Week 3 Pre-class
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Pennsylvania State University *
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BMES341
Subject
Mechanical Engineering
Date
Jan 9, 2024
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Briefly define what principal stresses and principal angles are.
Principal stresses are a specific pair of normal stress components where the transformed
normal stresses are equal to the maximum and minimum normal stresses.
Principle angles are the stress transformation angles which correspond to the principal stresses.
When bones break, they exhibit a variety of different fracture patterns: transverse,
oblique, spiral, etc. Briefly describe how stress transformations and principal stresses
may (in part) explain some of these fracture patterns.
Transverse fracture occurs when an external force acts directly perpendicular to the bone. The
principal stress responsible for the fracture is oriented along the direction of the applied force.
The bone breaks perpendicular to the applied force due to the high tensile or compressive
stress in that direction.
Oblique fracture happens when the applied force is at an angle to the bone. Stress
transformation results in both normal and shear stresses. The fracture occurs at an angle
because of the combination of principal normal stresses and shear stresses acting on the bone.
Spiral fracture exhibits a twisting or helical pattern, and the fracture pattern typically occurs
when a torsional force, such as twisting or rotating a bone, is applied. In this case, stress
transformation generates significant shear stresses along with the principal normal stresses,
resulting in a spiral fracture pattern along the bone.
What are the characteristics of a linear elastic material?
●
Directly proportional relationship between stress and strain
●
Small strains (infinitesimal strain) assumption
●
Linear superposition
Briefly describe one way you could apply something you learned in the pre-class
materials to your personal or professional life. Please be as specific as possible.
One way to apply the knowledge from the video lesson on Hooke's Law and linear elasticity to
my professional life is in the professional field is using the concepts of Hooke's Law to analyze
and select materials for specific applications. By understanding how stress and strain are
related through the elastic modulus (E) and how Poisson's ratio (ν) affects the material's
response to deformation, I can make informed decisions when choosing materials for various
engineering projects. This knowledge will help ensure that the materials I select can withstand
the required loads and deformations within specified limits, contributing to the success and
safety of the projects I'm involved in.
What is one question you would like answered during class this week?
Explain the other relevant material properties, such as the shear modulus, bulk modulus, or
Lamé’s first parameter, and when they might be more appropriate than the elastic modulus and
Poisson's ratio
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Related Questions
1. For the stress-strain curve shown below, please estimate the properties indicated.
(a) Fracture Strain
Please do your work on a separate sheet of paper,
and put your answers in the boxes on the right.
Be sure to include the proper symbol and units.
Stress Strain
70
60
50
Stress (ksi)
240
30
20
10
70
0
0.000
60
50
Stress (ksi)
40
20
10
KULL
0
0.000
0.010
0.050
0.100
Strain (in/in)
Stress Strain
0.020 0.030
Strain (in/in)
0.040
0.150
0.050
(b) Ultimate Tensile Stress
(c) Fracture Stress
(d) Proportional Limit
(e) Elastic Modulus
(1) Yield Stress
(g) Tensile Toughness
(Modulus of Toughness)
(h) Modulus of Resilience
arrow_forward
What is the magnitude of the maximum stress that exists at the tip of an internal crack
having a radius of curvature of 1.9 x 10" mm and a crack length of 3.8 x 10 mm when a
tensile stress of 140 MPa is applied?
4.
5. A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The
plane strain fracture toughness of the composite is 45 and the tensile strength is 550 MPa.
Will the flaw cause the composite to fail before the tensile strength is reached? Assume
that Y 1
A cylindrical rod 500 mm long, having a diameter of 12.7 mm, is to be subjected to a
tensile load. If the rod is to experience neither plastic deformation nor an elongation of
more than 1.3 mm. when the applied load is 29,000 N, which of the four met ls or alloys
listed below are possible cardidates? Justify your choice(s).
Materia!
Modulus
of
Yield
Tensiie
Strength
(MPa)
Elasticity(GPa)
Strength(MPa)
Aluminium alloy
70
255
420
Brass alloy
100
345
420
Сopper
110
210
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Mark the following points on an Engineering Stress-Strain curve.
Plastic limitElastic limitFracture strengthYield strengthUltimate tensile strength
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In the tensile test experiment, one of the
following is correct regarding the final
fracture:
Select one:
O The fracture occurs in the middle of
the specimen and is inclined with
45°.
O The fracture occurs anywhere in the
smallest cross sectional-area of the
specimen and is perpendicular to the
direction of the load applied.
O Fracture occurs randomly and we
can never predict the shape of it
O The fracture occurs anywhere in the
smallest cross sectional-area of the
specimen and is inclined with 45°.
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Based upon the information in the table below, which material is the stiffest?
Elastic
Strain at
Fracture
Fracture
Yield Strength
(MPa)
Tensile
Modulus
Strength
(MPa)
120
550
340
Fractures before yielding
850
Strength
(MPa)
105
500
265
650
720
Material
100
415
310
0.40
0.15
0.23
(GPa)
150
310
210
350
210
700
0.14
OE
OB
A.
OD
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A tensile test generates the stress-strain diagram shown here.
600
Stress (MPa)
500
400
300
200
100
T
1
T
0
0.00
Stress (MPa)
0.04
500
400
300
200
100
0.000
0.08
Young's modulus numeric value equal to
Young's modulus value obtain from
0.002 0.004
Yield strength numeric value equal to
Yield strength value obtain from
Tensile strength numeric value equal to
Tensile strength value obtain from
Strain
Strain
Assume a tensile specimen was originally 10 cm long with 2 cm x 2 cm square shape cross section.
0.12
1
0.006
0.16
0.20
The change in sample length while a stress of 480 MPa is applied to a tensile
specimen
[Choose ]
[Choose ]
[Choose ]
[Choose ]
[Choose ]
[Choose ]
[Choose ]
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QUESTION 1
Select from the following statements, those which describe the process and conditions required for brittle fracture.
(Note: Partial credit is not available for this question. Credit will only be awarded if all correct answers, and no incorrect answers, are selected).
A. Occurs along defined crystallographic planes, thus leaving smooth fracture surfaces
B. Occurs due to repeated loading and unloading below the yield strength of the material
OC. Occurs in materials where the bonding strength is high relative to the yield strength
D. The stress concentration at the crack tip exceeds the inter-atomic bonding energy
☐ E. Involves the nucleation, growth and coalescence of voids, thus resulting in rough fracture surfaces
OF. Is a slow process whereby there is warning prior to failure
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What is unit of shear strain?
arrow_forward
Strain energy due to direct stresses
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What is compressive principal stress?
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Name the 7 different points on a stress-strain diagram?
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9) An aircraft component is made of an aluminum alloy that has a fracture toughness of 60MPavm. When a crack of 2.5mm exists in the
component, the material fractures when a stress of 440MPA is reached. If a crack of 5.5mm exists in the material, what stress would need to
be reached to cause failure?
arrow_forward
According to Griffith's theory for brittle fracture, the stress required to propagate a crack is:
(This question has more than one correct answer)
0.5
a. Proportional to a'
where 2a is the crack length
b. More than the ideal fracture stress
-0.5
c. Proportional to a
d. Less than the ideal fracture stress
e. Proportional to the surface energy y0.5
f. Proportional to the surface energy y
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SIMPLE STRAIN
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You are called in to evaluate the failure of a component produced from a nickel superalloy (Inconel). The evaluator is not sure what they are looking at. You are sent the following micrograph. Educate this young engineer.
a. Detail everything that you can see in terms of the characteristics of failure.
b. Define if you can whether this is a ductile or brittle failure.
c. Can you interpret the direction of fracture progression?
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QUESTION 8
A fixation plate (thickness 3 mm, width 40 mm) has a circular hole of radius 5 mm. The plate contains a crack of length a=16 mm. If the material has a
fracture toughness of 90 MPa m^(1/2), calculate the maximum allowable tensile stress the plate can experience before it has uncontrollable crack
growth.
O 200.7 MPa
O 315.4 MPa
O 250.6 MPa
O 20.07 MPa
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If a specimen percentage elongation is observed to
be 30%. The original length of the specimen is 10 cm.
Calculate the final length of the specimen at the
fracture.
a. 13 mm
b. 1.3 mm
c. 1300 mm
d. 130 mm
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An Al2O3 part contains both internal cracks with a maximum length of 0.075 mm, and external cracks with a maximum length of 0.05 mm. At what applied stress will the internal cracks begin to propagate? At what applied stress will the external cracks begin to propagate? The fracture toughness KIC = 1.76 MPa x m1/2, Y=1.
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QUESTION 7
Which governing equations are required to solve a plane elasticity problem?
Boundary conditions
Equilibrium condition and boundary conditions
Equilibrium conditions, boundary conditions, and the compatibility equation
QUESTION 8
Which of the following conditions must a structure meet to produce a state of plane strain parallel to the x-y plane?
Thin-walled
Uniform in z-direction
Dimension in z-direction that is much smaller than x and y
QUESTION 9
What is the major advantage of using the Airy stress function for plane isotropic solids?
Reduces the governing equations to only the boundary conditions and compatibility equation
Reduces governing equations to a single function with a single unknown
QUESTION 10
Airy stress functions are applicable for solving all elasticity problems.
True
False
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Figure 2 shows the stress versus strain curve of four engineering materials.
Compare the ability of the materials to sustain stress without fracture and their ability to
absorb the energy before fracture between Material A and D.
C
B
D
Strain
Stress (MPa)
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Question 1
You are working on a design team at a small orthopaedic firm. You have been asked to select a cobalt-
chrome-molybdenum (CoCr) material that will not experience plastic deformation under a specific mechanical test, as follows...
A tensile stress is applied along the long axis of a solid cylindrical rod that has a diameter of 10 mm. An applied load of some
magnitude F produces a 7x10-³ mm change in diameter (see figure below, original shape is blue, elongated shape is unshaded).
Q1G: If your design required using the new material to create a wire, what is the largest diameter that would lead to ductile
behavior while still avoiding plastic deformation when exposed to the above loading conditions?
arrow_forward
Question 1
You are working on a design team at a small orthopaedic firm. You have been asked to select a cobalt-
chrome-molybdenum (CoCr) material that will not experience plastic deformation under a specific mechanical test, as follows...
A tensile stress is applied along the long axis of a solid cylindrical rod that has a diameter of 10 mm. An applied load of some
magnitude F produces a 7x10-³ mm change in diameter (see figure below, original shape is blue, elongated shape is unshaded).
Q1E: Of those two materials (F75 CoCr alloy (as cast) and F90 CoCr alloy (hot forged)), which materials would you select to assure
that the deformation is entirely elastic (No yield!)?
arrow_forward
No wrong answer please , i could downvote
The piece of suture is tested for its stress relaxation properties after cutting 3 cm long sample with a diameter of 1mm. The initial force recorded after stretching 0.1 cm between grips was 5 Newtons. Assume the suture material behave as if it has one relaxation time. The gage length was 1 cm.
a. Calculate the initial stress.
b. Calculate the initial strain.
c. Calculate the modulus of elasticity of the suture if the initial stretching can be considered as linear and elastic.
d. Calculate the relaxation time if the force recorded after 10 hours is 4 Newtons.
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When an external force acts on a body, the body tends to undergo some deformation. If the external force is removed and the body comes back to its origin shape and size, the body is known as elastic body. This property, by virtue of which certain materials return back to their original position after the removal of the external force, is called elasticity. We can explain these behaviours using following elastic constants.
Young’s modulus or Modulus of elasticity (E)
Shear modulus or Modulus of Rigidity (G)
Bulk modulus (K)
You are required to submit a report explaining these elastic constants and explaining the relations between these elastic constants. (You can use the relation with Poisson’s Ratio and get the relation between E, G and K.)
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Tensile engineering stress-strain curves through failure are depicted below.
Which specimen features the highest toughness?
Engineering Stress, MPa
1800-
1600-
1400-
1200-
1000-
800-
600-
400-
200-
Ol
0
Aluminum Alloy 2024-T81
0.04
0.08
Stainless Steel (18-8)
1340 Steel, Water-Quenched
& Tempered at 370 C
Magnesium
0.12
Structural Steel (Mild Steel)
Engineering Strain
0.16
1340 Steel, Water-Quenched & Tempered at 370°C
Aluminum Alloy 2024-T81
Structural Steel (Mild Steel)
Stainless Steel (18-8)
Magnesium
0.20
0.24
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Define the term Normal Strain?
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Question 6
What is the calculated fracture toughness, K, for the perspex specimen in MPa
root m? Give your answer to 2 decimal places.
P = 250 N, crack length = 5.5 mm W= 13.3 mm, B = 5.45 mm
Note that the span length (s) is 38mm. Take f(a/W) as 0.78
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Question 2
A machine component is made of Aluminium alloy having an ultimate tensile stress
Ou = 350MPa, a fracture toughness Kic= 40 MPa√m and a geometry factor Y=1.12.
The component is subject to cyclic tensile stress of 200 MPa and cyclic compressive
stress of 70 MPa. Before the operation, using an ultrasonic non-destructive testing
technique, it was found that there is an edge crack of length 2mm. The crack growth
C(AK)m, where is in m/cycle, AK ΔσΥ νπα , C ~ 2.2×10-12
da
rate is given by
da
dN
dN
and m=3.2. Find the critical crack length.
=
=
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1. What is the difference between direct and shear stress?
2. What type of fracture has occurred for the specimen?
3. Why single shear strength is greater than double shear?
4. How we can measure shear modulus?
arrow_forward
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Related Questions
- 1. For the stress-strain curve shown below, please estimate the properties indicated. (a) Fracture Strain Please do your work on a separate sheet of paper, and put your answers in the boxes on the right. Be sure to include the proper symbol and units. Stress Strain 70 60 50 Stress (ksi) 240 30 20 10 70 0 0.000 60 50 Stress (ksi) 40 20 10 KULL 0 0.000 0.010 0.050 0.100 Strain (in/in) Stress Strain 0.020 0.030 Strain (in/in) 0.040 0.150 0.050 (b) Ultimate Tensile Stress (c) Fracture Stress (d) Proportional Limit (e) Elastic Modulus (1) Yield Stress (g) Tensile Toughness (Modulus of Toughness) (h) Modulus of Resiliencearrow_forwardWhat is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 x 10" mm and a crack length of 3.8 x 10 mm when a tensile stress of 140 MPa is applied? 4. 5. A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the composite is 45 and the tensile strength is 550 MPa. Will the flaw cause the composite to fail before the tensile strength is reached? Assume that Y 1 A cylindrical rod 500 mm long, having a diameter of 12.7 mm, is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 1.3 mm. when the applied load is 29,000 N, which of the four met ls or alloys listed below are possible cardidates? Justify your choice(s). Materia! Modulus of Yield Tensiie Strength (MPa) Elasticity(GPa) Strength(MPa) Aluminium alloy 70 255 420 Brass alloy 100 345 420 Сopper 110 210arrow_forwardMark the following points on an Engineering Stress-Strain curve. Plastic limitElastic limitFracture strengthYield strengthUltimate tensile strengtharrow_forward
- In the tensile test experiment, one of the following is correct regarding the final fracture: Select one: O The fracture occurs in the middle of the specimen and is inclined with 45°. O The fracture occurs anywhere in the smallest cross sectional-area of the specimen and is perpendicular to the direction of the load applied. O Fracture occurs randomly and we can never predict the shape of it O The fracture occurs anywhere in the smallest cross sectional-area of the specimen and is inclined with 45°.arrow_forwardBased upon the information in the table below, which material is the stiffest? Elastic Strain at Fracture Fracture Yield Strength (MPa) Tensile Modulus Strength (MPa) 120 550 340 Fractures before yielding 850 Strength (MPa) 105 500 265 650 720 Material 100 415 310 0.40 0.15 0.23 (GPa) 150 310 210 350 210 700 0.14 OE OB A. ODarrow_forwardA tensile test generates the stress-strain diagram shown here. 600 Stress (MPa) 500 400 300 200 100 T 1 T 0 0.00 Stress (MPa) 0.04 500 400 300 200 100 0.000 0.08 Young's modulus numeric value equal to Young's modulus value obtain from 0.002 0.004 Yield strength numeric value equal to Yield strength value obtain from Tensile strength numeric value equal to Tensile strength value obtain from Strain Strain Assume a tensile specimen was originally 10 cm long with 2 cm x 2 cm square shape cross section. 0.12 1 0.006 0.16 0.20 The change in sample length while a stress of 480 MPa is applied to a tensile specimen [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ]arrow_forward
- QUESTION 1 Select from the following statements, those which describe the process and conditions required for brittle fracture. (Note: Partial credit is not available for this question. Credit will only be awarded if all correct answers, and no incorrect answers, are selected). A. Occurs along defined crystallographic planes, thus leaving smooth fracture surfaces B. Occurs due to repeated loading and unloading below the yield strength of the material OC. Occurs in materials where the bonding strength is high relative to the yield strength D. The stress concentration at the crack tip exceeds the inter-atomic bonding energy ☐ E. Involves the nucleation, growth and coalescence of voids, thus resulting in rough fracture surfaces OF. Is a slow process whereby there is warning prior to failurearrow_forwardWhat is unit of shear strain?arrow_forwardStrain energy due to direct stressesarrow_forward
- What is compressive principal stress?arrow_forwardName the 7 different points on a stress-strain diagram?arrow_forward9) An aircraft component is made of an aluminum alloy that has a fracture toughness of 60MPavm. When a crack of 2.5mm exists in the component, the material fractures when a stress of 440MPA is reached. If a crack of 5.5mm exists in the material, what stress would need to be reached to cause failure?arrow_forward
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