Stress Analysis

Question 1

What is the symbol for shear stress?

Shear strain?

Answer 1

Shear stress = tau

Shear strain = phi

Question 2

What is an isotropic material?

Answer 2

One which exhibits uniform mechanical properties in all directions

This is generally true for metals, alloys and plastics, but not for biological materials like bone, tendon or ligament

Question 3

Formula for shear stress?

SI Unit of shear stress?

Answer 3

Shear stress = shearing force/sheared area

tau = V/A

tau - shear stress
V - shearing force
A - sheared area

The SI unit is Pascal (N m^-2)

Question 4

What will happen to a material undergoing shear stress?

Answer 4

It will be subject to angular deformation, which is quantified using shear strain (phi)

(i.e. tilting an object)

Question 5

What is shear strain?

Answer 5

Shear strain = angle sheared = phi

It is the angle which the object has changed from its original position. Note the unit is RADIANS.

Question 6

Formula for sheared strain?

Answer 6

tan(phi) = x/d

x = distance sheared
d = distance between the 2 shearing forces

This is because a triangle is formed where phi is the angle between the origin and the top of the object, x is how far the top has travelled along the x axis, and d is the y axis.

However, in most cases the angle is so small it is less than 0.1 radians, so tan(phi) is pretty much equal to phi. Therefore:

phi = x/d

Question 7

What is shear strength?

Formula?

Answer 7

The maximum shear stress the material can withstand before fracturing.

Shear strength = shear force required to fracture material/sheared area

Question 8

What is the modulus of rigidity (G)?
Formula?
SI units?

Answer 8

The shear modulus - it is equal to the gradient of the shear-stress/shear-strain curve, up to a limiting stress

Modulus of rigidity = shear stress/shear strain

G = tau/phi

SI units = pascals

Question 9

Do axial forces (tensile and compressive loads) also give rise to shear stress?

Answer 9

yes - and the shear stress causes the planes on the material to slip relative to one another, ultimately resulting in failure.

Question 10

In axial loads, where does the largest shear stress occur?

How is it calculated?

Answer 10

At 45 degrees to the axial load

Since it is at an angle of 45 degrees, the shear stress is equal to HALF the axial stress:

tau(max) = sigma/2
Where tau - shear stress
and sigma - axial stress

Question 11

How does shear stress in axial loadings relate to real life?

Answer 11

Although shear stress is half the stress acting axially, the shear stress may actually be limiting the material if the material is less than half as strong in shear as it is axially

e.g. cortical bone is less than half as strong in shear than it is in compression, and therefore tends to break at 45 degrees if a compressive load is applied

Question 12

What is bending stress?

2 different types?

Answer 12

When a material is acted on by forces and moments that tend to bend or curve it. This will cause the material to be elongated in one side, and compressed in the other

2 types are Cantilever and 3-point

Question 13

Pattern of stress and strain in an object subject to a bending load?

Answer 13

The stress and strain vary along the length of the bar. The strain (and therefore the stress) is greatest at the surfaces. Between the elongated and compressed side of the bar there will be an axis (or plane in 3D) where there is neither compressive or tensile stresses - the neutral axis

The largest stress will occur at the furthest point from the applied load

Question 14

Equation to find the variation in stress for a segment subject to a bending load?

Answer 14

sigma = epsilonE = (y/r)E

sigma - stress in the segment under consideration
epsilon - strain in the segment
E - Young’s Modulus of the material
y - displacement of the segment from the neutral axis
r - radius of the circle containing the neutral axis

Question 15

What is the stress in any layer of a material being bent dependent on?

Answer 15

Its displacement from the neutral axis

The further away, the greater the stress - the maximum stress will therefore occur at the surfaces if the material, which is where failure will occur

Question 16

What is the bending moment?

Answer 16

An internal moment within a material which must balance the externally applied bending load in order to maintain static equilibrium. It can be calculated for any cross-section of a loaded bar by applying the principle of static equilibrium.

Question 17

Point C is on a metal beam, which is being subject to cantilever bending by downward force F. C is x distance away from F. Write out an expression for the Moment centred around C

Answer 17

Mc - Fx = 0

Mc = Fx

Question 18

What is the bending moment dependent on?

Answer 18

The bending force being applied and its displacement from the point of application of the bending force

Question 19

What will a bending moment diagram show?

When is the diagram positive and negative?

Answer 19

The magnitude of the bending moment as you travel along the length of the bar. The greatest magnitude is at the furthest point away from the force, F, thus it is greatest when x (distance) = length of bar

The magnitude of the bending moment is positive when the moment causes sagging

The bending moment is negative when the moment causes hogging/arching

Question 20

What is the bending strength of a beam dependent on?

Answer 20

The material, the cross-sectional area and the cross-sectional shape

Question 21

In the equation for finding stress in a layer of material in a bent object, what does the symbol y indicate?

Answer 21

sigma = epsilonE = (y/r)E

y = displacement of the layer from the neutral axis. It can be deduced that a material with its mass distributed may from its neutral axis will be able to better resist any bending moments applied to it.

If 2 beams from the same material have identical cross-sectional areas but different shapes, the beam arranged in such a way that the majority of its material is distributed as far as possible from the neutral axis will be able to best resist a bending moment

Question 22

General equation for the maximum bending moment a beam can resist?

Answer 22

Mmax = (sigmaI)/y(max)

Mmax - maximum bending moment
sigma - maximum bending stress
I(cap i) - second moment of area
y(max) - maximum displacement of the extreme layer of the beam from the neutral axis

(this can be altered depending on the material shape etc)

Question 23

Relationship between stress in a particular layer of a beam, and the maximum bending moment that the beam can resist?

Answer 23

M/I = sigma/y = E/R
(MISYER)

Moment/Second moment of area

stress in layer/displacement from neutral axis

Young’s Modulus/Rupture point?

Question 24

what is the second moment of area?

What is it dependent on?

Answer 24

Resistance to bending?

it is dependent on the cross-sectional shape of the beam - the further the material of a beam is concentrated away from its neutral axis, the greater its second moment of area

Question 25

What is torsional stress?

What is pure torsion?

Answer 25

Twisting due to application of a moment

A circular bar is in pure torsion when its cross section retains its shape (i.e. remains circular and the radius is unchanged)

Question 26

How does deformity vary in torsion?

How does angle of twist, shear strain and stress vary?

Answer 26

Deformation is zero at the central axis (line longitudinally through the bar), and is maximal at the outer surface

The angle of twist, and hence the shear stress and shear strain also increase towards the surface of the bar

Question 27

Equation to calculate shear stress in torsion?

Answer 27

tau = (Gθr(AB)) / L

τ = shear stress
G = modulus of rigidity
θ = angle of twist (rad)
r(AB) = radius of bar
L = length of bar

Question 28

Equation to calculare shear strain in torsion?

Answer 28

ϕ = θr(AB) / L

ϕ = shear strain
θ = angle of twist (rad)
r(AB) = radius of bar
L = length of bar

Question 29

What is the deformation at the centre of a twisted object?

Answer 29

Zero

Question 30

Why is it better to use hollow bars than solid bars for tortuous structures?

Answer 30

Since the deformation in the middle is zero, and therefore the stress and strain, the material is effectively redundant

Most of the deformation and stress occurs at the outer surface, therefore using a hollow bar will help save material, and increase the strength-to-weight ratio of the bar, as found in bones

Question 31

What is the polar second moment of area (J)?

Unit?

Answer 31

A measure of the distribution of the material about the central axis

Unit = m^4

Question 32

General expression for twisting moment?

Answer 32

M = JGθ/L

M = twisting moment
J = polar second moment of area
θ = angle of twist (rad)
L = Length of bar

Question 33

Combination of equation for shear stress and twisting moment?

Answer 33

M/J = τ/R = Gθ/L

M = twisting moment
J = polar second moment of area
τ = shear stress
R = radius of cross-section
G = modulus of rigidity
θ = angle of twist (rad)
L = length of bar

Question 34

Equation of polar second moment of area for a solid circular bar?

Answer 34

J = πd^4/32

Question 35

Equation of polar second moment of area for a hollow circular bar?

Answer 35

J = π(D^4 - d^4)/32

D = outer diameter
d = inner diameter

Question 36

What is shear stress inversely proportional to?

Answer 36

Length

Question 37

Why do most torsional fractures of the tibia occur distally?

Answer 37

The distal polar second moment of area is smaller than the proximal polar second moment of area

Question 38

An example of muscles minimising tensile load on the bones during reciprocal gait?

Answer 38

During reciprocal gait, loadings on the hip during the stance phase produce a tensile load on the superior aspect of the femoral neck, and compressive load on the inferior aspect

The gluteus medius, located superiorly to the femoral neck, contracts to pull up the superior aspect, therefore also bringing it into compression. This effectively neutralises the tensile stress.

Question 39

What are strain gauges?
How do they work?
Advantage/Disadvantages of strain gauges?

Answer 39

Devices used to measure the strain, and consequently the stress, at the surface of a structure

It consists of a very thin metal foil locate between 2 pieces of thin insulating film. In use, the gauge is cemented to the surface of the material where it is required to measure the strain.

As the material changes in dimension, the length of the gauge will also, altering the electrical resistance of the foil. The change is proportional to the strain and may be measured using an electronic circuit.

They can be applied to the actual structure under study, there is no need to build an experimental model. They are ideal for measuring strain in a few directions at a few specific locations, however are not really practical when an investigation of a complex structure is needed.

Question 40

What is photoelasticity?

How does it work?

Answer 40

An experimental technique for stress analysis which is useful for complex geometrical structures with complicated loadings, which are usually transparent e.g. certain plastics

When transparent materials are subjected to force, some of their optical properties change in direct proportion to the stresses developed. The material becomes birefringent and a colourful interference pattern is observed. Polarised light passing through the stressed material splits into 2 beams, creating a fringe pattern, which can be used to quantify and localise stress. The closed the fringes, the steeper the stress gradient, indicating an area of stress concentration

It is also possible to coat a structure in a special photo elastic resin, in this way the stress present at the surface can be quantified

Question 41

What is the finite element method?

Answer 41

A computing tool which creates a mathematical model of a structure and the applied loads to simulate the stress which would occur in real life

A large number of simplified problems are substituted for one very complex one. The structure is split up into several small sub-regions called elements, and assoc w each element are a number of points called nodes. The system generates a series of equations for each node, of which there may be tens of thousands. Certain assumptions can be made about the initial value of some of these nodes, which will allow the remainder of the equations to be solved.

It produces an illustration with arrows and colours. The arrows represent the point of application and direction of applied forces. Blue colour = compression, red colour = tension