

σmax = 492.26 kg/cm²

σ0,05 = 475.00 kg/cm²

εp = 1.05

εmax = 1.18

ε0,05 = 1.100

modulus elasticity = 442.857 kg/cm²

modulus kenyal = 224.125 kg/cm²

modulus secant = 431.818 kg/cm²

Camphor wood has absolute power σk = 47,826 kg/cm²

b. From the experimental results strongly urged meranti wood using a

hydraulic insisted, is obtained:

σp = 349.700 kg/cm²

σmax = 421.59 kg/cm²

σ0,05 = 391.800 kg/cm²

εp = 3.667

εmax = 4.650

ε0,05 = 4.100

modulus elasticity = 95.364 kg/cm²

modulus kenyal = 2564.699 kg/cm²

modulus secant = 95.561 kg/cm²

Meranti timber has absolute power σk = 41,159 kg/cm²

9. Suggestions

a. Wood urged to be tested must have a flat surface and angled so that the

entire cross section can receive urgent force evenly.

b. The accuracy in the measurement of length, width and height of timber

to be tested must be really careful, because it will affect the amount of

stress and strain that occurs.

c. In the installation of wood that will be tested in a hydraulic test

equipment urge to be completely straight.

d. In the notice of motion and the needle on the record when pressed

should be careful, because it supports the validity of data obtained

Figure V.5 Wood material after pressure strength test

C

⅔h

C. Flexural Strength Test of Wood

1. Objectives

To determine the flexural strength of wood so as to know the

strength of the wood very well.

2. Basic Theory

If the log is placed on two pedestal, burdened with force P, the top

edge fibers of the stem and urged each other on the bottom edge fibers will

each attraction. Because the fibers of the top edge of each emergency, there

will be a compressive stress. In contrast to the lower edge of the fiber tensile

stress will occur. The voltage is called the bending stress (σ lt). At the

boundary between the compressive stress and tensile stress, no fibers were

voltage = 0, meaning no voltage occurs. This tension lies in a straight line or

plane-called neutral line, as shown in the picture:

(a) Beams loaded (b) Looks beam (c) Diagram of stress

Figure V.6 Wood Flexure Strength Test

If the voltage is happening (σ LT) has reached the voltage permits (σ lt), it is

considered to be neutral line at half the height of the beam (½ h). At the

h

P

⅓h

TFiber tensile

Normal line

Fiber press

moment this is still a balance is: C compressive force equal to the tensile

force T.

The amount of compressive force or tensile force can be calculated by the

formula:

C = T = diagram area x wide beam C = ½ h . lt x b 2 C = ¼ . h . lt . b

Due to the tie style and compressive force, moment magnitude can cause

M = the force x distance between compressive force and tensile strength

= C x ⅔ h = ¼ . b . h lt . ⅔ h = 1/6 . b . h² . lt

Since the moment of detention W = 1/6 . b . h² , the stress obtained

permission σ lt

δ lt = M / V

Therefore in planning suffers beam or rod bending moment is not

connected, use the following formula:

δ = M / V lt

3. Materials Used

Table V.9 Data Materials Wood Flexure Strength Test

No Information Meranti Camphor1 Length (cm) 50 502 Wide (cm) 6,9 6,93 High (cm) 4,6 4,54 Weight (gr) 1275 11205 Volume (cm³) 1587 1552,56 Density (gr/cm³) 0,8 0,727 Area(cm2) 345 3458 P max (kg) 1900 24009 Stress (kg/cm2) 5,5072 6,9565

4. Apparatus

a. Scale

b. Ruler bracket

c. Calipers

d. Flexure test equipment

e. Desicator

f. Oven

g. Sandpaper

5. Steps Work

a. Timber is weighed and then measured the length, width and height.

b. Calculate wood density.

c. Wood put in flexural test equipment and record the maximum pressure

and measure the deflection of wood going up the wood cracked /

broken.

d. Draw a sketch of timber damage caused by the load acting on the wood

6. Workflow

Figure V.7 Workflow of Strong Flexure Wood

Start

Flexural Strength Test of Wood

Preparing Apparatus and Materials :a. Scale b. Ruler Bracketc. Calipersd. Flexure test equipmente. Desicatorf. Oveng. Sandpaper

Steps Work :a. Timber is weighed and then measured the length, width and

height.b. Calculate materials and tools.c. Wood put in flexural test equipment and record the

maximum pressure and measure the deflection of wood going up the wood cracked / broken.

d. Draw a sketch of timber damage caused by the load acting on the timber.

Observing Test Results

Data Analysis

Conclusion

Finish

7. Observation Results

a. Kind of Wood Camphor

Length = 50 cm

Wide = 6,9 cm

High = 4,5 cm

Weight of wood (S) = 1120 gram

Volume (V) = p x l x t

= 50 x 6,9 x 4,5

= 1552,5 cm³

Density (γ) = S / V = 1120 / 1552,5 = 0,72 gram/cm³

Pmaks = 24 kN = 2400 Kg

Area (A) = p x l = 50 x 6,9

= 345 cm²

Stress (σ) = 3xPmaksxL/ 2xbxh2

= 3x2400x50 / 2x6.9x4.5

= 5797.101 kg/cm²

b. Kind of Wood Meranti

Length = 50 cm

Wide = 6,9 cm

High = 4,6 cm

Weight of Wood (S) = 1275 gram

Volume (V) = p x l x t = 50 x 6,9 x 4,6

= 1587 cm³

Density (γ) = S / V = 1275 / 1587

= 0,8 gram/cm³

Pmaks = 19 kN = 1900 kg

Area (A) = p x l = 50 x 6,9

= 345 cm²

Stress (σ) = 3xPmaksxL/ 2xbxh2

= 3x1900x50 / 2x6.9x4.6

= 4489.603 kg/cm²

8. Conclusion

a. For camphor wood type:

1) Wood fractured at pressures of 2400 kg

2) Wood experiencing flexure stress 5797.101 kg/cm²

3) Density = 0,72 gr/cm2

According to the 1961 PKKI camphor wood are included in Class V.

b. For Meranti Wood Type:

1) Wood fractured at pressures of 1900 kg

2) Wood experiencing flexure stress 4489.603 kg/cm²

3) Density = 0,8 gr/cm2

According to the 1961 PKKI meranti are included in Class V.

9. Suggestions

a. In the watch and record the distance the needle must be flexible at

carefully, because it supports the truth of the data obtained.

b. Test materials cultivated in the dry state so as to obtain the maximum

value of flexure.