I. ObjectiveAt the end of the experiment, the student must able to perform a tensile test to destruction on two different tensile test pieces in order to determine their tensile properties.II. MaterialsThe materials and apparatus that were used in the experiment were, Mosanto (Hounsfiled) Tensometer, two tensile test pieces of ferrous material, load indicator and a digital dial test indictor.III. MethodologyIn testing for the tensile strength of a material, Mosanto (Hounsfiled) Tensometer was used.
The first part of the experiment was setting up of the said apparatus. And then, the load increment to be used in the experiment was determined and the following were used as load increment, 0.5N, 1N, 5N, 10N or 20N. The students monitored the tensile testing of the ferrous material, one student turning the handle of the tensometer, the other one keep an eye on the load in the load indicator while the third student checks the ferrous material. When the ferrous material reaches its fracture or becomes deformed, the load associated in the load indicator is the predetermined load.The steps above were repeated on different load increments.
The procedures were also done on the other ferrous material.IV. ResultsFigure 1. Stress-strain diagram of the two steel samples.
V. DiscussionFigure 1 shows the stress-strain diagram of the two ferrous materials used in the experiment. Stress is the force applied per unit area. It measures the capacity of the material to resist the tensile force, thus a material with high tensile stress can resist higher tensile force compared to other material.
While a strain is the deformation due to the application of some external forces like tensile force. In the figure, it can be seen that the graph of the stress-strain diagram is different on the two materials. From the figure, it suggest that high carbon steel will eventually need more tensile force to deform the material compared to the less carbon steel content. It shows that at the same deformation, the 0.8% carbon steel needs more tensile force compared to the 0.1% carbon steel.
Another interesting to the graph is that the rupture strength of the two materials was also different. 0.1% carbon steel has higher rupture strength than of the 0.8% carbon. This shows that the 0.8% carbon steel is more ductile than the other material.
VI. ConclusionForm the experiment, we can conclude that material with high carbon cnontent will have higher tensile strength but is prone to ductility. The strength of a material is not the only criterion that must be considered in designing structures. The stiffness of a material is frequently of equal importance. In a lesser degree, mechanical properties such as hardness, toughness, and ductility determine the selection of a material.
These properties are determined by making test on the materials.Exercise 2 ‘Forces on a Beam’I. ObjectiveAt the end of the experiment, the students must be able to determine the bending moment in a simply supported beam under the action of a series concentrated loads, and compare the bending moment obtained with that calculated theoretically.II. ApparatusIn determining the moment of a simply supported beam under the application of a concentrated load, the apparatus that were used in the experiment were, two weighing scale that are fixed at both ends.
An aluminium alloy channel section is supported between the scales. Three suspension hooks and load hangers enable loads to be applied at various points along the 1m span of beam. Various loads in Newton have been supplied.
III. MethodThe first step done in the experiment is the setting up of the beam at the two weighing scale that are fixed at both ends. 1 meter distance was allotted between the two weighing scales. Three load hangers were used in the experiment and were placed at different position depending on the choice of the students.
The distance from one end of each three hangers was recorded as well the load applied by each hanger. The mass shown in the weighing scale was then recorded. The steps were replicated for the second time and also for other three different distances of the three loads. .IV. ResultsFirst Beam 200mm = 0.2m100mm = 0.
1m100mm = 0.1m600mm = 0.6m RA = 1.125KgF = m x gF = 1.125 x 9.81F = 11.
04N RB = 425g = 0.425kgF = m x gF = 0.425 x 9.81F = 4.17N ?fx = 0?fy = 00 = 5.5 + 5.5 + 5.
5 _ RA _ RB0 = 16.5 _ RA _ RBB.M = 00 = (5.5 x 0.2) + (5.
5 x 0.3) + (5.5 x 0.4) _ (RA x 0) _ (RB x 1)0 = 1.1 + 1.65 + 2.
2 _ 0 _ RB0 = 4.95 _ RBRB = 4.95N0 = 16.5 _ RA _ RB0 = 16.5 _ RA _ 4.
95RA = 16.5 _ 4.95RA = 11.55N (0 < x < 0.2)S.F = 11.
55NWhen x = 0m S.F = 11.55NWhen x = 0.1m S.F = 11.
55NWhen x = 0.2m S.F = 11.
55N B.M = 11.55xWhen x = 0m B.
M = 0NmWhen x = 0.1m B.M = 1.155NmWhen x = 0.2m B.M = 2.
31Nm (0.2 < x < 0.3)S.F = 11.
55 _ 5.5 = 6.05NWhen x = 0.
2m S.F = 6.05NWhen x = 0.3m S.
F = 6.05N B.M = 11.55x _ 5.5(x _ 0.2)B.M = 11.55x _ 5.
5x + 1.1B.M = 6.05x + 1.1When x = 0.2m B.M = 2.
31NmWhen x = 0.3m B.M = 2.915Nm (0.
3 < x < 0.4)S.F = 11.
55 _ 5.5 _ 5.5 = 0.55NWhen x = 0.3m S.F = 0.55NWhen x = 0.
4m S.F = 0.55N B.M = 11.55x _ 5.5(x _ 0.
2) _ 5.5(x _ 0.3)B.
M = 11.55x _ 5.5x + 1.1 _ 5.5x + 1.
65B.M = 0.55x + 2.75When x = 0.3m B.
M = 2.915NmWhen x = 0.4m B.M = 2.97Nm (0.
4 < x < 1)S.F = 11.55 _ 5.5 _ 5.5 _ 5.5 = _ 4.
95NWhen x = 0.4m S.F = _ 4.95NWhen x = 0.5m S.F = _ 4.95NWhen x = 0.
6m S.F = _ 4.95NWhen x = 0.7m S.F = _ 4.95NWhen x = 0.8m S.
F = _ 4.95NWhen x = 0.9m S.F = _ 4.95NWhen x = 1m S.
F = _ 4.95N B.M = 11.55x _ 5.5(x _ 0.2) _ 5.5(x _ 0.
3) _ 5.5(x _ 0.4)B.M = 11.55x _ 5.5x + 1.1 _ 5.
5x + 1.65 _ 5.5x + 2.2B.M = _ 4.95x + 4.95When x = 0.
4m B.M = 2.97NmWhen x = 0.5m B.M = 2.475NmWhen x = 0.
6m B.M = 1.98NmWhen x = 0.7m B.
M = 1.485NmWhen x = 0.8m B.
M = 0.99NmWhen x = 0.9m B.
M = 0.495NmWhen x = 1m B.M = 0NmFigure 1: Shear Force Graph (First Beam)Figure 2: Bending Moment Graph (First Beam)Second Beam 200mm = 0.2m200mm = 0.
2m300mm = 0.3m300mm = 0.3m RA = 1.3KgF = m x gF = 1.
3 x 9.81F = 12.75N RB = 1kgF = m x gF = 1 x 9.
81F = 9.81N ?fx = 0?fy = 00 = 5.5 + 10.
5 + 7.5 _ RA _ RB0 = 23.5 _ RA _ RBB.M = 00 = (5.5 x 0.
2) + (10.5 x 0.4) + (7.5 x 0.7 _ (RA x 0) _ (RB x 1)0 = 1.1 + 4.2 + 5.25 _ 0 _ RB0 = 10.
55 _ RBRB = 10.55N0 = 23.5 _ RA _ RB0 = 23.5 _ RA _ 4.
95RA = 23.5 _ 10.55RA = 12.
95N (0 < x < 0.2)S.F = 12.95NWhen x = 0m S.F = 12.95NWhen x = 0.
1m S.F = 12.95NWhen x = 0.2m S.F = 12.
95N B.M = 12.95xWhen x = 0m B.M = 0NmWhen x = 0.1m B.M = 1.
295NmWhen x = 0.2m B.M = 2.59Nm (0.2 < x < 0.4)S.F = 12.
95 _ 5.5 = 7.45NWhen x = 0.2m S.F = 7.45NWhen x = 0.3m S.
F = 7.45NWhen x = 0.4m S.
F = 7.45NB.M = 12.
95x _ 5.5(x _ 0.2)B.M = 12.95x _ 5.
5x + 1.1B.M = 7.45x + 1.1When x = 0.2m B.M = 2.
59NmWhen x = 0.3m B.M = 3.335NmWhen x = 0.
4m B.M = 4.08Nm (0.4 < x < 0.7)S.
F = 12.95 _ 5.5 _ 10.
5 = _ 3.05NWhen x = 0.4m S.F = _ 3.05NWhen x = 0.5m S.F = _ 3.05NWhen x = 0.
6m S.F = _ 3.05NWhen x = 0.7m S.
F = _ 3.05N B.M = 12.95x _ 5.
5(x _ 0.2) _ 10.5(x _ 0.4)B.M = 12.95x _ 5.5x + 1.1 _ 10.
5x + 4.2B.M = _ 3.
05x + 5.3When x = 0.4m B.M = 4.08NmWhen x = 0.
5m B.M = 3.775NmWhen x = 0.6m B.M = 3.47NmWhen x = 0.
7m B.M = 3.165Nm (0.7 < x < 1)S.F = 12.
95 _ 5.5 _ 10.5 _ 7.5 = _ 10.55NWhen x = 0.7m S.F = _ 10.
55NWhen x = 0.8m S.F = _ 10.
55NWhen x = 0.9m S.F = _ 10.
55NWhen x = 1m S.F = _ 10.55N B.M = 12.95x _ 5.5(x _ 0.2) _ 10.5(x _ 0.
4) _ 7.5(x _ 0.7)B.M = 12.95x _ 5.5x + 1.1 _ 10.5x + 4.2 _ 7.5x + 5.25B.M = _ 10.55x + 10.55When x = 0.7m B.M = 3.165NmWhen x = 0.8m B.M = 2.11NmWhen x = 0.9m B.M = 1.055NmWhen x = 1m B.M = 0Nm Figure 3: Shear Force Graph (Second Beam) Figure 4: Bending Moment Graph (Second Beam) V. DiscussionFrom the result of the experiment, it shows that the force applied by the hangers on the beams will be equal to the reaction of the beam to the weighing scale. This shows that the system or the set up of the experiment is equilibrium. Another good evidence that shows this equilibrium condition is the result of the theoretical computation. From the computed values, the computed values were plotted in a graph to determine the shear and bending moment of the given problem. It shows from the graph that end of the curve returns to the zero value of the shear force just in figure 1 and 3. Also, the bending moment of the two beams shows that there is equilibrium in the moment applied to each beam.VI. ConclusionConsequently, 0.5N was the initial force on the load hangers which shows 0Kg at RA and RB on scale measurement. When a force is applied on the load hangers, it shows that at any moment when a force is applied on the load hangers the resultant force changes with respect of RA and RB. From the calculations, it proves that the beam is at equilibrium and which shows that ‘the bending moment on top of the beam equals the bending moment at the bottom of the beam’. The beam is also at equilibrium, when the shear force on top of the beam equals the shear force at the bottom of the beam. The experiment of force on a beam reveals the accuracy of the shear force and bending moment through calculation. Simply, mean that the experiment of force on a beam and calculation of the beam are the same.Despite calculating the force of RA and RB, the resultant force of the bending moment is slight different from the scale measurement when converting the mass (Kg) to force (N). Reference:Andrew Pytel, Ferdinand Singer, Strength of Materials (Fourth edn.: HarperCollins Publisher Inc. , 1987).