Axisymmetric stress

Question 1

Give the two definitions for integrity

Answer 1

1. An unimpaired condition
2. The quality of being whole and complete

Question 2

What are the dimensions of an axi-symmetric stress distribution?

Answer 2

3D stress distribution:
-Radial, Hoop, Axial

Question 3

What are the key components for which axi-symmetric stress is considered?

Answer 3

-Thick disks subjected to rotation
-Thick cylinders subjected to pressure loading

Question 4

What is represented by R(B), sigma(theta), sigma(r) and dr?

Answer 4

R(B): body force
sigma(theta): hoop stress
sigma(r): radial stress
dr: thickness of element

Question 5

What is the first step in setting up axi-symmetrical calculations?

Answer 5

Set up equilibrium equation on element

Question 6

What assumptions are made to simplify the axi-symmetric equilibrium equation?

Answer 6

-sin(dtheta/2) = dtheta/2
-Second order terms can be neglected
-Stress field is asymmetric
-Stresses vary only with radius

Question 7

What must be insured due to the deformation of the material of the element?

Answer 7

Compatibility of displacements

Question 8

How is compatibility of displacements ensured?

Answer 8

Considering the geometry of a typical displacement/deformation using ELASTIC HOOKE’S LAW relationships

Question 9

What is represented by:
u, du, r, dr?

Answer 9

u: radial displacement at r
du: deformation
r: distance from centre to inner edge
dr: original thickness

Question 10

Give the strains (not in terms of stresses) for E(r), E(A), E(theta)
(Where E is epsilon)

Answer 10

E(r) = du/dr
E(A) = dw/dA
E(theta) = u/r

Question 11

Give the equation and units for R(B) in relation to centrifugal force

Answer 11

R(B) = rhoomega^2r
In rad/s

Question 12

What assumption is made about E(A) in a THICK DISK and why?

Answer 12

The axial strain is constant with radius
The plane sections are assumed to remain plane

Question 13

How many boundary conditions are required to solve for A and B?

Answer 13

Two, as there are two unknowns

Question 14

What does the assumption of E(A) being constant lead to?

Answer 14

dE(A)/dr = 0

Question 15

What assumption is made about axial stress in a THIN DISK?

Answer 15

Sigma(A) = 0

Question 16

What does assuming sigma(A) = 0 lead to?

Answer 16

A different set of integration constant equations (Jim pls fix this wording I’ve forgotten maths)

Question 17

What boundary conditions are used for a solid rotating shaft?

Answer 17

sigma(Ro) = 0
B = 0

Question 18

Explain the boundary conditions for a solid rotating shafts

Answer 18

At Ro there is nothing to ‘push’ against, so there is no area for a stress to occur
In the centre, B/0 would be infinite if B=/= 0

Question 19

What are the boundary conditions for a hollow rotating shaft?

Answer 19

Sigma(Ro) and sigma(R1) are 0

Question 20

Explain the boundary conditions for a hollow rotating shaft

Answer 20

There is no stress as there is nothing to ‘push’ against- no area for force to be applied to

Question 21

How do you solve the equations for sigma(R1) and sigma(R0)?

Answer 21

Sigma(R1) - sigma(R0)

Question 22

What do the Lame equations apply to?

Answer 22

Thick cylinders subjected to internal and external pressure

Question 23

What happens to the body force term in the Lame equations?

Answer 23

Body forces are 0 as omega = 0

Question 24

What happens to the radial and hoop stresses as r decreases?

Answer 24

They both increase

Question 25

What sign do hoop and radial stresses have for internal pressure conditions?

Answer 25

Hoop: positive
Radial: negative

Question 26

What are the hoop and radial stresses also, due to the problems being axisymmetric?

Answer 26

They are principal stresses
sigma(1) = hoop stress
sigma(2) = radial stress

Question 27

How can max shear stress be calculated using the principal stress equivalents?

Answer 27

tau(max) = (sigma(1)-sigma(2)

Question 28

What does k represent in an internally pressurised cylinder?

Answer 28

r(o)/r(i)

Question 29

Why would an initial compressive hoop stress be induced in a cylinder?

Answer 29

Compression is negative, so initial pressurisation (which is tensile, so positive) overcomes the initial negative, bringing hoop stress to 0

Question 30

How can one contain a higher pressure in practice?

Answer 30

-Shrink fit
-Wind wire onto cylinder under tension
-Overstrain once with a pressure beyond yield, to leave compressive residual stress distribution near bore

Question 31

How does the appearance of the hoop strress vs r graph change from a single to compound cylinder?

Answer 31

Initially lower, then spikes at the interface to above

Question 32

What is needed to produce a shrink fit?

Answer 32

Outer diameter of inner cylinder must be slightly greater than inner diameter of outer cylinder
Called ‘diametral interference’

Question 33

What equation can be used to calculate diameter change in a shrink fit?

Answer 33

deltaD = D epsilon(theta)
Where epsilon(theta) is hoop strain

Question 34

What happens to the calculation for diametral interference when both cylinders are the same material?

Answer 34

Poisson’s ratio can be ignored and E can be factored out

Question 35

For an interference fit of a sleeve on a solid shaft, what are the initial conditions?

Answer 35

sigma(r) = sigma(theta) = C = -P (uniform across shaft)
B = 0 to avoid infinite stress at r = 0