Advanced Solid Mechanics: Creep and Fatigue
7388 words (30 pages) Essay in Mechanics
23/09/19 Mechanics Reference this
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Advanced Solid Mechanics
Table of Contents
Creep……………………………………………………………1
1. Creep 1
1.1 Stress relaxation 2
1.2 Creep rupture strength 4
1.3 Creep life 5
1.4 Factors aﬀecting creep 5
2. Creep Problem(MATLAB) 6
3. Creep Problem(MATLAB) 7
4. Creep Problem 8
Fatigue…………………………………………………………..9
5. Fatigue 9
5.1 Three major Fatigue life methods 9
5.2 Endurance limit modifying factor 10
5.3 Stress concentration and notch sensitivity 12
6. Fatigue Problem 14
7. Fatigue Problem 15
References……………………………………………………….18
1 Creep:
Creep occurs when the material is subjected to a constant tensile load at an evaluated temperature. In material sciences, creep is the tendency of a solid material to slowly moves or deforms permanently under the influence of stress.
1.1 Stress Relaxation:
Stress relaxation is the amount of stress observed in responds to the strain produced in the structure. The initial stress would be normal when the time is zero, although when the time increases the percentage of stress decreases. At higher temperature, the amount of stress will be more repaid. Additionally, high number stress will be inducing faster relaxation.
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Find out moreCreep is an increase in plastic strain under the constant stress. Stress relaxation is the decrease in stress under the constant strain. Creep got a rear important in some electrical contact design, since it can be an issue in the plastic housing in which most connectors are molded. Although creep is much easier to define understand than stress relaxation.
Creep is the time dependent deformation within a certain amount of applied load. In general it occurs in higher temperature, but can also at room temperature in certain materials (glass or lead), albeit much slower. As result the material undergoes a time dependent increase in length, which is dangerous while in service.
The rate of deformation is called creep rate. It is rate slop of the line in a Creep Strain vs. time curve.
Classical creep curve
Creep is divided into three stages as shown in Fig. In the primary stage the creep starts in rapid rates, and slows down with the time. Although in the secondary stage has relatively uniform rate. In the tertiary stage the creep rate accelerated and terminates when the materials ruptures or breaks. However at each stage depends on initial temperature and stress level. Higher the temperature depends on the stress level will accelerate the creep process. Similarly lower the temperature and the initial stress levels depends on the slower the process.
For example the first requirement is typically observed in the gas turbine blades whereas second standard is observed in those of stream turbine these occurs with respect to the time, its heavily depends on stress and temperature while creep is a thermally active process. Creep rate can be described by the fundamental by Arrhenius equation of the form.
ɛ= dɛ/dt
$=K{\sigma}^{n}$exp(-Q/RT); for T=const, dɛ/dt=K
${\sigma}^{n}$1.2 Creep rupture strength:
Creep rupture testing is the important adjunct for the creep testing for the selection size for preventing the rupture on the structure. It is noted that stress rupture value and the long term creep are extrapolated from the short term test. While these properties are determined directly or extrapolated, both of these methods provides the accurate results on higher temperature and operative life.
Stress rupture tests:
- Determine the time requires for the material to failure under the overload.
- Its helps in material section where the dimensional tolerance are acceptable and rupture cannot be tolerated
- These test generally performed at evaluated temperature.
- Flat specimen, notched, smooth or sample of any combination can be tested.
The original behavior of the material is difficult to predict accurately because of the service stress complexity, uniaxial loads in the standard test, variation in the relative application with the idealized conditions and also with the attenuating factors like cyclic loading, metal loosing from corrosion and temperature fluctuation. The metallurgist and the design engineer should be preparing to anticipate the synergistic effect of the variables.
One of the popular techniques for calculating the creep rupture data is Larson-miller temperature parameter. This parameter can be defined from the temperature and stress depended on the creep rate or time dependent rupture. Equation can be written as
PLM= T [21.577+ log (tr)].
1.3 Creep life:
Creep life reliable for the existing and new generation of materials being developed for ultra-super critical and super critical has a great challenge. The materials like low Cr-Mo Steel, low alloy has very long Creep database to induce high temperature structures which could survive at least 30-35years, whereas there temperature capacity limit is around 500°c. in order to increase the efficiency of cut down emission and power plants of
$\frac{{\mathit{co}}_{2}}{\mathit{MW}}$of power generated. Where it is required to increase the temperature to 600°c, the materials gets martensitic grades of steel having 10-12% Cr and 1% Mo with the addition of other alloys like V,B,W,NB,N etc. these materials are readily available from the last decades. Therefore creep database is limited. The data using for conventionally it cannot be reliable. These type of mechanical based model describing the behavior of such steel need to microscopic parameter and these type is often difficult to estimate.
1.4 Factors aﬀecting creep:
Creep is affected due to the fracture shows from the nucleation and growth of void from the boundaries of the grains. However at the lower temperature the structure usually failed by the crack formation at higher temperature and specifically after the creep and superplastic forming, intergranular cavity start forming these continuously growing and coalesce, leading to facture which is readily distinguishable which are observed by the electronic microscope reveals entire grains on the facture surface there are several reasons why cavities form at the grain boundaries.
Design consideration to avoid the creep
Grain boundaries effects need to be reduced
- Use large grain with single crystal material
- Solid solutions need to added to eliminate vacancies
Usage of materials which is having high melting temperatures.
Consult the creep test data during the selection of material
- Service type application
- Need regular inspection depend upon the life expectancy
2.
… Secondary creep rate obtained from creep tests
…-Aluminium alloy at 100Deg Centigrades
Scr=[6.2e-6 4.2e-5 1.42e-4 2.3e-4 4.7e-4 6.2e-4]; %secondary creep rate
stress=[120 160 210 250 280 300]; % Stress
p = polyfit(Scr,stress,2);
% Evaluate the fitted polynomial p and plot:
f = polyval(p,Scr);
figure(1)
plot(Scr,stress,’o’,Scr,f,’-‘)
legend(‘Aluminium Alloy’,’linear fit’)
title(‘Norton Power Law for Creep’)
xlabel(‘Stress’)
ylabel(‘Secondary creep rate’)
grid on
3.
Secondary creep rates for a series of tensile tests as various temperatures
… under the same load
T=[290 300 310 320 330];
str1=[4.8e-6 2.74e-5 1.4e-4 6.44e-4 2.7e-3];
p1=polyfit(T,str1,2);
%to determine the coefficients of the function formed
f1=polyval(p1,T);
figure(2)
plot(T,str1,’*’,T,f1,’-‘)
legend(‘Arrhenius Material’,’Quadratic Fit’)
title(‘Tensile Tests over various Temperatures’)
xlabel(‘Temp(Deg.C)’)
ylabel(‘Secondary creep rate’)
grid on
4)
Pressure vessel (p) = 1.5 mpa = 217.557 psi
Diameter of cover plate (D) = 450mm= 17.71 inches
Number of bolts= 30mm
Diameter (d) = 20 mm = 0.78 inches
Radius (r) =225mm
Elasticity (E)= 210 Gpa = 210*1000 mpa = 30457924.9 psi
Area ‘A’ =
$\frac{\pi}{4}*{d}^{2}$
A =
$0.7853981*{\left(450\right)}^{2}$A= 159043.128 mm^{2}
Secondary creep rate of =
$32.2*{10}^{\u201317}{\sigma}^{4}{h}^{\u20131}$1) Initial tightening stress in the bolts
$\sigma =\frac{{\mathit{pD}}^{2}}{\mathit{nd}}=\frac{1.5*450*450}{30*20}$
$=\frac{303750}{600}$
$\sigma =506.25\mathit{mpa}$
2) Time for re-tightened the bolt
$\sigma =\frac{\mathit{pr}}{t}$
$t=\frac{\mathit{pr}}{\sigma}=\frac{1.5*225}{506.25}$
$t=0.66\mathit{sec}$
2. FATIGUE
5. 1. Three major fatigue life methods and their brief and limitations:
Fatigue failure occurs when the material is subjected to fluctuating stress or repetitive and will fail at a stress much lower than its tensile strength.
The three major fatigue life methods used in design and analysis are
- Stress life (S-N)
- Strain life (ɛ-N)
- Linear Elastic fracture mechanics(LEFM)
These methods attempt to predict the life in number of cycles to failure, N, for specific level of loading. Life of 1 ≤ N ≤
${10}^{3}$cycles is generally classified as low-cycle fatigue, whereas higher cycle fatigue is considered to be N >
${10}^{3}$cycles.
Stress life (S-N):
Stress life method(S-N) depend on the stress levels only, these approach is the least accurate for the low cycle applications, whereas these method is most traditional method, hence this application is most simplest way for implementation for wide range of design application has represents high cycle applications adequately and ample supporting data.
Strain life (ɛ-N):
Strain life requires more detail analysis of plastic deformation at the local regions were the stain and stress is considered for the life estimates. This method has grater advantages for the low cycle fatigue application. However the application of the method, several idealizations to be compounded, and so some uncertainties will exit in the results. For this reason, it will be discussed only because of its value in adding to the natural fatigue.
Linear Elastic fracture mechanics (LEFM):
Fracture mechanics is a mechanical method which is assumes a crack on the structure which is already detected and present. It is then active to predict crack growth with the addition of stress intensity. This method is most applicable for larger assembles in contact with the computer codes and periodic inspection program.
5.2 Endurance limits with modifying factor.
Endurance limits:
The determination of the endurance is now simple by fatigue testing, though a lengthy procedure. In general for endurance stress testing is preferred to the strain testing.
For prototype, preliminary design and for some failure analysis as well, there is a need of quick method for estimating the endurance limits. There are some great quantities of the data in literature on the values of rotating beam tests and for specimen taken from same bar or ingot.
By plotting these as in Fig, it is possible to see whether there is any correlation between the two sets of results. The graph appears to suggest that the endurance limit ranges from about 40 to 60 percent of the tensile strength for steels up to about 210 kpsi (1450 MPa). Beginning at about
${S}_{\mathit{ut}}$= 210 kpsi (1450 MPa), the scatter appears to increase, but the trend seems to level off, as suggested by the dashed horizontal line at
${S}_{\mathit{e}}^{\u2018}=$105 kpsi
For steels, simplifying our observation of Fig, we will estimate the endurance limit as
Where
${S}_{\mathit{ut}}$= minimum tensile strength?
$\mathit{}{S}_{\mathit{e}}^{\u2018}$
= rotating beam specimen
${\mathit{S}}_{e}$
=endurance limit
Steel treated to give different microstructures have different
${S}_{\mathit{e}}^{\u2018}/{S}_{\mathit{ut}}$ratios. It appears that the more ratio have the higher ductile microstructure have a higher ratio. Martensite has a very brittle nature and is highly susceptible to fatigue-induced cracking; thus the ratio is low. When designs include detailed heat-treating specifications to obtain specific micro structures, it is possible to use an estimate of the endurance limit based on test data for the particular microstructure; such estimates are much more reliable and indeed should be used.
Endurance limits modifying factors:
Endurance limits are test under the controlled condition for the rotating specimen which is used in the laboratory determination. Whereas the endurance limit unrealistic to expect for the mechanical or structural member to compare the values obtained in the laboratory. So of the difference include.
- Manufacturing: heat treatment, method, fretting corrosion, stress concentration, surface condition.
- Design: shape, size, stress state, life, galling, speed, fretting
- Environment: temperature, relaxation times, corrosion, stresses state.
- Material: basic of failure, composition, variability
Marin identified factors that qualified the effect of size, surface condition, temperature, loading and miscellaneous items. The equation of Marin written as
${S}_{e}={K}_{a}{K}_{b}{K}_{c}{K}_{d}{K}_{e}{K}_{f}{S}_{\mathit{e}}^{\u2018}$
Where
${K}_{a}$
= Surface condition modification factor
${K}_{b}$
= Size modification factor
${K}_{c}$
= Load modification factor
${K}_{d}$
=Temperature modification factor
${K}_{e}$
= Reliability factor13
${K}_{f}$
= Miscellaneous-effects modification factor
${S}_{\mathit{e}}^{\u2018}$
= Rotary-beam test specimen endurance limit
${S}_{e}$
= Endurance limit at the critical location of a machine part in the geometry and condition of use
When endurance tests of parts are not available, estimations are made by applying Marin factors to the endurance limit.
5.3 Stress concentration and notch sensitivity.
Stress concentration and notch sensitivity:
In this stress concentration and notch sensitivity the existence of the discontinued or irregularities, which are Holes, notches, or grooves, in a part increases the theoretical stress significantly in the intermediate vicinity of the discontinuity. The equation defined as the stress concentration factor
${K}_{t}$which is used for nominal stress to obtain the maximum resulting stress due to the defects. Whereas for this a reduction value of
${K}_{t}$can be implemented. For these materials, the effective maximum stress in fatigue is,
${\sigma}_{\mathit{max}}$
=
${K}_{f}{\sigma}_{0}$The resulting factor is defined by the equation
${K}_{f}=\frac{\mathit{maximum\; stress\; in\; notch\; specimen}}{\mathit{stress\; in\; notch}\u2013\mathit{free\; specimen}}$
Notch sensitivity q is defined by the equation
q =
$\frac{{K}_{f}\u20131}{{K}_{t}\u20131}$- Where q=0,
$\mathit{}{K}_{f}$
=1 and the material has no sensitivity to notch
If q=1
${\mathit{K}}_{f}={K}_{t}$
in this case the material has full notch sensitivity
In design work
${K}_{f}=?q=?$${K}_{f}=$
1+q (
${K}_{t}\u20131$)
The sensitivity of some material can be obtained by experimentally. There are limited published values, although some values were available for aluminum and steel. Trends for notch sensitivity as function of ultimate strength and notch radius as shown in the fig.1 for axial loading or reversed bending
The fig.2 shows the reverse torsion. By using these chart we understand the actual testing results from which the curve may derive exhibit high amount of scatter. By the mean of the scatter it is important to use
Where
$\sqrt{a}=\mathit{Neuber\; constant}$
Correlating figs:
Bending or axial:
$\sqrt{a}$= 0.246 – 3.08(
${10}^{\u20133}$)
${S}_{\mathit{ut}}$+1.51(
${10}^{\u20135}$)
${S}_{\mathit{ut}}^{2}$– 2.67
${(10}^{\u20138}$)
${S}_{\mathit{ut}}^{3}$Torsion:
$\sqrt{a}$= 0.190 – 2.51(
${10}^{\u20133}$)
${S}_{\mathit{ut}}$+1.35(
${10}^{\u20135}$)
${S}_{\mathit{ut}}^{2}$– 2.67
${(10}^{\u20138}$)
${S}_{\mathit{ut}}^{3}$Where this equation applies to steel and
${S}_{\mathit{ut}}$is in kpsi. As per the graphs curve fits provides only approximate data to the experimental data.
For the cast iron the notch sensitivity is very low, varying from 0 to above 0.20, depending on the tensile strength. To be on conservative side, it is recommended for all cast iron grade the value of q is 0.20.
6)
Load on machine tool = 120cycle/day
Life at strain (
${\sigma}_{1}=\pm 130\mathit{mpa})$cycles
$\therefore $
No of days of life at
${\sigma}_{1}=\frac{8*{10}^{3}}{120}=66.67\mathit{days}$Life at strain (
${\sigma}_{2}=\pm 120\mathit{mpa})={10}^{4}$cycles
$\therefore $
No of days of life at
${\sigma}_{2}=\frac{{10}^{4}}{120}=82.33\mathit{days}$Life at strain (
${\sigma}_{3}=\pm 80\mathit{mpa})=4*{10}^{5}$cycles
$\therefore $
No of days of life at
${\sigma}_{3}=\frac{{4*10}^{5}}{120}=333.33\mathit{days}$Actual no.of cycles at each strain level in the sequence=
$250$Total cycles at all strain levels =250+250+250 = 750
%age of time spent at each strain level =
$\frac{250}{750}=\frac{1}{3}=0.33$1) According to miner’s rule sum of fractions of demand at each stress level is equal to total damage
$L=\frac{1}{3}$$L=\frac{1}{3}\left(66.67\right)+\frac{1}{3}\left(82.33\right)+\frac{1}{3}\left(333.33\right)$
L= Expected life =160.77 days.
2)
D
$=\sqrt{R}$D_{1} =
$\sqrt{8}*{10}^{3}=89.4\mathit{days}$D_{2}=
$\sqrt{{10}^{4}}=100\mathit{days}$D_{3}=
$\sqrt{4}*{10}^{5}=200\mathit{days}$_{ }
As the total is subjected to the different stresses for the same time of no.of days to
=
$\frac{89.4+100+200}{3}$= 129.8 days
7)
Endurance limit
${s}_{e}=0.5{s}_{\mathit{ut}}=300\mathit{mpa}$${\sigma}_{\mathit{max}}=+270\mathit{mpa}$
${\sigma}_{\mathit{min}}=\u2013270\mathit{mpa}$
${s}_{\mathit{ut}}=670\mathit{mpa};\mathit{}{s}_{\mathit{yt}}=260\mathit{mpa}$
${\sigma}_{a}=\frac{{\sigma}_{\mathit{max}\u2013{\sigma}_{\mathit{min}}}}{2}$
$=\frac{270+270}{2}$
${\sigma}_{a}=270\mathit{mpa}$
${\sigma}_{m}=\frac{{\sigma}_{\mathit{max}}+{\sigma}_{\mathit{min}}}{2}$
$=\frac{270\u2013270}{2}$
${\sigma}_{m}=0\mathit{mpa}$
1) According to Gerber’s criteria
${s}_{e}=0.5{s}_{\mathit{ut}}$
${s}_{e}=0.5*670$
${s}_{e}=335\mathit{mpa}$
$\frac{{\mathit{n\sigma}}_{a}}{\mathit{se}}$
+
${\left(\frac{{\mathit{n\sigma}}_{m}}{{s}_{\mathit{ut}}}\right)}^{2}=1$$\frac{{\mathit{n\sigma}}_{a}}{{s}_{e}}=1\mathit{}({\sigma}_{m}=0)$
Factor of safety
$n=\frac{{s}_{e}}{{\sigma}_{a}}=\frac{335}{270}=1.240$2) According to Goodman criteria
$\frac{{\sigma}_{a}}{{s}_{e}}+\frac{{\sigma}_{m}}{{s}_{\mathit{ut}}}=\frac{1}{n}$
$\left({\sigma}_{m}=0\right)\mathit{n}=\frac{{s}_{e}}{{\sigma}_{a}}$
$n=\frac{335}{270}=1.240$
3) Based on soderberg prediction
$\frac{{\sigma}_{a}}{{s}_{e}}+\frac{{\sigma}_{m}}{{s}_{\mathit{yt}}}=1$
${(\sigma}_{m}=0)$
$\frac{{\sigma}_{a}}{335}+\frac{185}{260}=1$
${\sigma}_{a}=96.634\mathit{mpa}$
Amplitude stress is,
${\sigma}_{a}=\frac{{\sigma}_{\mathit{max}}\u2013{\sigma}_{\mathit{min}}}{2}$
${\sigma}_{\mathit{max}}\u2013{\sigma}_{\mathit{min}}=2*{\sigma}_{a}$
$=2*96.6346$
${\sigma}_{\mathit{max}}\u2013{\sigma}_{\mathit{min}}=193.26$
………….(1)
Mean stress is,
${\sigma}_{m}=\frac{{\sigma}_{\mathit{max}}+{\sigma}_{\mathit{min}}}{2}$
${\sigma}_{\mathit{max}}+{\sigma}_{\mathit{min}}=2*{\sigma}_{m}$
${\sigma}_{\mathit{max}}+{\sigma}_{\mathit{min}}=370$
…………….(2)
From (1) and (2)
${\sigma}_{\mathit{max}}=281.6346\mathit{mpa}$
${\sigma}_{\mathit{min}}=88.3654\mathit{mpa}$
References:
Online Material
- http://www.msm.cam.ac.uk/phase-trans/mphil/MP6-14.pdf
- https://ocw.mit.edu/courses/materials-science-and-engineering/3-35-fracture-and-fatigue-fall-2003/lecture-notes/total_life.pdf
- https://ocw.mit.edu/courses/materials-science-and-engineering/3-35-fracture-and-fatigue-fall-2003/lecture-notes/fatigue_crack_growth.pdf
- http://www.sut.ac.th/Engineering/metal/pdf/MechMet/12_Fatigue%20of%20metals.pdf
- http://www.mse.berkeley.edu/groups/morris/MSE200/Lecture_Notes/lecture%20slides%20-%2034.pdf
- https://engkarrar.files.wordpress.com/2013/04/meyers-m-a-chawla-k-k-mechanical-behavior-of-materials-2.pdf
- https://ac.els-cdn.com/S1877705813006553/1-s2.0-S1877705813006553-main.pdf?_tid=59656f4c-3638-4e11-a005-90b48caddb9d&acdnat=1542786887_b30a503ec8cf2624eae2a7387800c288
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