Mechanics Of Materials PDF - Third Edition

Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Stress and Strain – Axial Loading Lecture Notes: J. Walt Ole...

Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Stress and Strain – Axial Loading Lecture Notes: J. Walt Oler Texas Tech University © 2002 The McGraw-Hill Companies, Inc. All rights reserved. Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Contents Stress & Strain: Axial Loading Generalized Hooke’s Law Normal Strain Dilatation: Bulk Modulus Stress-Strain Test Shearing Strain Stress-Strain Diagram: Ductile Materials Example 2.10 Stress-Strain Diagram: Brittle Materials Relation Among E, ν, and G Hooke’s Law: Modulus of Elasticity Sample Problem 2.5 Elastic vs. Plastic Behavior Composite Materials Fatigue Saint-Venant’s Principle Deformations Under Axial Loading Stress Concentration: Hole Example 2.01 Stress Concentration: Fillet Sample Problem 2.1 Example 2.12 Static Indeterminacy Elastoplastic Materials Example 2.04 Plastic Deformations Thermal Stresses Residual Stresses Poisson’s Ratio Example 2.14, 2.15, 2.16 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-2 Edition Third MECHANICS OF MATERIALS 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-3 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Normal Strain P 2P P P σ= = stress σ= = σ= A 2A A A δ δ 2δ δ ε= = normal strain ε= ε= = L L 2L L © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-4 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Stress-Strain Test © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-5 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Stress-Strain Diagram: Ductile Materials © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-6 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Stress-Strain Diagram: Brittle Materials © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-7 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Hooke’s Law: Modulus of Elasticity Below the yield stress σ = Eε E = Youngs Modulus or Modulus of Elasticity Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-8 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Elastic vs. Plastic Behavior If the strain disappears when the stress is removed, the material is said to behave elastically. The largest stress for which this occurs is called the elastic limit. When the strain does not return to zero after the stress is removed, the material is said to behave plastically. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2-9 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Fatigue Fatigue properties are shown on S-N diagrams. A member may fail due to fatigue at stress levels significantly below the ultimate strength if subjected to many loading cycles. When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 10 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Deformations Under Axial Loading From Hooke’s Law: σ P σ = Eε ε= = E AE From the definition of strain: δ ε= L Equating and solving for the deformation, PL δ = AE With variations in loading, cross-section or material properties, PL δ =∑ i i i Ai Ei © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 11 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.01 SOLUTION: Divide the rod into components at the load application points. Apply a free-body analysis on each −6 E = 29 × 10 psi component to determine the D = 1.07 in. d = 0.618 in. internal force Evaluate the total of the component Determine the deformation of deflections. the steel rod shown under the given loads. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 12 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf SOLUTION: Apply free-body analysis to each component to determine internal forces, Divide the rod into three components: P1 = 60 × 103 lb P2 = −15 × 103 lb P3 = 30 × 103 lb Evaluate total deflection, Pi Li 1 ⎛ P1L1 P2 L2 P3 L3 ⎞ δ =∑ = ⎜⎜ + + ⎟⎟ i Ai Ei E ⎝ A1 A2 A3 ⎠ = 1 ( ) ( ) ( ⎡ 60 × 103 12 − 15 × 103 12 30 × 103 16 ⎤ + + ) 6⎢ ⎥ 29 × 10 ⎢⎣ 0. 9 0. 9 0.3 ⎥⎦ = 75.9 × 10−3 in. L1 = L2 = 12 in. L3 = 16 in. δ = 75.9 × 10−3 in. A1 = A2 = 0.9 in 2 A3 = 0.3 in 2 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 13 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Sample Problem 2.1 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 The rigid bar BDE is supported by two and DC or the displacements of B links AB and CD. and D. Link AB is made of aluminum (E = 70 Work out the geometry to find the GPa) and has a cross-sectional area of 500 deflection at E given the deflections mm2. Link CD is made of steel (E = 200 at B and D. 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 14 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Sample Problem 2.1 SOLUTION: Displacement of B: PL Free body: Bar BDE δB = AE (− 60 × 103 N )(0.3 m ) = (500 ×10-6 m2 )(70 ×109 Pa ) = −514 × 10 − 6 m δ B = 0.514 mm ↑ ∑MB = 0 Displacement of D: 0 = −(30 kN × 0.6 m ) + FCD × 0.2 m PL FCD = +90 kN tension δD = AE ∑ MD = 0 (90 × 103 N )(0.4 m ) 0 = −(30 kN × 0.4 m ) − FAB × 0.2 m = (600 ×10-6 m2 )(200 ×109 Pa ) FAB = −60 kN compression = 300 × 10− 6 m δ D = 0.300 mm ↓ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 15 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Sample Problem 2.1 Displacement of D: BB′ BH = DD′ HD 0.514 mm (200 mm ) − x = 0.300 mm x x = 73.7 mm EE ′ HE = DD′ HD δE = (400 + 73.7 )mm 0.300 mm 73.7 mm δ E = 1.928 mm δ E = 1.928 mm ↓ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 16 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Static Indeterminacy Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate. A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium. Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. Deformations due to actual loads and redundant reactions are determined separately and then added or superposed. δ = δL +δR = 0 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 17 Edition Third MECHANICS OF MATERIALS 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. 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. Solve for the displacement at B due to the redundant reaction 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 reaction at A due to applied loads and the reaction found at B. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 18 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.04 SOLUTION: Solve for the displacement at B due to the applied loads with the redundant constraint released, P1 = 0 P2 = P3 = 600 × 103 N P4 = 900 × 103 N A1 = A2 = 400 × 10− 6 m 2 A3 = A4 = 250 × 10− 6 m 2 L1 = L2 = L3 = L4 = 0.150 m Pi Li 1.125 × 109 δL = ∑ = i Ai Ei E Solve for the displacement at B due to the redundant constraint, P1 = P2 = − RB A1 = 400 × 10 − 6 m 2 A2 = 250 × 10 − 6 m 2 L1 = L2 = 0.300 m δR = ∑ Pi Li =− ( 1.95 × 103 RB ) i Ai Ei E © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 19 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.04 Require that the displacements due to the loads and due to the redundant reaction be compatible, δ = δL +δR = 0 δ = − ( ) 1.125 × 109 1.95 × 103 RB =0 E E RB = 577 × 103 N = 577 kN Find the reaction at A due to the loads and the reaction at B ∑ Fy = 0 = R A − 300 kN − 600 kN + 577 kN R A = 323 kN R A = 323 kN RB = 577 kN © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 20 Edition Third MECHANICS OF MATERIALS 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. Treat the additional support as redundant and apply the principle of superposition. PL δ T = α (∆T )L δP = AE α = thermal expansion coef. The thermal deformation and the deformation from the redundant support must be compatible. δ = δT + δ P = 0 δ = δT + δ P = 0 P = − AEα (∆T ) PL α (∆T )L + =0 P AE σ= = − Eα (∆T ) A © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 21 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Poisson’s Ratio For a slender bar subjected to axial loading: σx εx = σy =σz = 0 E The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence), εy = εz ≠ 0 Poisson’s ratio is defined as lateral strain εy ε ν= =− =− z axial strain εx εx © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 22 Edition Third MECHANICS OF MATERIALS 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 With these restrictions: σ x νσ y νσ z εx = + − − E E E νσ x σ y νσ z εy = − + − E E E νσ x νσ y σz εz = − − + E E E © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 23 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Dilatation: Bulk Modulus Relative to the unstressed state, the change in volume is [ ( ) ] [ e = 1 − (1 + ε x ) 1 + ε y (1 + ε z ) = 1 − 1 + ε x + ε y + ε z ] = εx +ε y +εz 1 − 2ν = E ( σ x +σ y +σ z ) = dilatation (change in volume per unit volume) For element subjected to uniform hydrostatic pressure, 3(1 − 2ν ) p e = −p =− E k E k= = bulk modulus 3(1 − 2ν ) Subjected to uniform pressure, dilatation must be negative, therefore 0 < ν < 12 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. Edition Third MECHANICS OF MATERIALS 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, τ xy = f (γ xy ) 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, τ xy = G γ xy τ yz = G γ yz τ zx = G γ zx where G is the modulus of rigidity or shear modulus. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 25 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.10 SOLUTION: Determine the average angular deformation or shearing strain of the block. Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress. A rectangular block of material with modulus of rigidity G = 90 ksi is Use the definition of shearing stress to bonded to two rigid horizontal plates. find the force P. 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 26 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Determine the average angular deformation or shearing strain of the block. 0.04 in. γ xy ≈ tan γ xy = γ xy = 0.020 rad 2 in. Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress. ( ) τ xy = Gγ xy = 90 ×103 psi (0.020 rad ) = 1800 psi Use the definition of shearing stress to find the force P. P = τ xy A = (1800 psi )(8 in.)(2.5 in.) = 36 × 103 lb P = 36.0 kips © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 27 Edition Third MECHANICS OF MATERIALS 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. An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain. 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. Components of normal and shear strain are related, E = (1 + ν ) 2G © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 28 Edition Third MECHANICS OF MATERIALS 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 29 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf SOLUTION: Apply the generalized Hooke’s Law Evaluate the deformation components. to find the three components of normal strain. δB A ( ) = ε x d = + 0.533 × 10 −3 in./in. (9 in.) σ x νσ y νσ z δB A = +4.8 × 10−3 in. εx = + − − E E E δC D ( ) = ε z d = + 1.600 × 10 −3 in./in. (9 in.) 1 ⎡ 1 ⎤ = (12 ksi ) − 0 − (20 ksi ) 10 × 106 psi ⎢⎣ 3 ⎥⎦ δC D = +14.4 × 10−3 in. = +0.533 × 10−3 in./in. ( ) δ t = ε y t = − 1.067 ×10−3 in./in. (0.75 in.) νσ x σ y νσ z δ t = −0.800 ×10−3 in. εy = − + − E E E = −1.067 × 10−3 in./in. Find the change in volume νσ x νσ y σ εz = − − + z e = ε x + ε y + ε z = 1.067 × 10 −3 in 3/in 3 E E E = +1.600 × 10 −3 in./in. ∆V = eV = 1.067 × 10−3 (15 × 15 × 0.75)in 3 ∆V = +0.187 in 3 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 30 Edition Third MECHANICS OF MATERIALS 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. Normal stresses and strains are related by Hooke’s Law but with directionally dependent moduli of elasticity, σ σy σz Ex = x Ey = Ez = εx εy εz Transverse contractions are related by directionally dependent values of Poisson’s ratio, e.g., εy ε ν xy =− ν xz = − z εx εx Materials with directionally dependent mechanical properties are anisotropic. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 31 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Saint-Venant’s Principle Loads transmitted through rigid plates result in uniform distribution of stress and strain. Concentrated loads result in large stresses in the vicinity of the load application point. Stress and strain distributions become uniform at a relatively short distance from the load application points. 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 32 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Stress Concentration: Hole Discontinuities of cross section may result in σ max K= high localized or concentrated stresses. σ ave © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 33 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Stress Concentration: Fillet © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 34 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.12 SOLUTION: Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b. Determine the largest axial load P that can be safely supported by a Find the allowable average normal flat steel bar consisting of two stress using the material allowable portions, both 10 mm thick, and normal stress and the stress respectively 40 and 60 mm wide, concentration factor. connected by fillets of radius r = 8 Apply the definition of normal stress to mm. Assume an allowable normal find the allowable load. stress of 165 MPa. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 35 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b. D 60 mm r 8 mm = = 1.50 = = 0.20 d 40 mm d 40 mm K = 1.82 Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. σ max 165 MPa σ ave = = = 90.7 MPa K 1.82 Apply the definition of normal stress to find the allowable load. P = Aσ ave = (40 mm )(10 mm )(90.7 MPa ) = 36.3 × 103 N P = 36.3 kN © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 36 Edition Third MECHANICS OF MATERIALS 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 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 37 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Plastic Deformations σ A Elastic deformation while maximum P = σ ave A = max K stress is less than yield stress Maximum stress is equal to the yield σY A PY = stress at the maximum elastic K loading At loadings above the maximum elastic load, a region of plastic deformations develop near the hole As the loading increases, the plastic PU = σ Y A region expands until the section is at = K PY a uniform stress equal to the yield stress © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 38 Edition Third MECHANICS OF MATERIALS 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 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 Residual stresses also result from the uneven heating or cooling of structures or structural elements © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 39 Edition Third MECHANICS OF MATERIALS 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 Ar = 0.075 in.2 At = 0.100 in.2 b) determine the maximum Er = 30 × 106 psi Et = 15 × 106 psi elongation σY , r = 36 ksi σY ,t = 45 ksi c) determine the permanent set d) calculate the residual stresses in the rod and tube. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 40 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.14, 2.15, 2.16 a) draw a load-deflection diagram for the rod- tube assembly ( ) PY , r = σ Y , r Ar = (36 ksi ) 0.075 in 2 = 2.7 kips σ Y ,r 36 × 103 psi δY,r = εY , r L = L= 30 in. = 36 × 10-3 in. EY , r 30 × 106 psi ( ) PY ,t = σ Y ,t At = (45 ksi ) 0.100 in 2 = 4.5 kips σ Y ,t 45 × 103 psi δY,t = εY ,t L = L= 30 in. = 90 × 10-3 in. EY ,t 15 × 106 psi P = Pr + Pt δ = δ r = δt © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 41 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.14, b,c) 2.15,determine the maximum elongation and permanent set 2.16 at a load of P = 5.7 kips, the rod has reached the plastic range while the tube is still in the elastic range Pr = PY , r = 2.7 kips Pt = P − Pr = (5.7 − 2.7 ) kips = 3.0 kips Pt 3.0 kips σt = = = 30 ksi At 0.1in 2 σt 30 × 103 psi δ t = εt L = L= 30 in. δ max = δ t = 60 ×10−3 in. Et 15 × 106 psi the rod-tube assembly unloads along a line parallel to 0Yr 4.5 kips m= -3 = 125 kips in. = slope 36 × 10 in. Pmax 5.7 kips δ′ = − =− = −45.6 × 10−3 in. m 125 kips in. δ p = δ max + δ ′ = (60 − 45.6 )×10−3 in. δ p = 14.4 ×10 −3 in. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 42 Edition Third MECHANICS OF MATERIALS Beer Johnston DeWolf Example 2.14, 2.15, 2.16 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. δ′− 45.6 × 10−3 in. ε′ = = = −1.52 × 10 −3 in. in. L 30 in. ( )( ) σ r′ = ε ′Er = − 1.52 ×10−3 30 ×106 psi = −45.6 ksi σ t′ = ε ′Et = (− 1.52 ×10−3 )(15 ×106 psi ) = −22.8 ksi σ residual , r = σ r + σ r′ = (36 − 45.6 ) ksi = −9.6 ksi σ residual ,t = σ t + σ t′ = (30 − 22.8) ksi = 7.2 ksi © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 43