Nanyang Technological University Second Year Civil Engineering Course CV2712: Civil Engineering Laboratory 2B Group Laboratory Report For Experiment Lab CT1, CT2, CT3 Concrete Technology & Tensile Properties of Steel Reinforcements Name of Students : Goh Chung Hao Jeremy, Yeoh Chun Yen, Peh Xin Yun, Ang Jin Hao Group :4 Date : 24 March 2016 Testing on Hardened Concrete Discussion 1. Comment on the difference between designed and experimental values for workability and for the strength of hardened concrete. Comment on the importance (if any) of water. The workability and strength of hardened concrete are higher in their experimental values as compared to the design values. Water plays an important role in determining both the workability and strength of hardened concrete. For the workability part, generally a higher water to cement ratio would lead to increased workability for the concrete. As for the compressive strength of hardened concrete, we can see from the compressive strength results obtained from the tests done on our hardened concretes. Concretes that are water cured have a higher compressive strength compared to those that are air stored. From this we can conclude that water plays a crucial role in determining various properties of a concrete sample. 2. Comment on the effect of different curing condition on development of concrete strength. Based on the compressive strength test, we noticed that water cured concrete has a higher mean compressive strength of 34.2 MPa than that of air stored concrete with a mean compressive strength of 30.72 MPa. The reason behind the lower compressive strength for air stored concrete is due to water being lost to the surrounding and thus water to cement ratio will decrease leading to lower capillary porosity which in turn causes a decrease in the compressive strength. As for water cured concrete, the water content is being maintained or increased in the concrete and thus this leads to the water to cement ratio to increase and also ensure continuous hydration. This will lead to a decrease in capillary porosity and thus increase the compressive strength of the concrete. 3. Compare the apparent concrete strength of cube and cylinder specimen and comment on the differences. By comparing the mean compressive strength of air stored cube and cylinder, we noticed that the cube specimen has a higher mean compressive strength of 30.72 MPa as compared to cylinder’s mean compressive strength of 19.80 MPa. There is a big difference in the strength of 10.90 MPa. The reason behind this huge difference is mainly due to the loaded areas. For the cube is has a loaded area of 10000mm2 which is greater than that of the cylinder’s area of 7857mm2. The larger the area, the higher the confinement resist will be against the specimen expansion, which in turn will lead to higher compressive strength. 4. Comment on the reliability/ suitability of the mix design method. 5. What conclusion can you make from this experiment on mix designing, slump test, wet density and compressive tests? From the results obtained from the mixed design, and also with the values obtained from the actual concrete design, it is obvious that the actual design values are relatively higher as compared to the mixed design ones. This is not surprising as the mixed design values can only be obtained in an ideal world situation which obviously cannot be achieved in real life due to the many experimental errors and the conditions that we work in. One such factor that can affect the slump test would be the temperature that we work in. As Singapore is quite a hot country, the concrete might lose some of its water content early on, therefore losing some of its workability in its early stage. For this we may consider slightly increasing the water content to counter this issue. As from the compressive tests, the mean strength obtained from concretes that are water cured are relatively higher than the mean strength obtained from concretes that are air stored. From this we can see that it is more beneficial to water cure the concrete, so they can be stronger and thus can be used for more purposes. Tensile Splitting Strength of Concrete Discussion 1. Compare the compressive strength and the splitting tensile strength; and also the flexural strength and the splitting tensile strength for water cured concrete. The mean compressive strength of the air stored concrete cylinder is 19.8 MPa and the mean splitting tensile strength will be 2.60 MPa. By comparing both values, we notice that for the same dimensions of the concrete cylinder, the compressive strength is much higher than that of the splitting tensile strength by 86.9%. This is explain by the fact that concrete is stronger in compression as compared to tension. As for water cured concrete cube, it has a mean compressive strength of 34.2 MPa whereas the air stored concrete cube, it has a mean compressive strength of 30.72 MPa. This is mainly due to the water content as concrete that has gone through water curing will be able maintain the water in the concrete and then leading to higher compressive strength as compared to air stored concrete. As for air stored concrete prism, it has a mean flexural strength of 3.55 MPa. 2. Is the flexural strength higher than the splitting tensile strength? Why? Formula for flexural strength f cf = Formula for splitting tensile strength PL bd2 = 3P a bd2 2F f ct = πLD Yes, flexural strength is higher than that of splitting tensile strength. By comparing the formulas, we will notice that for flexural strength, it is being divided by bd2 whereby b is the width and d is the height. The values of b and d is considered just a small fraction of L which is the length of the beam. Thus this leads to the overall value of bd2 to be small as compared to the overall value of πLD . With this by dividing a big value, it will lead to splitting tensile strength having a smaller value as compared flexural strength. Moreover, for splitting tensile strength, it is making use of compressive force to cause the specimen to fail under tension. Since concrete itself is not able resist tension than that of compression due to low tensile strength and brittle nature, it will fail under tension first and thus a smaller value of splitting tensile strength. 3. Between flexural strength and splitting tensile strength, which one does give closer value to the “true tensile strength”? Why? The splitting tensile strength will give a closer value to the “true tensile strength”. 4. Conclusion Tensile Properties of Steel Reinforcement Discussion 1. Complete the data sheets and enclose the load-strain curves. Refer to the attached Data Sheet No. 5 and load-strain curves. 2. Compare the 2 load-strain diagrams obtained for specimens A and B. Describe the difference in the shape of the load-strain diagrams. Make reference to yielding behavior, ultimate tensile strength, elongation etc. Both the load-strain curves for specimens A and B has similar shape as both of them are made of the same material. 3. Explain what property of the material does modules of elasticity provide? Explain why the modulus of elasticity from the 2 specimens are similar or different? The property of the material that the modules of elasticity provide will be the stiffness of the material. As based from the formula, modules of elasticity is the ratio of stress over strain. The stiffer the material, the higher the elastic modules will be which mean the material has a high resistance to being deformed elastically. The modulus of elasticity of the 2 specimens 4. Of the strengths of materials (yield strength, ultimate strength, breaking strength etc.) determined, describe which property is most important for the design engineer, and explain why? Among all of the strengths of the materials, yield strength is the most important for the design engineer. Yield strength of the material is defined as the stress at which the material will start to deform physically. This means we will be able to know the maximum amount of load the steel can hold before it will start to deform plastically. By knowing the yield strength of the material, we will be able to select which material to use such that it will be able to overcome the applied force.
