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### Files 2 handouts-ch02

1. 1. 2. StrainCHAPTER OBJECTIVES • Define concept of normal strain • Define concept of shear strain • Determine normal and shear strain in engineering applications©2005 Pearson Education South Asia Pte Ltd 1
2. 2. 2. StrainCHAPTER OUTLINE 1. Deformation 2. Strain©2005 Pearson Education South Asia Pte Ltd 2
3. 3. 2. Strain 2.1 DEFORMATION Deformation • Occurs when a force is applied to a body • Can be highly visible or practically unnoticeable • Can also occur when temperature of a body is changed • Is not uniform throughout a body’s volume, thus change in geometry of any line segment within body may vary along its length©2005 Pearson Education South Asia Pte Ltd 3
4. 4. 2. Strain 2.1 DEFORMATION To simplify study of deformation • Assume lines to be very short and located in neighborhood of a point, and • Take into account the orientation of the line segment at the point©2005 Pearson Education South Asia Pte Ltd 4
5. 5. 2. Strain 2.2 STRAIN Normal strain • Defined as the elongation or contraction of a line segment per unit of length • Consider line AB in figure below • After deformation, Δs changes to Δs’©2005 Pearson Education South Asia Pte Ltd 5
6. 6. 2. Strain 2.2 STRAIN Normal strain • Defining average normal strain using εavg (epsilon) Δs − Δs’ εavg = Δs • As Δs → 0, Δs’ → 0 lim Δs − Δs’ ε= B→A along n Δs©2005 Pearson Education South Asia Pte Ltd 6
7. 7. 2. Strain 2.2 STRAIN Normal strain • If normal strain ε is known, use the equation to obtain approx. final length of a short line segment in direction of n after deformation. Δs’ ≈ (1 + ε) Δs • Hence, when ε is positive, initial line will elongate, if ε is negative, the line contracts©2005 Pearson Education South Asia Pte Ltd 7
8. 8. 2. Strain 2.2 STRAIN Units • Normal strain is a dimensionless quantity, as it’s a ratio of two lengths • But common practice to state it in terms of meters/meter (m/m) • ε is small for most engineering applications, so is normally expressed as micrometers per meter (μm/m) where 1 μm = 10−6 • Also expressed as a percentage, e.g., 0.001 m/m = 0.1 %©2005 Pearson Education South Asia Pte Ltd 8
9. 9. 2. Strain 2.2 STRAIN Shear strain • Defined as the change in angle that occurs between two line segments that were originally perpendicular to one another • This angle is denoted by γ (gamma) and measured in radians (rad).©2005 Pearson Education South Asia Pte Ltd 9
10. 10. 2. Strain 2.2 STRAIN Shear strain • Consider line segments AB and AC originating from same point A in a body, and directed along the perpendicular n and t axes • After deformation, lines become curves, such that angle between them at A is θ’©2005 Pearson Education South Asia Pte Ltd 10
11. 11. 2. Strain 2.2 STRAIN Shear strain • Hence, shear strain at point A associated with n and t axes is π lim γnt = − 2 B→A along n θ’ C →A along t • If θ’ is smaller than π/2, shear strain is positive, otherwise, shear strain is negative©2005 Pearson Education South Asia Pte Ltd 11
12. 12. 2. Strain 2.2 STRAIN Cartesian strain components • Using above definitions of normal and shear strain, we show how they describe the deformation of the body • Divide body into small elements with undeformed dimensions of Δx, Δy and Δz©2005 Pearson Education South Asia Pte Ltd 12
13. 13. 2. Strain 2.2 STRAIN Cartesian strain components • Since element is very small, deformed shape of element is a parallelepiped • Approx. lengths of sides of parallelepiped are (1 + εx) Δx (1 + εy)Δy (1 + εz)Δz©2005 Pearson Education South Asia Pte Ltd 13
14. 14. 2. Strain 2.2 STRAIN Cartesian strain components • Approx. angles between the sides are π π π − γxy − γyz − γxz 2 2 2 • Normal strains cause a change in its volume • Shear strains cause a change in its shape • To summarize, state of strain at a point requires specifying 3 normal strains; εx, εy, εz and 3 shear strains of γxy, γyz, γxz©2005 Pearson Education South Asia Pte Ltd 14
15. 15. 2. Strain 2.2 STRAIN Small strain analysis • Most engineering design involves applications for which only small deformations are allowed • We’ll assume that deformations that take place within a body are almost infinitesimal, so normal strains occurring within material are very small compared to 1, i.e., ε << 1.©2005 Pearson Education South Asia Pte Ltd 15
16. 16. 2. Strain 2.2 STRAIN Small strain analysis • This assumption is widely applied in practical engineering problems, and is referred to as small strain analysis • E.g., it can be used to approximate sin θ = θ, cos θ = θ and tan θ = θ, provided θ is small©2005 Pearson Education South Asia Pte Ltd 16
17. 17. 2. Strain EXAMPLE 2.1 Rod below is subjected to temperature increase along its axis, creating a normal strain of εz = 40(10−3)z1/2, where z is given in meters. Determine (a) displacement of end B of rod due to temperature increase, (b) average normal strain in the rod.©2005 Pearson Education South Asia Pte Ltd 17
18. 18. 2. Strain EXAMPLE 2.1 (SOLN) (a) Since normal strain reported at each point along the rod, a differential segment dz, located at position z has a deformed length: dz’ = [1 + 40(10−3)z1/2] dz©2005 Pearson Education South Asia Pte Ltd 18
19. 19. 2. Strain EXAMPLE 2.1 (SOLN) (a) Sum total of these segments along axis yields deformed length of the rod, i.e., 0.2 m z’ = ∫0 [1 + 40(10−3)z1/2] dz = z + 40(10−3)(⅔ z3/2)|00.2 m = 0.20239 m Displacement of end of rod is ΔB = 0.20239 m − 0.2 m = 2.39 mm ↓©2005 Pearson Education South Asia Pte Ltd 19
20. 20. 2. Strain EXAMPLE 2.1 (SOLN) (b) Assume rod or “line segment” has original length of 200 mm and a change in length of 2.39 mm. Hence, Δs’ − Δs 2.39 mm εavg = = = 0.0119 mm/mm Δs 200 mm©2005 Pearson Education South Asia Pte Ltd 20
21. 21. 2. Strain EXAMPLE 2.3 Plate is deformed as shown in figure. In this deformed shape, horizontal lines on the on plate remain horizontal and do not change their length. Determine (a) average normal strain along side AB, (b) average shear strain in the plate relative to x and y axes©2005 Pearson Education South Asia Pte Ltd 21
22. 22. 2. Strain EXAMPLE 2.3 (SOLN) (a) Line AB, coincident with y axis, becomes line AB’ after deformation. Length of line AB’ is AB’ = √ (250 − 2)2 + (3)2 = 248.018 mm©2005 Pearson Education South Asia Pte Ltd 22
23. 23. 2. Strain EXAMPLE 2.3 (SOLN) (a) Therefore, average normal strain for AB is, AB’ − AB 248.018 mm − 250 mm (εAB)avg = = AB 250 mm = −7.93(10−3) mm/mm Negative sign means strain causes a contraction of AB.©2005 Pearson Education South Asia Pte Ltd 23
24. 24. 2. Strain EXAMPLE 2.3 (SOLN) (b) Due to displacement of B to B’, angle BAC referenced from x, y axes changes to θ’. Since γxy = π/2 − θ’, thus γxy = tan−1 ( 3 mm 250 mm − 2 mm)= 0.0121 rad©2005 Pearson Education South Asia Pte Ltd 24
25. 25. 2. Strain CHAPTER REVIEW • Loads cause bodies to deform, thus points in the body will undergo displacements or changes in position • Normal strain is a measure of elongation or contraction of small line segment in the body • Shear strain is a measure of the change in angle that occurs between two small line segments that are originally perpendicular to each other©2005 Pearson Education South Asia Pte Ltd 25
26. 26. 2. Strain CHAPTER REVIEW • State of strain at a point is described by six strain components: a) Three normal strains: εx, εy, εz b) Three shear strains: γxy, γxz, γyz c) These components depend upon the orientation of the line segments and their location in the body • Strain is a geometrical quantity measured by experimental techniques. Stress in body is then determined from material property relations©2005 Pearson Education South Asia Pte Ltd 26
27. 27. 2. Strain CHAPTER REVIEW • Most engineering materials undergo small deformations, so normal strain ε << 1. This assumption of “small strain analysis” allows us to simplify calculations for normal strain, since first-order approximations can be made about their size©2005 Pearson Education South Asia Pte Ltd 27