2 axial loading
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This document provides an overview of Chapter 2 from the textbook "Mechanics of Materials" which covers stress and strain under axial loading. The chapter discusses normal strain, stress-strain diagrams for ductile and brittle materials, Hooke's law, elastic vs plastic behavior, fatigue, thermal stresses, Poisson's ratio, generalized Hooke's law, dilatation, shearing strain, and more. The document includes example problems and solutions related to determining deformations and stresses in structural members under various loading conditions.
![© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer • Johnston • DeWolf
Dilatation: Bulk Modulus
• Relative to the unstressed state, the change in volume is
( )( )( )[ ] [ ]
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21
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σσσ
ν
εεε
εεεεεε
• For element subjected to uniform hydrostatic pressure,
( )
( )
modulusbulk
213
213
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−
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−=
−
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ν
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• Subjected to uniform pressure, dilatation must be
negative, therefore
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2 axial loading
- 1. MECHANICS OF MATERIALS Third Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 Stress and Strain – Axial Loading
- 2. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Contents Stress & Strain: Axial Loading Normal Strain Stress-Strain Test Stress-Strain Diagram: Ductile Materials Stress-Strain Diagram: Brittle Materials Hooke’s Law: Modulus of Elasticity Elastic vs. Plastic Behavior Fatigue Deformations Under Axial Loading Example 2.01 Sample Problem 2.1 Static Indeterminacy Example 2.04 Thermal Stresses Poisson’s Ratio Generalized Hooke’s Law Dilatation: Bulk Modulus Shearing Strain Example 2.10 Relation Among E, ν, ανδ Γ Sample Problem 2.5 Composite Materials Saint-Venant’s Principle Stress Concentration: Hole Stress Concentration: Fillet Example 2.12 Elastoplastic Materials Plastic Deformations Residual Stresses Example 2.14, 2.15, 2.16
- 3. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Stress & Strain: Axial Loading • Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient. • Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate. • Determination of the stress distribution within a member also requires consideration of deformations in the member. • Chapter 2 is concerned with deformation of a structural member under axial loading. Later chapters will deal with torsional and pure bending loads.
- 4. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Normal Strain strainnormal stress == == L A P δ ε σ L A P A P δ ε σ = == 2 2 LL A P δδ ε σ == = 2 2
- 5. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Stress-Strain Test
- 6. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Stress-Strain Diagram: Ductile Materials
- 7. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Stress-Strain Diagram: Brittle Materials
- 8. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Hooke’s Law: Modulus of Elasticity • Below the yield stress ElasticityofModulus orModulusYoungs= = E Eεσ • Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.
- 9. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Elastic vs. Plastic Behavior • If the strain disappears when the stress is removed, the material is said to behave elastically. • When the strain does not return to zero after the stress is removed, the material is said to behave plastically. • The largest stress for which this occurs is called the elastic limit.
- 10. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Fatigue • Fatigue properties are shown on S-N diagrams. • When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles. • A member may fail due to fatigue at stress levels significantly below the ultimate strength if subjected to many loading cycles.
- 11. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Deformations Under Axial Loading AE P E E === σ εεσ • From Hooke’s Law: • From the definition of strain: L δ ε = • Equating and solving for the deformation, AE PL =δ • With variations in loading, cross-section or material properties, ∑= i ii ii EA LP δ
- 12. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Example 2.01 Determine the deformation of the steel rod shown under the given loads. in.618.0in.07.1 psi1029 6 == ×= − dD E SOLUTION: • Divide the rod into components at the load application points. • Apply a free-body analysis on each component to determine the internal force • Evaluate the total of the component deflections.
- 13. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf SOLUTION: • Divide the rod into three components: 2 21 21 in9.0 in.12 == == AA LL 2 3 3 in3.0 in.16 = = A L • Apply free-body analysis to each component to determine internal forces, lb1030 lb1015 lb1060 3 3 3 2 3 1 ×= ×−= ×= P P P • Evaluate total deflection, ( ) ( ) ( ) in.109.75 3.0 161030 9.0 121015 9.0 121060 1029 1 1 3 333 6 3 33 2 22 1 11 − ×= × + ×− + × × = ++=∑= A LP A LP A LP EEA LP i ii iiδ in.109.75 3− ×=δ
- 14. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Sample Problem 2.1 The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2 . Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2 ). For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E. SOLUTION: • Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC. • Evaluate the deformation of links AB and DC or the displacements of B and D. • Work out the geometry to find the deflection at E given the deflections at B and D.
- 15. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Displacement of B: ( )( ) ( )( ) m10514 Pa1070m10500 m3.0N1060 6 926- 3 − ×−= ×× ×− = = AE PL Bδ ↑= mm514.0Bδ Displacement of D: ( )( ) ( )( ) m10300 Pa10200m10600 m4.0N1090 6 926- 3 − ×= ×× × = = AE PL Dδ ↓= mm300.0Dδ Free body: Bar BDE ( ) ( ) ncompressioF F tensionF F M AB AB CD CD B kN60 m2.0m4.0kN300 0M kN90 m2.0m6.0kN300 0 D −= ×−×−= = += ×+×−= = ∑ ∑ SOLUTION: Sample Problem 2.1
- 16. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Displacement of D: ( ) mm7.73 mm200 mm0.300 mm514.0 = − = = ′ ′ x x x HD BH DD BB ↓= mm928.1Eδ ( ) mm928.1 mm7.73 mm7.73400 mm300.0 = + = = ′ ′ E E HD HE DD EE δ δ Sample Problem 2.1
- 17. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Static Indeterminacy • Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate. 0=+= RL δδδ • Deformations due to actual loads and redundant reactions are determined separately and then added or superposed. • Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. • A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium.
- 18. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Example 2.04 Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied. • Solve for the reaction at A due to applied loads and the reaction found at B. • Require that the displacements due to the loads and due to the redundant reaction be compatible, i.e., require that their sum be zero. • Solve for the displacement at B due to the redundant reaction at B. SOLUTION: • Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied loads.
- 19. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf SOLUTION: • Solve for the displacement at B due to the applied loads with the redundant constraint released, EEA LP LLLL AAAA PPPP i ii ii 9 L 4321 26 43 26 21 3 4 3 321 10125.1 m150.0 m10250m10400 N10900N106000 × =∑= ==== ×==×== ×=×=== −− δ • Solve for the displacement at B due to the redundant constraint, ( )∑ × −== == ×=×= −== −− i B ii ii R B E R EA LP δ LL AA RPP 3 21 26 2 26 1 21 1095.1 m300.0 m10250m10400 Example 2.04
- 20. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf • Require that the displacements due to the loads and due to the redundant reaction be compatible, ( ) kN577N10577 0 1095.110125.1 0 3 39 =×= = × − × = =+= B B RL R E R E δ δδδ • Find the reaction at A due to the loads and the reaction at B kN323 kN577kN600kN3000 = ∑ +−−== A Ay R RF kN577 kN323 = = B A R R Example 2.04
- 21. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Thermal Stresses • A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports. ( ) coef.expansionthermal= =∆= α δαδ AE PL LT PT • Treat the additional support as redundant and apply the principle of superposition. ( ) 0 0 =+∆ =+= AE PL LT PT α δδδ • The thermal deformation and the deformation from the redundant support must be compatible. ( ) ( )TE A P TAEP PT ∆−== ∆−= =+= ασ α δδδ 0
- 22. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Poisson’s Ratio • For a slender bar subjected to axial loading: 0=== zy x x E σσ σ ε • The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence), 0≠= zy εε • Poisson’s ratio is defined as x z x y ε ε ε ε ν −=−== strainaxial strainlateral
- 23. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Generalized Hooke’s Law • For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) strain is linearly related to stress 2) deformations are small EEE EEE EEE zyx z zyx y zyx x σνσνσ ε νσσνσ ε νσνσσ ε +−−= −+−= −−+= • With these restrictions:
- 24. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Dilatation: Bulk Modulus • Relative to the unstressed state, the change in volume is ( )( )( )[ ] [ ] ( ) e)unit volumperin volume(changedilatation 21 111111 = ++ − = ++= +++−=+++−= zyx zyx zyxzyx E e σσσ ν εεε εεεεεε • For element subjected to uniform hydrostatic pressure, ( ) ( ) modulusbulk 213 213 = − = −= − −= ν ν E k k p E pe • Subjected to uniform pressure, dilatation must be negative, therefore 2 10 <<ν
- 25. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Shearing Strain • A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the change in angle between the sides, ( )xyxy f γτ = • A plot of shear stress vs. shear strain is similar the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains, zxzxyzyzxyxy GGG γτγτγτ === where G is the modulus of rigidity or shear modulus.
- 26. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Example 2.10 A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate. SOLUTION: • Determine the average angular deformation or shearing strain of the block. • Use the definition of shearing stress to find the force P. • Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.
- 27. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf • Determine the average angular deformation or shearing strain of the block. rad020.0 in.2 in.04.0 tan ==≈ xyxyxy γγγ • Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress. ( )( ) psi1800rad020.0psi1090 3 =×== xyxy Gγτ • Use the definition of shearing stress to find the force P. ( )( )( ) lb1036in.5.2in.8psi1800 3 ×=== AP xyτ kips0.36=P
- 28. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Relation Among E, ν, and G • An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions. ( )ν+= 1 2G E • Components of normal and shear strain are related, • If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain. • An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain.
- 29. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Sample Problem 2.5 A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses σx = 12 ksi and σz = 20 ksi. For E = 10x106 psi and ν = 1/3, determine the change in: a) the length of diameter AB, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate.
- 30. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf SOLUTION: • Apply the generalized Hooke’s Law to find the three components of normal strain. ( ) ( ) in./in.10600.1 in./in.10067.1 in./in.10533.0 ksi20 3 1 0ksi12 psi1010 1 3 3 3 6 − − − ×+= +−−= ×−= −+−= ×+= −− × = −−+= EEE EEE EEE zyx z zyx y zyx x σνσνσ ε νσσνσ ε νσνσσ ε • Evaluate the deformation components. ( )( )in.9in./in.10533.0 3− ×+== dxAB εδ ( )( )in.9in./in.10600.1 3− ×+== dzDC εδ ( )( )in.75.0in./in.10067.1 3− ×−== tyt εδ in.108.4 3− ×+=ABδ in.104.14 3− ×+=DCδ in.10800.0 3− ×−=tδ • Find the change in volume ( ) 33 333 in75.0151510067.1 /inin10067.1 ×××==∆ ×=++= − − eVV e zyx εεε 3 in187.0+=∆V
- 31. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Composite Materials • Fiber-reinforced composite materials are formed from lamina of fibers of graphite, glass, or polymers embedded in a resin matrix. z z z y y y x x x EEE ε σ ε σ ε σ === • Normal stresses and strains are related by Hooke’s Law but with directionally dependent moduli of elasticity, x z xz x y xy ε ε ν ε ε ν −=−= • Transverse contractions are related by directionally dependent values of Poisson’s ratio, e.g., • Materials with directionally dependent mechanical properties are anisotropic.
- 32. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Saint-Venant’s Principle • Loads transmitted through rigid plates result in uniform distribution of stress and strain. • Saint-Venant’s Principle: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points. • Stress and strain distributions become uniform at a relatively short distance from the load application points. • Concentrated loads result in large stresses in the vicinity of the load application point.
- 33. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Stress Concentration: Hole Discontinuities of cross section may result in high localized or concentrated stresses. ave max σ σ =K
- 34. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Stress Concentration: Fillet
- 35. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Example 2.12 Determine the largest axial load P that can be safely supported by a flat steel bar consisting of two portions, both 10 mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r = 8 mm. Assume an allowable normal stress of 165 MPa. SOLUTION: • Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b. • Apply the definition of normal stress to find the allowable load. • Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor.
- 36. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf • Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b. 82.1 20.0 mm40 mm8 50.1 mm40 mm60 = ==== K d r d D • Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. MPa7.90 82.1 MPa165max ave === K σ σ • Apply the definition of normal stress to find the allowable load. ( )( )( ) N103.36 MPa7.90mm10mm40 3 ×= == aveAP σ kN3.36=P
- 37. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Elastoplastic Materials • Previous analyses based on assumption of linear stress-strain relationship, i.e., stresses below the yield stress • Assumption is good for brittle material which rupture without yielding • If the yield stress of ductile materials is exceeded, then plastic deformations occur • Analysis of plastic deformations is simplified by assuming an idealized elastoplastic material • Deformations of an elastoplastic material are divided into elastic and plastic ranges • Permanent deformations result from loading beyond the yield stress
- 38. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Plastic Deformations • Elastic deformation while maximum stress is less than yield stressK A AP ave maxσ σ == • Maximum stress is equal to the yield stress at the maximum elastic loading K A P Y Y σ = • At loadings above the maximum elastic load, a region of plastic deformations develop near the hole • As the loading increases, the plastic region expands until the section is at a uniform stress equal to the yield stress Y YU PK AP = = σ
- 39. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Residual Stresses • When a single structural element is loaded uniformly beyond its yield stress and then unloaded, it is permanently deformed but all stresses disappear. This is not the general result. • Residual stresses also result from the uneven heating or cooling of structures or structural elements • Residual stresses will remain in a structure after loading and unloading if - only part of the structure undergoes plastic deformation - different parts of the structure undergo different plastic deformations
- 40. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf Example 2.14, 2.15, 2.16 A cylindrical rod is placed inside a tube of the same length. The ends of the rod and tube are attached to a rigid support on one side and a rigid plate on the other. The load on the rod-tube assembly is increased from zero to 5.7 kips and decreased back to zero. a) draw a load-deflection diagram for the rod-tube assembly b) determine the maximum elongation c) determine the permanent set d) calculate the residual stresses in the rod and tube. ksi36 psi1030 in.075.0 , 6 2 = ×= = rY r r σ E A ksi45 psi1015 in.100.0 , 6 2 = ×= = tY t t σ E A
- 41. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf a) draw a load-deflection diagram for the rod- tube assembly ( )( ) in.1036in.30 psi1030 psi1036 kips7.2in075.0ksi36 3- 6 3 , , , 2 ,, ×= × × === === L E Lδ AP rY rY rYY,r rrYrY σ ε σ ( )( ) in.1009in.30 psi1015 psi1045 kips5.4in100.0ksi45 3- 6 3 , , , 2 ,, ×= × × === === L E Lδ AP tY tY tYY,t ttYtY σ ε σ tr tr PPP δδδ == += Example 2.14, 2.15, 2.16
- 42. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf b,c) determine the maximum elongation and permanent set • at a load of P = 5.7 kips, the rod has reached the plastic range while the tube is still in the elastic range ( ) in.30 psi1015 psi1030 ksi30 in0.1 kips0.3 kips0.3kips7.27.5 kips7.2 6 3 t 2t , × × === === =−=−= == L E L A P PPP PP t t t t t rt rYr σ εδ σ in.1060 3 max − ×== tδδ • the rod-tube assembly unloads along a line parallel to 0Yr ( ) in.106.4560 in.106.45 in.kips125 kips7.5 slopein.kips125 in.1036 kips5.4 3 maxp 3max 3- − − ×−=′+= ×−=−=−=′ == × = δδδ δ m P m in.104.14 3− ×=pδ Example 2.14, 2.15, 2.16
- 43. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSThird Edition Beer • Johnston • DeWolf • calculate the residual stresses in the rod and tube. calculate the reverse stresses in the rod and tube caused by unloading and add them to the maximum stresses. ( )( ) ( )( ) ( ) ( ) ksi2.7ksi8.2230 ksi69ksi6.4536 ksi8.22psi10151052.1 ksi6.45psi10301052.1 in.in.1052.1 in.30 in.106.45 , , 63 63 3 3 =−=′+= −=−=′+= −=××−=′=′ −=××−=′=′ ×−= ×− = ′ =′ − − − − tttresidual rrrresidual tt rr . E E L σσσ σσσ εσ εσ δ ε Example 2.14, 2.15, 2.16