Stress AND Strain IN PURE Shear PDF

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Engineering Mechanics Statics...

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MOM120S NUST Lecture11

OkorieM.E

Objectives: • • • • •

Revisit Hook’s law Poisson’s ratio Generalized Hook’s Law Stress and strain in Homogenous Isotropic materials (properties are same at all material point) Stress and strain in pure shear (irrotational strain in which a body is elongated in one direction while being shortened perpendicularly)

ElasticityandHooke’sLaw • • • •

All solid materials deform when they are stressed, and as stress is increased, deformation also increases. If a material returns to its original size and shape on removal of load causing deformation, it is said to be elastic. If the stress is steadily increased, a point is reached when, after the removal of load, not all the induced strain is removed. This is called the elastic limit.

Hooke’sLaw • States that providing the limit of proportionality of a material is not exceeded, the stress is directly proportional to the strain produced. • If a graph of stress and strain is plotted as load is gradually applied, the first portion of the graph will be a straight line. • The slope of this line is the constant of proportionality called modulus of Elasticity, E or Young’s Modulus. • It is a measure of the stiffness of a material.

Note: For normal stress   Modulus of Elasticity,    



For shear stress  Modulus of rigidity or shear modulus    For volumetric strain   is proportional to hydrostatic stress  within the elastic range. 

Therefore   



Where  is called the bulk modulus What is hydrostatic stress? Stress that acts along the x, y, and z which brings about change in volume of the shape

Lateral Strain and Poisson’s Ratio • Under the action of a longitudinal stress, a body will extend in the direction of the stress and contract in the transverse or lateral direction (see Fig. below). • The reverse occurs under a compressive load.

Poisson’s Ratio

• If force P is directed along the x  axis, the stress developed equals   ⁄ (1) • From Hooke’s law, the strain along    is    ⁄ (2) • A and E is the modulus of elasticity of the materials respectively Fig1b: Note:     0     0

Poisson’s ration represented by the Greek word  (nu) is given              Note: the negative value used in the formula is to obtain a positive value of   because the lateral and axial stress has opposite value in every engineering material Combining equation 2 and 3 gives            Example 1: A 500 mm long, 16 mm diameter rod made of a homogenous, isotropic material is observed to increase in length by 300 and a decrease in dimeter by 2.4 when subjected to an axial 12 load. Determine the modulus of elasticity and Poisson’s ratio of the material.

MultiaxialLoading:GeneralizedHooke’sLawor GeneralStress‐StrainRelationship Using superposition methodto solvestrainon eachdirection andcombing themgivesthe generalized stressequation

General Stress-Strain Relationship

General Stress-Strain Relationship

General Stress-Strain Relationship For an element subjected to triaxial stresses,

x, y

and  z , the total strain in x direction will be

due to  x and lateral strains due to  y

and  z .

Using the principle of superposition, the resultant strain in x-direction is:

x 

 x  y  z 

E



E

i. e.  x 

E

1 {   ( y   z )} E x

y 

1 {   ( x   z )} E y

z 

1 {   ( x   y )} E z

Generalised Hooke’s Law in three dimensions

General Stress-Strain Relationship Note: In the case of shear strain, there is no lateral strain, hence the shear stress/shear strain relationship is the same for both uniaxial and complex strain systems.

Plain Stress and Plain Strain Note: • • • •

A plain stress condition is said to exist when stress in the z direction is zero. The above equations may be applied for but strain in the z direction is not zero. Also plain strain condition exists when the strain in z direction is zero. Using strain in Z direction as zero in this case does not mean that stress in the z direction is zero.

Strain Caused by Stress and Temperature In addition to strain caused by stress, there may also be thermal strain due to change in temperature. The general form of the stress/strain relations is

Example2: Aplateofuniformthickness1cmanddimension3  2isacted uponbytheloadsshown.Taking  2  10 ⁄ ,determine and .Poison’sratiois0.3

Example 2 The steel block shown below is subjected to a uniform pressure on all its faces. Knowing that the change in length of edge AB is 1.2  10 in. , determine (a) the change in length of the other two edges, (b) the pressure  applied to the faces of the block. Assume E  29  10 Psiandν  0.29

Note: Solution • If a body is subjected to a uniform hydrostatic pressure , each of the stress components is then equal to –  and equation

Volumetric Strain • Hydrostatic stress refers to tensile or compressive stress in all dimensions within or external to a body. • Hydrostatic stress results in change in volume of the material. • Consider a cube with sides x, y, z. Let dx, dy, and dz represent increase in length in all directions. • i.e. new volume = (x + dx) (y + dy) (z + dz) • Neglecting products of small quantities: • New volume = x y z + z y dx + x z dy + x y dz Original volume = x y z = z y dx + x z dy + x y dz • Volumetric strain, = z y dx + x z dy + x y dz xyz = dx/x + dy/y + dz/z 







Note: By similar reasoning, on area x y 





To do Try to show that

Relationship between Elastic Modulus (E) and Bulk Modulus, K • Remember that        1  x   ( y   z ) E For hydrostatic stress,  x   y   z  

x 

i .e .

x 

 1   2    1 2  E E

Similarly ,  y and  z are each

v   x   y   z



1 2  E  Volumetric strain

3 1 2  E 3 E 12 

v 

v

Volumetric or hydrostatic stress   v Volumetric strain E i .e . E  3 K 1  2  and K  3 1 2  Bulk Modulus, K 

Dilation A dilation is a transformation (notation ) that produces an image that is the same shape as the original, but is a different size. A dilation stretches or shrinks the original figure.      = hydrostatic pressure The constant  is the bulk modulus or modulus of compression of the material.  is a positive quantity because the dilation  is always negative. (Note: hydrostatic pressure can only decrease the volume, hence   )

Maximum Value For Poisson’s Ratio From the equation, if v = 0.5, the value of K becomes infinitely large. Hence the body is incompressible. If v > 0.5, K becomes negative i.e. the body will expand under hydrostatic pressure which is inconceivable. It may be concluded that the upper limit of Poisson’s ratio is 0.5. Note:

K

2 G 1  3 1 2 

Where: G is Shear Modulus

and

E  2 G 1 

Shear Strain •

In our previous discussion, we assumed that shear stress does not exist but in more representative manner, shear stress exists as shown in fig.1 below

Fig.1 Fig.2

• Theshearingstressesshownwilltendtodeformacubicelementof materialintoanobliqueparallelepiped.Fig.2 • τ andτ areappliedtofacesoftheelementrespectively perpendicularto and axes.Sinceτ   τ ,theelementistend todeformtorhombusofsideequaltooneshowninfig.3below.

• Graph of    shows similarity to the normal stressstrain diagram discussed earlier • For any homogeneous isotropic material    and    also    The generalized Hooke’s law for a homogeneous isotropic material

Note: Topredictthedeformationscausedina givenmaterialbyanarbitrary combinationofstresses,wehaveto knowthevaluesofE, ν and G.

How does The are related as shown previously. Here are the relationship again   2 1 and   2 1 G is the shear modulus of rigidity. It is the coefficient of elasticity for shear force

Example 3: A circle diameter d  225 mm is scribed on an unstressed aluminium plate of thickness t  18mm. Forces acting in the plane of plates later cause normal stresses σ  84MPa and σ  140MPa. For E  70GPa  and   , 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.

Question 1 A 500 mm long bar has rectangular cross-section 20 mm  40 mm. The bar is subjected to:40kN tensile force on 20  40 face; 200kN compressive force on 20mm  500mm face; 300kN tensile force on 40mm  500mm face. Find the change in dimensions and volume, if E  2  10 N⁄mm and Poisson ratio   0.3

Question 2 Consider a cubical element subjected to volumetric stress which acts simultaneously along the mutually perpendicular axes ,  directions such that       . Show that   3 1  2

Question 3 A 20 mm square has been scribed on the side of a large steel pressure vessel. After pressurization, the biaxial stress condition of the square is as shown. Determine the percentage change in the diagonal DB due to the pressurization of the vessel. For the structural steel   200  77.2.

Question 4 A fabric used in air-inflated structures is subjected to biaxial loading that result in normal stresses σ  120MPaandσ  160MPa. Knowing that the properties of the fabric can be approximated as E  87GPaandν  0.34, determine the change in length of (4.1) side AB, (4.2) side BC, (4.3) diagonal AC

Question 5 5.1. For the axial loading shown, determine the change in height and the change in volume of the brass cylinder shown. (5.2) Solve part 5.1, assuming that the loading is hydrostatic with        70...