Weekly Laboratory Report Formatting GuideAuthor/Student Name

Group Name

ABSTRACT

The purpose of this lab was to test the compression strength of the concrete and obtained the E value for the modulus of the elasticity. Two sample mix concrete has been place under compression stress and the results for the maximum capacity of the stress vs strain has been obtained in this lab report. The slope of the curve will indicated the experimental value for modulus of elasticity for the concrete. Therefore relative value for the each sample has been compared with the theoretically value and the results have been included in this report.

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INTRODUCTION

Concrete has played an important role in the development of the material of the structure and buildings. From the time of the post revolution in the industry, Concrete has been introduced into the market. Since concrete has higher capability of the compression strain, there has been many researches based on the experiment which has already developed variety method of the mixing in the design of a

METHODS

Sample crushing method in this laboratory work has been done. Prior to the testing, in previous lab work it was intended to make a well graded aggregate mixed based on the calculation of the aggregated mix lab work. So after the mix was made, it was poured into cylindrical molds with known height and diameter. To obtain the certain level of the healing, the concrete specimens were left for certain days till it reaches the relative strength of the compression. In today’s lab work, two sample of the concreted intended for testing the sample were removed from molds and the compression machine has been set up. The machine has already been connected to the computer system which operate and obtaining readings for test. After the collaboration and placing the sample into the machine as shown in figure 1, the machine starts compression. Based on the strength of the sample, the compression strength will vary just before the crushing happens. For the first sample, the data for the maximum load and extension has been recorded which has been shown in table 1. Similar to the first sample, the second specimen of the concreted sample was tested and the results have been recorded into table 2. After all the data have been obtained from, calculation has been done and the result as below have found.

Figure 1: showing the adjuster operating system .

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Figure 2: sample just before starting the compression.

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Figure 3: showing the concrete cylindrical specimen has failed at its maximum compression load.

Formula has been used for the calculation:

Equations:

σ=E∗ϵ (1)

σ= LoadArea (2)

ϵ (longitudenal)=∆lL

(3)

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Area=(π d2)/ 4(4)

f 'c=0.85∗σ (max ) (5 )

E=0.57∗(√ f c ' ) (6)

Area=π∗(D2 )2

4(7)

E=slope=σϵ

(8)

Table1: data and result for the specimen number1 .

Time Extension Load deflection strain Stres(s) (in) (lbf) (In) in/in psi

71.914 0.084 42274.42700 0.01381 0.00173 3364.0972.014 0.084 42268.658 0.01384 0.00173 3363.6372.114 0.084 42269.203 0.01391 0.00174 3363.6872.214 0.084 42245.503 0.01394 0.00174 3361.7972.314 0.084 42218.329 0.01400 0.00175 3359.6372.414 0.084 42169.690 0.01403 0.00175 3355.7672.514 0.084 42159.654 0.01409 0.00176 3354.9672.614 0.084 42190.997 0.01416 0.00177 3357.4572.714 0.084 42228.575 0.01422 0.00178 3360.4473.614 0.085 42465.854 0.01481 0.00185 3379.3373.714 0.085 42484.931 0.01491 0.00186 3380.8473.814 0.085 42517.392 0.01497 0.00187 3383.4373.914 0.085 42555.918 0.01506 0.00188 3386.4974.014 0.085 42599.594 0.01513 0.00189 3389.9774.114 0.085 42651.799 0.01522 0.00190 3394.1274.214 0.085 42685.493 0.01531 0.00191 3396.8074.314 0.085 42716.629 0.01541 0.00193 3399.2874.414 0.086 42765.584 0.01550 0.00194 3403.1874.514 0.086 42845.943 0.01556 0.00195 3409.5774.614 0.086 42965.524 0.01566 0.00196 3419.0974.714 0.086 43076.281 0.01575 0.00197 3427.90

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Table2 : data and results for the specimen number 2.

Time Extension Load

Compressive strain (Extension)

Compressive stress deflection

(s) (in) (lbf) (in/in) (psi)0 0 0 0 0 0

7.1 0.0954 495.5089 0.0954 0.011925 39.431347.2 0.095833 643.4692 0.095833 0.011979 51.20565

27.91 0.1032 16733.89 0.1032 0.0129 1331.64128.01 0.1032 16796.62 0.1032 0.0129 1336.63367.51 0.108133 40439.25 0.108133 0.013517 3218.05467.61 0.108133 40499.16 0.108133 0.013517 3222.82167.71 0.108167 40556 0.108167 0.013521 3227.34470.51 0.108667 42029.33 0.108667 0.013583 3344.58870.61 0.108667 42069.27 0.108667 0.013583 3347.76670.71 0.1087 42104.38 0.1087 0.013588 3350.5670.81 0.108733 42139.58 0.108733 0.013592 3353.36170.91 0.108767 42168.75 0.108767 0.013596 3355.68371.01 0.108767 42196.32 0.108767 0.013596 3357.87671.11 0.1088 42219.44 0.1088 0.0136 3359.71671.21 0.108833 42238.65 0.108833 0.013604 3361.24571.31 0.108867 42254.09 0.108867 0.013608 3362.47471.41 0.1089 42261.43 0.1089 0.013613 3363.05871.51 0.108933 42260.32 0.108933 0.013617 3362.9771.61 0.108967 42237.9 0.108967 0.013621 3361.185

Graph showing the LVD baseline

.

RESULTS:

From the data and calculation as indicated below, it has been found that the maximum stress and strain for the specimen number one are 3427.9 psi and 0.00154∧theEexp and Etho for the same sample are 2.81 ksi3.083 ksi, therefore the % errors is 8.0 %.

For the sample number due same calculation as for sample one has been done and the max stress found to be 3363. 058 psi and strain equal to0.00154. Also, the values of the modulus of elasticity for excremental found to be 2.18 and for theoretical it was found as 3.01 ksi and the % Errors from comparison of these valuefind to be 27 %

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0.055 0.06 0.065 0.07 0.075 0.08 0.085 0.090

5000100001500020000250003000035000400004500050000

Load vs Extention Diaghram

Extention (inches)

Load

(lb)

Figure 4: the value for the baseline is 0.0715 for specimen number 1

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.0180.00

5,000.0010,000.0015,000.0020,000.0025,000.0030,000.0035,000.0040,000.0045,000.0050,000.00

Load vs extention (deflection) Specimen # 1

Deflection( extention) (in)

Load

(lb)

Figure 5: load verses defelction ghraph showing for sample 1.

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LVCD_Baseline 0.0715

Max load = 43076.3 lb

0 0.0005 0.001 0.0015 0.002 0.00250

5001000150020002500300035004000

Stress vs strain specimen #1

strian

stre

ss (

psi) max stress=

3427.9 psi

Figure 6: stress versus strain ghrpah for sample 1.

0.08 0.085 0.09 0.095 0.1 0.105 0.11 0.115 0.12-5000

0

5000

10000

15000

20000

25000

30000

35000

40000

45000

Load vs extention for specimen # 2

extention (inches)

Axis Title

Figure 7: load versus strain graph for sample 2.

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LVCD_Baseline = 0.088 inch

0 0.005 0.01 0.015 0.02 0.0250

50001000015000200002500030000350004000045000

Load vs corrected Deflection specimen #2

Extention (corrected deflection) (in)

Load

(lb)

Figure 8: load versus corrected correction for sample 2.

0 0.0005 0.001 0.0015 0.002 0.0025 0.0030

500

1000

1500

2000

2500

3000

3500

4000 Stress vs strain for specimen #2

Strain

Stre

ss

(lb

)

Max stress = 3363.058 psi

Figure 9: showing the stress vs strain for sample 2.

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Max load = 42261.40 lb

Calculatioan:

A. Calculation of the sample one.

Area=π d2

4=( 3.14∗42

4 )=12.57 i n2 (4 )

σ (max )= parea

= 43076.28lb12.566∈2¿=3428 psi ¿

f 'c=0.85∗σ (max )=(0.85 )∗(3427.9∨3428 )=2913.72 psi (5 )

E( theo)=57∗(√ f c' ) = 57000 * (√2913.72=3.083 ksi (6)

E(exp)=slope= σϵ= 4327.9

0.00154=2.81ksi (8 )

Errors%=3.083−2.813.083 *100 = 8.0 %

Calculation for specimen #2:

σ (max )= parea

= 42261.4412.566∈2¿=3363.058 psi¿

f 'c=0.85∗σ (max )=(0.85 )∗(3363.058 )=2858.68 psi

E theo ¿57000∗(√ f c ' ) = 57 000 * (√2858.68=3 .01ksi

Eexp=slopeof t h e stress vs stain gh raph=¿

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3363.0580.00154

=(2.18 ) ksiErrors%=3.01−2.183.01

*100 = 27 %

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DISCUSSION

Here is where you should discuss whether the findings are consistent or inconsistent with what was previously believed (or some theoretical model). Why is consistent or not? What are some sources of discrepancy, inaccuracy, or potential error?

CONCLUSION

APPENDIX

.

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