Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr Chapter 7 Analysis of Stress and Strain 7.1 Introduction 1. Flexure and shear formula (ππ = −ππππ/πΌπΌ and ππ = ππππ/πΌπΌπΌπΌ), torsion formula (ππ = ππππ/πΌπΌππ ), etc. help determine the stresses on cross sections 2. However, larger stresses may occur on inclined sections 3. Example 1: Uniaxial stress (Section 2.6) – maximum shear at 45° and maximum normal at cross sections 4. Example 2: Pure shear (Section 3.5) – maximum tensile and compressive stresses occur on 45° 5. Generalization of these examples ο need theories for “Plane Stress” 6. Transformation equations help determine the stresses in any general direction from the given state of stress 7.2 Plane Stress 1. Stress element under “plane stress” condition, e.g. in the π₯π₯π₯π₯ plane: only the π₯π₯ and π¦π¦ faces of the element are subjected to stresses, and all stresses act parallel to the π₯π₯ and π¦π¦ axis. 2. Normal stress (ππ) - Subscript identifies the face on which the - stress acts, e.g. πππ₯π₯ and πππ¦π¦ For e__________, equal normal stresses act on the opposite faces - Sign convention: ________ is positive while ___________ is negative 3. Shear stress (ππ) - Two subscripts: the first indicates the face, and the second direction - Sign convention: positive for plus(face)plus(direction), and negative otherwise Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. - Junho Song junhosong@snu.ac.kr The sign convention described above is consistent with the shear stress pattern discussed in Section 1.7 (derived from the equilibrium equation) - Thus, πππ₯π₯π₯π₯ = πππ¦π¦π¦π¦ ο€ Stresses on Inclined Sections 1. To express the stresses acting on the inclined π₯π₯1 π¦π¦1 element in terms of those on the π₯π₯π₯π₯ element, consider the e__________ of the forces on the wedge-shaped element 2. Equilibrium equation in π₯π₯1 direction: πππ₯π₯1 π΄π΄0 sec ππ − πππ₯π₯ π΄π΄0 cos ππ − πππ₯π₯π₯π₯ π΄π΄0 sin ππ − πππ¦π¦ π΄π΄0 tan ππ sin ππ − πππ¦π¦π¦π¦ π΄π΄0 tan ππ cos ππ = 0 3. Equilibrium equation in π¦π¦1 direction: πππ₯π₯1 π¦π¦1 π΄π΄0 sec ππ + πππ₯π₯ π΄π΄0 sin ππ − πππ₯π₯π₯π₯ π΄π΄0 cos ππ − πππ¦π¦ π΄π΄0 tan ππ cos ππ + πππ¦π¦π¦π¦ π΄π΄0 tan ππ sin ππ = 0 4. Using the relationship πππ₯π₯π₯π₯ = πππ¦π¦π¦π¦ , and also simplifying and rearranging, we obtain πππ₯π₯1 = πππ₯π₯ cos2 ππ + πππ¦π¦ sin2 ππ + 2πππ₯π₯π₯π₯ sin ππ cos ππ πππ₯π₯1 π¦π¦1 = −οΏ½πππ₯π₯ − πππ¦π¦ οΏ½ sin ππ cos ππ + πππ₯π₯π₯π₯ (cos2 ππ − sin2 ππ) 5. For ππ = 0, πππ₯π₯1 = 6. For ππ = 90°, πππ₯π₯1 = and πππ₯π₯1 π¦π¦1 = and πππ₯π₯1 π¦π¦1 = Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr ο€ Transformation Equations for Plane Stress 1. Using the following trigonometric identities: 1 1 1 cos2 ππ = (1 + cos 2ππ), sin2 ππ = (1 − cos 2ππ), sin ππ cos ππ = sin 2ππ 2 2 2 2. The transformation equation is expressed in a more convenient form πππ₯π₯1 = πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ + cos 2ππ + πππ₯π₯π₯π₯ sin 2ππ 2 2 πππ₯π₯1 π¦π¦1 = − πππ₯π₯ − πππ¦π¦ sin 2ππ + πππ₯π₯π₯π₯ cos 2ππ 2 ο transformation equation for plane stress 3. Normal stress on the π¦π¦1 face – can be obtained by substituting ππ + 90° πππ¦π¦1 = πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ − cos 2ππ − πππ₯π₯π₯π₯ sin 2ππ 2 2 4. It is noted that πππ₯π₯1 + πππ¦π¦1 = πππ₯π₯ + πππ¦π¦ ο€ Special Cases of Plane Stress 1. Uniaxial stress, i.e. πππ¦π¦ = πππ₯π₯π₯π₯ = πππ₯π₯1 = πππ₯π₯ (1 + cos 2ππ) 2 πππ₯π₯1 π¦π¦1 = − πππ₯π₯ (sin 2ππ) 2 2. Pure shear, i.e. πππ₯π₯ = πππ¦π¦ = πππ₯π₯1 = πππ₯π₯π₯π₯ sin 2ππ πππ₯π₯1 π¦π¦1 = πππ₯π₯π₯π₯ cos 2ππ 3. Biaxial stress, i.e. πππ₯π₯π₯π₯ = πππ₯π₯1 = πππ₯π₯ + πππ¦π¦ 2 πππ₯π₯1 π¦π¦1 = − + πππ₯π₯ − πππ¦π¦ πππ₯π₯ − πππ¦π¦ 2 2 sin 2ππ cos 2ππ Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr ο€ Example 7-1: Internal pressure results in longitudinal stress πππ₯π₯ = 6,000 psi and circumferential stress πππ¦π¦ = 12,000 psi. Differential settlement after an earthquake ο rotation at support B ο shear stress πππ₯π₯π₯π₯ = 2,500 psi. Find the stresses acting on the element when rotated through angle ππ = 45 ° 7.3 Principal Stresses and Maximum Shear Stresses ο€ Principal Stresses 1. Principal stresses: maximum and minimum stresses (ο occurs at every 90°) 2. Setting the derivative to be zero, i.e. πππππ₯π₯1 = −οΏ½πππ₯π₯ − πππ¦π¦ οΏ½ sin 2ππ + 2πππ₯π₯π₯π₯ cos 2ππ = 0 ππππ from which we get the principal angle ππππ tan 2ππππ = 2πππ₯π₯π₯π₯ πππ₯π₯ − πππ¦π¦ Note: the principal angles for minimum and maximum stresses are perpendicular to each other (why?) 3. Substituting ππππ into the transformation formula via (ο ) Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. πππ₯π₯ − πππ¦π¦ 2 2 , π π = οΏ½οΏ½ οΏ½ + πππ₯π₯π₯π₯ 2 cos 2ππππ = πππ₯π₯ − πππ¦π¦ , 2π π sin 2ππππ = 4. The larger of the two principal stresses, ππ1 ππ1 = πππ₯π₯1 οΏ½ππππ οΏ½ = = = Junho Song junhosong@snu.ac.kr πππ₯π₯π₯π₯ π π πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ + cos 2ππππ + πππ₯π₯π₯π₯ sin 2ππππ 2 2 πππ₯π₯π₯π₯ πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ πππ₯π₯ − πππ¦π¦ + οΏ½ οΏ½ + πππ₯π₯π₯π₯ οΏ½ οΏ½ 2 2 2π π π π πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ 2 2 + οΏ½οΏ½ οΏ½ + πππ₯π₯π₯π₯ 2 2 5. The smaller of the principal stresses, ππ2 ο From the property ππ1 + ππ2 = ππ2 = πππ₯π₯ + πππ¦π¦ − ππ1 = πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ 2 2 − οΏ½οΏ½ οΏ½ + πππ₯π₯π₯π₯ 2 2 6. A single formula for the principal stresses ππ1,2 = πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ 2 2 ± οΏ½οΏ½ οΏ½ + πππ₯π₯π₯π₯ 2 2 ο€ Principal Angles 1. Principal angles ππππ1 and ππππ2 (corresponding to ππ1 and ππ2 , respectively) are roots 2πππ₯π₯π₯π₯ of the equation tan 2ππππ = ππ ππππ2 π₯π₯ −πππ¦π¦ ο Check the normal stresses to determine ππππ1 and 2. Alternatively, ππππ1 is the root that satisfies both cos 2ππππ1 = Then ππππ2 is 90° larger or smaller than ππππ1 πππ₯π₯ −πππ¦π¦ 2π π and sin 2ππππ1 = πππ₯π₯π₯π₯ π π ο€ Shear Stresses on the Principal Planes 1. If we set πππ₯π₯1 π¦π¦1 = 0 for the transformation equation πππ₯π₯1 π¦π¦1 = − 2πππ₯π₯π₯π₯ πππ₯π₯π₯π₯ cos 2ππ, we get the equation tan 2ππ = ππ for having principal stresses π₯π₯ −πππ¦π¦ πππ₯π₯ −πππ¦π¦ 2 sin 2ππ + , which is the same as the condition 2. “The shear stresses are zero on the principal planes” . Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr ο€ Maximum Shear Stresses 1. Setting the derivative to be zero, i.e. πππππ₯π₯1 π¦π¦1 = −οΏ½πππ₯π₯ − πππ¦π¦ οΏ½ cos 2ππ − 2πππ₯π₯π₯π₯ sin 2ππ = 0 ππππ tan 2πππ π = − πππ₯π₯ − πππ¦π¦ 2πππ₯π₯π₯π₯ tan 2πππ π = − 1 tan 2ππππ 2. Relationship between πππ π and ππππ : 3. Can show (See textbook for the derivation) πππ π = ππππ ± 45° 4. Maximum shear stress πππ₯π₯ − πππ¦π¦ 2 ππ − ππ2 2 = 1 ππmax = οΏ½οΏ½ οΏ½ + πππ₯π₯π₯π₯ 2 2 ο€ Example 7-3: The state of stress in the beam web at element C is πππ₯π₯ = 86 MPa, πππ¦π¦ = −28 MPa, and πππ₯π₯π₯π₯ = −32 MPa. (a) Determine the principal stresses and show them on a sketch of a properly oriented element; and (b) Determine the maximum shear stresses and show them on a sketch of a properly oriented element. Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr 7.4 Mohr’s Circle for Plane Stress ο€ Equations of Mohr’s Circle 1. Recall the transformation equation πππ₯π₯1 = πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ + cos 2ππ + πππ₯π₯π₯π₯ sin 2ππ 2 2 → πππ₯π₯1 − πππ₯π₯1 π¦π¦1 = − πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ = cos 2ππ + πππ₯π₯π₯π₯ sin 2ππ 2 2 πππ₯π₯ − πππ¦π¦ sin 2ππ + πππ₯π₯π₯π₯ cos 2ππ 2 2. It can be shown that οΏ½πππ₯π₯1 − πππ₯π₯ + πππ¦π¦ 2 οΏ½ + πππ₯π₯21 π¦π¦1 = 2 3. Note from the previous note that ππaver = πππ₯π₯ + πππ¦π¦ 2 πππ₯π₯ − πππ¦π¦ 2 2 π π = οΏ½οΏ½ οΏ½ + πππ₯π₯π₯π₯ 2 4. Now the equation above becomes 2 οΏ½πππ₯π₯1 − ππaver οΏ½ + πππ₯π₯21 π¦π¦1 = π π 2 5. In words, (πππ₯π₯1 , πππ₯π₯1 π¦π¦1 ) is located on a circle whose center is ( is ____ , ) and the radius ο€ Construction of Mohr’s Circle 1. What’s required: _____, _____ and _____ 2. Sign conventions (consistent with the transformation formula, etc.) - Positive shear stress: d________ - Positive normal stress: to the r______ - Positive rotation: c___________ 3. Construction procedure 1) Draw a set of coordinate axes with πππ₯π₯1 as abscissa and πππ₯π₯1 π¦π¦1 as ordinate Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. 2) Locate the center C, ( , Junho Song junhosong@snu.ac.kr ) 3) Locate point A representing the stress condition ( , ) shown in Figure (a), ππ = 4) Locate point B representing the stress condition on the π¦π¦ face, i.e. ( , ), ππ = 5) Draw a line AB. This goes through C (why?), i.e. opposite ends of the diameter of the circle 6) Using Point C as the center, draw Mohr’s circle through points A and B. The radius is the length of the line segments AC and BC, which is π π = πππ₯π₯ −πππ¦π¦ 2 οΏ½οΏ½ 2 2 οΏ½ + πππ₯π₯π₯π₯ ο€ Stresses on an Inclined Element 1. Consider the new axes π₯π₯1 and π¦π¦1 after rotation ππ 2. From the point A representing the original state (πππ₯π₯ , πππ¦π¦ ), rotate by 2ππ c__________wise to locate the point D representing the inclined element 3. The coordinates of D are the normal and shear stresses of the inclined element Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr 4. Proof: available in the textbook ο€ Principal Stresses 1. The points ππ1 and ππ2 on Mohr’s circle represent m________ and m_________ normal stresses, respectively ο principal stresses 2. Principal stresses: ππ1 = ππππ + πΆπΆππ1 = ππ2 = ππππ − πΆπΆππ2 = + 2 − 2 3. The angle of rotation to achieve the principal planes = the angle between A and ππ1 (or ππ2 ) divided by _____ 4. The angle ππππ1 can be obtained from cos 2ππππ1 = πππ₯π₯ − πππ¦π¦ 2π π sin 2ππππ1 = 5. From Mohr’s circle, it is clear that ππππ2 = ππππ1 + πππ₯π₯π₯π₯ π π ° ο€ Maximum Shear Stresses 1. The points ππ1 and ππ2 on Mohr’s circle represent maximum positive and negative shear stresses, respectively. 2. The angle between these points and ππ1 and ππ2 (on Mohr’s circle) = 3. This confirms once again that the angle between principal stress and maximum shear stresses is 4. The normal stresses under maximum shear stresses = Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. ο€ Example 7-4: At a point on the surface of a hydraulic ram on a piece of construction equipment, the material is subjected to biaxial stresses πππ₯π₯ = 90 MPa and πππ¦π¦ = 20 MPa. Using Mohr’s circle, determine the stresses acting on an element inclined at an angle ππ = 30°. Junho Song junhosong@snu.ac.kr Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. ο€ Example 7-5: An element in plane stress on the surface of an oil-drilling pump arm is subjected to stresses πππ₯π₯ = 100 MPa, πππ¦π¦ = 34 MPa, and πππ₯π₯π₯π₯ = 28 MPa. Using Mohr’s circle, determine (a) the stresses acting on an element inclined at an angle ππ = 40°, (b) the principal stresses, and (c) the maximum shear stresses. Junho Song junhosong@snu.ac.kr Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. ο€ Example 7-6: At a point on the surface of a metalworking lathe the stresses are πππ₯π₯ = −50 MPa, πππ¦π¦ = 10 MPa, and πππ₯π₯π₯π₯ = −40 MPa. Using Mohr’s circle, determine (a) the stresses acting on an element inclined at an angle ππ = 45°, (b) the principal stresses, and (c) the maximum shear stresses. Junho Song junhosong@snu.ac.kr Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr 7.5 Hooke’s Law for Plane Stress ο€ Hooke’s Law for Plane Stress 1. Conditions (in addition to being in “Plane Stress”) 1) Material properties uniform throughout the body and in all directions (h__________ and i_________) 2) The material is l______ e______, i.e. follows Hooke’s law 2. “Resultant” strains 1 1 ππ οΏ½πππ₯π₯ − πππππ¦π¦ οΏ½ πππ¦π¦ = οΏ½πππ¦π¦ − πππππ₯π₯ οΏ½ πππ§π§ = − οΏ½πππ₯π₯ + πππ¦π¦ οΏ½ πΈπΈ πΈπΈ πΈπΈ πππ₯π₯ = 3. Shear strain: πΎπΎπ₯π₯π₯π₯ = πππ₯π₯π₯π₯ πΊπΊ 4. Solving the equations of resultant strains simultaneously for πππ₯π₯ and πππ¦π¦ , we can obtain “Hooke’s law for plane stress” as πππ₯π₯ = πππ¦π¦ = πΈπΈ οΏ½ππ + πππππ¦π¦ οΏ½ 1 − ππ 2 π₯π₯ πΈπΈ οΏ½ππ + πππππ₯π₯ οΏ½ 1 − ππ 2 π¦π¦ πππ₯π₯π₯π₯ = πΊπΊπΎπΎπ₯π₯π₯π₯ 5. Hooke’s law for plane stress contain three material constants: πΈπΈ, πΊπΊ, and ππ, but only two are independent because of the relationship πΊπΊ = πΈπΈ 2(1 + ππ) ο€ Volume Change (for Plane Stress) 1. Original volume: ππ0 = ππππππ 2. Final volume: ππ1 = (ππ + πππππ₯π₯ )οΏ½ππ + πππππ¦π¦ οΏ½(ππ + πππππ§π§ ) 3. Volume change: π₯π₯π₯π₯ = ππ1 − ππ0 ≅ ππ0 (πππ₯π₯ + πππ¦π¦ + πππ§π§ ) 4. Unit volume change (“dilatation”): ππ = π₯π₯π₯π₯/ππ0 = Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr ο€ Strain-Energy Density (for Plane Stress) 1. Work done by the force on π₯π₯-face and π¦π¦-face: 1 1 (πππ₯π₯ ππππ)(πππππ₯π₯ ) + οΏ½πππ¦π¦ πππποΏ½οΏ½πππππ¦π¦ οΏ½ 2 2 ππππππ = οΏ½πππ₯π₯ πππ₯π₯ + πππ¦π¦ πππ¦π¦ οΏ½ 2 2. Strain energy density by the normal stresses: π’π’1 = 1 οΏ½ππ ππ + πππ¦π¦ πππ¦π¦ οΏ½ 2 π₯π₯ π₯π₯ π’π’2 = πππ₯π₯π₯π₯ πΎπΎπ₯π₯π₯π₯ 2 3. Strain energy density associated with the shear stresses: 4. Strain energy density in plane stress: π’π’ = 1 οΏ½ππ ππ + πππ¦π¦ πππ¦π¦ + πππ₯π₯π₯π₯ πΎπΎπ₯π₯π₯π₯ οΏ½ 2 π₯π₯ π₯π₯ π’π’ = 2 πππ₯π₯π₯π₯ 1 οΏ½πππ₯π₯2 + πππ¦π¦2 − 2πππππ₯π₯ πππ¦π¦ οΏ½ + 2πΈπΈ 2πΊπΊ 5. Using Hooke’s law, the strain energy density can be alternatively described as π’π’ = 2 πΊπΊπΎπΎπ₯π₯π₯π₯ πΈπΈ 2 2 οΏ½ππ + ππ + 2ππππ ππ οΏ½ + π¦π¦ π₯π₯ π¦π¦ 2 2(1 − ππ 2 ) π₯π₯ ο€ Example 7-7: Consider a rectangular plate with thickness π‘π‘ = 7mm under plane stress (biaxial) condition. The readings of the gages A and B give πππ₯π₯ = −0.00075 and πππ¦π¦ = 0.00125. πΈπΈ = 73 GPa and ν = 0.33. Find the stresses πππ₯π₯ and πππ¦π¦ and the change Δπ‘π‘ in the thickness. Find the volume change (or dilatation) ππ and the strain energy density π’π’ for the plate. Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. 7.7 Plane Strain ο€ Plane Strain Versus Plane Stress 1. Recall “plane stress” condition: πππ§π§ = 0 πππ₯π₯π₯π₯ = 0 πππ¦π¦π¦π¦ = 0 2. “Plane strain” condition: πππ§π§ = 0 πΎπΎπ₯π₯π₯π₯ = 0 πΎπΎπ¦π¦π¦π¦ = 0 Junho Song junhosong@snu.ac.kr Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr 3. Under ordinary conditions, plane stress and strain (do/do not) occur simultaneously. 4. The following exceptional cases of plane stress = plain strain (why?) 1) Plane stress with πππ₯π₯ = −πππ¦π¦ 2) Zero Poisson effect, i.e. ν = 0 ο€ Transformation Equations for Plane Strain 1. Transformation equations (Proof in Textbook) πππ₯π₯1 = πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ πΎπΎπ₯π₯π₯π₯ + cos 2ππ + sin 2ππ 2 2 2 πΎπΎπ₯π₯1 π¦π¦1 πππ₯π₯ − πππ¦π¦ πΎπΎπ₯π₯π₯π₯ =− sin 2ππ + cos 2ππ 2 2 2 2. Sum of strains is conserved, i.e. πππ₯π₯ + πππ¦π¦ = const 3. Principal strains tan 2ππππ = ππ1,2 = πΎπΎπ₯π₯π₯π₯ πππ₯π₯ − πππ¦π¦ πππ₯π₯ + πππ¦π¦ πππ₯π₯ − πππ¦π¦ 2 πΎπΎπ₯π₯π₯π₯ 2 οΏ½ +οΏ½ οΏ½ ± οΏ½οΏ½ 2 2 2 4. Maximum shear strains and corresponding normal strains πππ₯π₯ − πππ¦π¦ 2 πΎπΎπ₯π₯π₯π₯ 2 πΎπΎmax οΏ½ +οΏ½ οΏ½ = οΏ½οΏ½ 2 2 2 ππaver = πππ₯π₯ + πππ¦π¦ 2 5. Mohr’s circle for strains (ο ) 6. Applications of transformation equations: one can use transformation equations for plane strain for plane stress conditions (and vice versa) because πππ§π§ does not affect the strains Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. Junho Song junhosong@snu.ac.kr ο€ Strain Measurements 1. Electrical-resistance strain gage measures n_________ strain in t______ directions with 45° angle differences (why?) 2. How it works: electrical resistance of wire is altered when it stretches or shortens ο€ Example 7-8: Consider a plane strain condition with πππ₯π₯ = 340 × 10−6 , πππ¦π¦ = 110 × 10−6 , πΎπΎπ₯π₯π₯π₯ = 180 × 10−6 . Determine (a) strains at ππ = 30°, (b) principal strains, and (c) maximum shear strains. Dept. of Civil and Environmental Engineering, Seoul National University 457.201 Mechanics of Materials and Lab. ο€ Example 7-9: Explain how to obtain strains πππ₯π₯1 , πππ¦π¦1 and πΎπΎπ₯π₯1 π¦π¦1 associated with an angle ππ from the gage readings ππππ , ππππ and ππππ . Junho Song junhosong@snu.ac.kr
