CIVE302 Fall 2013 Dr. Dowell T. Johnson Lab 1. Tensile Test of Steel A computer-controlled Tinius-Olsen testing machine will be used to perform experiments on steel coupons by loading them to failure in tension. This lab project will demonstrate typical stress-strain responses of hot-rolled (Figure 1-1) and cold-rolled (Figure 1-2) steels, starting with the initial linear-elastic behavior and going through yield, the yield plateau, strain-hardening, and finally rupture. Failure occurs when the single axial member fractures into two pieces following necking – a strain localization signified by a visible reduction of the cross-sectional area. Strain is defined as the change in length L of a member, for a given load, divided by the member length. While there are several methods for defining strain, engineering strain is the most common and is very useful for structural engineering, and herein will be referred to simply as strain. Engineering strain  considers the original member length L prior to being loaded and may thus be written as  L L (1-1) For an axially loaded member, stress is defined as the force divided by the cross-sectional area. Engineering stress considers the member force F and the original, un-stressed, cross-sectional area A when determining axial stress. As with strain, there are varying definitions of stress that allow for the changing cross-sectional area as the member is being loaded. However, engineering stress  is the most useful definition for practical applications and from here on will be referred to simply as stress. Thus stress is given as  F A (1-2) For the axially loaded member discussed here the stresses and strains are the same at all locations within a cross-section; however, it is important to realize that the concepts of stress and strain are not limited to axial loading as in this lab project, but can be used for 2-D and 3-D objects. In 3-D analysis these values can vary from point to point, with stress and strain quantities defined at a single point rather than over a member length as in Eq. (1-1). In the finite element approach, for example, solid objects of arbitrary shape can be modeled by many interconnected elements, with stress and strain contours developed that provide an understanding of locations of high stress and strain. Stress Useful range of stress-strain curve Necking su Fracture sy E ey eh eu Figure 1-1. Typical stress-strain diagram for hot-rolled steel 2 Strain In Eq. (1-1) the member length can be replaced by a gage length (shorter than the member length), providing a different measured displacement while retaining the same strain. For example, if the elongation were measured over half of the member length (gage length = 1/2 of the member length) the measured elongation would also be 1/2 of the full member length value, resulting in the same strain. This is an important concept that regardless of the chosen gage length the measured strain will be the same, with one exception as discussed below. As the member begins to experience significant deformation, local imperfections in material structure cause a large increase in strain over a short length of the member and typically only over a portion of a chosen gage length. When this localization takes place the measured strain will directly depend on the gage length and, therefore, the very concept of strain will lose meaning and any engineering significance. This is why the useful range of the stress-strain curve for steel ends at peak stress (as indicated in Figures 1-1 and 1-2). Note that peak stress is equal to the ultimate stress. Thus, regardless of the gage length the stress-strain curve will be consistent until the peak stress and associated strain are reached. The shape of the stressstrain curve at strains larger than the ultimate strain (necking) will depend on the chosen gage length and is, therefore, not unique – that is, the measured strain value will vary for a given stress dependent upon a selected gage length. Figure 1-3 shows a coupon fixed at the base and loaded at the top. The dark lines on the left figure indicate the initial geometry of the coupon with length L. With load F applied, an equal and opposite reaction develops at the fixed base and the member elongates L as shown with the light gray outline on the same figure on the left. In addition to elongating, the cross-section reduces in size as shown in the figure. It is assumed here that the fixed base allows the crosssection to reduce size (fixed axially only). It is clear that the figure on the left is below the ultimate stress because there is no localization with the same strains throughout the member length. The figure in the middle shows necking (localization) leading to the failure and fracture 3 shown on the right figure. Ductile materials develop extensive necking while brittle materials fracture abruptly. Stress Useful range of stress-strain curve Necking su Fracture sy E E 0.002 ey eu Strain Figure 1-2. Typical stress-strain diagram for cold-rolled steel 4 F F F DL L R=F R R Figure 1-3. Constant strain, necking and fracture of steel coupon Experiment Tension tests will be conducted for hot-rolled and cold-rolled steels using circular test coupons that have two threaded ends. The threaded end regions are larger diameter than the center portion of interest, resulting in lower stresses due to the larger cross-sectional area (stress = force/area) and preventing yield in the attachment regions. Proper attachment of the specimen to the machine grips will ensure most of the deformations will occur over the member length that has constant cross-section. Displacements will be recorded by computer, measuring the elongation of a machined length using a potentiometer. Since most of the 5 specimen deformations occur between the threaded ends, and because the machine is stiff, use this distance as the gage length (member length with constant cross-section). Axial force will also be recorded throughout the experiment and saved to computer. Using a digital caliper, the diameter of the specimen at the center of the gage length should be measured before testing. Force and Displacement data will be posted on Blackboard for each test, with one column providing measured axial displacement and a second column giving measured force. Students will be responsible for changing the force and displacement data to stress and strain and plotting the axial stress-strain curve for each experiment. From the recorded data the following is required. REQUIRED CALCULATIONS: Hot-Rolled Steel Plot of stress versus strain Plot of force versus displacement (this should be presented in data section) Member-end-stiffness K (kips/inch) at yield stress and at ultimate stress* Modulus of elasticity E (ksi) Yield stress (ksi) Yield strain Strain at first hardening Ultimate stress (ksi) Ultimate strain (strain associated with ultimate stress) 6 Strain ductility Modulus of toughness (ksi) Cold-Rolled Steel Plot of stress versus strain Plot of force versus displacement (this should be presented in data section) Initial member-end-stiffness K (kips/inch) Modulus of elasticity E (ksi) Yield stress (ksi) – from 0.2% offset Yield strain (strain associated with yield stress) – from 0.2% offset Ultimate stress (ksi) Ultimate strain (strain associated with ultimate stress) Strain ductility Modulus of toughness (ksi) REQUIRED DISCUSSION: From the measured results, discuss any differences and similarities between hot-rolled steel and cold-rolled steel (numerically and/or conceptually). Comment on the overall behavior of the materials in the force-displacement and stress-strain curves (be specific). Provide example applications in the real world for both types of steel. 7 Definitions: Modulus of elasticity E (ksi). This is the initial, linear slope of the stress-strain diagram. In this range the material (and any structure that is made of this material) will be linear-elastic, meaning that if the load is doubled the deformations will double (linear), and upon removal of the load it will return to its original position (elastic) with no permanent deformation. Yield stress (ksi). For hot-rolled steel the yield stress is the stress at the end of the proportional limit, which is followed by the yield plateau. For cold-rolled steel there is no well-defined yield stress due to the nonlinear nature of the stress-strain curve. Therefore the yield stress is found by drawing a line on the stress-strain diagram with the same slope as the initial E, but offset by a strain of 0.002 (0.2%), and finding where it intersects the measured stress-strain diagram. Yield strain. Yield strain is the strain associated with the yield stress (at the same point on the stress-strain diagram). Strain at first hardening. For hot-rolled steel, strain hardening occurs at the end of the yield plateau. Beyond yield strain, the stress remains approximately constant (equal to the yield stress) for quite a while until strain hardening begins. This does not apply to cold-rolled steel due to the shape of the stress-strain curve. Ultimate stress (ksi). This is the maximum stress achieved. At strains beyond this point the stresses decrease and necking develops, followed soon after by fracture and failure of the specimen. 8 Ultimate strain (ksi). This is the strain associated with ultimate stress. It is not the strain at rupture as this strain is not unique, with different values for different arbitrary gage lengths. Strain ductility. This is the ratio of ultimate strain to yield strain. Modulus of toughness (ksi). This is the area under the stress-strain curve to ultimate stress and strain (not to rupture). Member-end-stiffness (kips/inch). Axial force divided by displacement, which is the slope of the force-displacement curve. In the plastic region stiffness is defined as the tangent slope to the stress-strain curve. 9 CIVE302 Spring 2013 Dr. Dowell T. Johnson Lab 4. Torsion of Member with Circular Cross-Section: Principal Strains and Stresses Thus far our only focus has been on axial deformations which cause elongation or compression of a member along their axis of application. Other types of deformation exist, though, such as shear deformations which translate or slide the face on which they act. To help visualize this behavior, in this lab a solid steel shaft with circular cross-section is loaded under pure torsion. Torsion is caused when a moment is applied in-plane with the cross-section of the member, twisting it and causing it to deform in shear. A visualization of this is provided below in Figure 4-1. Figure 4-1: Deformation of a body under shear conditions Recall that the measurement devices used thus far only measure elongation or contraction along their primary axis. Meaning that strain gages cannot directly measure these values. Looking at more detail at Figure 4.1, it is clear that the shear strain, γ , is not a linear quantity but an angle of distortion. We thus need additional techniques to be able to somehow determine shear strain from linear-based measurement equipment. For a bar in pure torsion, each section along the longitudinal axis rotates strictly in a plane; that is, all points within the member follow a strictly a circular arc in the plane and do not translate in either direction along the member axis. Note that this is not the case for non-circular cross- sections, as torsion develops warping deformations which displace points axially in addition to rotationally. This behavior is shown below in Figure 4-2, where it can be seen that point B at the outermost fiber of the cross-section travels along the circumference of the circle to B’ at an applied torsion of To. To B A B’ To L Figure 4-2. Member deformation from applied torsion To Torsional loading at a section causes shear stresses that are zero at the center of the member and increase linearly with radius r to a maximum at the surface of the cylinder at r = R. Shear stress is found from the section torsion T, radius r, and polar moment of inertia J as  Tr J (4-1) For a circular section, the polar moment of inertia is equal to the sum of moments of inertia about the x and y axes as 2 J  Ix  I y  r 4 4  r 4 4  R 4 (4-2) 2 Shear strain is found by dividing the shear stress by the shear modulus G, where the shear modulus is found for a given material from material testing as the slope of the shear stress versus shear strain plot. Thus shear strain is found from   Tr  G JG (4-3) Note the similarities in this equation to those seen in the previous labs. To calculate axial stresses, we had a driving axial load P that was divided by the cross-sectional area – a geometric property of the member. Then, to calculate strains, we divided this stress by the modulus of elasticity – a material property of the section – and concluded that the deformation is a function of the driving load P, the geometrical resistance of the member, and the material resistance of the member. In the case of torsional strain, we see a similar pattern: a torque T drives the deformation of the member, a geometric resistance forms via r and J, and a material resistance forms via the shear modulus G. Fixing a member on one end and applying a torsion To to the end free to rotate yields the deformed shape given in Figures 4-2 and 4-3. At the left end of the member all points of the cross-section are restrained from rotation and remain in the same place before and after torsion is applied. At the right end of the member, however, torsional loading causes the crosssection to rotate through an angle indicated by the movement of point B to B’. This point starts at the top of the section but shifts along the perimeter of the cross-section. All points on the surface at the right end of the member move through the same angle as point B, though, generating a rotation of the full cross-section. 3 For the example shown in Figure 4-3, the torsion is constant along the member length which results in a linear variation of deformation from Points A to B. In other words, the amount of rotation at half the member length is one-half of the maximum rotation at the right end of the member. To demonstrate this, two lines are given in Figure 4-3 that were originally parallel to the member geometry and have deformed from load application. A stress element shows that the original rectangular shape has been deformed due to shear stresses; the length of the sides remain unchanged, but the amount of deformation of the corners on the left face and the corners on the right face differ in magnitude. Applied and reacting torsions are indicated with double arrows: arrows pointing away from the member indicate positive torsion while arrows facing toward the member indicate negative torsion, with the rotational direction following that defined by the right hand rule. Positive torsion definition is given for any slice of the beam in Figure 4-4 and the torsion diagram for the example provided in Figure 4-3 is given in Figure 4-5. To Stress Element A B B’ L Figure 4-3. Side view of member under torsion with stress element indicated 4 To T + T Figure 4-4. Definition of positive torsion for a short length of the member (slice) T To A B X Figure 4-5. Torsion diagram over member length When the torsion is constant along a member length as in Figure 4-4 the rotation at its right end, called the angle of twist, is given as  Tl JG (4-4) 5 If the torsion varies along the member length, however, rotation needs to integrated over the length of the member similar to how distributed loads must be integrated for beam displacements. To do this the twist of the member, which is the rotation per unit length, is calculated for a given cross-section  T JG (4-5) The rotation at this cross-section is then calculated by integrating over the length l   0 T dx JG (4-6) An example of torsion that varies along the member length is given in Figure 4-6, with evenly distributed torsion of To/L applied for the entire member length. There is no torsion in the member at its right end and the torsion increases linearly to a maximum at the left of the member. From statics, the torsion reaction at the fixed end is To and the torsion diagram can be developed as shown in Figure 4-6. The torsion in the member at any section along its length is given as T  To  To  x x  To 1   L  L (4-7) 6 Figures 4-6. Distributed torsion along member length with torsion diagram Torsion given as a function of x in Eq. (4-7) is substituted into Eq. (4-6) to find member rotation. Thus  x To 1   T L   dx    dx JG JG 0 0 l l (4-8) 7 If the material is the same for the entire member and the cross-section is constant then J and G are constant and can be moved out of the integral, as can the constant To, giving B  To l  Tl x 1  dx  o  JG 0  L  2 JG (4-9) This is half the rotation of a member that has constant torque along its length with the same reacting torsion at its left end of To. The stress element shown in Figure 4-3 distorts due to an upward shear stress on its left side and, from statics, an equal and opposite shear stress on its right side (down) similar to how the block in Figure 4-1(b) deforms. However, while this satisfies equilibrium of forces in the vertical direction, equilibrium of the stress block as a whole is not satisfied: by themselves these shear stresses form a couple and thus give a tendency for the block to rotate. To correct for this, moments can be taken about one of the bottom corners of the stress block to generate a balancing horizontal stress on the top face that imparts an equal but opposite rotation to the vertical shear stresses. It is found that this horizontal shear stress is equal in magnitude to the two vertical shear stresses, and equilibrium in the horizontal direction shows that an equal but opposite shear stress forms on the bottom face. To help visualize this process, an element is also shown at the top right of Figure 4-7 with these shear stresses indicated. The stress element has dimensions of dx and dy, and when it has a square or rectangular shape it vanishes to a single point. With shear stresses defined at a stress element from torsion, it is possible to determine maximum compressive and tensile stresses of the stress element that act on different faces of 8 the element (at different angles). These are called principal stresses and can be determined through the use of Mohr’s circle as shown in Figure 4-7. For a given set of shear and normal stresses on two planes of a stress element, Mohr’s circle is constructed. The horizontal axis is for normal stresses on a plane and the vertical axis is for shear stresses on a plane. X and Y planes are defined by passing the x and y axes through the element and the faces that are intersected are defined, as indicated at the bottom right of Figure 4-7. There are two X planes and two Y planes, and either one can be used in the following discussion as they are completely interchangeable. t t dy X (0, t ) t dx f 1 s1 s2 f s 2 y Plane x Plane x Plane Y (0, -t ) y Plane y x Figure 4-7. Mohr’s circle with stress element and plane definitions 9 To construct Mohr’s circle an X coordinate and Y coordinate are plotted, representing the normal and shear stresses on X and Y faces, respectively, of the stress element being considered. On the X plane of our stress element there is a positive shear and no normal stress, resulting in the X point given in Figure 4-7. As discussed before, the X plane on either left or right sides of the element can be used to determine the shear and normal stresses to plot in Mohr’s circle. If the shear stress on the face being considered wants to turn the element in the clockwise direction we call this positive shear. The stress element in Figure 4-7 shows that the upward shear stress on the left of the element wants to rotate the element in the clockwise direction, and so does the downward shear stress on the right of the element, thus they both give the same positive shear stress and no normal stress. On the Y plane (top or bottom of the element) the shear stress wants to turn the element in the counterclockwise direction and this is defined as negative shear. There is no normal stress on the Y plane, yielding an X coordinate of zero and a Y coordinate equal to the negative shear stress found from the torsion expression given in Eq. (4-1). Once two points are given on a circle then the entire circle is completely defined as shown in Figure 4-7. In Mohr’s circle we are working in stress space where all values are stresses. Thus, if a ruler is put on a graph of Mohr’s circle to scale a result, the values will be in stress units such as psi or ksi. Only results that lie on the circle have meaning, providing normal and shear stresses at all different angles within the stress element. One can imagine a plane cut through the stress element at some arbitrary angle and asking what the shear and normal stresses are on this exposed new plane. The construction of Mohr’s circles provides the answer to this question in an elegant and efficient manner. It can be shown from solid mechanics that the angle swept in Mohr’s circle is twice the angle within the element. Or, equivalently, the angle within the stress element is one-half of the angle swept in Mohr’s circle. The direction of rotation, clockwise or counterclockwise, is the same within the element and in Mohr’s circle. 10 Maximum tensile and compressive stresses 1 and 2, called principal stresses, occur where the shear stresses are zero, as indicated in Figure 4-7. The angle rotated to the principal stresses from the X and Y points on Mohr’s circle is 90 degrees in the clockwise direction. Therefore the angle rotated within the physical element from the X and Y planes to the principal planes is 45 degrees. Also from Mohr’s circle it is clear that for this case the magnitude of the principal stresses is equal to the magnitude of the shear stresses found from torsion (radius of Mohr’s circle). A stress element showing the principal planes is given in Figure 4-8, occurring on a surface at 45 degrees from the element in pure shear that has normal stresses but no shear stresses. Figure 4-9 breaks this down mechanically, showing how tensile and compressive stresses develop internally in an element to enforce equilibrium in the element. Note that rotating 45 degrees clockwise from either X plane gives the same values on the principal tensile stress plane, and rotating 45 degrees clockwise from either Y plane gives the same values on the principal compressive plane. It is interesting to note that if the whole element is rotated 45 degrees to the principal planes then only normal stresses are seen on the surfaces, with no shear stresses. Thus, even though the element deforms in pure shear, pure tensile and compressive stresses develop at an angle 45 degrees from this axis. 45 Figure 4-8. Element rotated to principal planes 11 t t x Plane dy t 45 45 t dx x Plane t t y Plane t y Plane x Plane x Plane 45 t 45 y Plane y x t t y Plane Figure 4-9. Stresses on principal planes from Mohr’s circle The reverse is true when considering a wire in tension. From Mohr’s circle for a stress element within the wire a normal stress is found in the vertical direction and no normal stress in the transverse direction. A rotation of 90 degrees in Mohr’s circle shows a maximum shear stress at this angle which is an element rotation of 45 degrees. Yielding of the wire in tension is caused by shear of the crystals at 45 degrees resulting in Luder’s lines. This is clearly shown on a polished coupon tensile test. Experiments A steel bar with ¾-inch diameter is tested in torsion. The steel material has modulus of elasticity E and Poisson’s ratio  of 30,000 ksi and 0.25, respectively. Two strain gages are applied to the surface of the specimen one positioned in the principal tensile stress direction and the other in the principal compressive stress direction. Thus they are both positioned 45 degrees from the longitudinal member axis and 90 degrees from reach other. The strain gages utilize a half bridge. 12 Apply a torsion of 1,000 lb-in. at intervals of 250 lb-in. and record the principal strains at each load increment. Compare theoretical and measured principal strains. Principal strains are related to principal stresses in Eq. (4-10) and Eq. (4-11). With measured principal strains, these equations can be solved for the principal stresses. 1  1   2 E E (4-10) 2  2   1 E E (4-11) Provide the following plots of theory versus measured (or calculated from measured): Torsion diagram (torsion along member length) Torsion versus principal tensile stress Principal tensile stress versus principal compressive stress Principal tensile strain versus principal compressive strain Torsion versus shear stress REQUIRED DISCUSSION: How do the measured results in the above plots compare to theoretical results? Provide detailed explanation for any variations. 13 Also, if an axial load was applied to the rod would the orientation of the strain gages need to change or would the orientation of 45 degrees still be valid? Justify your answer. Definitions: Shear Modulus of elasticity G (ksi). This is the initial, linear slope of the stress-strain diagram in shear. In this range the material (and any structure that is made of this material) will be linearelastic, meaning that if the load is doubled the deformations will double (linear), and upon removal of the load it will return to its original position (elastic) with no permanent deformation. Strain gage. An electrical strain gage is a resistor that changes its resistance as a function of strain and temperature. By calibrating the strain gage (gage factor), this change in resistance that is associated with strain in the gage is accurately measured as a voltage change and converted to a strain value. The change in resistance of the strain gage due to temperature effects is removed from the measured results by use of a Wheatstone bridge. Polar moment of inertia. For circular sections the polar moment of inertia is equal to the sum of the section moment of inertias about the x and y axes that run through the section centroid. 14 CIVE302 Spring 2013 Dr. Dowell T. Johnson Lab 4. Torsion of Member with Circular Cross-Section: Principal Strains and Stresses Thus far our only focus has been on axial deformations which cause elongation or compression of a member along their axis of application. Other types of deformation exist, though, such as shear deformations which translate or slide the face on which they act. To help visualize this behavior, in this lab a solid steel shaft with circular cross-section is loaded under pure torsion. Torsion is caused when a moment is applied in-plane with the cross-section of the member, twisting it and causing it to deform in shear. A visualization of this is provided below in Figure 4-1. Figure 4-1: Deformation of a body under shear conditions Recall that the measurement devices used thus far only measure elongation or contraction along their primary axis. Meaning that strain gages cannot directly measure these values. Looking at more detail at Figure 4.1, it is clear that the shear strain, γ , is not a linear quantity but an angle of distortion. We thus need additional techniques to be able to somehow determine shear strain from linear-based measurement equipment. For a bar in pure torsion, each section along the longitudinal axis rotates strictly in a plane; that is, all points within the member follow a strictly a circular arc in the plane and do not translate in either direction along the member axis. Note that this is not the case for non-circular cross- sections, as torsion develops warping deformations which displace points axially in addition to rotationally. This behavior is shown below in Figure 4-2, where it can be seen that point B at the outermost fiber of the cross-section travels along the circumference of the circle to B’ at an applied torsion of To. To B A B’ To L Figure 4-2. Member deformation from applied torsion To Torsional loading at a section causes shear stresses that are zero at the center of the member and increase linearly with radius r to a maximum at the surface of the cylinder at r = R. Shear stress is found from the section torsion T, radius r, and polar moment of inertia J as  Tr J (4-1) For a circular section, the polar moment of inertia is equal to the sum of moments of inertia about the x and y axes as 2 J  Ix  I y  r 4 4  r 4 4  R 4 (4-2) 2 Shear strain is found by dividing the shear stress by the shear modulus G, where the shear modulus is found for a given material from material testing as the slope of the shear stress versus shear strain plot. Thus shear strain is found from   Tr  G JG (4-3) Note the similarities in this equation to those seen in the previous labs. To calculate axial stresses, we had a driving axial load P that was divided by the cross-sectional area – a geometric property of the member. Then, to calculate strains, we divided this stress by the modulus of elasticity – a material property of the section – and concluded that the deformation is a function of the driving load P, the geometrical resistance of the member, and the material resistance of the member. In the case of torsional strain, we see a similar pattern: a torque T drives the deformation of the member, a geometric resistance forms via r and J, and a material resistance forms via the shear modulus G. Fixing a member on one end and applying a torsion To to the end free to rotate yields the deformed shape given in Figures 4-2 and 4-3. At the left end of the member all points of the cross-section are restrained from rotation and remain in the same place before and after torsion is applied. At the right end of the member, however, torsional loading causes the crosssection to rotate through an angle indicated by the movement of point B to B’. This point starts at the top of the section but shifts along the perimeter of the cross-section. All points on the surface at the right end of the member move through the same angle as point B, though, generating a rotation of the full cross-section. 3 For the example shown in Figure 4-3, the torsion is constant along the member length which results in a linear variation of deformation from Points A to B. In other words, the amount of rotation at half the member length is one-half of the maximum rotation at the right end of the member. To demonstrate this, two lines are given in Figure 4-3 that were originally parallel to the member geometry and have deformed from load application. A stress element shows that the original rectangular shape has been deformed due to shear stresses; the length of the sides remain unchanged, but the amount of deformation of the corners on the left face and the corners on the right face differ in magnitude. Applied and reacting torsions are indicated with double arrows: arrows pointing away from the member indicate positive torsion while arrows facing toward the member indicate negative torsion, with the rotational direction following that defined by the right hand rule. Positive torsion definition is given for any slice of the beam in Figure 4-4 and the torsion diagram for the example provided in Figure 4-3 is given in Figure 4-5. To Stress Element A B B’ L Figure 4-3. Side view of member under torsion with stress element indicated 4 To T + T Figure 4-4. Definition of positive torsion for a short length of the member (slice) T To A B X Figure 4-5. Torsion diagram over member length When the torsion is constant along a member length as in Figure 4-4 the rotation at its right end, called the angle of twist, is given as  Tl JG (4-4) 5 If the torsion varies along the member length, however, rotation needs to integrated over the length of the member similar to how distributed loads must be integrated for beam displacements. To do this the twist of the member, which is the rotation per unit length, is calculated for a given cross-section  T JG (4-5) The rotation at this cross-section is then calculated by integrating over the length l   0 T dx JG (4-6) An example of torsion that varies along the member length is given in Figure 4-6, with evenly distributed torsion of To/L applied for the entire member length. There is no torsion in the member at its right end and the torsion increases linearly to a maximum at the left of the member. From statics, the torsion reaction at the fixed end is To and the torsion diagram can be developed as shown in Figure 4-6. The torsion in the member at any section along its length is given as T  To  To  x x  To 1   L  L (4-7) 6 Figures 4-6. Distributed torsion along member length with torsion diagram Torsion given as a function of x in Eq. (4-7) is substituted into Eq. (4-6) to find member rotation. Thus  x To 1   T L   dx    dx JG JG 0 0 l l (4-8) 7 If the material is the same for the entire member and the cross-section is constant then J and G are constant and can be moved out of the integral, as can the constant To, giving B  To l  Tl x 1  dx  o  JG 0  L  2 JG (4-9) This is half the rotation of a member that has constant torque along its length with the same reacting torsion at its left end of To. The stress element shown in Figure 4-3 distorts due to an upward shear stress on its left side and, from statics, an equal and opposite shear stress on its right side (down) similar to how the block in Figure 4-1(b) deforms. However, while this satisfies equilibrium of forces in the vertical direction, equilibrium of the stress block as a whole is not satisfied: by themselves these shear stresses form a couple and thus give a tendency for the block to rotate. To correct for this, moments can be taken about one of the bottom corners of the stress block to generate a balancing horizontal stress on the top face that imparts an equal but opposite rotation to the vertical shear stresses. It is found that this horizontal shear stress is equal in magnitude to the two vertical shear stresses, and equilibrium in the horizontal direction shows that an equal but opposite shear stress forms on the bottom face. To help visualize this process, an element is also shown at the top right of Figure 4-7 with these shear stresses indicated. The stress element has dimensions of dx and dy, and when it has a square or rectangular shape it vanishes to a single point. With shear stresses defined at a stress element from torsion, it is possible to determine maximum compressive and tensile stresses of the stress element that act on different faces of 8 the element (at different angles). These are called principal stresses and can be determined through the use of Mohr’s circle as shown in Figure 4-7. For a given set of shear and normal stresses on two planes of a stress element, Mohr’s circle is constructed. The horizontal axis is for normal stresses on a plane and the vertical axis is for shear stresses on a plane. X and Y planes are defined by passing the x and y axes through the element and the faces that are intersected are defined, as indicated at the bottom right of Figure 4-7. There are two X planes and two Y planes, and either one can be used in the following discussion as they are completely interchangeable. t t dy X (0, t ) t dx f 1 s1 s2 f s 2 y Plane x Plane x Plane Y (0, -t ) y Plane y x Figure 4-7. Mohr’s circle with stress element and plane definitions 9 To construct Mohr’s circle an X coordinate and Y coordinate are plotted, representing the normal and shear stresses on X and Y faces, respectively, of the stress element being considered. On the X plane of our stress element there is a positive shear and no normal stress, resulting in the X point given in Figure 4-7. As discussed before, the X plane on either left or right sides of the element can be used to determine the shear and normal stresses to plot in Mohr’s circle. If the shear stress on the face being considered wants to turn the element in the clockwise direction we call this positive shear. The stress element in Figure 4-7 shows that the upward shear stress on the left of the element wants to rotate the element in the clockwise direction, and so does the downward shear stress on the right of the element, thus they both give the same positive shear stress and no normal stress. On the Y plane (top or bottom of the element) the shear stress wants to turn the element in the counterclockwise direction and this is defined as negative shear. There is no normal stress on the Y plane, yielding an X coordinate of zero and a Y coordinate equal to the negative shear stress found from the torsion expression given in Eq. (4-1). Once two points are given on a circle then the entire circle is completely defined as shown in Figure 4-7. In Mohr’s circle we are working in stress space where all values are stresses. Thus, if a ruler is put on a graph of Mohr’s circle to scale a result, the values will be in stress units such as psi or ksi. Only results that lie on the circle have meaning, providing normal and shear stresses at all different angles within the stress element. One can imagine a plane cut through the stress element at some arbitrary angle and asking what the shear and normal stresses are on this exposed new plane. The construction of Mohr’s circles provides the answer to this question in an elegant and efficient manner. It can be shown from solid mechanics that the angle swept in Mohr’s circle is twice the angle within the element. Or, equivalently, the angle within the stress element is one-half of the angle swept in Mohr’s circle. The direction of rotation, clockwise or counterclockwise, is the same within the element and in Mohr’s circle. 10 Maximum tensile and compressive stresses 1 and 2, called principal stresses, occur where the shear stresses are zero, as indicated in Figure 4-7. The angle rotated to the principal stresses from the X and Y points on Mohr’s circle is 90 degrees in the clockwise direction. Therefore the angle rotated within the physical element from the X and Y planes to the principal planes is 45 degrees. Also from Mohr’s circle it is clear that for this case the magnitude of the principal stresses is equal to the magnitude of the shear stresses found from torsion (radius of Mohr’s circle). A stress element showing the principal planes is given in Figure 4-8, occurring on a surface at 45 degrees from the element in pure shear that has normal stresses but no shear stresses. Figure 4-9 breaks this down mechanically, showing how tensile and compressive stresses develop internally in an element to enforce equilibrium in the element. Note that rotating 45 degrees clockwise from either X plane gives the same values on the principal tensile stress plane, and rotating 45 degrees clockwise from either Y plane gives the same values on the principal compressive plane. It is interesting to note that if the whole element is rotated 45 degrees to the principal planes then only normal stresses are seen on the surfaces, with no shear stresses. Thus, even though the element deforms in pure shear, pure tensile and compressive stresses develop at an angle 45 degrees from this axis. 45 Figure 4-8. Element rotated to principal planes 11 t t x Plane dy t 45 45 t dx x Plane t t y Plane t y Plane x Plane x Plane 45 t 45 y Plane y x t t y Plane Figure 4-9. Stresses on principal planes from Mohr’s circle The reverse is true when considering a wire in tension. From Mohr’s circle for a stress element within the wire a normal stress is found in the vertical direction and no normal stress in the transverse direction. A rotation of 90 degrees in Mohr’s circle shows a maximum shear stress at this angle which is an element rotation of 45 degrees. Yielding of the wire in tension is caused by shear of the crystals at 45 degrees resulting in Luder’s lines. This is clearly shown on a polished coupon tensile test. Experiments A steel bar with ¾-inch diameter is tested in torsion. The steel material has modulus of elasticity E and Poisson’s ratio  of 30,000 ksi and 0.25, respectively. Two strain gages are applied to the surface of the specimen one positioned in the principal tensile stress direction and the other in the principal compressive stress direction. Thus they are both positioned 45 degrees from the longitudinal member axis and 90 degrees from reach other. The strain gages utilize a half bridge. 12 Apply a torsion of 1,000 lb-in. at intervals of 250 lb-in. and record the principal strains at each load increment. Compare theoretical and measured principal strains. Principal strains are related to principal stresses in Eq. (4-10) and Eq. (4-11). With measured principal strains, these equations can be solved for the principal stresses. 1  1   2 E E (4-10) 2  2   1 E E (4-11) Provide the following plots of theory versus measured (or calculated from measured): Torsion diagram (torsion along member length) Torsion versus principal tensile stress Principal tensile stress versus principal compressive stress Principal tensile strain versus principal compressive strain Torsion versus shear stress REQUIRED DISCUSSION: How do the measured results in the above plots compare to theoretical results? Provide detailed explanation for any variations. 13 Also, if an axial load was applied to the rod would the orientation of the strain gages need to change or would the orientation of 45 degrees still be valid? Justify your answer. Definitions: Shear Modulus of elasticity G (ksi). This is the initial, linear slope of the stress-strain diagram in shear. In this range the material (and any structure that is made of this material) will be linearelastic, meaning that if the load is doubled the deformations will double (linear), and upon removal of the load it will return to its original position (elastic) with no permanent deformation. Strain gage. An electrical strain gage is a resistor that changes its resistance as a function of strain and temperature. By calibrating the strain gage (gage factor), this change in resistance that is associated with strain in the gage is accurately measured as a voltage change and converted to a strain value. The change in resistance of the strain gage due to temperature effects is removed from the measured results by use of a Wheatstone bridge. Polar moment of inertia. For circular sections the polar moment of inertia is equal to the sum of the section moment of inertias about the x and y axes that run through the section centroid. 14 Statement of Objective State the objective(s) of the experiment concisely in two to three sentences maximum. The objective should state the topic of the experiment and primary goal of the experiment. This section should inform the reader precisely why the experiment is being conducted and contain a minimal amount of explanation. Theory Provide a brief description of the relevant theory behind each experiment. Accurately summarize the material theory material discussed in the lab write-up and in the lecture. Writing in this section should be technical and demonstrate understanding of the purpose behind each experiment and its following calculations. Does not need to be extensively long (i.e. re-deriving equations done in class) but should be thorough enough to demonstrate a general understanding of the material. Description of Experimental Setup Provide a technical description of the setup for the experiment and include pictures, schematic drawings, or other visual aids whenever possible. If you use pictures, do not simply Google random items which may or may not be related to the lab; use either the provided pictures of the experiment or your own. Any diagrams included must be referenced in the text. This section should explain only the physical setup of the experiment and list any equipment used. Your description should include enough information for a third party to physically set up the experiment using only your report as a reference. Procedure Detail the procedure used to carry out the experiment step-by-step. This should describe how the experiment was performed from start to finish, including relevant information such as measurement intervals, the order in which the tests were performed (if multiple), and other such chronological information. Organizing this information in a bullet point or list format is recommended, and a proper procedure allows for a third party to read each step and reproduce the experiment exactly as it was performed. Data All the pertinent experimental/raw data obtained during the experiment are presented in this section. This section should contain only raw information, not results from manipulation of data. If data and calculations are intermixed for any reason, each should be clearly labeled. For experiments with excessive amount of point data, plots should be presented in place of large tables: never provide a printout of 8000 data points. All numerical data should be tabulated carefully, and each table, figure and graph in the report must have a caption or label and a number that is referenced in the written text. Variables tabulated or plotted should be clearly identified by a symbol or name. Units, if any, should always be clearly noted. Data presented should be neat, organized, and readily interpreted by the reader. Analysis of Data and Calculations This section tabulates any calculations performed in the experiment. While using computer aides to perform calculations is often helpful and/or required, it is important to submit sample hand calculations describing the computational procedure and showing how the computer program or spreadsheet you are using arrives at its final calculated values. Any data manipulation - such as converting loads to stresses and displacements to strains should be included in this portion of the report. In addition, calculation of percentage errors – if applicable – should be listed next to final results. Theoretical and experimental calculations should be clearly labeled Discussion of Results and Conclusions This section is devoted to your interpretation of the outcome of the experiment or project. The information from the data analysis is examined and explained. You should describe, analyze, and explain (not just restate) all your results. This section should answer the question “What do the experiment and accompanying results tell me?” Logically evaluate your results and rationally explain any errors. If your results seem very far off, this should be noted and explanations provided. Unless expressly told that this is valid reasoning, generic explanations such as “something was wrong” or “the machines were old” are not acceptable. Additionally, any discussion questions presented in the accompanying lab manual should be answered in detail here. References If sources outside the lab lecture or lab manual are used, they must be included here. Use a standardized bibliographical format of your choosing, but be consistent.
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