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Solid Mechanics-I Stress-strain Relationship 1

SHEAR STRESS-STRAIN DIAGRAM

Solid Mechanics-I Stress-strain Relationship 2

SHEAR STRESS-STRAIN DIAGRAM

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Solid Mechanics-I Stress-strain Relationship 3

SHEAR STRESS-STRAIN DIAGRAM

POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 4

• In all engineering materials, the elongation produced by an axial tensile force P in the direction of the force is accompanied by a contraction in transverse direction

• All materials considered will be assumed to be both homogeneous and isotropic, i.e., their mechanical properties will be assumed independent of both position and direction

• It follows that the strain must have the same value for any transverse direction, referred to as the lateral strain

• Poisson’s ratio is an important constant for a given material and is defined as:

𝜈 = −lateral strain

axial strain

𝝂 = −𝝐𝒚

𝝐𝒙= −

𝝐𝒛

𝝐𝒙

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POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 5

• Consider an element of material subjected to uniaxial stress σx the corresponding strain system is shown.

• In x-direction the strain is εx and in y and z-directions the strains are –νεy and –νεz, respectively.

• Using Hooke’s law (σ=Eε), 휀𝑥 =𝜎𝑥

𝐸, 휀𝑦 = −𝜈

𝜎𝑥

𝐸 and 휀𝑧 = −𝜈

𝜎𝑥

𝐸

POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 6

• Element subjected to triaxial stresses σx, σy, and σz, total strain in x-direction is therefore composed of a strain due to σx, and lateral strains due to σy and σz.

휀𝑥 =𝜎𝑥

𝐸− 𝜈

𝜎𝑦

𝐸− 𝜈

𝜎𝑧

𝐸

• Similarly, 휀𝑦 =𝜎𝑦

𝐸− 𝜈

𝜎𝑥

𝐸− 𝜈

𝜎𝑧

𝐸

휀𝑧 =𝜎𝑧

𝐸− 𝜈

𝜎𝑦

𝐸− 𝜈

𝜎𝑥

𝐸

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POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 7

• Equations giving strains can be solved to determine stress components:

𝜎𝑥 =𝜈𝐸

1+𝜈 1−2𝜈휀𝑥 + 휀𝑦 + 휀𝑧 +

𝐸

1+𝜈휀𝑥

• Similarly, 𝜎𝑦 =𝜈𝐸

1+𝜈 1−2𝜈휀𝑥 + 휀𝑦 + 휀𝑧 +

𝐸

1+𝜈휀𝑦

𝜎𝑧 =𝜈𝐸

1+𝜈 1−2𝜈휀𝑥 + 휀𝑦 + 휀𝑧 +

𝐸

1+𝜈휀𝑧

• Lame constant, λ =𝜈𝐸

1+𝜈 1−2𝜈, μ =

𝐸

2 1+𝜈

• Shear Modulus, 𝐺 =𝐸

2(1+𝜈) Bulk Modulus, 𝐾 =

𝐸

3(1−2𝜈)

Plane stress • In many practical situations stress component in z-direction is zero and is referred as plane stress

휀𝑥 =𝜎𝑥

𝐸− 𝜈

𝜎𝑦

𝐸, 휀𝑦 =

𝜎𝑦

𝐸− 𝜈

𝜎𝑥

𝐸

POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 8

Plane strain • If the strain in z-direction is zero (휀𝑧 = 0), this referred to a plane

strain condition

• Zero strain does not mean zero stress in that direction

• Owing to the Poisson’s ratio effect, changes will occur in z-direction

• To keep strain zero in z-direction, it is necessary to have stress in the z-direction

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POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 9

POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 10

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POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 11

Problem 3-33 Plug has diameter of 30mm and fits within a rigid sleeve having an inner diameter of 32mm. Both are 50mm long. Determine the axial pressure that must be applied at the top of plug to cause it contact the sides of sleeve. Also, how far must the plug be compressed downward to do this? Young’s modulus is 5MPa and Poisson’s ratio is 0.45. Solution: Plug should be compressed with the particular amount of load in order to touch the sleeve. That load is used to determine the axial pressure. Sleeve

PLUG

POISSON’S RATIO

Solid Mechanics-I Stress-strain Relationship 12

That load will do the compression and resulted in lateral strain, which can be calculated using,

𝜖𝐿𝑎𝑡𝑒𝑟𝑎𝑙 =𝑑𝑠 − 𝑑𝑝

𝑑𝑝=

32 − 30

30= 0.06667

𝜖𝐴𝑥𝑖𝑎𝑙 = −𝜈𝜖𝐿𝑎𝑡𝑒𝑟𝑎𝑙 → 𝜖𝐴𝑥𝑖𝑎𝑙 = −0.1481

𝑃 = 𝜎 = 𝐸𝜖𝐴𝑥𝑖𝑎𝑙

𝑃 = 741𝑘𝑃𝑎 Answer

𝛿 = 𝜖𝐴𝑥𝑖𝑎𝑙L

𝛿 = 0.1481 × 50

𝛿 = 7.41𝑚𝑚 Answer