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# Principle Stresses for Elements

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The figure shows a schematic of typical steel (E=200GPa, v=0.3) kneed support. The shaft has a diameter of 16mm. The dimensions are a=150mm and b=300mm and the forces are F=100N, P=75N and P2=50N. At the wall, Kt=1.75 and Kts=2. Considering the Generalized Hooke's law:

a. What are the principal stresses at the critical stressed element located at the wall and on the outer surface of the beam, due to the applied forces?

b. Is the element in plane strain or plane stress?
c. What are the principal strains?

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The figure shows a schematic of a typical steel (E = 200 GPa,  = 0.3) kneed support.
The shaft has a diameter of 16 mm. The dimensions are a = 150 mm and b = 300 mm and the forces are F = 100 N, P = 75 N, P2 = 50 N. At the wall, Kt = 1.75 and Kts = 2.
Considering the generalized Hooke's law:

a) What are the principal stresses at the critical stressed element located at the wall and on the outer surface of the beam, due to applied forces?
b) Is the element in plane strain or plane stress?
c) What are the principal strains?

Solution:

a) The cross section at the wall is loaded as follows:

1. traction due to force P2, which determines the normal stress on (x) direction:
(1)
where A = cross sectional area, (2)
D = diameter of the shaft, D = 16 mm = 0.016 m, R = 8 mm = 0.008 m
2. bending around (y) axis due to forces P and P2, which yield the moment:
(3)
The stresses determined by this bending moment are:
(4)
where = moment of inertia of the cross section with respect to (y) axis
3. bending around (z) axis due to force F, which yield the moment:
(5)
The normal stresses determined by this bending moment are:
(6)
where = moment of inertia of the cross section with respect to (z) axis
For a circular cross section, we have (7)
From (1), (4) and (6), the total normal stress pattern will be found:

 (8)
4. cutting on (y) and (z) directions, due to cutting forces F and P, which yield the shear stresses:
(9)
These will be neglected in the following analysis.
( unit vectors on the corresponding directions)
5. torsion around (x) axis due to the torque produced by the force F, which yield the shear stresses:
(10)
where = polar moment of inertia of the cross section with respect to (x) axis:
(11)
r = actual radius at which the shear stress is considered,
unit vector on tangent direction to the circle of radius (r):
(see the figure)

The maximum normal and shear stresses occur somewhere on the outer diameter, that is
r = R. In order to find the location of critical stress, we will consider in (8) that (y) and (z) are on the circumference, so that:
(12)

If we denote then

(13)

(14)
and also, the position of minimum stress will be:
(15)
Numerically, this means:
(16)
(measured from y axis in <yz> frame - see the figure) - this is the position of the critical element.
The value of maximum normal stress will be found using (8), (16) and applying the concentration factor (Kt):
(17)

The maximum torsional shear stress occurs on the outer diameter too and can be computed by applying (10), (11) and the shear concentration factor:

It is constant on the circumference.
Therefore, we can attach to the critical element the following stress tensor:
(18)
The principal stresses are the eigenvalues of the above tensor:

b) Since the critical element is submitted only to 1 = (x)max and 12 = (t)max, there is a plane state of stress.

c) The generalized Hooke's law consists in following equations:

which in our case simplify to:

The strain tensor attached to the critical element will be:
(19)
Computing the eigenvalues of the above matrix, the principal strains will be found:
(20)

(21)
The numerical values of the principal strains are:

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com September 29, 2022, 2:39 pm ad1c9bdddf>
https://brainmass.com/physics/torques/principle-stresses-elements-304817