Tensile Loadings and Elastic and Plastic Regions - Lab Report Example

Tensile test The objective of the experiment was subject specimens of materials to tensile loadings and to compare the results in elastic and plastic regions. The prepared specimens were placed in the test ring and loaded to a point of failure. While loading there was recording of the load with the corresponding change in length. The measurements of cross section area were also recorded. Results for 0.4 % steel The load at yield point = 1200 N Cross section area at yield point= 20mm2 Yield stress =  = 675 N/mm2 Change in length at yield point = 2mm Strain at yield point = 2/25.4 = 0.0786 Maximum load = 17171 N Cross area at maximum load = 18mm2 Stress at maximum =  = 953.96 N/mm2 Change in length at point of maximum loading = 2.7 mm Strain at point of maximum loading = 2.7/25.4 = 0.106 Load at fracture point = 11700 Cross section area at fracture point = 7 mm2 Stress at fracture point =  = 1671 N/mm2 Change in length at fracture point = 4.5 Strain at fracture point = 4.5/25.4 = 0.177 The stress strain values are as summarized in TABLE 1 The graphic representation is as shown in line graph TABLE 1 Strain 0.05 0.0787 0.106 0.177 Stress 400 675 953.96 1671 Figure 1 From the figure A= Is the yield stress = 675 N/mm2 B= Is the ultimate tensile strength = 953.96N/mm2 C= Is fracture stress = 1671 N/mm2 Results for 0.15 % steel The load at yield point = 1200 N Cross section area at yield point= 20mm2 Yield stress =  = 600 N/mm2 Change in length at yield point = 2mm Strain at yield point = 2/25.4 = 0.0786 Maximum load = 16176 N Cross area at maximum load = 18mm2 Stress at maximum =  = 899 N/mm2 Change in length at point of maximum loading = 3.4 mm Strain at point of maximum loading = 3.4/25.4 = 0.1339 Load at fracture point = 13 800 N Cross section area at fracture point = 12 mm2 Stress at fracture point =  = 1150 N/mm2 Change in length at fracture point = 5.5 Strain at fracture point = 5.5/25.4 = 0.2165 The stress strain values are as summarized in TABLE 2 TABLE 2 Strain 0.03 0.0786 0.1339 0.2165 Stress 200 600 899 1151 Figure 2 From the figure A= Is the yield stress = 600N/mm2 B= Is the ultimate tensile strength =899 N/mm2 C= Is fracture stress = 1152 N/mm2 Hardness test Hardness is the measure of material resistance to scratching, abrasion, indentation or wear. The Brinell test and the Vickers tests were used in the testing of the hardness of 0.4 % steel and 0.15 % steel. The Brinell hardness test involves of indenting the material under test with hardened steel or carbide ball which is hardened and subjected to a load of 3000 kg. In case the materials are softer there can be reduction in loading to 1500 kg or 500 kg so as to avoid excessive indentation. During the test the full load is applied for 10 to 15 seconds in the case of iron and steel and for at least 30 seconds in the case of other metals. A low powered microscope is used to find the diameter of the indentation left in the test material. To find the Brinell harness number the load applied is divided by surface area of the indentation. The value for the diameter is an average value for the two readings which are at right angles. In Vickers test there are two distinct force ranges, micro (10g to 1000g) and macro (1kg to 100kg),that carters for testing requirements. In both ranges the indenter is the same thus the values (Vickers hardness values) are continuous for the various metals with the values typically ranging from HV100 to HV1000. Vickers values are considered to be independent of the test force used with the exception of test forces that are below 200g in the experiment a force of 10kg was used. The results 0.4% steel 0.15 % steel BHN Number BHN Number Reading 1 = 304 Reading 1 = 262 Reading 2 = 300 Reading 2 = 255 Average = 302 Average = 258 Vickers test = 279 HV Vickers test = 201HV Discussion From the tensile test and the hardness test different characteristics are observed for the two type of steel. 0.4% steel has high yield strength of 670N/mm2 compared to that of 0.15% steel. The values of the ultimate tensile strength and fracture stress were also in the 0.4% steel with values being 953.96 N/mm2 and 1671 N/mm2 respectively while values foe 0.15% steel were 899 N/mm2 and 1151N/mm2 respectively. The strain values for 0.4% steel were lower compared to those of 0.15% steel. This is an indication that 0.15% steel has high elasticity and high plasticity than 0.4% steel. In the hardness test 0.4% recorded higher values both for the Vickers test and the Brinell test. Thus it can be concluded that from the experiment it was proved that a higher carbon level increases the strength of the steel and also the hardness but reduces elasticity and plasticity (Nagpa G. R. 2009). Slump test The slump test was performed using the following procedure The dimensions of the slump cone were taken in order to ensure that it was clean Concrete was filled in three equal depth layers while tamping each layer using 25 stokes of a bullet nosed 16mm diameter rod. The concrete was heaped above the top of the cone before roding of the third layer was done The was topping up and then by use of rod with a sawing and rolling motion in order to strike the concrete level with the top of the cone The cone was carefully lifted straight up and clear to a count of between 5 and 10 seconds The rode was laid across the upturned slump cone the distance between the underside of the rode and the highest point of the concrete was measured ( true slump) The results obtained for the concrete at different w/c ratios are as summarized in TABLE 3 TABLE 3 Slump test results for the different mixes Water/Cement ratio (W/C) 0.8 0.75 0.7 0.65 0.6 0.5 Slump 125 116 82 65 50 45 Compacting factor test The following are the procedure that was used in performing the compacting factor test The bottom cylinder was weighed empty (WE) and concrete (not concrete used in the slump test) was filled in the upper hopper. The bottom door was released so that concrete fell into the lower hopper. The door of the lower hopper was released so that to allow the concrete to fall into a cylindrical mould underneath The surplus concrete was then removed, the mould wiped and the mould was then weighed to with the concrete the value being recorded and this was used to obtain weight of un compacted concrete as WU The cylinder was refilled with concrete in layers of depth 50mm where each layer was consolidated by vibration in order to obtain full compaction The weight of the cylinder with compacted soil was recorded and this was used in calculating the weight of compacted concrete WC From WC and WU compaction ratio (CF) was calculated as The results of the experiment are as summarized in the TABLE 4 W/C 0.8 0.75 0.7 0.65 0.6 0.55 WE+WU 17.1 17.11 17.29 17.3 17.35 17.4 WU 11.16 11.18 11.29 11.3 11.35 11.4 WC 11.09 11.1 11.2 11.4 11.6 11.7 CF = WC - WU 0.07 0.08 0.09 0.1 0.25 0.3 Discussion From the result of the slump test it can be seen that the higher the W/C the higher the value of the slump test. The slump test value of W/C of 0.75 and 0.8 were above 115 meaning that there was excess water. Concrete with excess water has low strength when dry even though it maybe of good workability. On the other hand concrete with low W/C value will results into a stronger concrete as long as the water content is too low to inhibit complete chemical reaction in the mix. From the compaction test it is observed that samples with high W/C had a low CF value. This is an indication that in most of the mixes the water was excess thus occupying some space after compaction. It should be noted that continued reduction of W/C does not guarantee increased compaction. This is due to the fact that when the mix is too dry compaction is hindered thus leading to lower values of CF. References Nagpal G.R. (1997). Machine design. Khanna Publishers, Dehli Read More