MECH 101 Tutorial 5

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MECH 101 Tutorial 5

• 9th March, 2007

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Outline
1.Lecture review about concepts 2.Example about
normal stress ,normal strain, Hookes Law, and
Poissions ratio 3.Example about shear stress
and bearing stress 4. Example about shear
stress, shear strain, and Hookes law in shear
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Normal stress and Normal strain
Normal stress force per unit area
This equation is valid only if the stress is
uniformly distributed over the cross section of
the bar.
Normal strain elongation per unit length
Remind strain is a dimensionless quantity
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Hooks law and Poissons ratio
Hooks law
In which E is a constant, called Youngs modulus
or modulus of elasticity, depends on different
materials.
Note A permanent strain exists in the specimen
after unloading from the plastic region.
Poissons ratio
Poissons ratio is also a constant, a property of
the material, and dimensionless
Dashed means the original shape with out P
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Shear stress and Bearing stress
Shear stress acts tangential to the surface of
the material.
Average shear stress
Where
Average bearing stress
Where
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Shear strain and Hookes law in shear
Shear strain change in the shape of the
element
is small
When
Hooks law in shear
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Sign conventions for shear stresses and strains
Positive surface the outward normal direct in
the positive direction of a coordinate axis
Surfaces with red outward normal is positive
surface
Surfaces with blue outward normal is negative
surface
Y
Shear stress in red is positive
Shear stress in blue is negative
X
Z
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Example about normal stress ,normal strain,
Hookes Law, and Poissions ratio
Solution
Which means linearly elastic behavior occurs,
then we can use Hookes Law
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Hookes Law
From the definition of Poissions ratio
From the definition of strain
Substituting numerical values into the above two
equations
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Example about shear stress and bearing stress
The lap joint shown is fastened by four rivets of
¾-in. diameter. Find the maximum load P that can
be applied if the working stresses are 14 ksi for
shear in the rivet and 18 ksi for bearing in the
plate. Assume that the applied load is
distributed evenly among the four rivets, and
neglect friction between the plates.
Solution From Fig. (b) we can obtain
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The value of P that would cause the shear stress
in the rivets to reach its working value is found
as follows
The value of P that would cause the bearing
stress to reach its working value is found as
follows
Comparing the above solutions, we conclude that
the maximum safe load P that can be applied to
the lap joint is
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Example about shear stress, shear strain, and
Hookes law in shear
Two 1.75-in.-thick rubber pads are bonded to
three steel plates to form the shear mount shown
in Figure. Find the displacement of the middle
plate when the 1200-lb load is applied. Consider
the deformation of rubber only. Use E500 psi and
Solution
When the load is applied, the grid deforms as
shown in the figure.
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Each rubber pad has a shear area of A5x945
in-square. that carries half the 1200-lb load.
Hence, the average shear stress in the rubber is
From the relationship among E, G and v, we get
Shear stress acting on the sides of a grid
element is shown in Figure. The corresponding
shear strain is
So the displacement of the middle plate is