Final Project

Joseph D Hawley

Principles of Mechanical Design (MEE 342)

Dr. Fard

7 August 2018

Problem Statement

Figure 1 shows the engine mounted on a base with its output shaft connected via a clutch

to the input shaft of a gearbox. The gearbox contains a single gearset to reduce the high engine

speed. The output shaft has the larger torque which is defined as shown in Figure 2(a). In

addition to the torque on the shafts, there are loads from gears that apply bending moments to the

shaft.

The torque-time function on the output shaft is given in figure 2(a). The required gear ratio is a

3:1 reduction in velocity from the input to the output shaft. The diameters of the input and output

gears are 4 in and 12 in, respectively. Both gears have the same thickness and 20° pressure angle

as shown in Figure 2(b). We assume that the gears are centered between the simply supported

bearings that are set 6 in apart. A stress concentration factor of 3 (for static force) for both

bending and torsion at the critical location where both moment and torque are largest is assumed.

Low – carbon cold-rolled steel SAE 1018 (Sut = 64 kpsi and Sy = 54 kpsi) is considered for the

design. Also assume the followings: (i) Machined finish; (ii) 50% reliability; (iii) notch radius

of 0.01 in; and (iv) safety factor of 2.5 to account for the uncertainties.

(a) Determine reasonable sizes for the input and output shafts of the gearbox considering the

following failure criteria (i) Goodman (ii) Gerber (iii) ASME elliptic, and (iv) Soderberg.

Summarize the results in a table and discuss.

(b) Draw the fatigue diagram for input and output shafts with various criteria of failure

including the load lines (use the diameters of the input and output shafts from Soderberg

criterion in part a)

Stress Analysis First, let us draw free-body diagram of the of the pinion and gear.

In the figure above we can see our system with forces acting on the two gears. The forces acting on the outside of the gear have a pressure angle of 20 degrees. The forces at points a and b are reaction forces acting on the gears. Now let us draw a free-body diagram for pinion 2 and gear 3.

In the free-body diagram, 𝑊32

𝑡 is the transmitted force from gear 3 to pinion 2. We can express the transmitted force in terms of torque by,

𝑇2 = 𝑊32𝑡 ⋅ (

𝑑2

2) → 𝑊32

𝑡 =2 ⋅ 𝑇2

𝑑2

Here, 𝑇2 is the torque from the input shaft and 𝑑2 is the diameter of the pinion. From the torque-time function of the output shaft, we have a maximum of 585 lb-in and a minimum of -175 lb-in. Because of the 3:1 gear ratio, we know that the input torque will be 3 times less than the output torque. Therefore, the maximum input torque is 195 lb-in and the minimum is -58.33 lb-in. Substituting these values is along with a diameter of 4 inches we get,

𝑊32𝑡 =

2 ⋅ 195

4= 97.50 𝑙𝑏𝑓 → 𝑚𝑎𝑥𝑖𝑚𝑢𝑚

𝑊32𝑡 =

2 ⋅ −58.33

4= −29.17 𝑙𝑏𝑓 → 𝑚𝑖𝑛𝑖𝑚𝑢𝑚

To the get force in the radial direction, we can take the transmitted force and multiply by the tangent of the pressure angle. Doing this gives us,

𝑊32𝑟 = 97.50 ⋅ tan(20) = 35.49 𝑙𝑏𝑓 → 𝑚𝑎𝑥𝑖𝑚𝑢𝑚

𝑊32

𝑟 = −29.17 ⋅ tan(20) = −10.62 𝑙𝑏𝑓 → 𝑚𝑖𝑛𝑖𝑚𝑢𝑚

Then the resultant force, which will give us our maximum and minimum force for bending, is found by,

𝑊32 = √(97.50)2 + (35.49)2 = 103.72 𝑙𝑏𝑓 → 𝑚𝑎𝑥𝑖𝑚𝑢𝑚

𝑊32 = √(−29.17)2 + (−10.62)2 = −31.04 𝑙𝑏𝑓 → 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 Similarly, the resultant force can also be found by,

𝑊32 =2 ⋅ 𝑇

cos(𝜙) ∗ 𝑑

If we sum the forces in both the x and y directions, we see that

𝑊32𝑡 = 𝐹𝑎2

𝑡

𝑊32𝑟 = 𝐹𝑎2

𝑟 Because of this, we can also say that,

𝑊32 = 𝐹𝑎2 where 𝐹𝑎2 will cause maximum or minimum bending in the shaft. Looking at a free-body diagram of gear 3, we can use the same approach to solve for the forces,

𝑊23𝑡 =

2 ⋅ 585

12= 97.50 𝑙𝑏𝑓

𝑊23𝑡 =

2 ⋅ −175

12= −29.17 𝑙𝑏𝑓

Instead of continuing with the rest of the calculations, we should notice that the forces on gear 3 are equal to the forces on pinion 2 because the system is n static equilibrium. This means that both the input and output shaft will experience the same magnitude of bending moment but different magnitudes of torsional stress. To find the maximum and minimum bending moment, we can take the resultant force and multiply by the distance between the gear/pinion and the bearings.

𝑀{𝑚𝑎𝑥} = 103.72 ⋅ 3 = 311.27 𝑙𝑏𝑓 ⋅ 𝑖𝑛

𝑀{𝑚𝑖𝑛} = −31.04 ⋅ 3 = −93.12 𝑙𝑏𝑓 ⋅ 𝑖𝑛

Let us also define our maximum and minimum torque in both the input and output shafts.

𝑇{𝑚𝑎𝑥,𝑖} = 195.00 𝑙𝑏𝑓 ⋅ 𝑖𝑛

𝑇{𝑚𝑖𝑛,𝑖} = −58.33 𝑙𝑏𝑓 ⋅ 𝑖𝑛

𝑇{𝑚𝑎𝑥,𝑜} = 585.00 𝑙𝑏𝑓 ⋅ 𝑖𝑛

𝑇{𝑚𝑖𝑛,𝑜} = −175 𝑙𝑏𝑓 ⋅ 𝑖𝑛

From here, we can find our midrange and alternate stresses from,

𝑀𝑚 =𝑀{𝑚𝑎𝑥} + 𝑀{𝑚𝑖𝑛}

2, 𝑇𝑚 =

𝑇{𝑚𝑎𝑥} + 𝑇{𝑚𝑖𝑛}

2

𝑀𝑎 =|𝑀{𝑚𝑎𝑥} − 𝑀{𝑚𝑖𝑛}|

2, 𝑇𝑎 =

|𝑇{𝑚𝑎𝑥} − 𝑇{𝑚𝑖𝑛}|

2

Substituting values in to these equations yield,

𝑀𝑚 = 109.08 𝑙𝑏𝑓 ⋅ 𝑖𝑛, 𝑀𝑎 = 202.19 𝑙𝑏𝑓 ⋅ 𝑖𝑛

𝑇𝑚,𝑖 = 68.33 𝑙𝑏𝑓 ⋅ 𝑖𝑛, 𝑇𝑎,𝑖 = 126.67 𝑙𝑏𝑓 ⋅ 𝑖𝑛

𝑇𝑚,𝑜 = 205 𝑙𝑏𝑓 ⋅ 𝑖𝑛, 𝑇𝑎,𝑜 = 380 𝑙𝑏𝑓 ⋅ 𝑖𝑛

Endurance Limit and Modification Factors The material of the shaft is low – carbon cold-rolled steel SAE 1018 with a yield strength of 54 ksi and an ultimate strength of 64 ksi. Our endurance limit under standard conditions can then be approximated as 50 % of the ultimate strength,

𝑆𝑒′ = 0.5 ⋅ 64 = 32 𝑘𝑠𝑖

Our shaft however is not under standard conditions and modification factors are needed to find the endurance limit under non-standard conditions. The problem states the shaft has a machine finish, 50 % reliability, and notch radius of 0.01 inches. Since the diameter is also unknown, we will have to add a sizing factor. The surface condition modification factor can be found by,

𝑘𝑎 = 𝑎𝑆𝑢𝑡𝑏

where the constants ‘a’ and ‘b’ are to be found in table 6-2. For a machined finish ‘a’ is 2.7 and be is -0.265 if the ultimate strength is expressed in ksi.

𝑘𝑎 = 2.7(64)−0.265 = 0.8969

Since the diameter is unknown, we will start with an assumption that the diameter of the shaft is 1 inch. The sizing factor is then found by,

𝑘𝑏 = (𝑑

0.3)

−0.107

= (1

0.3)

−0.107

= 0.8791

This equation assumes that the diameter of the shaft is between 0.11 and 2 inches. If the diameter of the shaft exceeds 2 inches, a different equation will need to be used. The sizing factor will need to be updated as we go through iterations till a solution has converged. Since we have both torsion and bending, we can use 𝑘𝑐 = 1 for the loading factor. The reliability factor is found using the relation,

𝑘𝑒 = 1 − 0.08𝑧𝑎

where 𝑧𝑎 is a transformation variate and can be found in table 6-5. For 50 % reliability the transformation variate is equal to zero which yields 𝑘𝑒 = 1. Using these modification factors, we can now find the new endurance limit under non-standard conditions by,

𝑆𝑒 = 𝑘𝑎𝑘𝑏𝑘𝑐𝑘𝑒 ⋅ 𝑆𝑒′ = 0.8969 ⋅ 0.8791 ⋅ 1 ⋅ 1 = 25.231 𝑘𝑠𝑖

This problem has both a stress concentration of 3 for both bending and torsion at the critical area where torque and bending is maximum. To find the notch sensitivity factor, we need to use figures 6-20 and 6-21. The notch sensitivity factor is about 0.5 for bending and torsion. The stress concentration is then,

𝐾𝑓 = 1 + 𝑞(𝐾𝑡 − 1)

𝐾𝑓𝑠 = 1 + 𝑞𝑠(𝐾𝑡𝑠 − 1)

Since 𝑞 = 𝑞𝑠 & 𝐾𝑡 = 𝐾𝑡𝑠, our stress concentrations will both be equal.

𝐾{𝑓,𝑓𝑠} = 1 + 0.5(3 − 1) = 2

Static Failure Before we analyze the fatigue, it is necessary to check for static failure. We can do this by finding the maximum stress. Since we do not have a diameter, we will solve for the smallest diameter possible before static failure. Then when we check for fatigue, we will know if our solutions will result in static failure.

𝜎′ =𝐾𝑡𝑀{𝑚𝑎𝑥}𝑐

𝐼, 𝜏′ =

𝐾𝑡𝑠𝑇{𝑚𝑎𝑥}𝑟

𝐽

These can reduce to,

𝜎′ =𝐾𝑡32𝑀{𝑚𝑎𝑥}

𝜋𝑑3, 𝜏′ =

𝐾𝑡𝑠16𝑇{𝑚𝑎𝑥}

𝜋𝑑3

The maximum stress is then,

𝜎{𝑚𝑎𝑥} = 𝜎′ + 𝜏′ =𝐾𝑡16

𝜋𝑑3(2 ⋅ 𝑀{𝑚𝑎𝑥} + 𝑇{𝑀𝑎𝑥})

Assuming here that 𝐾𝑡 = 𝐾𝑡𝑠. Setting this equal to 1 and solving for the diameter we get,

𝑛 =𝑆𝑦

𝜎{𝑚𝑎𝑥}→ 𝑛 =

𝑆𝑦

𝐾𝑡16𝜋𝑑3 (2𝑀{𝑚𝑎𝑥} + 𝑇{𝑚𝑎𝑥})

Set the design factor equal to 1 and solve for d,

1

𝜋𝑑3=

𝑆𝑦

𝐾𝑡16(2𝑀{𝑚𝑎𝑥} + 𝑇{𝑚𝑎𝑥})→ 𝑑{𝑚𝑖𝑛} = [

(𝐾𝑡16(2𝑀{𝑚𝑎𝑥} + 𝑇{𝑚𝑎𝑥}))

𝜋𝑆𝑦]

13

Plugging in our values we get the minimum allowable diameter to be 0.6991 inches for the output shaft and 0.6139 inches for the input shaft. Shaft Stresses & Failure Criteria We have both alternating and midrange stresses as well as bending and torsional stress. Axial stresses may be present but are considered negligible. The stresses are given by the following,

𝜎𝑎 = 𝐾𝑓 ⋅𝑀𝑎𝑐

𝐼, 𝜎𝑚 = 𝐾𝑓 ⋅

𝑀𝑚𝑐

𝐼

𝜏𝑎 = 𝐾𝑓𝑠 ⋅𝑇𝑎𝑟

𝐽, 𝜏𝑚 = 𝐾𝑓𝑠 ⋅

𝑇𝑚𝑟

𝐽

Since we have a solid shaft with a circular cross-section, we can express these terms as,

𝜎𝑎 = 𝐾𝑓 ⋅32𝑀𝑎

𝜋𝑑3, 𝜎𝑚 = 𝐾𝑓 ⋅

32𝑀𝑚

𝜋𝑑3

𝜏𝑎 = 𝐾𝑓𝑠 ⋅16𝑇𝑎

𝜋𝑑3, 𝜏𝑚 = 𝐾𝑓𝑠 ⋅

16𝑇𝑚

𝜋𝑑3

If we use the distortion energy failure theory with a rotating, round shaft, we can combine the alternating and midrange stresses as,

𝜎𝑎′ = (𝜎𝑎

2 + 3𝜏𝑎2)

12 = [(

32𝐾𝑓𝑀𝑎

𝜋𝑑3)

2

+ 3 (16𝐾𝑓𝑠𝑇𝑎

𝜋𝑑3)

2

]

12

𝜎𝑚′ = (𝜎𝑚

2 + 3𝜏𝑚2 )

12 = [(

32𝐾𝑓𝑀𝑚

𝜋𝑑3)

2

+ 3 (16𝐾𝑓𝑠𝑇𝑚

𝜋𝑑3)

2

]

12

DE-Goodman The fatigue failure criteria for the modified Goodman line is given as,

1

𝑛=

𝜎𝑎′

𝑆𝑒+

𝜎𝑚′

𝑆𝑢𝑡

Substituting 𝜎𝑚

′ & 𝜎𝑎′ into the equation will yield,

1

𝑛=

16

𝜋𝑑3{

1

𝑆𝑒[4(𝐾𝑓𝑀𝑎)

2+ 3(𝐾𝑓𝑠𝑇𝑎)

2]

12

+1

𝑆𝑢𝑡[4(𝐾𝑓𝑀𝑚)

2+ 3(𝐾𝑓𝑠𝑇𝑚)

2]

12

}

Since our goal is to find the appropriate diameter, we can rearrange the equation and solve for the diameter.

𝑑 = ((16𝑛

𝜋) {

1

𝑆𝑒[4(𝐾𝑓𝑀𝑎)

2+ 3(𝐾𝑓𝑠𝑇𝑎)

2]

12

+1

𝑆𝑢𝑡[4(𝐾𝑓𝑀𝑚)

2+ 3(𝐾𝑓𝑠𝑇𝑚)

2]

12

})

13

DE-Soderberg

𝜎𝑎′

𝑆𝑒+

𝜎𝑚′

𝑆𝑦=

1

𝑛

1

𝑛=

16

𝜋𝑑3{

1

𝑆𝑒[4(𝐾𝑓𝑀𝑎)

2+ 3(𝐾𝑓𝑠𝑇𝑎)

2]

12

+1

𝑆𝑦[4(𝐾𝑓𝑀𝑚)

2+ 3(𝐾𝑓𝑠𝑇𝑚)

2]

12

}

𝑑 = (16𝑛

𝑝𝑖{

1

𝑆𝑒[4(𝐾𝑓𝑀𝑎)

2+ 3(𝐾𝑓𝑠𝑇𝑎)

2]

12

+1

𝑆𝑦[4(𝐾𝑓𝑀𝑚)

2+ 3(𝐾𝑓𝑠𝑇𝑚)

2]

12

})

13

DE-Gerber

𝑛𝜎𝑎′

𝑆𝑒+ (

𝑛𝜎𝑚′

𝑆𝑢𝑡)

2

= 1

1

𝑛=

8𝐴

𝜋𝑑3𝑆𝑒{1 + [1 + (

2𝐵𝑆𝑒

𝐴𝑆𝑢𝑡)

2

]

12

}

𝑑 = (8𝑛𝐴

𝜋𝑆𝑒{1 + [1 + (

2𝐵𝑆𝑒

𝐴𝑆𝑢𝑡)

2

]

12

})

13

where

𝐴 = √4(𝐾𝑓𝑀𝑎)2

+ 3(𝐾𝑓𝑠𝑇𝑎)2

𝐵 = √4(𝐾𝑓𝑀𝑚)2

+ 3(𝐾𝑓𝑠𝑇𝑚)2

DE-ASME Elliptic

(𝑛𝜎𝑎

′

𝑆𝑒)

2

+ (𝑛𝜎𝑚

′

𝑆𝑦)

2

= 1

1

𝑛=

16

𝜋𝑑3[4 (

𝐾𝑓𝑀𝑎

𝑆𝑒)

2

+ 3 (𝐾𝑓𝑠𝑇𝑎

𝑆𝑒)

2

+ 4 (𝐾𝑓𝑀𝑚

𝑆𝑦)

2

+ 3 (𝐾𝑓𝑠𝑇𝑚

𝑆𝑦)

2

]

12

𝑑 = {16𝑛

𝜋[4 (

𝐾𝑓𝑀𝑎

𝑆𝑒)

2

+ 3 (𝐾𝑓𝑠𝑇𝑎

𝑆𝑒)

2

+ 4 (𝐾𝑓𝑀𝑚

𝑆𝑦)

2

+ 3 (𝐾𝑓𝑠𝑇𝑚

𝑆𝑦)

2

]

12

}

13

Solution Method As stated earlier, due to an unknown diameter we are not able to directly find the sizing factor. A diameter of 1-inch was assumed to obtain a sizing factor, where the diameter, sizing factor, and endurance limit would be reevaluated every iteration until the change in diameter from one iteration to the next was less than 1%. Solution The results from the analysis are detailed in Table 1.

Failure Criteria Input Shaft Diameter (inches) Output Shaft Diameter (inches)

DE-Goodman 0.8211 0.9809

DE-Soderberg 0.8303 0.9920

DE-Gerber 0.7792 0.9314

DE-ASME Elliptic 0.7760 0.9276

Table 1: Input and output shaft diameters for various criteria of fatigue failure. The results are consistent with what is expected regarding the relationships between different criteria. Looking at a fatigue diagram for the four methods, Soderberg is the most conservative and therefore should have the largest diameters. At the other end, ASME Elliptic is the least conservative and should have the smallest diameters. Gerber is expected to yield a similar diameter size to ASME Elliptic but should yield a larger. Goodman is expected to yield a similar size as Soderberg but should yield a smaller value. The values in the table are found analytically and are not necessarily realistic for an actual shaft size. The actual shaft size may depend on sizing that is readily available from a manufacturer, and if a different sizing is needed than what the analytical results yield, the next size up should be chosen to avoid violating the factor of safety requirement.

Also note that none of the diameters found will result in static failure. Check for Yielding By using the Langer static yield, we can check for first cycle yielding by,

𝜎𝑎′ + 𝜎𝑚

′ =𝑆𝑦

𝑛

Solving for n and inputting in our values yields the following:

Failure Criteria Input Shaft Factor of Safety Output Shaft Factor of Safety

DE Soderberg 4.29 4.35

DE Goodman 4.14 4.21

DE Gerber 3.54 3.60

DE ASME Elliptic 3.50 3.56

Table 2: First cycle yielding check. Fatigue Diagrams The fatigue diagrams can be obtained by using the failure criteria equations. For the Soderberg line, this equation is

𝑆𝑎

𝑆𝑒+

𝑆𝑚

𝑆𝑦= 1

where 𝑆𝑎 & 𝑆𝑚 are 𝑛𝜎𝑎

′ & 𝑛𝜎𝑚′ , respectively. For the modified Goodman, the line equation is,

𝑆𝑎

𝑆𝑒+

𝑆𝑚

𝑆𝑢𝑡= 1

ASME elliptic is,

(𝑆𝑎

𝑆𝑒)

2

+ (𝑆𝑚

𝑆𝑦)

2

= 1

And lastly, Gerber,

𝑆𝑎

𝑆𝑒+ (

𝑆𝑚

𝑆𝑢𝑡)

2

= 1

Also on the fatigue diagram is the Langer first-cycle-yielding criterion given by,

𝑆𝑎 + 𝑆𝑚 = 𝑆𝑦

The load lines can also be plotted by taking 𝑆𝑎/𝑆𝑚 as the slope with a y-intercept of zero. Using these equations, we can then plot the fatigue diagrams.

References

Books

Budynas, Richard G., and Nisbett, J. K., Shigley’s Mechanical Engineering Design, 10th ed., Ch. 6, 7, 13., McGraw-Hill 2015.

Juvinall, Robert C., and Marshek, Kurt M., Fundamentals of Machine Component Design, 5th ed., Ch. 6, 8, 15., Wiley 2012 Collins, Jack A., Busby, Henry, and Staab, George, Mechanical Design of Machine Elements and Machines, 2nd ed., Ch. 8., Wiley 2010

MATLAB Code

clc; close; clear;

format long

mg = 3;

d_1 = 4; % in

d_2 = 12; % in

phi = 20; % degrees

l = 6; % in

k_t = 3;

S_ut = 64e3;% psi

S_y = 54e3; % psi

Fs = 2.5;

T_max_o = 585;

T_min_o = -175;

T_max_i = 195;

T_min_i = (-175/3);

F_max = (T_max_o*2)/(cosd(phi)*d_2);

M_max = F_max*l/2;

F_min = (T_min_o*2)/(cosd(phi)*d_2);

M_min = F_min*l/2;

M_m = (M_max + M_min)/2;

M_a = abs((M_max - M_min)/2);

% static failure

d_min = (((k_t*16*(2*M_max+T_max_i))/(pi*S_y)))^(1/3);

D_min = (((k_t*16*(2*M_max+T_max_o))/(pi*S_y)))^(1/3);

T_m_i = (T_max_i + T_min_i)/2;

T_a_i = abs((T_max_i - T_min_i)/2);

T_m_o = (T_max_o + T_min_o)/2;

T_a_o = abs((T_max_o - T_min_o)/2);

Se_prime = (0.5*S_ut);

k_a = 2.7*(S_ut*10^-3)^-0.265; % machine finish

k_e = 1; % R = 50%

q = 0.5; % notch radius 0.01 inch

qs = 0.5;

k_f = 1+q*(k_t-1);

k_fs = 1+qs*(k_t-1);

% goodman

d_s_goodman = 1;

k_b_goodman_d = (d_s_goodman/0.3)^-0.107;

Se_goodman_d(1) = Se_prime*k_a*k_b_goodman_d*k_e;

% gerber

d_s_gerber = 1;

k_b_gerber_d = (d_s_gerber/0.3)^-0.107;

Se_gerber_d = Se_prime*k_a*k_b_gerber_d*k_e;

% soderberg

d_s_soderberg = 1;

k_b_soderberg_d = (d_s_soderberg/0.3)^-0.107;

Se_soderberg_d = Se_prime*k_a*k_b_soderberg_d*k_e;

% asme

d_s_asme = 1;

k_b_asme_d = (d_s_asme/0.3)^-0.107;

Se_asme_d = Se_prime*k_a*k_b_asme_d*k_e;

error_d_goodman = 100;

% loop

i = 1;

while error_d_goodman > 1

d_s_goodman(i+1) = ((16*Fs)/pi*(1/Se_goodman_d(i)*(4*(k_f*M_a)^2+3*(k_fs*T_a_i)^2)^0.5 +

1/S_ut*(4*(k_f*M_m)^2+3*(k_fs*T_m_i)^2)^0.5))^(1/3);

error_d_goodman(i) = abs((d_s_goodman(i) - d_s_goodman(i+1))/(d_s_goodman(i)))*100;

k_b_goodman_d(i+1) = (d_s_goodman(i+1)/0.3)^-0.107;

Se_goodman_d(i+1) = Se_prime*k_a*k_b_goodman_d(i+1)*k_e;

i = i +1;

end

error_d_gerber = 100;

i = 1;

A = sqrt(4*(k_f*M_a)^2+3*(k_fs*T_a_i)^2);

B = sqrt(4*(k_f*M_m)^2+3*(k_fs*T_m_i)^2);

while error_d_gerber > 1

d_s_gerber(i+1) =

((8*Fs*A)/(pi*Se_gerber_d(i))*(1+(1+((2*B*Se_gerber_d(i))/(A*S_ut))^2)^0.5))^(1/3);

error_d_gerber(i) = abs((d_s_gerber(i) - d_s_gerber(i+1))/(d_s_gerber(i)))*100;

k_b_gerber_d(i+1) = (d_s_gerber(i+1)/0.3)^-0.107;

Se_gerber_d(i+1) = Se_prime*k_a*k_b_gerber_d(i+1)*k_e;

i = i+1;

end

error_d_soderberg = 100;

i = 1;

while error_d_soderberg > 1

d_s_soderberg(i+1) =((16*Fs)/pi*(1/Se_soderberg_d(i)*(4*(k_f*M_a)^2+3*(k_fs*T_a_i)^2)^0.5 +

1/S_y*(4*(k_f*M_m)^2+3*(k_fs*T_m_i)^2)^0.5))^(1/3);

error_d_soderberg(i) = abs((d_s_soderberg(i) - d_s_soderberg(i+1))/(d_s_soderberg(i)))*100;

k_b_soderberg_d(i+1) = (d_s_soderberg(i+1)/0.3)^-0.107;

Se_soderberg_d(i+1) = Se_prime*k_a*k_b_soderberg_d(i+1)*k_e;

i = i+1;

end

error_d_asme = 100;

i = 1;

while error_d_asme > 1

d_s_asme(i+1) =

((16*Fs/pi)*(4*(k_f*M_a/Se_asme_d(i))^2+3*(k_fs*T_a_i/Se_asme_d(i))^2+4*(k_f*M_m/S_y)^2+3*(k_fs*T

_m_i/S_y)^2)^0.5)^(1/3);

error_d_asme(i) = abs((d_s_asme(i) - d_s_asme(i+1))/(d_s_asme(i)))*100;

k_b_asme_d(i+1) = (d_s_asme(i+1)/0.3)^-0.107;

Se_asme_d(i+1) = Se_prime*k_a*k_b_asme_d(i+1)*k_e;

i = i+1;

end

% goodman

D_s_goodman = 1;

k_b_goodman_D = (D_s_goodman/0.3)^-0.107;

Se_goodman_D(1) = Se_prime*k_a*k_b_goodman_D*k_e;

% gerber

D_s_gerber = 1;

k_b_gerber_D = (D_s_gerber/0.3)^-0.107;

Se_gerber_D = Se_prime*k_a*k_b_gerber_D*k_e;

% soderberg

D_s_soderberg = 1;

k_b_soderberg_D = (D_s_soderberg/0.3)^-0.107;

Se_soderberg_D = Se_prime*k_a*k_b_soderberg_D*k_e;

% asme

D_s_asme = 1;

k_b_asme_D = (D_s_asme/0.3)^-0.107;

Se_asme_D = Se_prime*k_a*k_b_asme_D*k_e;

error_D_goodman = 100;

% loop

i = 1;

while error_D_goodman > 1

D_s_goodman(i+1) = ((16*Fs)/pi*(1/Se_goodman_D(i)*(4*(k_f*M_a)^2+3*(k_fs*T_a_o)^2)^0.5 +

1/S_ut*(4*(k_f*M_m)^2+3*(k_fs*T_m_o)^2)^0.5))^(1/3);

error_D_goodman(i) = abs((D_s_goodman(i) - D_s_goodman(i+1))/(D_s_goodman(i)))*100;

k_b_goodman_D(i+1) = (D_s_goodman(i+1)/0.3)^-0.107;

Se_goodman_D(i+1) = Se_prime*k_a*k_b_goodman_D(i+1)*k_e;

i = i +1;

end

error_D_gerber = 100;

i = 1;

A = sqrt(4*(k_f*M_a)^2+3*(k_fs*T_a_o)^2);

B = sqrt(4*(k_f*M_m)^2+3*(k_fs*T_m_o)^2);

while error_D_gerber > 1

D_s_gerber(i+1) =

((8*Fs*A)/(pi*Se_gerber_D(i))*(1+(1+((2*B*Se_gerber_D(i))/(A*S_ut))^2)^0.5))^(1/3);

error_D_gerber(i) = abs((D_s_gerber(i) - D_s_gerber(i+1))/(D_s_gerber(i)))*100;

k_b_gerber_D(i+1) = (D_s_gerber(i+1)/0.3)^-0.107;

Se_gerber_D(i+1) = Se_prime*k_a*k_b_gerber_D(i+1)*k_e;

i = i+1;

end

error_D_soderberg = 100;

i = 1;

while error_D_soderberg > 1

D_s_soderberg(i+1) =((16*Fs)/pi*(1/Se_soderberg_D(i)*(4*(k_f*M_a)^2+3*(k_fs*T_a_o)^2)^0.5 +

1/S_y*(4*(k_f*M_m)^2+3*(k_fs*T_m_o)^2)^0.5))^(1/3);

error_D_soderberg(i) = abs((D_s_soderberg(i) - D_s_soderberg(i+1))/(D_s_soderberg(i)))*100;

k_b_soderberg_D(i+1) = (D_s_soderberg(i+1)/0.3)^-0.107;

Se_soderberg_D(i+1) = Se_prime*k_a*k_b_soderberg_D(i+1)*k_e;

i = i+1;

end

error_D_asme = 100;

i = 1;

while error_D_asme > 1

D_s_asme(i+1) =

((16*Fs/pi)*(4*(k_f*M_a/Se_asme_D(i))^2+3*(k_fs*T_a_o/Se_asme_D(i))^2+4*(k_f*M_m/S_y)^2+3*(k_fs*T

_m_o/S_y)^2)^0.5)^(1/3);

error_D_asme(i) = abs((D_s_asme(i) - D_s_asme(i+1))/(D_s_asme(i)))*100;

k_b_asme_D(i+1) = (D_s_asme(i+1)/0.3)^-0.107;

Se_asme_D(i+1) = Se_prime*k_a*k_b_asme_D(i+1)*k_e;

i = i+1;

end

% check for yielding

sig_a_d =

((32*k_f*M_a/(pi*d_s_soderberg(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_soderberg(end)^3))^2)^(0.5);

sig_m_d =

((32*k_f*M_m/(pi*d_s_soderberg(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_soderberg(end)^3))^2)^(0.5);

sig_max_d = sig_a_d+sig_m_d;

n_d_soderberg = S_y/sig_max_d;

sig_a_D =

((32*k_f*M_a/(pi*D_s_soderberg(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_soderberg(end)^3))^2)^(0.5);

sig_m_D =

((32*k_f*M_m/(pi*D_s_soderberg(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_soderberg(end)^3))^2)^(0.5);

sig_max_D = sig_a_D+sig_m_D;

n_D_soderberg = S_y/sig_max_D;

sig_a_d =

((32*k_f*M_a/(pi*d_s_goodman(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_goodman(end)^3))^2)^(0.5);

sig_m_d =

((32*k_f*M_m/(pi*d_s_goodman(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_goodman(end)^3))^2)^(0.5);

sig_max_d = sig_a_d+sig_m_d;

n_d_goodman = S_y/sig_max_d;

sig_a_D =

((32*k_f*M_a/(pi*D_s_goodman(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_goodman(end)^3))^2)^(0.5);

sig_m_D =

((32*k_f*M_m/(pi*D_s_goodman(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_goodman(end)^3))^2)^(0.5);

sig_max_D = sig_a_D+sig_m_D;

n_D_goodman = S_y/sig_max_D;

sig_a_d = ((32*k_f*M_a/(pi*d_s_asme(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_asme(end)^3))^2)^(0.5);

sig_m_d = ((32*k_f*M_m/(pi*d_s_asme(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_asme(end)^3))^2)^(0.5);

sig_max_d = sig_a_d+sig_m_d;

n_d_asme = S_y/sig_max_d;

sig_a_D = ((32*k_f*M_a/(pi*D_s_asme(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_asme(end)^3))^2)^(0.5);

sig_m_D = ((32*k_f*M_m/(pi*D_s_asme(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_asme(end)^3))^2)^(0.5);

sig_max_D = sig_a_D+sig_m_D;

n_D_asme = S_y/sig_max_D;

sig_a_d =

((32*k_f*M_a/(pi*d_s_gerber(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_gerber(end)^3))^2)^(0.5);

sig_m_d =

((32*k_f*M_m/(pi*d_s_gerber(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_gerber(end)^3))^2)^(0.5);

sig_max_d = sig_a_d+sig_m_d;

n_d_gerber = S_y/sig_max_d;

sig_a_D =

((32*k_f*M_a/(pi*D_s_gerber(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_gerber(end)^3))^2)^(0.5);

sig_m_D =

((32*k_f*M_m/(pi*D_s_gerber(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_gerber(end)^3))^2)^(0.5);

sig_max_D = sig_a_D+sig_m_D;

n_D_gerber = S_y/sig_max_D;

% Soderberg

S_a_d =

Fs*((32*k_f*M_a/(pi*d_s_soderberg(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_soderberg(end)^3))^2)^(0.5)

;

S_a_D =

Fs*((32*k_f*M_a/(pi*D_s_soderberg(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_soderberg(end)^3))^2)^(0.5)

;

S_m_d =

Fs*((32*k_f*M_m/(pi*d_s_soderberg(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_soderberg(end)^3))^2)^(0.5)

;

S_m_D =

Fs*((32*k_f*M_m/(pi*D_s_soderberg(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_soderberg(end)^3))^2)^(0.5)

;

sig_m = linspace(0, S_y, 1000);

r = S_a_d/S_m_d;

ll = sig_m.*r;

sig_a_yield = S_y - sig_m;

figure

plot([0 S_m_d S_y], [Se_soderberg_d(end) S_a_d 0],'r',[0 S_m_D S_y], [Se_soderberg_D(end) S_a_D

0],'--k', sig_m, ll, sig_m, sig_a_yield)

grid on

ylim([0 60e3]);

legend('Input Shaft','Output Shaft','Load Line','Yield Line')

title 'Soderberg Fatigue Diagram'

xlabel 'Midrange Stress (psi)'

ylabel 'Alternating Stress (psi)'

% Goodman

S_a_d =

Fs*((32*k_f*M_a/(pi*d_s_goodman(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_goodman(end)^3))^2)^(0.5);

S_a_D =

Fs*((32*k_f*M_a/(pi*D_s_goodman(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_goodman(end)^3))^2)^(0.5);

S_m_d =

Fs*((32*k_f*M_m/(pi*d_s_goodman(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_goodman(end)^3))^2)^(0.5);

S_m_D =

Fs*((32*k_f*M_m/(pi*D_s_goodman(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_goodman(end)^3))^2)^(0.5);

sig_m = linspace(0, S_ut, 1000);

r = S_a_d/S_m_d;

ll = sig_m.*r;

sig_a_yield = S_y - sig_m;

figure

plot([0 S_m_d S_ut], [Se_goodman_d(end) S_a_d 0],'r',[0 S_m_D S_ut], [Se_goodman_D(end) S_a_D

0],'--k', sig_m, ll, sig_m, sig_a_yield)

grid on

ylim([0 60e3]);

legend('Input Shaft','Output Shaft','Load Line','Yield Line')

title 'Goodman Fatigue Diagram'

xlabel 'Midrange Stress (psi)'

ylabel 'Alternating Stress (psi)'

% ASME

S_a_d = Fs*((32*k_f*M_a/(pi*d_s_asme(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_asme(end)^3))^2)^(0.5);

S_a_D = Fs*((32*k_f*M_a/(pi*D_s_asme(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_asme(end)^3))^2)^(0.5);

S_m_d = Fs*((32*k_f*M_m/(pi*d_s_asme(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_asme(end)^3))^2)^(0.5);

S_m_D = Fs*((32*k_f*M_m/(pi*D_s_asme(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_asme(end)^3))^2)^(0.5);

sig_m = linspace(0, S_y, 1000);

r = S_a_d/S_m_d;

ll = sig_m.*r;

sig_a_yield = S_y - sig_m;

sig_alt_d = sqrt((1-(sig_m./S_y).^2))*Se_asme_d(end);

sig_alt_D = sqrt((1-(sig_m./S_y).^2))*Se_asme_D(end);

figure

plot(sig_m, sig_alt_d,'r',sig_m, sig_alt_D,'--k', sig_m, ll, sig_m, sig_a_yield)

grid on

ylim([0 60e3]);

legend('Input Shaft','Output Shaft','Load Line','Yield Line')

title 'ASME Elliptic Fatigue Diagram'

xlabel 'Midrange Stress (psi)'

ylabel 'Alternating Stress (psi)'

% Gerber

S_a_d =

Fs*((32*k_f*M_a/(pi*d_s_gerber(end)^3))^2+3*(16*k_fs*T_a_i/(pi*d_s_gerber(end)^3))^2)^(0.5);

S_a_D =

Fs*((32*k_f*M_a/(pi*D_s_gerber(end)^3))^2+3*(16*k_fs*T_a_o/(pi*D_s_gerber(end)^3))^2)^(0.5);

S_m_d =

Fs*((32*k_f*M_m/(pi*d_s_gerber(end)^3))^2+3*(16*k_fs*T_m_i/(pi*d_s_gerber(end)^3))^2)^(0.5);

S_m_D =

Fs*((32*k_f*M_m/(pi*D_s_gerber(end)^3))^2+3*(16*k_fs*T_m_o/(pi*D_s_gerber(end)^3))^2)^(0.5);

sig_m = linspace(0, S_ut, 1000);

r = S_a_d/S_m_d;

ll = sig_m.*r;

sig_a_yield = S_y - sig_m;

sig_alt_d = (1-(sig_m./S_ut).^2)*Se_gerber_d(end);

sig_alt_D = (1-(sig_m./S_ut).^2)*Se_gerber_D(end);

figure

plot(sig_m, sig_alt_d,'r',sig_m, sig_alt_D,'--k', sig_m, ll, sig_m, sig_a_yield)

grid on

ylim([0 60e3]);

legend('Input Shaft','Output Shaft','Load Line','Yield Line')

title 'Gerber Fatigue Diagram'

xlabel 'Midrange Stress (psi)'

ylabel 'Alternating Stress (psi)'