We were discussing the concept of stress
and strain and also we have discussed the different
types of stress and also different
types of strain in our previous posts.

Now we are going further to Derive the expression
for normal stress and tangential stress with the help of this post. We will
recommend you to first review the principal planes and principal stresses
before going ahead.

###
**Let
us see here the derivation of normal stress and tangential stress**

Let us consider one rectangular section ABCD of
uniform cross-sectional area and of unit thickness as displayed here in
following figure.

Ïƒ

_{X}is the tensile stress acting in X direction over face AB and CD
Ïƒ

_{Y}is also the tensile stress acting in Y direction over face AD and BC
Ï„ is the shear stress acting on plane or rectangular
section ABCD as displayed here in above figure

Let us consider one oblique or inclined section AE which is at angle Î¸ with the normal face of plane AB. We will derive here an expression for normal stress and tangential or shear stress on inclined section AE.

Let us concentrate here wedge or triangle ABE. Now
we will figure out here the forces acting on rectangular bar or plane ABCD due
to above discussed stresses.

If we analyze here the stresses acting over the
triangle ABE, we will conclude that there will be 6 forces acting over the
triangle ABE. Now, we will write each force here for each face of triangle ABE.

####
*Forces
acting on face AB of triangle ABE*

*Forces acting on face AB of triangle ABE*

Force due to tensile stress Ïƒ

_{x}= Ïƒ_{x}. AB
Force due to shear stress Ï„ = Ï„. AB

We have considered above that rectangular plane ABCD
is having unit thickness and hence area of face AB will be AB x 1 = AB

####
*Forces
acting on face AE of triangle ABE*

*Forces acting on face AE of triangle ABE*

Force due to normal stress Ïƒ

_{n}= Ïƒ_{n}. AE
Force due to tangential stress Ïƒ

_{t}= Ïƒ_{t}. AE
We have considered above that rectangular plane ABCD
is having unit thickness and hence area of face AE will be AE x 1 = AE

####
*Forces
acting on face BE of triangle ABE*

*Forces acting on face BE of triangle ABE*

Force due to tensile stress Ïƒ

_{Y}= Ïƒ_{Y}. BE
Force due to shear stress Ï„ = Ï„. BE

We have considered above that rectangular plane ABCD is having unit thickness and hence area of face BE will be BE x 1 = BE

####
*Now
we will resolve the above forces along the oblique section AE and normal to the
oblique section AE*

*Now we will resolve the above forces along the oblique section AE and normal to the oblique section AE*

We can refer here the following figure where forces
are resolved and displayed along the oblique section AE and normal to the oblique
section AE.

Let us consider here first the forces acting normal
to the oblique section and we will have following equation as mentioned here

Ïƒ

_{n}. AE = Ïƒ_{x}. AB Cos Î¸ + Ï„. AB Sin Î¸ + Ïƒ_{Y}. BE Sin Î¸ + Ï„. BE Cos Î¸
Ïƒ

_{n}= Ïƒ_{x}. (AB/AE) Cos Î¸ + Ï„. (AB/AE) Sin Î¸ + Ïƒ_{Y}. (BE/AE) Sin Î¸ + Ï„. (BE/AE) Cos Î¸
Ïƒ

_{n}= Ïƒ_{x}. Cos Î¸ Cos Î¸ + Ï„. Cos Î¸ Sin Î¸ + Ïƒ_{Y}. Sin Î¸ Sin Î¸ + Ï„. Sin Î¸ Cos Î¸
Let us consider here now the forces acting along the
oblique section and we will have following equation as mentioned here

Ïƒ

_{t}. AE = Ïƒ_{x}. AB Sin Î¸ - Ï„. AB Cos Î¸ - Ïƒ_{Y}. BE Cos Î¸ + Ï„. BE Sin Î¸
Ïƒ

_{t}= Ïƒ_{x}. (AB/ AE) Sin Î¸ - Ï„. (AB/ AE) Cos Î¸ - Ïƒ_{Y}. (BE/ AE) Cos Î¸ + Ï„. (BE/ AE) Sin Î¸
Ïƒ

_{t}= Ïƒ_{x}. Cos Î¸ Sin Î¸ - Ï„. Cos Î¸ Cos Î¸ - Ïƒ_{Y}. Sin Î¸ Cos Î¸ + Ï„. Sin Î¸ Sin Î¸
Ïƒ

_{t}= (Ïƒ_{x}- Ïƒ_{Y}) Sin Î¸ Cos Î¸ - Ï„. Cos^{2}Î¸ + Ï„. Sin^{2}Î¸
Ïƒ

_{t}= (Ïƒ_{x}- Ïƒ_{Y}) Sin Î¸ Cos Î¸ + (- Ï„ ) (Cos^{2}Î¸ - Ï„. Sin^{2}Î¸)
Ïƒ

_{t}= (Ïƒ_{x}- Ïƒ_{Y}) Sin Î¸ Cos Î¸ - Ï„ Cos2Î¸###
**Location
of principal plane**

A plane which is subjected with only normal stresses i.e. tensile stresses or compressive stresses will be termed as principal planes. Principal planes will not be subjected with shear stress i.e. there will be zero shear stress or tangential stress.

Ïƒ

_{t}= 0
(1/2) x (Ïƒ

_{x}- Ïƒ_{Y}) Sin 2Î¸ - Ï„ Cos2Î¸ = 0
(1/2) x (Ïƒ

_{x}- Ïƒ_{Y}) Sin 2Î¸ = Ï„ Cos2Î¸

*Tan 2Î¸ = 2 Ï„ / (Ïƒ*_{x}- Ïƒ_{Y})
Do you have suggestions? Please write in
comment box.

We will now discuss the Slope and deflection of beam and expression of shear stress developed in a circular shaft subjected to torsion in the category of strength of material, in our
next post.

###
**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google

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