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Monday, 4 September 2017

NORMAL STRESS AND SHEAR STRESS ACTING ON THE INCLINED PLANE

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.

σX is the tensile stress acting in X direction over face AB and CD
σY is also the tensile stress acting in Y direction over face AD and BC
τ 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 θ 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

Force due to tensile stress σx = σx. AB
Force due to shear stress τ = τ. 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

Force due to normal stress σn = σn. AE
Force due to tangential stress σt = σt. 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

Force due to tensile stress σY = σY. BE
Force due to shear stress τ = τ. 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

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

σn. AE = σx. AB Cos θ + τ. AB Sin θ + σY. BE Sin θ + τ. BE Cos θ
σn = σx. (AB/AE) Cos θ + τ. (AB/AE) Sin θ + σY. (BE/AE) Sin θ + τ. (BE/AE) Cos θ
σn = σx. Cos θ Cos θ + τ. Cos θ Sin θ + σY. Sin θ Sin θ + τ. Sin θ Cos θ


Let us consider here now the forces acting along the oblique section and we will have following equation as mentioned here

σt. AE = σx. AB Sin θ - τ. AB Cos θ - σY. BE Cos θ + τ. BE Sin θ
σt = σx. (AB/ AE) Sin θ - τ. (AB/ AE) Cos θ - σY. (BE/ AE) Cos θ + τ. (BE/ AE) Sin θ
σt = σx. Cos θ Sin θ - τ. Cos θ Cos θ - σY. Sin θ Cos θ + τ. Sin θ Sin θ
σt = (σx - σY) Sin θ Cos θ - τ. Cos2 θ + τ. Sin2 θ
σt = (σx - σY) Sin θ Cos θ + (- τ ) (Cos2 θ - τ. Sin2 θ)
σt = (σx - σY) Sin θ Cos θ - τ Cos2θ

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.
σt = 0
(1/2) x (σx - σY) Sin 2θ - τ Cos2θ = 0
(1/2) x (σx - σY) Sin 2θ = τ Cos2θ
Tan 2θ = 2 τ / (σx - σY)

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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