Civil lab Essay Example

Civil Lab Report

civil lab

College of Engineering

Civil Engineering Lab Report

Module: Civil Engineering Laboratory 1 (EG107) Student Name:

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The Lab Report discusses numerous experiments including:

  • Equilibrium of forces test.

  • Deflection of Beam.

  • Concrete mix and casting Design.

  • Hardened Concrete Testing.

  • Steel and Timber Testing.

Practical 1: Equilibrium

The purpose of this experiment is to determine the equilibrium position of a metal plate that is attached onto through the use of different loads.

  • Method Used and Results

A cord ring was attached to a board through the use of strings that were later hooked to four pulleys, which were located at the plate’s corners. The four pulleys were held in tension because of the numerous weights acting on the strings, which were hooked to the pulleys.

With the help of drawing materials, the direction of resultant forces were drawn on a paper indicating their respective vertical component through the use of two cord rings that are attached to the plate.

1) The following are the results of string attached to three forces:

Angle with horizontal (x-axes)

The equilibrium conditions are obtained based on the following calculations:

1.5cos54º+0.5cos76º=1.5cos48º

1.003=1.003

2) The following are the results obtained through attaching a single string to six forces

Vertical forces component (y-axes)

1.5sin42º

1.5cos48º

1.5cos38º

1.5sin77º

1.5cos13º

1.5sin52

The equilibrium conditions are obtained based on the following calculations:

1.5sin42º+1.5sin52º+1.5sin77º=1.5cos48º+1.5cos38º+1.5cos13º

3.6N=3.6N

Conclusion

The experiment was successful because no problems were witnessed while the errors were within the acceptable limits. In addition, the sources of the errors are traceable and it includes apparatus imprecision and lack of incorporation of friction.

Practical 2: Deflection of Beam

  • Aim:-

from two symmetrical loads are measured to determine the modulus. 2 and δ1 The aim of the practical is to calculate the Young’s modulus (E) through measuring material deflection of the beams. The displacements δ

  • Method

    • The following image represents the method used during the experiment:

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    • In determining the beam cross section, a micrometer was used

    • The displacement gauges were adjusted to read zero

    • was measured 1 δAt each end of the 1 m long beam, loads were loaded on the two symmetrical points (P) and the vertical displacement indicated by

    • The weights were then increased and the load forced the steel beam to bend because of the weight. The displacement resulting from the weights was measured through the use of gauges.

    • The procedure of changing weights, analysis the symmetrical load requirements and determining the spacing (d) were repeated for numerous times. The Young’s modulus of the beam was then calculated through the use of the varying loads scenarios.

  • Results and Calculations:

The following table summarizes the results that were obtained after the experiment:

Thickness of Beam

oCalculations:

Based on the results obtained from set 1 of the data and utilization of the largest load scenario, it is possible to calculate the bending moment indicated by M through the use of the formula: M = Pa. P is the external load (N) and (a) is the perpendicular distance.

In the calculation, it is important to convert any weights to Newton and it is possible through using the following approach:

M=Pa= 10.1 * 0.1 = 1.01Nm-2

The second moment of area I was obtained through using the formula (I= (1/12) bh3) whereby the length was changed from centimeters to meters — from the formulae:

b= width of the beam h= height of beam.

Therefore I = (1/12)*(0.01266) (0.00531)3 = 1.57955962*10-10


civil lab 3In determining the curvature of the deflected beam, the methodology used was through the Pythagoras theorem indicated above – the following are the meaning of the letters:

D is the distance between the two displacement gauges

Using the formulae, we obtain:

R= ((0.1982)2 + (8.2*10) / 2(3*10-4) =22.82

In calculating the Young’s modulus, the following equation is used:

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Using the equation, we obtain

E = (1.01 * 22.82) / (1.55-10) = 1.48*1011GPa

  • Conclusion:

The value of Young’s Modulus is 128 GPA. The value is high but based on the material used, the value is appropriate. In carrying out the research, numerous errors were evident that may affect the results. For instance, the displacement gauges that were used to obtain the measures were old meaning that the springs were unable to release appropriate tension. Therefore, the measurement may have been inappropriate because of the limitations of the measuring equipment. To minimize the errors from the old measuring gauge, we were forced to tap the gauge to ensure the right readings were obtained. The manipulation of the gauge meant that the obtained data was inaccurate and may affect the credibility of the entire experiment. In future, new instruments and equipments or tools that were working appropriately should be used to minimize the errors. The hooks were also sliding along the bar meaning that the loads may have not been entirely symmetrical. To solve the problem, the solution is to use materials that enable to fix the weights appropriately in place.

Practical 3: Concrete mix and Casting

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  • Aim:-

The aim of the practical is to design a concrete mix to enable carrying out the workability test on the wet concrete. In addition, the concrete is hardened and concrete testing is done to determine the effectiveness of the mixture.

  • Method:

  • (V) was designed 3A concrete mixture of 0.035 m

  • 2 in order to fulfill the requirements of cube strength of 24 N/mm2The average cube strength of 28 days was used. It was 30 N/nm

  • The ratio of water and cement was 0.6

  • It is imperative to note the ratio since excess water can create problems. Some of the problems include reducing its durability and creates channels on the concrete mix surface

  • Results and Calculations:

The Mass of the concrete mix can be obtained using this formula:-

The following formulae can be used to obtain concrete mix mass:

M = M (cement) + M (water) + M (aggregate)

M = C + 0.6C + 5C = 6.6C

The C stands for the mass of the cement

The following equitation is used to calculate the mass of the cement:

)3(assume the density of mixed concrete = 2300 kg/mMass = Density * Volume

  • V=0.035m3

Therefore;

M= 2300V

6.6C = 2300 * 0.035 C = 12.197Kg

M (cement) = 9 Kg M (water) = 5Kg

M (aggregate) = 65Kg

Fine Aggregate = 22.7 kg

Course Aggregate = 42.1 kg

  • It is also important to note that all the water was not added and around 2kg was left. The weight of the bucker was 0.675 kg.

  • After obtaining the right data, a mixing machine was used to mix the concrete

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  • The following section discusses the tests used to find the workability of the concrete mix

  • The Slump Test:

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    • The mixed concrete was placed in a cone shaped mould

    • The mould was poured into the cone shaped object in three stages and in each stage, the mould was hit twenty times through the use of cylindrical shape object

    • After the mould had hardened, the cone was removed

    • Measurement was done to determine the difference in height between the concrete

    • Theoretically, the greater the slump means that it is easier to work with it and the right measurement was 74 mm

  • Compact factor test:

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The section discusses the second test for the sample:

    • The sample of the concrete is placed on the upper hoper of the brim

    • The trap door is then opened so that the concrete can flow downwards into the lower hopper

    • The trap door of the lower hopper is opened to enable the flow of concrete to the cylinder

    • Through the use of plane blades, the excess concrete remaining on the top level of the cylinder is removed or trimmed

    • The concrete in the cylinder is then weighed; the weight obtained is the partially compacted concrete

    • The cylinder is then again filled with fresh concrete, which is then vibrated to obtain full compaction. The cylinder concrete is weighted again to obtain the fully compacted concrete

  • Calculations:

The following equations and procedure are used to calculate the CF

CF = (Partial Compact) / (Fully Compact)

M1= Partially Compacted Mass = 16.8 kg M2 = Fully Compacted Mass =17.4 kg Mc = Mass of the Cylinder = 6.2kg

CF = (M1 – Mc) / (M2- Mc) = (16.8 – 6.2) / (17.4 – 6.2) =0.946

  • Conclusion:

From the workability tests, the mix design was appropriate because the tests indicated that the concrete was mixed accordingly. The slump test obtained was within the specified range while the compaction factor test was below 1. The data indicates that the overall concrete was high in workability capability. Based on the results and analysis, the workability was high even to high the target was medium workability. The reason may be attributed to excessive water that was added during the mix.

Errors are inherent in anty experioenment and during the mixing and adding the different components contribute to increase in number of errors. The errors affect the reliability of the results and influence the conclusion. The level of compaction also determines the level of usability and workability of the concrete and associated errors. To minimize complications of compactions, the compaction process and procedure should be effective. The unique and common during compaction that may occur is use of excessive water that results in collapse of the slump test. Too much water makes the mixture too fluid and wet raising additional complications. It results in segregation in which the larger aggregate sink to the bottom of the mixture during compaction. On the other hand, when a single section of the concrete collapses, it means that it is too dry making it unworkable. To prevent these challenges, all the water should not be added in an instant but small amounts of water should be added until the testing confirms the workability of the slump.

Practical 4: Hardened Concrete Testing

  • Aim:

The aim of the experiment is to measures tensile and compressive strength through testing beam, cylinder and cube samples. In addition, the experiment measures the stiffness of the concrete through static means and the outcome of the tests is discussed extensively in the report. One beam, two cylinders and two cubes are used and also the concrete was left to dry for seven days.

  • Methods and Results:

o Direct Compression Test:

The purpose of the test is to measure the strength of the concrete. The testing process started with the 150 mm cubes and the load was applied through the use of compression testing machine. The procedure was repeated for the 150 mm diameter cylinder.

Force(N)

)2Area(m

Stress(Pa)

Stress(MPa)

33333333

Cylinder

30337087

And through conducting numerous calculations as indicated in the table, the results indicate that the cube has a higher compressive stress value compared to the cylinder. The higher compressive stress value translates to higher compressive strength. In addition, the cube has a higher capacity to resist crushing compared with the cylinder meaning that the cube has a largest surface area for compressor to impact pressure.

o Splitting Test:

    • The split test was accomplished through placing the cube block on steel packers. The placement of the cube provided a knife edge load across the center of the cube

    • The compression testing machine was used to load the cube to the systems

    • The data is recorded based on the changes evidenced

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The tensile stress is achieved through the use of the following formula:

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Whereby:

  • P represents the splitting load (P = 10.1 KN)

  • D is the cube’s length (d = 0.150 m)

  • W is the packer’s width (w. 0.0125)

Therefore: 3.43MPa00

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From the calculation, it is evident that concrete has a higher compressive strength but lower tensile capability. Therefore, concrete is stronger when compressed.

o Static Stiffness measurement:

  • The cylinder that had metal studs in four quadrants was loaded into the compression testing machine

  • The load was increased to a failure load that was estimated based on the first cylinder

  • With the Demec strain gauge, the strain was measured and recorded after load increment. The Demec strain gauge measures the movement between the studs and the following table summarizes the data recorded:

% Of Destruction Load Applied

Quadrant

0

0

The following picture illustrates the use of tools to measure the strain and record the load added

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  • Bending Test:

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    • A concrete load was placed onto the failure equipment and the load was documented

    • The beam was supported by supports at each end

    • With the use of compression testing machine, a load is applied on the beam’s center

    • With the help of Euler-Bernoulli beam theory, the tensile strength was calculated. The result of calculation was 12.7 KN

The following formulae was used to calculate the results

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P is the split load

A is the distance from the applied load to support

B and h is the dimensions of the cross sectional area of the beam used.

Applying the formula

= (6 * 12700 * 0.15) / (0.098 * 0.0104) = 11.2MPa

Practical 5: Steel and Timber Testing

    • Aims:

      • The aim of the testing is to determine the behavior of timber and steel under load until failure occurs

      • The Young’s modulus and the ductility of steel from a simple tensile test is done to accomplish the requirements of tensile strength

    • Method of Steel Testing:

      • The Dartec Testing Machine was used to accomplish the requirements of the experiment

      • The steel specimens were obtained through subjecting same steel to different heat treatments

      • The two samples are then placed into the testing machine and compressed until failure is reported

      • The sample from the same steel should be of the same dimensions

      • The load is increased gradually until formation is reported

      • The strain gauge equipment plays an important role in the entire process

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Nm)Kess(rtS

  • Results and Graphs:

  • Normal Steel (non-annealed):

civil lab 19Diameter: 9.99 mm so the radius (R1) = 4.995mm

  • Annealed Steel:

Diameter: 9.96 mm so the radius (R2) = 4.98mm

  • Gauge length = 25mm

  • Young’s Modulus = (STRESS) /(STRAIN)

  • 2RInitial Cross sectional area =

  • =78.3832 * (4.995)Cross sectional area of the normal (non-annealed) steel=

  • =77.9132 * (4.98)Cross sectional area of the annealed steel=

The following table summarizes the information analyzed:

Tensile Strength

Young’s Modulus

Elongation at failure

0.881KNm

110.75GPa

Annealed

0.629KNm

118.70GPa

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The following graph illustrate the stress-strain for the normal tested steel versus the Annealed tested steel

Nm)Kess(rtS

Nm)Kess(rtS

  • Timber Testing:

  • Aim and Method:-

Through the use of compression, two specimens of the same timber type were tested. One of the timber would be loaded transversely while the other axially. The aim is to incorporate the aspect of fibers in the testing process. Data was collected and the amount of load was also recorded.

  • civil lab 21civil lab 22Graphs: