We were discussing “Determination of volumetric strain for a rectangular bar subjected with an axial load in the direction of length of the rectangular bar, “Total elongation of the bar due to its own weight” and “Thermal stress and strain” in our previous posts.

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Let us determine the

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Today we will see here the determination of
volumetric strain for a rectangular bar subjected with three forces mutually
perpendicular with each other with the help of this post.

Before going ahead we must recall the basic concept of volumetric strain which is explained here briefly.

Before going ahead we must recall the basic concept of volumetric strain which is explained here briefly.

When an object will be subjected with a system of
forces, object will undergo through some changes in its dimensions and hence,
volume of that object will also be changed.

Volumetric strain will be defined as the ratio of
change in volume of the object to its original volume. Volumetric strain is
also termed as bulk strain.

Ԑ

_{v}= Change in volume /original volume
Ԑ

_{v}= dV/V
Let us consider one rectangular bar as displayed in
following figure. x, y and z are the dimensions of the rectangular bar which is
subjected with three forces and hence three direct tensile stresses mutually
perpendicular with each other.

We can also say that we have following initial
dimensions of the rectangular bar

x = Length of the rectangular bar

y = Width of the rectangular bar

z = Thickness or depth of the rectangular bar

Volume of the rectangular bar, V = xyz

Δx=Change in length of the rectangular bar

Δy=Change in width of the rectangular bar

Δz= Change in thickness or depth of the rectangular
bar

ΔV= Change in volume of the rectangular bar

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*Let
us determine the final dimensions of the rectangular bar *

*Let us determine the final dimensions of the rectangular bar*

Final length of the rectangular bar = x+ Δx

Final width of the rectangular bar = y + Δy

Final thickness or depth of the rectangular bar = z
+ Δz

Final volume of the rectangular bar = (x+ Δx). (y +
Δy). (t + Δt)

Let us ignore the product of small quantities and we
will have

Final volume of the rectangular bar = xyz + y. z. Δx
+ x. z. Δy + x. y. Δz

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Let us determine the **change in volume** of the rectangular bar

Change in volume of the rectangular bar = Final
volume – initial volume

ΔV= (xyz + y. z. Δx + x. z. Δy + x. y. Δz) - xyz

ΔV= y. z. Δx + x. z. Δy + x. y. Δz

Volumetric strain also known as bulk strain will be
determined as following

Ԑ

_{v}= Change in volume /original volume
Ԑ

_{v}= dV/V
Ԑ

_{v}= (y. z. Δx + x. z. Δy + x. y. Δz)/ (xyz)
Ԑ

_{v}= (Δx/x) + (Δy/y) + (Δz/z)
Ԑ

_{v}= Ԑ_{x}+ Ԑ_{y}+ Ԑ_{z}
Let us consider the stresses in various directions
and young’s modulus of elasticity

σ

_{x}= Tensile stress in x-x direction
σ

_{y}= Tensile stress in y-y direction
σ

_{z}= Tensile stress in z-z direction
E= Young’s modulus of elasticity

Let us brief here first lateral strain before going
ahead; lateral strain will be the strain at perpendicular or right angle to the
direction of applied force.

Therefore we can define here lateral strain such as
lateral strain will be basically defined as the ratio of change in breadth of
the body to the original breadth of the body.

Tensile stress acting in x-x direction i.e. σ

_{x}will produce a tensile strain in x-x direction and lateral strain in y-y direction and z-z direction.
Let us brief here first “

**Poisson ratio**” before going ahead
Poisson ratio = Lateral strain /Linear strain

We can also say that, Lateral strain = Poisson ratio
(ν) x Linear strain

As we have already seen that, lateral strain will be
opposite in sign to linear strain and therefore above equation will be written
as following

Lateral strain = - Poisson ratio (ν) x Linear strain

Therefore tensile stress acting in x-x direction
i.e. σ

_{x}will produce a tensile strain (σ_{x}/E) in x-x direction and lateral strain (ν. σ_{x}/E) in y-y direction and z-z direction.
Similarly, tensile stress acting in y-y direction
i.e. σ

_{y}will produce a tensile strain (σ_{y}/E) in y-y direction and lateral strain (ν. σ_{y}/E) in x-x direction and z-z direction.
Similarly, tensile stress acting in z-z direction
i.e. σ

_{z}will produce a tensile strain (σ_{z}/E) in z-z direction and lateral strain (ν. σ_{z}/E) in x-x direction and y-y direction.####
**Strain
produced in x-x direction**

σ

_{x}will produce a tensile strain (σ_{x}/E) in x-x direction
σ

_{y}will produce a lateral strain i.e. compressive strain here (ν. σ_{y}/E) in x-x direction
σ

_{z}will produce a lateral strain i.e. compressive strain here (ν. σ_{z}/E) in x-x direction####
**Total
strain produced in x-x direction**

Ԑ

_{x}= (σ_{x}/E) - (ν. σ_{y}/E) - (ν. σ_{z}/E)
Ԑ

_{x}= (σ_{x}/E) - ν [(σ_{y }+ σ_{z})/E]####
**Total
strain produced in y-y direction**

Ԑ

_{y}= (σ_{y}/E) - ν [(σ_{x }+ σ_{z})/E]####
**Total
strain produced in z-z direction**

Ԑ

_{z}= (σ_{z}/E) - ν [(σ_{x }+ σ_{y})/E]
Let us add total strain produced in
each direction to find out the volumetric strain for a
rectangular bar subjected with three forces mutually perpendicular with each
other

Volumetric strain, Ԑ

_{v}= Ԑ_{x}+ Ԑ_{y}+ Ԑ_{z}
Volumetric strain, Ԑ

_{v}= [(σ_{x}+ σ_{y}+ σ_{z})/E-2 ν (σ_{x}+ σ_{y}+ σ_{z})/E]###
*Volumetric strain, **Ԑ*_{v}= (σ_{x}
+ σ_{y}+ σ_{z}) (1-2 ν)/E

*Volumetric strain,*

*Ԑ*

_{v}= (σ_{x}+ σ_{y}+ σ_{z}) (1-2 ν)/E
Do you have any suggestions? Please write in comment
box

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**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google

We will see another important topic i.e. Stress
analysis of bars of varying sections in the category of strength of
material.

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