We were discussing the “derivation of relationship between young’s modulus of elasticity (E) and bulk modulus of elasticity (K)”, “Elongation
of uniformly tapering rectangular rod” and also we have seen “Volumetric strain for a cylindrical rod” and “Volumetric strain of a rectangular body”
with the help of previous posts.
Now we are going further to start our discussion to understand
the basic principle of complementary shear stresses with the help of this post.
Principle of complementary shear stresses
Let us consider a rectangular block ABCD as
displayed in following figure. Let us assume that a set of shear stresses (τ)
of opposite direction, as displayed in following figure, is applied over the
opposite surfaces of rectangular block i.e. AB and CD.
Now if we consider the effect of this set of shear
stresses (τ) of opposite direction, we can easily say that there will be zero
net force acting over the rectangular block but there will be one couple acting
over the rectangular block in clockwise direction.
In order to balance the rectangular block, there
must be one more couple of similar intensity acting over the rectangular block in
opposite direction i.e. in anti clockwise direction. Therefore, there will be
one more set of shear stresses (τ’) of same intensity acting over the rest two
opposite surfaces of rectangular block and this set of shear stresses will be
termed as complementary shear stress.
Therefore we can say that according to principle of
complementary shear stresses,
“A set of shear stresses acting across a plane will
always be accompanied by a set of balancing shear stresses of similar intensity
across the plane and acting normal to it”
Let us consider that thickness of the rectangular
block ABCD, normal to the plane of paper, is one or unity.
We have already seen that applied set of shear
stresses is τ and set of complementary shear stresses is τ’.
Let us determine the force acting on surface AB and CD of rectangular block
Force acting on surface AB = Shear stress x area
Force acting on surface AB = τ x AB x 1 = τ x AB
Direction of force (τ x AB) acting on surface AB will
be towards left.
Similarly, force acting on surface CD = τ x CD x 1 =
τ x CD
Direction of force (τ x CD) acting on surface CD will
be towards right.
As these two forces i.e. (τ x AB) and (τ x CD) are
equal in magnitude and acting in opposite direction and therefore these two
forces will develop one couple which will be acting in clockwise direction.
Couple developed by set of shear stresses (τ) = τ x
AB x BC
Similarly, Couple developed by set of complementary shear
stresses (τ’) = τ’ x BC x AB
In order to balance the rectangular block, couple developed by applied set of shear stresses (τ) and complementary set of shear stresses (τ’) must be equal.
τ x AB x BC = τ’ x BC x AB
τ = τ’
Therefore we can say from above equation that A set
of shear stresses acting across a plane will always be accompanied by a set of balancing
shear stresses of similar intensity across the plane and acting normal to it.
Do you
have any suggestions or any amendment required in this post? Please write in
comment box.
Reference:
Strength of material, By R. K. Bansal
Image Courtesy: Google
We will see another important topic i.e. Elongation
of uniformly tapering circular rod, in the category of strength of
material, in our next post.
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