1.  2005 Pearson Education South Asia Pte Ltd 2. Strain 1 CHAPTER OBJECTIVES To define the concept of normal strain To define the concept of shear strain To determine the normal and shear strain in engineering applications

2.  2005 Pearson Education South Asia Pte Ltd 2. Strain 2 CHAPTER OUTLINE 1.Deformation 2.Strain

3.  2005 Pearson Education South Asia Pte Ltd 2. Strain 3 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 2.1 DEFORMATION

4.  2005 Pearson Education South Asia Pte Ltd 2. Strain 4 Assumptions 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 2.1 DEFORMATION

5.  2005 Pearson Education South Asia Pte Ltd 2. Strain 5 Normal strain 2.2 STRAIN Defined as the elongation or contraction of a line segment per unit of length Undeformed body Deformed body After deformation, Δs changes to Δs ’ Consider line AB in figure above

6.  2005 Pearson Education South Asia Pte Ltd 2. Strain 6 Defining average normal strain using  avg (  = epsilon) 2.2 STRAIN  avg = Δs’ − Δs ΔsΔs  = = Δs − Δs’ ΔsΔs lim B → A along n As Δs → 0, Δs’ → 0 Undeformed body Deformed body

7.  2005 Pearson Education South Asia Pte Ltd 2. Strain 7 If normal strain  is known, use the equation to obtain approx. final length of a short line segment in direction of n after deformation. 2.2 STRAIN Δs’ ≈ (1 +  ) Δs Hence, when  is positive, initial line will elongate, if  is negative, the line contracts

8.  2005 Pearson Education South Asia Pte Ltd 2. Strain 8 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)  is small for most engineering applications, so it is normally expressed as micrometers per meter (μm/m) where 1 μm = 10 −6 m Also expressed as a percentage, e.g., 0.001 m/m = 0.1 %

9.  2005 Pearson Education South Asia Pte Ltd 2. Strain 9 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).

10.  2005 Pearson Education South Asia Pte Ltd 2. Strain 10 2.2 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’ Undeformed body Deformed body

11.  2005 Pearson Education South Asia Pte Ltd 2. Strain 11 2.2 STRAIN Shear strain Hence, shear strain at point A associated with n and t axes is γ nt = lim B→A along n C→A along t ’’ 22 − Deformed body Shear strain positive if  ’ <  /2, and Shear strain negative if  ’ >  /2. Undeformed body t n

12.  2005 Pearson Education South Asia Pte Ltd 2. Strain 12 Cartesian strain components 2.2 STRAIN A point in the un-deformed body is regarded as a small element having dimensions of Δx, Δy and Δz Using above definitions of normal and shear strain, we show how to describe the deformation of the body Undeformed point

13.  2005 Pearson Education South Asia Pte Ltd 2. Strain 13 Undeformed point Since element is very small, deformed shape of element is a parallelepiped (1 +  x )Δx (1 +  y )Δy (1 +  z )Δz 2.2 STRAIN Approx. lengths of sides of parallelepiped are Deformed point (1+  x )Δx (1+  y )Δy (1+  z )Δz

14.  2005 Pearson Education South Asia Pte Ltd 2. Strain 14 22 − γ xy Deformed point (1+  x )Δx (1+  y )Δy (1+  z )Δz 2.2 STRAIN Approx. angles between the sides are 22 − γ yz 22 − γ xz 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 of  x,  y,  z and 3 shear strains of γ xy, γ yz, γ xz

15.  2005 Pearson Education South Asia Pte Ltd 2. Strain 15 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.,  << 1. 2.2 STRAIN 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 sinθ = θ, cosθ = θ and tanθ = θ, provided θ is very small

16.  2005 Pearson Education South Asia Pte Ltd 2. Strain 16 EXAMPLE The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displaced 10 mm downward, determine the normal strain developed in each wire.

17.  2005 Pearson Education South Asia Pte Ltd 2. Strain 17 EXAMPLE (CONT.) A B C 10 mm  CE  BD Free-body diagram  BD = change in length of BD  CE = change in length of CE  CE = 7.14 mm  BD = 4.28 mm

18.  2005 Pearson Education South Asia Pte Ltd 2. Strain 18 Normal strain developed in each wire EXAMPLE (CONT.)

19.  2005 Pearson Education South Asia Pte Ltd 2. Strain 19 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 State of strain at a point is described by six strain components: a)Three normal strains:  x,  y,  z b)Three shear strains: γ xy, γ xz, γ yz c)These components depend upon the orientation of the line segments and their location in the body

20.  2005 Pearson Education South Asia Pte Ltd 2. Strain 20 CHAPTER REVIEW Strain is a geometrical quantity measured by experimental techniques. Stress in body is then determined from material property relations Most engineering materials undergo small deformations, so normal strain  << 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