TORSION TEST. Figure 1 Schematic view of torsion test

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1 TORSION TEST 1. THE PURPOSE OF THE TEST The torsion test is performed for determining the properties of materials like shear modulus (G) and shear yield stress ( A ). 2. IDENTIFICATIONS: Shear modulus: Shear modulus is the slope of shear stress shear deformation curve s line section. Also known as stifness modulus.. Torque: The moment in the act of torsion. Angle of twist: : If a shaft of length L is subjected to a constant twisting moment T along its length, than the angle q through which one end of the bar will twist relative to the other is known is the angle of twist. 3. THE IMPORTANCE OF TORSION TEST AND FIELDS OF APPLICATION The torsion test of metalic metarials is generally used to determine yielding characteristics in high plastic zone stress. This test method does not have a broad use of application like tensile test and it is completely standardized. It is rarely used for determining general mechanic properties of materials. However, it is a test method required in theorical studies about plastic deformation and enginering applications to determine the extrusion and forging characteristics of metals. Torsion test is a method to determine forging characteristics of brittle materials like tool steels also in high temperatures. At the same time, it is a test which can be applied directly to machine parts like shaft axle and drill which torsion stress is highly important. 4. THEORETICAL KNOWLADGE: 4.1 Mechanical Properties in Torsion In torsion test, a torsion sample was stably mounted to the testing device and one end of the sample was fixed while other end was exposed to torsion torque (Fig.1). Shear stress occur in the sample with the effect of torsion. Torque (T) angle of twist () is obtained during the test (Fig.2). Figure 1 Schematic view of torsion test

2 Figure 2 Torque (T) angle of twist () 4.2 SHEAR STRESS In a cylinder which has a Radius c, the shear stress () occurred in Radius (less then c) was formulated: T In formula: T: Torque : Radius at where the shear stress occured : Polar moment of inertia 1 c 2 4 The moment of inertia in solid cylinders: (c: Radius) The moment of inertia in hallow cylinders: 1 c c (c 2 : outer Radius, c 1 : inner radius) Maximum shear stress occurs in the surface of sample, and we can calculate such as: Tc

3 4.3 SHEAR YIELDING STRENGTH Shear yielding strength is calculated from torsion test torque (T) angle of twist () diagram. It is clear to find shear strength in materials exhibiting yield point behavior (like steels). For other materials it is shown in Fig. 2. To find it: A line was drawn parallel to torsion test torque (T) angle of twist () curve s line section from the point = The intersection point of line and graph is the torque which the shear strength occurred. 4.4 SHEAR STRAIN The deformation occurred in specimen with the effect of shear stress is shear strain. It is calculated with the given formula: c L : Angle of twist (Radian) c: Radius of sample (mm) L: Length of sample (mm) 4.5 SHEAR ELASTICITY MODULUS Shear elasticity modulus (G) was calculated from the linear zone of torque twist diagram. Shear stress increases proportionally with shear strain in the elastic zone of torque twist diagram. In elastic zone, the ratio of shear stress () over shear strain () gives the shear elastic modulus. G G: Shear Elasticity modulus : Shear stress : Shear strain 4.6 FAILURE TYPES IN TORSION State of stress in torsion on the surface of a bar occurs on two mutually perpendicular planes (longitudinal yy and transverse xx). The principal stresses σ 1, (longitudinal) σ 3 (compressive) make an angle of 45 o and σ 3 = -σ 1. (intermediate stress σ 2 = 0). Torsion failures are different from tensile failures in that there is little localized reduction of area or elongation.

4 a) Shear (ductile) failure is along the maximum shear plane. b) Tensile (brittle) failure is perpendicular to the maximum tensile stress (at 45 o ), resulting in a helical fracture 5. DEVICES AND SPECIMEN Torsion specimen (aluminum 6061, copper and polymer cylinder which have 10 mm diameter and 2 mm thickness) has circular section area which is the easiest shape to calculate. Shear stress occurs in the specimen by the effect of torque. This shear stess increases linearly from center of specimen to surface so the maximum shear stress occurs in the surface while it is zero in the center. So stress distribution is not homogenous in the specimen. It can be provided a more uniform stress distribution by using thin pipe shaped specimen in torsion test. In torsion test, specimen should have a thin thickness, but very thick specimen may be distorted before twisting. The specimen used to determine the shear yield strength ( A ) and shear modulus (G) should have a ratio of specimen length (L) per specimen s outer radius (D), L/D10. In the specimen used to determine modulus of rupture this ratio should be L/D0.5. Also the ratio of specimen s outer Radius per thickness should be D/t Figure 4 Bar torsion test device

5 7. THE EXPERIMENT: 7.1 Calibration Before the test, the device should be calibrated as follows: - Calibration bar is mounted - The cursor is checked if the angle is zero. - Calibration bar is loaded from one side only (e.g kg.) - The moment is calculated as the formula: T = mass (kg) x x distance x cos() angle (from the cursor on device). 8. DESIRED FROM STUDENTS - Draw the twist angle- torque curve of specimen tested. - Calculate the shear modulus by using the drawn curve. - Calculate the Shear yield strength. 9. Referances [1]. KAYALI, E.S., ENSARI, C., DİKEÇ, F., 1990, "Metalik Malzemelerin Mekanik Deneyleri" İ.T.Ü. Chemistry-Metallurgy Faculty, İstanbul [2]. KAYALI, E.S., 1983, "Malzeme Teknolojisinde Burma Deneyi", Malzeme Teknolojisinde Deneysel Verilerin Değerlendirilmesi Semineri, Boğaziçi University Engineering Faculty, İstanbul [3]. DIETER, G.E., 1986, Mechanical Metallurgy, McGraw-Hill, Singapour [4]. ASTM E143-87, 1987, Shear Modulus at Room Temperature [5]. MARIN.,., 1966, Mechanical Behavior Of Engineering Materials,Prentice Hall, New Delhi