Change in dial gage Torque Reading reading (thousandths (in-lb) of an inch) 0 480 850 1160 1280 1580 1940 2220 2400 2580 2660 2740 2760 2820 2860 2880 2920 2960 2970 2985 3000 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 angle of twist from Conversion testing conversion of column machine of column B C into (degrees) into radians Radians 0 0.004 0.008 0.012 0.016 0.02 0.024 0.028 0.032 0.036 0.04 0.044 0.048 0.052 0.056 0.06 0.064 0.068 0.072 0.076 0.08 Divide Columns D-E by the gage length to get radians per inch: write all values in this column 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0.022 0.024 0.026 0.028 0.03 0.032 0.034 0.036 0.038 0.04 Torque vs Unit twist 3500 Torque (in-lb) 3000 2500 2000 1500 1000 500 0 0 0.01 0.02 0.03 0.04 0.05 unit twist (radians per Inch) Torque Vs unit twist for the straight line portion 1400 y1200 = 192500x + 45 1000 800 600 400 200 0 0 0.2 0.4 0.6 0.8 1 1.2 Mechanics of Materials Laboratory Torsion of a Steel Bar 1 A. Introduction and Objective The objectives of this experiment are to perform a torsion test of a steel bar, examine the strain and stress distribution due to torsion in a bar with a circular cross-section, experimentally determine the shearing modulus of steel, and observe the failure mode of a ductile material in torsion. B. Theory 1.0 Equilibrium Shearing stresses develop in a material when torque is applied to a bar. Our goal in this section is to derive an equation that relates applied toque to the shearing stresses that develop in our bar. The torques and the shear stresses are related through equilibrium. Using a polar coordinate system in terms of (ρ, θ) centered at the center of the circular cross-section, a small, or infinitesimal, amount of torque from the stress at a point on the cross-section is given by dT =  dF =  dA . 2 dA The total torque is found by integrating (essentially adding up) the infinitesimal torques over the cross-sectional area: T =  dT = A  dA . The preceding statement of equilibrium is valid whether or not the material remains linear elastic. However, when the material is linear elastic, and shear stress and strain are related through Hooke’s law for shear:  = G , which leads to T = A G dA . Unfortunately, the preceding equation is not very useful until we can find an expression for the shear strain, γ. The shearing strain in a twisted bar that has a circular cross-section can be determined based on some simple observations and assumptions of elastic deformation: 3 -the bar does not elongate, -plane cross-sections remain plane, and -radial lines on the cross-sections remain radial lines after deformation. To illustrate these observations, consider a bar that is clamped on one end. Before the torsional load is applied, points A and C lie on the outside radius at the fixed end of the bar. Point B lies on the free end of the bar, and, on a side view, the angle BAC creates a right angle. T A B C After a torque is applied to the end of the bar, point B moves to B*. The reduction in angle from the right angle is the shearing strain, γ. 2r T A B C B 4 L r Geometrically, tan  = BB * . L Since the deformation of elastic metals are usually very small, the tangent of a small angle and the angle (when expressed in radians) are approximately equal. Additionally, the distance BB* is approximately equal to the arc length, rφ:  = r . L This equation gives the shearing strain on the outside edge of a bar of length L, that has been twisted by an angle, φ, and its free end. What is needed in the expression containing the integral, however, is the shearing strain as a function of ρ and θ, the polar coordinates on the crosssection. Since our basic observation is that radial lines remain radial lines after deformation, the shearing strain will not depend on its angular position, θ. However, using the same concepts that were used to find the strain on the outside of the bar, we can find the shearing strain as a function of position, ρ, an arbitrary point radially outward from the center of the cross-section. We can visualize this by examining a small length of the bar that has been removed from the core of the bar, and examining the relative movement of points on each end of the segment. 5 X R S X S’ R* S* R* X S’ S* 6 In the same way as presented previously,  = SS *  = , x x which is the shearing stress over a length of the segment of the bar, x . To get the strain at a point, we can take the limit as this length goes to zero:  = lim  x   =d . x → 0  dx   d is the twist rate of the bar, and is a constant. Then, the preceding dx equation implies that the shearing strain varies linearly with the distance from the center of the bar (ρ). i.e. that the shearing strain is zero at the center of the bar, and is maximum on the outer edge of the bar. For a linear elastic material that obeys Hooke’s law in shear ( = G ), the shearing stress is also maximum on the outside of the bar. The term An expression for the shearing stress in terms of the twist rate is d  = G . dx With an equation for the shearing strain determined in terms of the polar coordinate variable, ρ, the torque on the bar can now be determined by evaluating the integral T =  G dA =  G A A d d dA = G A  2 dA . dx dx The integral in the expression above contains only terms that involve the crosssection of the bar. It has a special name, and it is called the polar moment of inertia, Ip. This results in an equation for the torque, T, as 7 T =G d I. p dx Eliminating the twist rate (which can be difficult to measure) by replacing it with the expression containing the shearing stress results in:  T =G I, p G and finally upon solving for the shearing stress in terms of the torque results in  = T . Ip The polar moment of inertia can be quickly found by evaluating the integral, and is an exercise typically assigned in Calculus class. For a solid circular crosssection, the polar moment of inertia is: I p=  d 4, 32 where d is the diameter of the cross-section. If the angle of twist of the bar, φ, is required, then torque-twist rate equation can be separated and integrated: d = 8 T dx , I pG  = T L 0 I pG dx = TL I pG if the torque, shear modulus, and polar moment of inertia are constant over the length of the bar for which the twist angle is determined. Note also that the units of the twist angle are in radians. Given any four of the five variables in the twist equation, the remaining unknown term can be found. The data that will be collected is the torque applied to the bar, and the angle of twist between a section of the bar. The dial needle of the torsion machine will indicate the value of torque. The initial angle of twist is determined by the use of a device called a troptometer and then later by the angular indictor on the testing machine. Neither of these two instruments are connected to a data acquisition system, so you will carefully determine the value of torque at a specified troptometer reading. 9 10 It is essential to understand how the troptometer operates in order to calculate the relative angle of twist. The troptometer consists of a dial gage, wire, a left and right side, and a connecting bar as shown in the figure. The troptometer is installed on the testing machine with the connecting bar in place. This establishes a precise 2.0 inch spacing between the left and right sides of the troptometer. Prior to loading the specimen, the connecting bar is removed. As the bar is loaded, both the left and right sides of the troptometer rotate with the specimen. However, because of the deformation of the material, one side of the troptometer rotates more than the other. This relative rotation (over the 2.0 inches) is measured by dial gage as a thin wire moves along the arc length of a wheel that is precisely 2.5 inches in radius. If s is the arc length, and R is the radius of the wheel (2.5 inches in this case), the relative angle of twist is determined by the arc length relation: s = R , or  = s / R. Since the dial gage reads in thousandths of an inch, to determine the radians of relative twist over 2.0 inches, you will use  (radians) = (dial gage reading / 1000)in. . 2.5in. After a certain amount of twist, the amount of rotation will not be able to be measured with the troptometer, and the dial gage will be disconnected. At this point, the angle of twist will be given approximately by the angular degree markings on the testing machine. The torque will also need to be recorded for the corresponding values of angle of twist in degrees. The angle of twist in degrees will also need to be converted into radians, by the simple conversion factor that 180º = π radians, or 11  (radians) = (degree reading) * . 180. C. Lab Procedure The testing machine should be on 20-30 minutes prior to testing. Verify the diameter of the steel bar. Place the troptometer onto the specimen. Mount the specimen into the torsion machine, being careful to ensure that the flat of the grips align with the flat ends of the specimen. Tighten the chuck on the specimen and remove the chuck key prior to testing. Verify the operation of the troptometer and wire system. (Note: the way the troptometer is set up, the dial gage decreases when the angle of twist changes.) Remove the troptometer bar. Apply the torque in the forward testing machine mode, at a relatively slow rate. Call out the values of dial gage reading listed in the table and the specimen is loaded. Use a dry-erase marker to mark the torque values from the testing machine needle directly on the glass face of the testing machine indicator. When the troptometer has reached it’s useful limit repeat this process for every 90º interval based on the testing machine indicator, until the specimen fractures. Record your test data on the following data sheet. After the specimen fails, photograph or sketch the shape of the specimen and failure surface. 12 A Torque Reading (in-lb) B C D E F Change in dial gage reading (thousandths of an inch) angle of twist from testing machine (degrees) Conversion of column B into radians Conversion of column C into radians Divide columns D-E by the gage length to get radians per inch; write all values in this column 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 *200 45 90 180 270 360 450 540 630 *release dial gage wire 13 D. Lab Report Complete the data table columns D-F. Construct a graph with torque values on the y-axis (column A) and radians per inch on the x-axis (column F). These data points should be plotted as open symbols (Do not connect the points with a line plot). Draw a best fit curve through all of the data. Indicate the ultimate torque value on the graph. Construct a second graph that has a maximum x-axis value of 0.04 radians/inch which corresponds to the dial gage reading of 200 thousandths of an inch. This graph is expected to have a relatively straight-line portion and some initial curving in it. The deviation from a straight line indicates the yield torque of the material. Note this value on the graph. Determine the slope, m, of the best fit line from the initial portion of the data. 14 Ultimate torque value is 3000in-lb Yield torque is 1500 in-lb E. Questions Determine the shear yield stress of the material from the yield torque and the shear stress equation. 𝜏= 𝑇𝜌 𝐼𝑝 T= 1500in-lb . 627 1500 𝑥 ( 2 ) 391.875 𝜏= 𝜋 = = 30998.68𝑝𝑠𝑖 = 30.999 𝑘𝑠𝑖 0.01517 (. 627)4 𝑥 32 15 Is it valid to use the shear stress equation to determine ultimate shearing stress of the material? Explain. No, the ultimate stress occurs in the elastic region of the graph Sketch the elastic shearing stress distribution for an elastic bar with a round cross-section subjected to torsion. 16 Determine the shear modulus for this material, by using the slope determined from the second graph (Show Calculation). Hints: m= T ( / L)  T = L I pG G= T I p ( / L) = m . Ip 192500 𝐺= 𝜋 = 12687051.63𝑝𝑠𝑖 = 12687.05𝑘𝑠𝑖 (. 627)4 32 Shear yield stress from measurements 30.999 ksi Accepted value of shear stress 36.000 ksi Percent difference 16.1% 17 F. Discussion The graphical representation of the torque (in-lb) versus Unit twist in (rads/in) was able to shore the Ultimate shear stress and the yield shear stress. The Elastic region produced a straight line curve of the following equation. the Ultimate torque was found to be 3000 in-lb, while the Yield Torque was found to be 1500in-lb. The shear yield stress was found to be 30.999 KSI, this value had a percentage difference of 16.1% against the accepted value of shear stress 36.000KSI. these could be due to errors such as: • use of approximations instead of exact measurements. • human errors such as parallax • error in measuring equipment recommendations: • use of more accurate equipmens • minimize human errors 18 B. Lab Report Complete the data table columns Dr. Construct a graph with que values on the yaxis com Aanspre the x-axis (column. These data peints should be plotesohet connect the points with a lineplex Dewable while Indicate the time values the pa Construct a second that has vehich corresponds to the dialading of 200 thenis en inch This is deviation from a straight line indicates the id torque of the material Note this vale on the graph. Determine the slopes of the best line from the initial portice of the data Torque vs Unit twist 00 pelrably not motor 1000 your Somebodene luntat o ex diferite de Most forque 2000 ㅚ Utore value is on Yield toqueis 1500 in-lb Torque Vs unit twist for the same portion 100 600 400 200 O 0.005 0002 60 9004 BOS C. Questions Determine the shear yield stress of the material from the yield torque and the shear stress equation. Тр -- T-1500 627 391.875 - 309986830999 1500 x *(627) 0.01517 changes sincet calculation chargy 1160 1280 20 30 angel video Change in dial Cute tone conversion of temper Torque Reading reading the machine of an inch) of con depresenterade in 0 0 40 10 350 D. 1580 0 0.000 0.00 0.02 0.004 0.008 10 1940 2220 2.400 2580 2660 2740 2760 2820 2860 2880 2920 0. 40 SO 60 70 BO 90 100 110 120 150 140 150 160 170 180 190 200 0.032 0.036 0.04 0.044 0.048 0.052 0.056 0.06 0.04 0.068 0.092 0.096 0.08 0012 0014 0.016 0.00 0.022 0.024 0.096 0.028 0.03 0.032 0.054 0.036 0.038 2960 2970 2985 3000 Torque vs Unit twist Torque finib) 4200 3000 2000 1000 0.01 0.04 0.00 0.05 untis radians per inch از هم