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FATIGUE

(Life Calculations and Sizing)

12.

Damage caused by repeated loading

( tvids: 12.a, 12.b , 12.c)

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• If we want to lift a package that we can easily lift once, we can only lift it up to a certain number of times. Because this repetitive process tires us.
• Solid objects also behave similarly against repetitive loads.
• Even if solid objects can withstand a statically applied load, if we apply the same load repeatedly, they can withstand a certain number of cycles and then become damaged.
• This phenomenon is called fatigue.

Yorulma / Ömür Hesaplamaları

12.2- Our Aims in This Subject :

• Our first goal is that when there is a stress change depending on time at any point of an object subject to

cyclic (repeated) loading, to calculate how many cycles this point can withstand, that is, its life; or

in other words, is to calculate the number of load cycles required for the first crack (damage) to form at this point.

In addition, the concept of infinite life, which is taken as a design criterion in solid systems, will be explained.

12.3- Importance of the Topic:

Many machines, mechanisms and all moving solid parts that we use in industry or in our daily lives are exposed to repeated loads and some of them become damaged over time. It may be misleading to make stress calculations and dimensions of these parts only when they are in a static state. At the same time, it is extremely important to make life calculations and safe dimension determinations in terms of fatigue.

12.1- What is Fatigue??

Figure 12.1.a

Figure 12.1.b

Figure 12.1.c

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12.4- Important Definitions used in Fatigue Calculations: In fatigue calculations, normal stresses are shown with the symbol S. As in Figure 12.2.a, we apply a time-varying force F(t) to the midpoint of the BC beam, which has two jointed ends (simply supported). The change of the F force with respect to time is shown in Figure 12.2.b. Bending loading will occur in the beam due to the effect of the F force.

Figure 12.2.b

Figure 12.3

(a)

(b)

(12.1)

Figure 12.4

Now the first question we are looking for an answer to is this: At a point subjected to cyclic loading, after how many cycle does a crack form?

In other words, what is the fatigue life of this point?..>>

(12.4)

(12.2)

(12.3)

(12.5)

Maximum Stress:

Minimum Stress:

Mean Stress:

Alternating Stress:

Fatigue

We know from the subject of bending number 5.1 that the normal stress distribution in any D section of the beam (at any instant) will be as in Figure 12.3.b (We consider the beam section as symmetrical, in which case simple bending will occur). The stress at a point d in the D section is found from equation 12.1.

Since F(t) changes with time, the stress S will also vary with time. Depending on the type of loading, the F-t and S-t diagrams may be similar or different in form.

Also notice that S-t diagrams that are the same shape but have different limiting values ​​will be formed at different cross sections and points of the beam.

Examine the different stress definitions in the S-t diagram in Figure 12.4. Note that this diagram is drawn for any point d.

After a certain cycle in the beam, the first crack occurs at the points where maximum stress will occur (for this example, at the outermost points of section A).

Since the stress equation S(t) (equation 12.1) is a first-order (linear) equation, it is clear that the S-t diagram for this example will be similar to the F-t diagram.

(or Stress Amplitude)

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S_m

t (time)

S_m1

S_m2

S_m3

Sa_d =

Try to understand the following points from the diagrams above:

• If the alternating stress (or stress amplitude) S_a increases for the same mean stress (S_m) value, the material fatigues in fewer cycles.
• If S_m decreases for the same S_a value, the material fatigues in more cycles.
• A logarithmic scale is used because there is a large difference between the values ​​on the N axis. Note that the intervals on this scale are not equal.
• The Concept of Infinite Life: The S - N curve generally shows an asymptotic state to the horizontal axis after 10⁶ cycles, that is, it can be assumed that the curve is parallel to the horizontal axis at this number of cycles. This means infinite life. This concept is explained in more detail on the next page.

Fatigue Tests

12.5- Material properties required for life calculation

In order to calculate how many cycles will cause a crack at a point, the fatigue characteristic curves (S-N diagrams) of the material must first be known. These curves are obtained experimentally. As follows: Samples made of the same material are subjected to cyclic (repeated) loading.

Figure 12.5

Figure 12.6

Fatigue

The mean stress (S_m) or alternating stress (S_a) of each sample is different. Three different cyclic loading tests (b, c, d tests) are shown in Figure 12.5.

Each of these tests continues until damage (cracks) occur in the sample. The number of cycles (N) reached when cracks occur is determined for each test.

In the logarithmic diagram, N is placed on the horizontal axis and Sa on the vertical axis, the points corresponding to each test are marked, and the points with the same mean stress Sm are connected by a curve. All of these curves are called S-N diagrams. It is clear that a large number of tests must be performed to obtain S-N diagrams.

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12.6- The Concept of Infinite Life :

• If we pay attention to the S-N diagrams in Figure 12.7, after N=10⁶ cycles, it can be assumed that all the curves in the graph continue parallel to the horizontal axis and coincide with the tangents at these points (points a, b, c).
• The stress amplitudes (alternating stress) at these points where the tangent is horizontal are specially indicated with the symbol Sao, each of which we can call the «stress amplitude limit value».
• The N value corresponding to a Sa_o value is not only 10⁶, but also all values ​​greater than 10⁶ and theoretically continuing forever (because the curve continues horizontally in that region).

• In summary: Points such as a, b, c shown above with coordinates 10⁶; S_ao ) and the horizontal tangents passing through these points are infinite life limits. For the material in this example, the minimum number of cycles in the infinite life limits is 10⁶. However, this value may differ depending on the material type.

• This means that: At S_ao (or lower) alternating stresses, there is infinite cycle resistance, which is called infinite life. (For example, in the S_m1 curve, when the amplitude is S_ao1, the N value is 10⁶, 10⁷ and all values ​​larger, and the point c is the infinite life limit). If the point on the solid body whose fatigue is being examined has infinite life, a fatigue crack will never occur at that point, and cyclic loading can continue at that point forever.

Fatigue

Figure 12.7

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12.7-Fatigue Test Setups

Fatigue test setups (or mechanisms) can have different designs. What is important in each setup is that the Smax - Smin stresses occur repeatedly at the critical point where the crack will first form and a stress-time diagram such as in Figure 12.4 is obtained. The aim is to measure how many cycles the crack will form after, which is necessary to obtain the fatigue characteristic curves (S-N diagrams) used in theoretical calculations (explained on the previous page). 4 different fatigue test setups are shown below as examples:

Using your strength information, try to predict the point where a crack will first occur in each of these mechanisms and draw S-t diagrams at these points.

• In mechanisms, variable loads should not be applied impulsively (like hitting with a hammer), but slowly. Because instantaneous stresses such as Smax -Smin are calculated for instantaneous (as if the mechanism were at rest at that instant) static loading situations.

• In the mechanisms, the stress of the most critical point should be calculated from the Strength formulas. This may vary depending on the loading condition. For example, for a beam subjected to simple bending, the stress can be calculated from equation 12.4.

If the mechanism is in tension-compression style, the formula S=P/A is used.

Depending on the type of material, a single sample fatigue test can sometimes take days or even weeks. In order to determine the fatigue characteristic behavior of a material, a large number of tests must be performed. For this reason, fatigue tests for a material type can require a very long process. Instead of taking on this burden, it may be a much more practical solution to first use the fatigue curves previously found experimentally for our material in the literature and various sources (with reference). In addition, theoretical fatigue strength curves were developed from experiments performed only for S_m=0, which will be explained later in this topic.

Figure 12.8.a

Figure 12.8.c

Figure 12.8.d

Figure 12.8.b

• In order to calculate the fatigue life, the normal stresses Smax and Smin must occur at the same point in order and this situation must be repeated in each period. This can only happen under bending or tension/compression loading. For this reason, mechanisms are designed according to one of these loading types. Devices 12.8a-c are designed for bending, and device 12.8.d is designed for tension/compression.

Fatigue

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• Since the test setups (or mechanisms) examples on the previous page are aimed at finding material characteristic fatigue curves, bar-shaped test samples are used.
• In some test mechanisms, final products are directly tested and whether the product has the desired life is directly measured.
• Although theoretical calculations are not required thanks to product fatigue tests, which are very useful for businesses, it should not be forgotten that computer-aided fatigue analyses for R&D activities and product development processes are indispensable in today's technology.2 examples of product fatigue tests are shown below.

Figure 12.9.a (Shoe fatigue test setup)

Figure 12.9.b (Rim fatigue test setup)

Fatigue

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2- Tip: Life calculations can be made for a point and show us how many cycles will cause a crack at that point. The important thing is to correctly determine the stress-time change at that point.

1- Due to the design of the experimental setup, loading types may differ and therefore external load-time and stress-time graphs may be similar or different to each other.

12.8 Let's repeat the Important Points in Fatigue Life Calculations:

4- Fatigue occurs when the external load is variable and the system is fixed. (Example: Water discharge grid on the roads)

6- Before fatigue calculations, the stresses at the critical point created by the maximum and minimum loads should be calculated separately for the static loading case.

3- The fatigue life of an object is possible by determining the fatigue life of its most critical point, because the first crack formation occurs at this point.

5- However, sometimes external loads are fixed, the system is mobile, a point is constantly changing position and is subject to fatigue because the stresses on it change over time. (Example: Rim)

Figure 12.10

Figure 12.11

Fatigue

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Axial force P is applied to a wooden block with a section of 6cm x 5cm, fixed from the bottom, repeatedly, varying between 60kN and -30kN in Figure. If the S-N diagram of the material is as shown on the side, how many cycles can the wooden block withstand this cyclic load? (i.e. what is its fatigue life?)

Example 12.1

This block can withstand this repetitive load up to 400 repetitions. (It is said that the fatigue life is 400 cycles)

Maximum Stress:

, Minimum Stress:

Mean Stress:

(or stress amplitude):

Alternating Stress:

Solution:

Since there is tension or compression loading, stresses of the same intensity occur at all points of the block at every t instant. Theoretically, all points are critical points and stresses are calculated with the formula S=P/A.

Number of cycles until fatigue crack occurs

Fatigue

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Example 12.2

The durability criteria for a high-heeled women's shoe have been determined as follows: "A 70kg woman who walks 1 meter in 2 steps should not have any damage to the heel of the shoe when she walks an average of 2km a day for 2 years."

We can assume the heel part of the shoe as cylindrical and its diameter is d=7.5mm and its height is h=5cm. If you pay attention while taking a step, you will see that all the body load comes to one shoe, while the other foot is out of contact with the ground. It can be assumed that half of the body load is carried by the heel and the other half by the front of the foot during walking. The compressive fracture stress of the shoe material is S_{compression-breaking}=S_oc=-40MPa and the S-N diagrams are as shown in the figure below. Accordingly;

a-) Is it appropriate to use the wooden material whose properties are given for the heel?

b-) If not, what kind of precautions would you consider?

S-N diagrams for shoe material

Fatigue

(g = 10kgm/s²)

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Solution:

First, we will calculate how much repeated load a shoe will be exposed to in 2 years by establishing a proportion:

During 1meter

She touches the ground once.

A Shoe

Since she walks 2000m a day, in 2 years

She touches the ground n times.

In one step (one cycle), when she steps on the ground, her entire body weight falls on a shoe. However, half of her entire body weight falls on her heel (the other half is carried by the front of her shoe).

Force-time diagram for the shoe heel:

Stress-time diagram at a point on the shoe heel

Fatigue

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Does damage occur when static (standing on one leg or taking the first step)?

no damage occurs.

Since the desired number of cycle is n =1460000 and n>N

Mean stress while walking:

Alternating stress while walking:

S_m=-4MPa

In the question, the S_m=-4MPa curve is not given in the S-N diagrams. However, by using the existing curves, this curve is drawn as an average as in the figure on the side.

(The number of cycle corresponding to 𝑆_𝑎=4MPa on the curve of S_m=-4MPa) :

That means the heel of the shoe can withstand this load for up to 40000 repetitions.

This material is not suitable for use as heel material in these sizes.

How can we use the same material with any changes?..>>

Number of cycles until fatigue crack occurs

Fatigue

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b-) Change 1 :

Let's take the diameter of the heel as d=10mm and use the same calculation method:

• The S_m=-2.23MPa curve is not given in the S-N diagrams. However, by using the existing curves, this curve was also drawn as an average as in the figure.

• Therefore, the heel will also withstand n = 1460000 repetitions. For these reasons, we can say that Change 1 is correct.

If S_m=-2.23MPa and S_a<2.23MPa, we would remain in the lower region of the tangent. In this case, infinite life would still be provided and a safer situation would be in question.

• However, another limitation is that the diameter d should not be enlarged too much in this change, which would damage the originality of the shoe and cause the desired aesthetics to be lost.

Question 2: Another risk for the shoe heel is buckling. Calculate whether there will be a problem with buckling in the redesigned heel according to Change 1.

Question 1: What other changes could there be for the shoe heel? Think about it and prove the accuracy of these changes with numerical calculations.

Fatigue

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Example 12.3

A force of P=50N is applied repeatedly up and down to a metallic wire with a radius of r=2mm, whose left end is fixed to a wall, by means of pliers from its right end. In this case, after how many repetitions (cycles) can the first crack be formed in the wire? (The crushing and cutting effect in the jaws of the pliers and the possibility of the wire coming out of the wall will be neglected.)

Properties of Wire Materials

(S_yieid= 250MPa)

S-N diagrams

Number of cycles until fatigue crack occurs

Solution:The wire is subjected to bending loading. The most critical cross section is the built-in section and the most critical points are points b and c, and at these points, at a time t, the stresses are equal but with opposite signs. The stress-time diagrams are symmetrical. It is enough to examine one of b or c. If we examine b:

x

M_z-max

In the S_m=0 curve, the number of cycles corresponding to the amplitude of S_a = 79.6MPa is approximately N= 2x102 =200. Try to see this yourself from the graph. Then a crack occurs after 200 cycles. The pliers are in the upper position and then return to the upper position after one cycle.

c

b

Fatigue

(mean stress)

(alternating stress)

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• Because when the fatigue tests were stopped from time to time, a recovery was observed in the smooth test pieces and longer lives were obtained compared to the tests carried out without interruption..

12.9 Effect of Interruption of Repetitive Loads

• While we carry a pile of loads from one place to another with our own physical strength, we take breaks from time to time, rest, recover ourselves, and then we can continue working.

• We can say that such a situation is also the case for solid objects.

• We should also not forget that interrupting the repetitive load would create safer situations than in our calculations, would provide us with an advantage in terms of strength and would not risk our calculations.

• As engineers, we must always take the most critical situation into consideration in our calculations.

• However, when and for how long to interrupt the repetitive load on an industrial product or part is generally not standardized and can vary depending on the user.

• Therefore, using S-N diagrams and other fatigue data obtained from continuous tests without interruption would be a more accurate approach as it would mean examining the most critical situation.

Figure 12.12

Fatigue

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Our aim in this section:

Our aim in this section is to determine the minimum dimensions required for parts exposed to repeated loads to have an infinite life.

But first of all, fatigue strength curves developed by interpreting S-N diagrams in different ways and different fatigue theories (criteria) need to be understood thoroughly…>>

φ D_emn=?

12.10 Fatigue Criteria and Sizing

Figure 12.13

Fatigue

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12.10.1 Smith Diagram

Smith Diagram is a derived diagram that is a different interpretation of the S-N diagrams we obtain from fatigue tests. The Smith diagram is obtained with the help of S-N diagrams in the following steps:

c

d

e

S_m

S_max,min

45^o

Sa_0-1

Sa_0-1

S_m2

Sa_0-2

.

.

.

S_max0-1

S_max0-2

.

.

Sa_0-2

S_m1

S_min0-1

Smith Diagrams

S_m1

f

g

S_ult =S_Breaking

S_e

-S_e

Figure 12.14

Figure 12.15

17

Fatigue

• The horizontal axis shows the mean stress (S_m), the vertical axis shows the maximum and minimum stresses. A line with a slope of 45⁰ is drawn between the vertical and horizontal axes.

• At any S_m1 value, a line parallel to the vertical axis is drawn. The point e where this line intersects the 45⁰ line is determined.

• From point e, go up and down as much as the amplitude Sa_0-1 (the limiting amplitude for infinite life) and mark points c and d.

• The y coordinate of point e (because of 45⁰) is also equal to the S_m1 value.

• For S_m1 there is an infinite life at the S_a0-1 amplitude. In this case, the maximum and minimum stresses can be calculated from equations 12.2 and 12.3 (or from the geometry of the figure) as S_max0-1 = S_m1 + S_a0-1 ; S_min0-1 = S_m1- S_a0-1

(These values ​​are also the vertical coordinates of points c and d.) In other words, we

can say that the infinite life limits for Sm1 are points c and d.)

• The same operations above are performed for different mean stresses such as S_m2, S_m3 and the limit points are obtained for each. (For example, notice that the limit points obtained for S_m2 are f and g.)

• When all these limit points are combined, the Smith diagram is obtained. As a result, in order to obtain infinite life at any point of a material, it is necessary not to go beyond the Smith diagram. For this reason, the Smith diagram can also be called the «fatigue strength limit diagram».
• The S_max value cannot exceed the breaking stress (in other words tensile strength S_ult). (in that case, a crack will occur in a single repetition, which means that there can be no cyclic loading above it). For this reason, the upper limit of the Smith diagram is the breaking strength of the material. For S_m=0, the stress amplitude is specifically indicated by Se and is called the fatigue strength amplitude (endurunce stress) or simply the fatigue limit.

.

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Reinforce the topic and concepts by thoroughly examining the Smith diagram below and the cyclic load diagram next to it.

Smith diagram also includes the Negative region.

Since the compressive strength of brittle materials is significantly greater than the tensile strength, the negative part of the diagram is larger.

Figure 12.17

Figure 12.16

(a)

Fatigue

(b)

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We explained that S-N curves for different Sm values ​​are necessary to draw the original Smith diagram. In order to obtain these S-N curves, we may need to enter into a very burdensome experimental study in terms of time, labor and cost. Instead of the original Smith diagram, the Modified Smith diagram, which can be drawn with only S_e and S_yield values, is frequently used in the field of engineering and provides correct solutions, in which case it is sufficient to perform fatigue tests only for Sm=0.

12.10.2 Modified Smith Diagram and safety factor (n)

Modified Smith diagram is drawn with these steps:

1-) For S_m=0, fatigue tests are performed at different Sa amplitudes and the S-N curve is obtained. The amplitude Sa0 at the fatigue limit is read on the curve, which is equal to the value Se (endurance stress).

4-) The intersection point c of the line drawn at an angle of 40⁰ from the Se value and the line drawn parallel to the horizontal from Sakma is determined.

5-) The points d, e, f where the horizontal and vertical lines drawn from c intersect the 45⁰ line are determined so that ce = ef.

6-) The Modified Smith diagram is obtained by connecting the points Se, c, d, f, -Se by lines, respectively.

7-) If there is a safety factor (n), the safer inner diagram is obtained by using the S_e / n and S_yield / n points with a similar drawing.

Modified Smith Diagram (n=1)

Original Smith Diagram

Safer Modified Smith Diagram (n>1)

c

d

e

f

Figure 12.18

Fatigue

40⁰

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12.10.3 Different Fatigue Criteria and Limit Diagrams:

Different fatigue criteria have also been developed for fatigue strength calculations. The positive regions of the type 1 fatigue strength limit diagrams of these criteria are shown in Figure 12.19. The type 2 limit diagrams shown in Figure 12.20 are derived from the type 1 diagrams. While a large number of fatigue tests are required for the original Smith diagram, in other diagrams (in order to determine the Se value) it is sufficient to perform fatigue tests only for Sm = 0. The S_Breaking (=S_B) and S_yield and b values ​​are taken from the literature or found by tension/compression tests. Not going beyond the curve (or line) of a criterion diagram means staying in the safe region in terms of fatigue according to that criterion and having an infinite life. Examine the graphs below carefully and try to understand the criteria and all the concepts.

Figure 12.20 –2nd Type Limit Diagrams

(12.6)

(12.7.a)

(12.7.b)

(12.8.a)

(12.8.b)

Soderberg Criterion

In the 1st Type, it is created from S_ed and -S_ed lines.

Goodman and Modifiye Goodman Criteria

Goodman is created from S_eh and -S_eh lines in the 1st type graph.In Modified Goodman, the upper S_eh line is finished at point i, which is the intersection point of the S_yieldd line.j point is determined so that ik = k j. Modified Goodman diagram is obtained by connecting S_e, i, d, j, -S_e points with lines, respectively.

Gerber Criterion

• All curves for both types are drawn for n=1.
• Drawing of Smith charts was explained earlier.
• There is no equation for Smith diagrams.

n: safety factor

Fatigue

Knowing the S_e and S_B values, it is drawn in accordance with equation 12.6.

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• When the two diagrams in Figure 12.19 and 12.20 are examined;
• We see that Goodman, Gerber and Original Smith Criteria do not take into account yielding, that is, plastic deformation. In other words, the life calculated as a result of cyclic loading is the number of repetitions at the instant of crack formation, and plastic deformation up to this repetition (cycle) is not important for these criteria. For this reason, the upper limit is the breaking stress (S_B).
• Modified Smith, Modified Goodman, Soderberg criteria take into account the instant of plastic deformation in the calculation of fatigue life. In other words, if yielding occurs before a crack forms during cyclic loading, it is considered that damage has occurred at that point. For this reason, the upper limit of these criteria in the graphs is the yield stress (S_yield).

12.10.4 Importance of plastic deformation for fatigue criteria

• Although Smith, Goodman and Gerber criteria can be used for all types of materials,
• Goodman is a better choice for brittle materials, while Gerber is a better choice for ductile materials.
• Soderberg, on the other hand, is useful for materials with low ductility (with a narrow plastic area).

12.10.5 Which fatigue criterion should be preferred for which material type?

Fatigue

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• Examine the diagram on the side carefully and try to understand which area is damaged and for what reason, and which areas will not be damaged at all.

Fatigue

Figure 12.20

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Total Correction Factor : k = k_a.k_b.k_c.k_d.k_e

k_a: Surface factor (depending on manufacturing method)

k_b : Dimension factor depending on geometry

k_c : Load factor depending on the loading type

k_d : temperature factor depending on operating temperature

k_e: stress concentration factor =1/K_f

(K_f :stress concentration factor for fatigue.)

12.10.6 Correction Factors

For S_m = 0

Fatigue

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Example 12.4

Solution:

Fatigue

A variable force P acts on a steel rod in the axial direction, the graph of which is given below. Determine the safe cross-section of the bar according to

a-) Modified Smith,

b-) Gerbercriteria.

From diagram P-t:

Stress limit values :

Mean Stress:

n=3

Modified Smith Diagram

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Solution:

For the intersection point b of the lines (𝐼𝐼) and (𝐼𝐼𝐼), the equations of both lines are equal to each other.

Safe (allowable) cross-sectional area :

Slope of the Ob line:

Fatigue

b

n=3

Modifiye Smith Diagram

Let the infinite life limit point be point b for this loading case. We will first find the A section value corresponding to the critical section value for n=1, which will allow us to be at point b.

If we find the value of 𝑆_𝑚𝑎𝑥 at point b, we can calculate area A from equation (I). As follows:

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b-) Solution according to Gerber criterion:

Let h be the point on the curve for the given loading situation. For this point h:

45⁰

h

The Gerber curve drawn for n=1 is as shown in the figure.

Alternating Stress:

(stress amplitude)

Mean Stress:

Substituting it into the Gerber Equation:

(For n=1 )

From the solution of this second degree equation:

Safe (allowable) cross-sectional area:

• As can be seen, we do not necessarily need to draw the Gerber curve for the solution, it is enough to know the curve equation and its limits.
• We should always keep in mind that the limit curves of these and other criteria are infinite life limits, that if we stay below the curves, we will stay in the safe zone, and if we exceed the curves, finite life and fatigue cracks will occur, so that we do not break away from the solution logic.

Gerber curve

If we rearrange the last equation:

Fatigue

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Example 12.5

Determine the safe diameter of the solid cylindrical foot area of ​​a crutch that is in contact with the ground according to the Goodman and Soderberg criteria. The crutch must be able to carry a 100 kg person within the safety limits.

S_yield = 120MPa,

S_e = 60MPa,

S_m’ =110MPa , S_Breaking=S_B=220MPasafety factor n=2,

g=10m/s²

​

According to

Goodman Criterion:

Solution:

When the crutch is lifted up (no contact with the ground), there is no load on it:

A single crutch carries half of the body weight when in contact with the ground:

Mean Stress:

Alternating Stress:

According to

Soderberg Criterion:

Fatigue

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Example 12.6: Vertical P force will be applied to the AB shaft made of low ductility material from point C of the BC arm. The shaft is fixed at the A end and the change of P force over time is shown in the graph. Accordingly, check the safety of the shaft in terms of fatigue.

Due to the P force, bending and torsion occur at the same time. We carry the force P to point B together with its torsion moment T. We see Bending + Torsion occurring.

Bending

Solution:

T_A=P.L₂

M_z-A=-P.L₁

T=P.L₂

x

M_z

T

x

Torsion

S_Breaking=S_B=360MPa, S_yield =240MPa,

S_e = 80MPa,

L₁ =50cm,

L₂ 40cm, d=6cm

Bending + Torsion

Fatigue

Since the section is symmetrical ρ_c = ρ_b = d/2 and y_c = -y_b = d/2 and therefore the stresses are equal in intensity at both points. Therefore, it is sufficient to examine only point c in terms of fatigue.

If we examine the moment diagrams at a moment for each loading, it is understood that the most critical section is the fixed section A; the most critical points are the top and bottom points (c and b) where the normal and shear stresses are at their highest values.

Strength of Materials - Lecture Notes / Mehmet Zor

29

23 Agust 2024

If we look at point c from the x-z plane, we see that it is a plane stress state:

Let's find the limiting values ​​of the stresses that occur over time at point c:

Normal stress limit values ​​

from equation (I):

Shear stress limit values ​​

from equation (II):

The limiting values ​​of the variable force are read from the graph given in the question:

Fatigue

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Strength of Materials - Lecture Notes / Mehmet Zor

30

23 Agust 2024

At point c, there is a combined stress situation. Since steel is a ductile material, we can use the Von-Mises yield criterion, which we use in static loading, in fatigue calculations. (In brittle materials, it is more accurate to use principal stresses and especially the maximum principal stress.)

(from equation 7.2.a )

Fatigue

Now we will check whether there will be fatigue at point c according to a suitable criterion.

Since the material has low ductility, it is a more accurate approach to check fatigue according to the Soderberg criterion…>>

Strength of Materials - Lecture Notes / Mehmet Zor

31

23 Agust 2024

Fatigue control for point c according to Soderberg

Alternating stress:

45⁰

Current stresses at point c due to force P :

Alternating stress :

Maximum Stress:

Minimum Stress:

c

.

Soderberg

Soderberg:

c

Fatigue

Since these two stresses are known,

the location of point c is now determined and marked on the diagrams..>>

Drawing of Soderber Diagrams: It is drawn as shown on the side using the material properties 𝑆_𝑒 and 𝑆_yield stresses (see topic 12.10.3).

For fatigue control we must determine the location of the critical point c on the diagrams.

The limit stresses that will bring point c to the fatigue limit can also be calculated:

Mean Stress:

If you pay attention, point c remains within the boundary lines. For this reason, we conclude that fatigue does not occur at point c.

* One of these two diagrams is sufficient for fatigue control.

Or we can draw the 2nd type of diagram.

Strength of Materials - Lecture Notes / Mehmet Zor

32

23 Agust 2024

In the load lifting machine shown in the figure, the contact force between the gears is calculated as 4 kN. Accordingly, determine whether the selection of the radius r = 4 cm of the shaft connected to the motor will be sufficient for the safety of the system and if it is safe, calculate the safety factor according to the Soderberg criterion. (Radius of the gear connected to the motor R = 20 cm, small shaft length L = 0.6 m)

For Shaft Materials:

• Experimental Endurance Stress: Se’ =60MPa
• Total correction factor k: 0.9,
• Yield Stress S_yield =330MPa

Answer:

Fatigue

Example 12.7