UKCustomPapers | This is the analysis of the unloaded teeth:

This is the analysis of the unloaded teeth:

This is the analysis of the unloaded teeth:
. anova bs t if l == 0

Number of obs = 30 R-squared = 0.3255
Root MSE = 34.2935 Adj R-squared = 0.2756

Pairwise comparisons of marginal linear predictions

Margins : asbalanced

—————————
| Number of
| Comparisons
————-+————-
t | 3
—————————

—————————
| Number of
| Comparisons
————-+————-
t | 3
—————————

Incisor Group Maximum Compressive Load (N)
Control 1.5 N 3 N 6 N 9 N 12 N
1 55.48 51.68 55.19 52.73 20.41 84.26
2 81.82 44.08 59.19 141.87 33.12 13.16
3 72.02 78.91 97.55 12.68 32.8 55.85
4 44.59 76.36 83.86 80.52 43.4 46.28
5 65.47 69.9 66.43 83.4 58.37 48.29
6 81.54 68.55 42.96 71.09 114.86 97.3
7 67.18 71.67 58.71 51.99 36.02 27.59
8 68.17 49.72 102.98 43.27 58.31 36.11
9 71.99 32.12 15.29 45.59 47.77 54.31
10 59.73 59.32 20.43 16.8 74.79 54.36
Mean 66.8 60.23 60.26 59.99 51.99 51.75
Standard Deviation 11.41 15.41 29.34 37.27 27.12 24.77

Premolars Group Maximum Compressive Load (N)
Control 1.5 N 3 N 6 N 9 N 12 N
1 133.83 75.24 40.07 46.4 46.17 20.47
2 157.26 46.13 36 15.86 29.86 47.62
3 76.38 173.22 61.51 66.27 14.78 32.48
4 162.92 73.63 49.25 20.02 39.38 39.56
5 88.77 35.35 54.62 30.31 50.2 51.43
6 75.83 152.8 59.56 5.01 44.81 44.36
7 88.23 96.02 44.3 48.55 37.07 21.48
8 108.74 21.76 43.96 68.64 30.64 14.8
9 39.26 37.42 32.93 27.19 39.71 65.69
10 97.43 75.12 43.34 37.06 41.51 54.21
Mean 102.86 78.67 46.55 36.53 37.41 39.21
Standard Deviation 38.7 50.18 9.58 21.02 10.20 16.60

Molars Group Maximum Compressive Load (N)
Control 1.5 N 3 N 6 N 9 N 12 N
1 126.66 155.63 58.33 59.78 63.09 28.34
2 90.29 69.47 167.23 169.34 40.44 38.98
3 227.72 56.9 126.72 108.09 95.15 89.09
4 87.67 92.14 91.92 82.36 65.05 22.67
5 139.12 82.99 130.85 140.43 45.46 41.63
6 92.78 125.03 79.65 148.29 39.11 63.58
7 83.52 103.81 92.66 158.54 202.02 130.07
8 100.44 176.81 150.89 52.89 133.67 35.24
9 146.7 85.08 98.04 125.76 63.5 52.68
10 117.21 58.45 143.64 75.67 122.12 68.67
Mean 121.21 100.63 113.99 112.12 86.96 57.10
Standard Deviation 43.59 40.45 34.95 42.43 52.18 32.66

Incisor
. tabstat bs if t == 1, stats(n mean sd min max) f(%7.2f) by(l) notot

Summary for variables: bs by categories of: l

l | N mean sd min max
——–+————————————————–
Control | 10.00 66.80 11.41 44.59 81.82
1.5 N | 10.00 60.23 15.41 32.12 78.91
3 N | 10.00 60.26 29.34 15.29 102.98
6 N | 10.00 59.99 37.27 12.68 141.87
9 N | 10.00 51.98 27.12 20.41 114.86
12 N | 10.00 51.75 24.77 13.16 97.30
250

Incisor
200
150
100
50
0
Control 1.5 N 3 N 6 N 9 N 12 N

. anova bs l if t == 1

Number of obs = 60 R-squared = 0.0442 Root MSE = 25.7139 Adj R-squared = -0.0443

Source | Partial SS df MS F Prob > F
———–+—————————————————-

Model | 1651.90045 5 330.38009 0.50 0.7751
|
l | 1651.90045 5 330.38009 0.50 0.7751
|
Residual | 35705.1629 54 661.206719
———–+—————————————————- Total | 37357.0633 59 633.170564

No significant effect of load on bond strength.

Molar
. tabstat bs if t == 2, stats(n mean sd min max) f(%7.2f) by(l) notot

Summary for variables: bs by categories of: l

l | N mean sd min max
——–+————————————————–
Control | 10.00 121.21 43.59 83.52 227.72
1.5 N | 10.00 100.63 40.44 56.90 176.81
3 N | 10.00 113.99 34.95 58.33 167.23
6 N | 10.00 112.12 42.43 52.89 169.34
9 N | 10.00 86.96 52.18 39.11 202.02
12 N | 10.00 57.10 32.66 22.67 130.07
250

Molar
200
150
100
50
0
Control 1.5 N 3 N 6 N 9 N 12 N

. anova bs l if t == 2

Number of obs = 60 R-squared = 0.2307 Root MSE = 41.5271 Adj R-squared = 0.1595

Source | Partial SS df MS F Prob > F
———–+—————————————————-

Model | 27930.8583 5 5586.17166 3.24 0.0125
|
l | 27930.8583 5 5586.17166 3.24 0.0125
|
Residual | 93122.808 54 1724.49644
———–+—————————————————- Total | 121053.666 59 2051.75706

A significant effect of load. All pairwise comparisons:
. pwcompare l, mcomp(sidak) groups eff

Pairwise comparisons of marginal linear predictions Margins : asbalanced
—————————
| Number of
| Comparisons
————-+————-
l | 15
—————————
———————————————-
————-+——————————–
l |

Control | 121.211 13.13201 A
1.5 N | 100.631 13.13201 AB
3 N | 113.993 13.13201 A
6 N | 112.115 13.13201 AB
9 N | 86.961 13.13201 AB
12 N | 57.095 13.13201 B
———————————————-
Note: Margins sharing a letter in the group label are not significantly different at the 5% level.

—————————
| Number of
| Comparisons
————-+————-
l | 15
—————————

————————————————————————————

1.5 N vs Control | -20.58 18.57146 -1.11 0.992 -77.46478 36.30478
3 N vs Control | -7.218 18.57146 -0.39 1.000 -64.10278 49.66678
6 N vs Control | -9.096 18.57146 -0.49 1.000 -65.98078 47.78878
9 N vs Control | -34.25 18.57146 -1.84 0.667 -91.13478 22.63478
12 N vs Control | -64.116 18.57146 -3.45 0.016 -121.0008 -7.231216
3 N vs 1.5 N | 13.362 18.57146 0.72 1.000 -43.52278 70.24678
6 N vs 1.5 N | 11.484 18.57146 0.62 1.000 -45.40078 68.36878
9 N vs 1.5 N | -13.67 18.57146 -0.74 1.000 -70.55478 43.21478
12 N vs 1.5 N | -43.536 18.57146 -2.34 0.292 -100.4208 13.34878
6 N vs 3 N | -1.878 18.57146 -0.10 1.000 -58.76278 55.00678
9 N vs 3 N | -27.032 18.57146 -1.46 0.915 -83.91678 29.85278
12 N vs 3 N | -56.898 18.57146 -3.06 0.050 -113.7828 -.0132163
9 N vs 6 N | -25.154 18.57146 -1.35 0.950 -82.03878 31.73078
12 N vs 6 N | -55.02 18.57146 -2.96 0.066 -111.9048 1.864784
12 N vs 9 N | -29.866 18.57146 -1.61 0.836 -86.75078 27.01878
————————————————————————————

Adjusted Predictions of l with 95% CIs
Control 1.5 N 3 N 6 N 9 N 12 N l

There is a complicated distribution of significant comparisons.

Sequential comparison of loads:
. contrast a.l, mcomp(sidak) eff Contrasts of marginal linear predictions Margins : asbalanced
——————————————————————

| Sidak
| df F P>F P>F
——————–+———————————————
l |

—————————
| Number of
| Comparisons
————-+————-
l | 5
—————————

————————————————————————————-

(Control vs 1.5 N) | 20.58 18.57146 1.11 0.797 -28.86094 70.02094
(1.5 N vs 3 N) | -13.362 18.57146 -0.72 0.960 -62.80294 36.07894
(3 N vs 6 N) | 1.878 18.57146 0.10 1.000 -47.56294 51.31894
(6 N vs 9 N) | 25.154 18.57146 1.35 0.632 -24.28694 74.59494
(9 N vs 12 N) | 29.866 18.57146 1.61 0.453 -19.57494 79.30694

There are no sequential differences.

Premolar
. tabstat bs if t == 3, stats(n mean sd min max) f(%7.2f) by(l) notot

Summary for variables: bs by categories of: l

l | N mean sd min max
——–+————————————————–
Control | 10.00 102.87 38.70 39.26 162.92
1.5 N | 10.00 78.67 50.18 21.76 173.22
3 N | 10.00 46.55 9.58 32.93 61.51
6 N | 10.00 36.53 21.02 5.01 68.64
9 N | 10.00 37.41 10.20 14.78 50.20
12 N | 10.00 39.21 16.60 14.80 65.69
250

Premolar
200
150
100
50
0
Control 1.5 N 3 N 6 N 9 N 12 N l

. anova bs l if t == 3

Number of obs = 60 R-squared = 0.4615 Root MSE = 28.6614 Adj R-squared = 0.4116

Source | Partial SS df MS F Prob > F
———–+—————————————————-

Model | 38012.8157 5 7602.56314 9.25 0.0000
|
l | 38012.8157 5 7602.56314 9.25 0.0000
|
Residual | 44359.6489 54 821.474979
———–+—————————————————- Total | 82372.4646 59 1396.14347

A significant effect of load.

. pwcompare l, mcomp(sidak) groups eff

Control | 102.865 9.063526 C
1.5 N | 78.669 9.063526 BC
3 N | 46.554 9.063526 AB
6 N | 36.531 9.063526 A
9 N | 37.413 9.063526 A
12 N | 39.21 9.063526 A
———————————————-
Note: Margins sharing a letter in the group label are not significantly different at the 5% level.

—————————
| Number of
| Comparisons
————-+————-
l | 15
—————————

————————————————————————————

| Sidak Sidak
| Contrast Std. Err. t P>|t| [95% Conf. Interval]
——————-+—————————————————————-
1.5 N l
vs Control |
|
-24.196
12.81776
-1.89
0.632
-63.45707
15.06507
3 N vs Control | -56.311 12.81776 -4.39 0.001 -95.57207 -17.04993
6 N vs Control | -66.334 12.81776 -5.18 0.000 -105.5951 -27.07293
9 N vs Control | -65.452 12.81776 -5.11 0.000 -104.7131 -26.19093
12 N vs Control | -63.655 12.81776 -4.97 0.000 -102.9161 -24.39393
3 N vs 1.5 N | -32.115 12.81776 -2.51 0.206 -71.37607 7.14607
6 N vs 1.5 N | -42.138 12.81776 -3.29 0.026 -81.39907 -2.87693
9 N vs 1.5 N | -41.256 12.81776 -3.22 0.032 -80.51707 -1.99493
12 N vs 1.5 N | -39.459 12.81776 -3.08 0.048 -78.72007 -.1979302
6 N vs 3 N | -10.023 12.81776 -0.78 1.000 -49.28407 29.23807
9 N vs 3 N | -9.141 12.81776 -0.71 1.000 -48.40207 30.12007
12 N vs 3 N | -7.344 12.81776 -0.57 1.000 -46.60507 31.91707
9 N vs 6 N | .882 12.81776 0.07 1.000 -38.37907 40.14307
12 N vs 6 N | 2.679 12.81776 0.21 1.000 -36.58207 41.94007
12 N vs 9 N | 1.797 12.81776 0.14 1.000 -37.46407 41.05807
——————————————————————————–

Adjusted Predictions of l with 95% CIs
Control 1.5 N 3 N 6 N 9 N 12 N l

Sequential load comparison
. contrast a.l, mcomp(sidak) eff Contrasts of marginal linear predictions Margins : asbalanced
——————————————————————

| Sidak
| df F P>F P>F
——————–+———————————————
l |

—————————
| Number of
| Comparisons
————-+————-
l | 5
—————————

————————————————————————————-

(Control vs 1.5 N) | 24.196 12.81776 1.89 0.283 -9.927438 58.31944
(1.5 N vs 3 N) | 32.115 12.81776 2.51 0.074 -2.008438 66.23844
(3 N vs 6 N) | 10.023 12.81776 0.78 0.944 -24.10044 44.14644
(6 N vs 9 N) | -.882 12.81776 -0.07 1.000 -35.00544 33.24144
(9 N vs 12 N) | -1.797 12.81776 -0.14 1.000 -35.92044 32.32644
————————————————————————————-

There no sequential differences.
Introduction to the report . The topic is Shear bond strength of orthodontic bracket .

Assessment of the bond strength is performed on shear,tensile, and torsional load. In shear testing ,the bonded bracket is loaded by a blade in tension or compression or by a wire loop , therefore the force is parallel to the tooth’s long axis and the bracket slides parallel to the enamel surface. In tensile testing, the bracket is pulled perpendicularly from the enamel surface .I attach photo to understand it .

the aim of the project is to determine the relationship between the horizontal force on orthodontic bracket ( tensile
force) and shear force on the orthodontic bracket ( vertical force ) . ? is there relation or no

However the project Idea came from the clinical situation when orthodontist put orthodontic bracket for patient and his teeth irregular .In this case there is two force on bracket the first one shear force which is in vertical direction when patient bite on the bracket and the second force horizontal which called tensile force from the wire which ligated to bracket to aling the
teeth .

The Methods

The study consisted of 180 human teeth . The experiment consisted of 60 incisors, 60 premolars and 60 molars. The teeth were stored in normal saline (0.9% NaCl) in a closed plastic box. Teeth were divided into 6 group each one (10 incisors, 10 premolars and 10 molars) .

Each group was tested at a separate time. On the day of the test, teeth were placed into acrylic block with facial surface of the teeth expose and parallel to the chisel of the introns machine . Following the acrylic sit, the surface of teeth etched with phosphoric acid 37% for 10 – 15 sec. The teeth are subsequently washed and dried. Then, the XB bond resin is applied followed by light curing for 10- 20 sec, each tooth bond with corresponding bracket APC II which has preload transbond XT light cure adhesive .

The first group is the control group were the shear load measure with no horizontal force is applied. The second, third, fourth, fifth and sixth group were the shear load measure with horizontal forces applied of 1.5N, 3N, 6N, 9N, and 12N respectively. The acrylic block with bonded tooth is positioned in such way in instron machine so that the chisel of the instron is perpendicular to the bracket. Apart from control group, we used wire 016 which is 90 mm in length with two ended loop, one attached to the bracket with elastic module (for the incisors , premolars brackets and in molars the wire pass through the tubes) and the other side attached to electronic calibration holed manually to apply specific tension force in horizontal direction according to each group load( second group apply 1.5 N . third group I apply 3N ……………)
. Total shear load is recorded with instron machine with cross head speed of 1mm/min when the bracket came off.
The result

In the sheet attach i wrote the result
in result you will find the force i used and the type of tooth used e.g first one control incisor , it mean i did not use horizontal force and the shear force( compress load ) register by sliding the blade of the instron on the bracket of incisor , i did for ten incisors and the mean calculated and the stander deviation.
In the second group is written in 1.5 N incisors which indicate in this group i applied horizontal force ( tensile force ) 1.5 N for the incisors teeth then i slide the blade of the instron machine vertical on the bracket of the incisors
and the shear force ( compress load ) register , i did for ten incisors and the mean calculated and the stander deviation.
Similar for other group i apply the horizontal force on the bracket ,then i apply the vertical force by blade of the instron machine to evaluate the vertical force which applied by the blade of the instron machine on bracket is effected by the amount of horizontal force .
Theoretically if we increase the horizontal force on the bracket the force required by vertical force ( shear /compress load ) . to cause bracket off should be reduce
Statistic analysis
Used one way a nova and it show the shear force ( compress load ) reduce withe increase the horizontal force but in incisors not significant reduce . In Premolars and molar reduce and statistically significant but there are no sequential differences as the sequential comparison done .The reason in the result it may because the flat surface of the incisors , more force on the posterior teeth , nonuniform thickness of adhesive , the location of the blade of the instron machine on the bracket ,different shape of the teeth tested …. or other reason the writer need to find it from previous study on the factors effect the shear bond strength .
I start to write about the first table in result please add what you can see better :
In the incisors groups, the first group that is control group, ten incisors were applied to shear load only by the instron machine without horizontal force (tensile load) was applied, the mean and stander deviation were 66.8 and 11.41(table) respectively.
In the second group, ten incisors were applied to horizontal load 1.5 N. and their mean shear load( compress load ) and standard deviation were 60.23 and 29.34 respectively, when compare with the control group the shear load is reduced according to the mean of the shear load .
In the third group, also ten incisors were applied to horizontal (tensile) force 3N, and shear (compressive) load mean was 60.26 which approximately similar to compressive load of the second group where 1.5 N used as horizontal force. The maximum shear load in the third group was 102.98, which is higher than the maximum shear load in control group, which is 81.82. However, the stander deviation in the second and the third group were 15.41 and 29.34 respectively.
From fourth group of incisors groups the mean shear load reduce gradually. In fourth group ten incisors were applied to 6 N horizontal force, in the fifth group where 9 N horizontal force and in sixth group 12 N used as horizontal force , the mean shear load are 59.99 , 51.99 and 51.75 respectively . Variable results were available from the ten teeth members of the fourth group of incisor groups where 6N used as horizontal force as stander deviation is 37.27
From the above result, there is gradually reduction in compressive load with increase of horizontal (tensile load). Moreover, based on statistical analysis for the sixty incisors, there is no significant effect of horizontal (tensile) load on the shear load p=0.7751.

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This is the analysis of the unloaded teeth:

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