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BMSc - Applied Ortho Semester 1 > Mechanics unit 4 - stress analysis deck 2 > Flashcards

Mechanics unit 4 - stress analysis deck 2 Flashcards

1
Q

What is the bending strength of a beam dependant on ?

A
  1. The strength of the material
  2. The cross-sectional area
  3. The cross-sectional shape
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2
Q

Why does the cross-sectional shape of a beam affect the bending strength ?

A

We know that the bending stess at any particular layer in the beam is dependent on the layers displacement from the neutral axis, and considering the equation to calculate the stress (fig.)

The variable y indicates that a beam with its mass distributed away from its neutral axis will be better able to resist any applied moments.

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3
Q

State the equation for calculating the maximum bending moment that a beam can resist

A
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4
Q

What equation combines the equations to calculate bending strength and the max bending moment into 1 to make it easier to remember ?

A

MISYER

  • M = bending moment
  • I = second moment area
  • S = stress
  • Y = displacement from neutral axis
  • E = youngs modulus
  • the R = radius
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5
Q

What is factor ɪ the second moment of area dependent on?

A

Upon the cross-sectional shape of a beam - the further the material of a beam is concentrated away from its neutral axis the larger the second moment of area

Pic shows 5 different cross-sections and their corresponding second moment of area and bending strength values. All the beams have the same cross-sectional area

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6
Q

State the equation for calculating the second moment of area (ɪ) and Ymax for a rectangular cross-section

A
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7
Q

State the equation for calculating the second moment area and Ymax for a circular cross-section

A
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8
Q

State the equation for calculating the second moment area (ɪ) and Ymax for a hollow cicular cross-section

A
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9
Q

This pic shows what each of the symbols for the equations for second moment areas of the different shapes stands for

A
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10
Q

Do SAQ 6 pg. 13 stress analysis unit 4

A

Ans in workbook

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11
Q

Describe how the typical boot top fracture arises and state the type of bending that causes it

A
  • It is a fracture as a result of 3-point bending. As the skier falls forward over the top of the ski boot a force is exerted on the proximal end of the tibia
  • As the distal end of the tibia is fixed in the boot, the tibia is bent over the top of the rigid ski boot, if the bending load is large enough, the tibia will fracture.
  • Hence quick release ski bindings
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12
Q

What are torsional stresses caused by ?

A

Twisting due to the application of a moment.

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13
Q

State what is meant when a bar is said to be under torsion

A

This is when a bar is under the action of a twisting moment

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14
Q

State what is meant when a bar is said to be in pure torsion

A

This is when the cross-section of the bar retains its shape (it remains circular and its radius is unchanged)

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15
Q

Describe how the deformation, angle of twist, shear strain and shear stress all vary with the distance from the central axis ?

A

They all increase from 0 at the central axis to a max at the outer surface

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16
Q

How does the angle of twist vary along the length of the bar acted upon by a twisting moment ?

A

It increases along the length of the bar

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17
Q

What is shear stress in a bar under the action of a twisting moment equal to ?

A

It is equal to the angle of shear

18
Q

How does the shear strain vary along the length of a bar under the action of a twisting moment ?

A

It remains constant along the length of the bar

19
Q

State the equation and units for the modulus of rigidity (G) of a bar under torsion

A
20
Q

State the equation used to calculate the shear stress of a bar under torsion

A
  • G = modulus of rigidity
  • θ = angle of twist
  • rAB = radius - distance of the point of interest from the centre of the bar
  • L = length of bar
  • τ = shear stress

Units are the same for all stress = N m-2 or Pa

21
Q

State the equation and units for shear strain in a bar under torsion

A
  • Φ = shear strain
  • θ = angle of twist
  • rAB = distance of the point of interest from the centre of the bar (radius)
  • L = length of the bar

Units of shear strain = rad NOT degrees

22
Q

Do SAQ 7 pg. 15 stress analysis unit 4

A

Ans in workbook

23
Q

Why is a hollow section (e.g. those found in bones) in a bar the best way to maximise the stress to weight ratio ?

A
  1. Because as we know the deformation, shear stress and strain is greatest on the outer surfaces of the bar and decreases to 0 as you move inwards to the centre of the bar.
  2. This means that a lot of the material of a bar will be stressed below the max tolerable (i.e. the material closer to the centre of the bar) therefore when weight reduction and material saving is important, hollow bars will maximise the strength to weight ratio as there ability to resist torsion will be similar to a solid bar of similar dimensions since the majority of stress is at the outer surface of the bar.
  3. The material near the centre of the bar is effectively redundant
24
Q

In equilibrium what will the applied twisting moment on the bar be countered equally by ?

A

The moments arising due to the shear stresses within the twisted rod.

25
Q

State the equation for calculating the twisting moment applied to a bar

A

Units = N m

26
Q

State the equation which is used to combine and simplify the equations for shear stress in torsion and twisting moment

A

MJTRG(O)L

Michael jackson thought of really good opening lines

27
Q

State the equations for calculating the polar second moment of area (J) for:

  1. A circular cross-section
  2. A hollow circular cross-section
A

Units = m4

Easiest way to remember these two is that both equations are just divided by half the number the second moment area is divided by for the same shape

28
Q

Do worked example pg. 16 stress analysis unit 4

A

Ans in workbook

29
Q

Do SAQ 8 pg 17. stress analysis unit 4

A
30
Q

How are long bones in the human body designed to resist torsional loads efficiently ?

A

Because they are hollow with strong cortical bone forming the outer layer which is the most efficient way of distributing bone tissue to maximise their strength to weight ratio to resist torsional loads.

If the same quantity of bone tissue was used to construct a completely solid bone rather than a hollow one the bone would be smaller in diameter and would be less able to resist torsional loads (break more easily). If the bone was the same size but completely solid compared to a hollow one of the same size then the weight would dramatically increase with only a small increase in its capacity to resist torsional loads

31
Q

Why are fractures of the tibia caused by torsional loads more common distally ?

A

Because the distal polar second moment of area of the tibia is smaller than the proximal one, given by the equation M = J (Gθ / L) and because the twisting moment (M) will be countered equally by the moments arising due to the shear stresses within the twisted rod

32
Q

What is the other function of muscle contraction apart from to cause movements and state why this function is beneficial?

A
  • Muscles will often contract not to cause movement but to alter the stress distribution within a bone.
  • If a muscle contracts, producing a compressive load on a bone (don’t believe the can produce tensile loads), it can eliminate any tensile loading and produce an overall compressive loading on the bone.
  • This is beneficial because bones are stronger in compression than in tension therefore the action of muscles can reduce the likelihood of fractures
33
Q

Describe how the action of the soleus muscle reduces the chances of tibial fractures

A
  1. When 3 point bending is applied to the tibia the ant. surface is in compression and post. surface is in tension.
  2. if the soleus muscle contracts it will produce a compressive load on the tibia by pulling downwards on the proximal end of the tibia, this will superimpose the bending load on the tibia reducing or eliminating the tensile load ==> producing an overall compressive load throughout the cross-section of the tibia
  3. This will however result in a higher compressive load on the ant. surface of the tibia but since bones are stronger in compression than tension the risk of fracture is still reduced
34
Q

Describe how the gluteus medius muscle allows the femoral neck to be able to withstand higher loads than otherwise would be possible during reciprocal gait

A
  1. During reciprocal gait the loading on the hip joint during stance phase produces a tensile stress on the superior aspect and a compressive stress on the inferior aspect of the femoral neck (fig. A)
  2. The gluteus medius muscle which lies superior to the femoral neck produces a compressive stress that effectively neutralises the tensile stress (fig. B) resulting in an overall compressive stress across the cross-section of the femoral neck allowing it to therefore withstand much higher loads
35
Q

Why are tired athletes more likely to fracture a bone than when they are fresh?

A

Because their muscles are fatigued and they are therefore less able to control the distribution of stres within their bones

36
Q

Do SAQ 9 pg. 19 stress analysis unit 4

A

Ans in workbook

37
Q

When we need to fully investigate the stresses present within complex structures e.g. bones etc what is usually used instead of simplified assumptions covered in the previous sections?

A

Experimental or computation methods

38
Q

What are the 3 main experimental or computation methods used to fully investigate the stresses within complex structures?

A
  1. Strain gauges
  2. Photoelasticity
  3. The finite element method
39
Q

What does a strain gauge allow us to measure?

A

The strain and therefore the stress at the surface of a structure

40
Q

How does a strain gauge work?

A
  • It consists of a thin metal foil located between 2 pieces of thin insulating film, it is then attached on to the surface of the material for which it is to measure the strain.
  • As the material changes in dimension due to changes in stress, the length of the gauge will also change, altering the electrical resistance of the foil. The change in resistance is proportional to the strain which is measured using an electronic circuit
41
Q

What are the advantages of strain gauges ?

A
  1. They can be applied to the actual structure being studied, there is no need to build a model of the structure
  2. They are ideal for measuring strain in a few directions and at a few specific locations
42
Q

What is the main disadvantages of strain gauges ?

A

They are not really practical when an investigation of a complete structure is needed

BMSc - Applied Ortho Semester 1 (52 decks)

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