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[Audio] EXPERIMENT NO. – 01 TENSILE TEST ON MILD STEEL Aim: To conduct Tension test on ferrous and non-ferrous alloys (mild steel / cast iron /aluminium etc.) and determine the following. 1. Yield stress 2. Ultimate stress 3. Breaking stress 4. Percentage elongation 5. Percentage reduction in area Reference: IS 1608 Method for tensile testing of steel products. Apparatus: Universal testing machine, test specimen, steel rule, vernier calipers..

[Audio] THEORY: The tensile test is most applied one of all mechanical tests. In this test, a test specimen is fixed into grips connected to a straining device and to a load-measuring device. (One end in stationary grips and the others are in movable grips). If the applied load is small enough, the deformation of any solid body is entirely elastic. An elastically deformed solid will return to its original form as soon as load is removed. However if the load is too large, the material can be deformed permanently. The initial part of the tension curve, which represents the manner in which solid undergoes plastic deformation is termed as plastic. The stress below which the deformation is essentially entirely elastic is known as the elastic limit of the material. In some materials like mild steel a sudden drop in load indicating both an upper and lower yield point denotes the onset of plastic deformation. However some materials do not exhibit a sharp yield point. During plastic deformation at lager extensions, strain hardening cannot compensate for the decrease in section and thus the load passes through a maximum and then begins to decrease. At this stage the ultimate strength, which is defined as the ration of the load on the specimen to the original cross-section area, reaches a maximum value..

[Audio] Until this point the deformation is uniform at all the sections of the specimen. Further loading will eventually cause neck formation and rupture follows. Usually a tension test is conducted at room temperature; the tensile load is applied slowly. During this test either round or flat specimens may be used. The load on the specimen is applied mechanically or hydraulically depending on the type of testing machine. OBSERVATION: 1. Nominal stress and Nominal strain: Original C/S area (A0) = 143.139 mm2 Original gauge length (L0) = 300 mm Weight of the bar = 0.3375 Kg Increase in gauge length (δL0) = 15 (mm) Nominal stress = Nominal Load / Original C/S area = 120 x 103 / 143.139 = 838.35 (N/ mm2) Nominal strain = δLA/L0 = 8 / 300 = 0.0267.

[Audio] These are diagram of universal testing machine And stress strain diagram for mild steel .A Universal Testing Machine (UTM) is a device used to test the mechanical properties of materials under tension, compression, or bending. It consists of two main parts — the loading unit, which applies the force, and the control unit, which measures and records load and deformation. The UTM helps determine important material properties like yield strength, ultimate strength, and elongation. The stress-strain diagram for mild steel starts with a linear portion, where stress is directly proportional to strain, obeying Hooke's law. Beyond this, the curve reaches the yield point, where the material begins to deform permanently with little increase in stress. After yielding, the stress increases again until it reaches the ultimate stress point, representing the maximum load the material can bear. Beyond this point, the curve drops slightly due to necking, where the cross-sectional area reduces significantly. Finally, the material fractures at the breaking point, marking the end of the test..

[Audio] 2. Limit of Proportionality: Stress is proportional to strain up to this point. Normal Stress = Normal Load / Original C/S area = 114 x 103 / 143.139 = 796.43 (N/ mm2) Normal Strain = (δ LO)A /LO = 7 / 300 = 0.023 3. Elastic limit: When the load is removed at A, the specimen will go back to original Dimensions Nominal stress = PA/AO = Normal Load / Original C/S area = 114 x 103 / 143.139 = 796.43 (N/ mm2) Normal Strain = (δ LO) B/LO = 7 / 300 = 0.023.

[Audio] If the specimen is loaded beyond elastic limit it will undergo permanent strain i.e. Plastic deformation. 4. Upper yield point: Nominal stress = PB/AO Nominal stress = Upper Yield Load / Original C/S area = 120 x 103 / 143.139 = 838.35 (N/ mm2) Nominal strain = (δ LO) D/LO = 8 / 300 = 0.0267 5. Lower yield point: Nominal stress = PC/AO = Lower Yield Load / Original C/S area = 118 x 103 / 143.139 = 824.37 (N/ mm2) Nominal strain = (δ LO) C/LO = 7.5 / 300 = 0.025.

[Audio] 6. Ultimate load or maximum load point: Nominal ultimate stress = PD/AO = Maximum Load / Original C/S area = 124 x 103 / 143.139 = 866.29 (N/ mm2) Nominal strain = (δ LO) D/LO = 20.60 / 300 = 0.068 7. Fracture Load point F: Nominal fracture stress = PE/AO = Fracture Load / Original C/S area = 122 x 103 / 143.139 = 852.32 (N/ mm2) Nominal strain at fracture = (δ LO) E/LO = 27 / 300 = 0.09.

[Audio] 8. Young's modulus (E): Young's modulus (E) = Stress/Strain =120 x 103 / 0.0267 = 44.9438202 x 105 N/mm2 9. Yield point Elongation: Elongation taking place in the specimen from C to D this is taking place without increase in stress. Percentage Elongation = (Final Length – Initial Length) x 100 Initial Length = (315-300) x 100 300 = 5 % Percentage reduction in area = (Original area –Reduced area) x100 Original C/S area = (143.139 – 78.539) x 100.

[Audio] PROCEDURE: 1. Measure the original gauge length, weight and diameter of the specimen. 2. Insert the specimen into grips of the test machine. 3. Begin the load application and record load vs elongation data. 4. Take the readings more frequently as yield point is approached. 5. Measure elongation values. 6. Continue the test till fracture occurs. 7. By joining the two broken halves of the specimen together measure the final length and diameter of specimen at fracture. RESULTS: Tension test for given specimen was conducted and the results are as follows 1. Yield stress = 838.35 N/ mm2 2. Ultimate stress = 866.29 N/ mm2 3. Breaking stress = 852.32 N/ mm2 4. Young`s modulus = 44.9438202 x 105 N/mm2.