Volumetric strain and Bulk Modulus

Volumetric strain

The ratio of change in volume to original volume is known as volumetric strain.

Volumetric strain  = Ԑv

Ԑv = Change in volume /original volume

Ԑv= dV/V

Where dV is change in volume of the body and V is original volume.

Volumetric strain in a body

Let us consider a rectangular body of length L, breadth b and thickness t as shown in figure.
L = Length of the rectangular bar
b = Width of the rectangular bar
t  = Thickness or depth of the rectangular bar
Original Volume of the rectangular bar is = V

                          V= L x b x t
ΔL= Change in length of the rectangular bar
Δb = Change in width of the rectangular bar
Δt  = Change in thickness of the rectangular bar

The final volume of the bar is = V'

V' = (L+ ΔL).(b+Δb).(t+Δt) 

    = Lbt + btΔL + LtΔb + LbΔt 

{The small quantities of squared and cubed terms such as ΔL.Δb, ΔL.Δt etc. are neglected}

Change in volume of the rectangular bar is = ΔV

ΔV = V' - V 

    = Lbt + bt.ΔL + Lt.Δb + Lb.Δt - Lbt

      =  btΔL + LtΔb + LbΔt 

The Volumetric strain  = Ԑv

             Ԑv = dV/V 

          = [btΔL + LtΔb + LbΔt]/ Lbt

                  =  ΔL/L + Δb/b + Δt/t

             Ԑv =  Ԑx + Ԑy + Ԑz

Where 

Ԑx = Strain in x-axis (Length direction)

Ԑy = Strain in y-axis (breadth direction)

Ԑy = Strain in z-axis (thickness direction)

Volumetric strain in uniaxial stress body

Ԑx = Longitudinal Strain in x-axis = σx /E

Ԑy = Lateral Strain in y-axis 

     = - Poisson's ratio(n) x Longitudinal strain

     =  - n.(σx /E)

Ԑz =  Lateral Strain in z-axis

     = - Poisson's ratio(n) x Longitudinal strain

     =  - n.(σx /E)

Volumetric strain Ԑv =  Ԑx + Ԑy + Ԑz

Ԑv =  (σx /E) - n.(σx /E) - n.(σx /E)

        = σx /E (1- 2 n)

Volumetric strain (Ԑv) for a rectangular bar subjected to an axial load P, is given by 

Ԑv = σx /E (1- 2 n) or

     = Ԑy (1- 2n)

Note: Volumetric strain of a cylindrical rod, subjected to an axial tensile load is given by

Ԑv = longitudinal strain – 2 x strain of diameter

Ԑv = ԐL - 2xԐd

Volumetric strain in Triaxial stress body

Let us consider a rectangular bar as shown in the figure. x, y and z are the three dimensional axises of the rectangular bar. 

The bar is subjected with three forces and hence three direct tensile stresses σx, σy, σz are induced in three mutually perpendicular axes x,y,z with each other. The poisson's ratio is n and  elastic modulus is E for the material

The strains in the bar Ԑx, Ԑy, Ԑz corresponding x, y, z axes are given by

The Volumetric strain = Ԑv

                          Ԑv =  Ԑx + Ԑy + Ԑz

substitute the Ԑx,Ԑy,Ԑz in the above equation,the volumetric strain will be,

      Ԑv = (σx + σy+ σz)/E - 2 n (σx + σy+ σz)/E

      Ԑv = (σx + σy+ σz) (1-2 n)/E

Example:

1. A rectangular block 250 mmx100 mmx80mm is subjected to axial loads as follows.
480 kN (tensile in direction of its length)
900 kN ( tensile on 250mm x 80 mm faces)
1000kN (comp. on 250mm x100mm faces)
taking E=200 GN/m2 and Poisson's ratio is 0.25, find  Change in volume of the block

Solution:

sx =480x103/(0.1*0.08)=60 *106N/m2 (tens.)
sy=1000x103/(0.25*0.1)=40*106N/m2(comp)
sz=900x103/(0.25*0.08)=45*106N/m2(tens.)
ex= (60 *106/E)+(0.25* 40*106/E) 

      - (0.25* 5*106/E) = (58.75* 106/E)
ey= -(40 *106/E) - (0.25* 45*106/E) 

       - (0.25* 60*106/E) = (- 66.25* 106/E)
ez= (45 *106/E)-(0.25* 60*106/E) 

       + (0.25* 40*106/E) =(40* 106/E)
Volumetric strain = ev = ex + ey + ez
=(58.75* 106/E)- (66.25* 106/E)+ (40* 106/E) 

= 32.5*106/E
ev = dV/V
so dV= ev V

=32.5*106*[(0.25*0.10*0.08)/(200*109)]*109

=325 mm3(increase)

Volumetric Stress

When a body is subjected to three mutually perpendicular stresses, of equal intensity the stress is called volumetric stress or direct stress.

The most common example of volumetric stress is an object under hydrostatic pressure.

 sx = sy = sz = constant and equal to s 

 s - is called as Volumetric stress 

Volumetric stress produces a change in volume of the body without producing any distortion to the shape of the body.

Note:

If the body is subjected to volumetric stress, the corresponding volumetric strain will be,

Ԑv  =  (σx + σy+ σz) (1-2 n)/E

      =  3σ (1-2 n)/E

Bulk Modulus

Bulk modulus is a ratio of volumetric stress to volumetric strain and constant within the elastic limit of the material. It is denoted by K.

The bulk modulus,

K = Volumetric stress/Volumetric strain

The Units of Bulk Modulus are newtons per square meter (N/m2) or MPa (N/mm2) in the metric system

Characteristics of Bulk Modulus of Elasticity:

• Within the elastic limit, it is the ratio of volumetric stress to volumetric strain.
• It is associated with the change in the volume of a body.
• It exists in solids, liquids, and gases.
• It determines how much the body will compress under a given amount of external pressure.

The relationship between the elastic constants of  Modulus of Elasticity E and Bulk Modulus K  and Poisson's ratio n 

The bulk modulus,

 K = Volumetric stress/Volumetric strain 

     K = σ /Ԑv  = σ /[3σ (1-2 n)/E]

     K = E / [3.(1-2 n)]

     E = 3K (1-2n)

Example:

1. A bar subjected to a tensile load of 55 KN. The diameter of the bar is 31 mm. In a Gauge length of 300mm,the bar was extended by115mm and its diameter was reduced by 0.00367mm. Find the Poisson’s ratio, Young’s modulus, Bulk modulus of the bar material.

Solution:

Given data:
Tensile load(P) = 55KN
Diameter(d) = 31 mm
Gauge Length (L) =300 mm
Extension (ΔL)=0.115mm
Change in diameter (Δd) = 0.00367 mm

a) Poisson's ratio

 n =lateral strain /longitudinal strain

lateral strain = Δd/d = 0.00367/31

longitudinal strain = ΔL/L = 0.115/300
n = 0.308

b) Young's modulus E= stress / strain
Stress = P/A = 55000/[(π/4).31x31]

Strain  = ΔL/L = 0.115/300
E=190.09x103N/mm2

3)Bulk Modulus

 E=3K(1−2n)
190.09∗103 = 3K (1−2x0.308)

Bulk modulus(K)=165x103N/mm2

Assignment:

1. A material has Young's modulus of 1.8x105 N/mm2 and Bulk modulus 1.2x105 N/mm2 .Find the Poisson’s ratio of the material.(Ans:0.25)

2. For a certain material E=2.8K. Calculate the Poisson’s ratio. (Ans: 0.033)

3. For a certain material E=2K, determine the Poisson’s ratio (Ans: 0.165)

4.A material has Young's Modulus of 2.1x105 N/mm2 and Poisson’s ratio 0.29. Calculate the Bulk modulus (Ans:81.4e3,)

5. In an experiment an alloy bar of 1m long and 20mm20mm in section was tested to increase through 0.1 mm, when subjected to an axial tensile load of 6.4 kN. If the value of bulk modulus of the bar is 133 GPa, find the value of Poisson's ratio. (Ans: Poisson's ratio=0.30)

6.A steel cube block of 100 mm side is subjected to a stress of 50 N/mm2 (tensile) in x direction, a stress of 40 N/mm2 (compressive) in y direction, a stress of 30 N/mm2 (tensile) in z direction, calculate strain in each direction and dv. E=200 Gpa, and Poisson ratio =0.25

(Ans:dv=100 mm3 )

7.In a tri-axial stress system the stresses along the three directions are 100 N/mm2 (tensile) in x direction, 60 N/mm2 (tensile) in y direction, 30N/mm2 (comp) in z direction, calculate strain in each direction and change in volume take E=200Gpa and Poisson ratio as 0.25 take x=400,y=150,z=300

(ans dv= 5850 mm3 )

8. A cube of 100 mm side is subjected to a tensile force of 200 kN on all faces (tensile). Find the strains in each direction. Also find the change in volume of the cube.Take E as 200 Gpa and m =4.(ans dv=150mm3)