Contents

CASE STUDY 7B

% Programmer: Aaron Klapheck
% 26-Nov-08
clear, clc, home
fprintf('The date and time: %s \n \n', datestr(now))

fprintf('CorrectedStrengthEnglishSh function: \n \n')
help CorrectedStrengthEnglishSh
fprintf('\n')
The date and time: 29-Dec-2008 12:46:24 
 
CorrectedStrengthEnglishSh function: 
 
  Calculates correction factors and uses these vaues to find the corrected
  strength (yeild or endurance). Shows C values.
 
  Synopsis: [S] = CorrectedStrengthSh(load, size, surf, temp, reliab,
  S_prime)
 
  input:    load = (required) 'bending' or 'axial'. Must be a string. Use
                    bending case for all cases that are non-axial.
            size = (required) Must be a structured array. size must contain
                    size.shape: 'rectangle' or 'circle'
                    size.rotation: 'nonrotating' or 'rotating'
                    size.dim: 1x1 (radius of circle) or 1x2 (height - the
                    cubed length, and width); must be in inches.
            surf = (required) 'ground' or 'machined' or 'hot-rolled' or
                    'as-forged' or 1.
            temp = (required) 0 to 1020 degrees F
            reliab = (required) 50, 90, 99, 99.9, 99.99, or 
                    99.999 percent
            mat = (required) 'steel', 'iron', or 'aluminum'. Must be a
                   sting.
            S_ut = (required) The ultamate tensile strength. Must be in
                    kpsi.
 
  output:   S = corrected strength value in kpsi.
 
  Note: For the following variables 1 can be assigned to them:
        load, surf, temp, and reliab. When this occurs the C value for
        theses variables is simply made equal to one.
 
 
  All values must be constants, no matricies allowed. Use the following to
  get around this rule though:
  
  use matix for the size, mat, or other variable, for example:
 
  Reliability = [50, 90, 99];
  len = length(Reliability);
  S = ones(1,len);
  for coun = 1:len
    reliab = Reliability(coun);
    S(1,coun) = CorrectedStrengthEnglish(load, size, surf, temp, reliab, mat, S_ut);
  end


Given and Assumes

% given
fprintf('----------------------- \n \n')
fprintf('----- Given Data: ----- \n \n')
fprintf('----------------------- \n \n')
fprintf('d_p, diameter of pinion (in)')
d_p = 4            % in
fprintf('r_g, radius of the gear (in). ')
r_g = 5.000        % in
fprintf('T_max, maximum Torque (lb*in)')
T_max = 800        % lb*in
fprintf('T_min, minimum Torque (lb*in)')
T_min = -250       % lb*in
fprintf('Sut, ultamate tensile strength (psi)')
Sut = 64*1000      % psi
S_ut = 64;          % kpsi
fprintf('Sut, yield strength (kpsi)')
Sy = 54            % kpsi
fprintf('Nsf, safety factor')
Nsf = 3

fprintf('PA, pressure angle (degrees)')
PA = 20        % degrees. Pressure angle. For all three gears
AGMA = 'Full-Depth'; % For all three gears.
Loading = 'HPSTC'; % Loading conditions for all three gears.

Phi = PA*(pi/180);  % rad


% Correction factors
fprintf('load, type of load applied')
load = 'bending'
fprintf('o.shape, shape of cross section for output shaft \n')
size_o.shape = 'circle';
fprintf('o.rotation, rotate value for output shaft \n')
size_o.rotation = 'rotating';
fprintf('o.dim, radius dimension for output shaft (in)')
size_o.dim = 1.1423 % in
fprintf('i.shape, shape of cross section for input shaft \n')
size_i.shape = 'circle';
fprintf('i.rotation, rotation value for input shaft \n')
size_i.rotation = 'rotating';
fprintf('i.dim, radius dimension for input shaft (in)')
size_i.dim = 0.8686 % in
fprintf('surf, surface condition')
surf = 'machined'
fprintf('temp, temperature (degrees F)')
temp = 70          % degrees F
fprintf('reliab, reliability (percent)')
reliab = 50        % percent
fprintf('mat, material used')
mat = 'steel'      % SAE 1018

% Assumptions
fprintf('Kt, goemetric stress consentration facter - assumed. ')
Kt = 3
----------------------- 
 
----- Given Data: ----- 
 
----------------------- 
 
d_p, diameter of pinion (in)
d_p =

     4

r_g, radius of the gear (in). 
r_g =

     5

T_max, maximum Torque (lb*in)
T_max =

   800

T_min, minimum Torque (lb*in)
T_min =

  -250

Sut, ultamate tensile strength (psi)
Sut =

       64000

Sut, yield strength (kpsi)
Sy =

    54

Nsf, safety factor
Nsf =

     3

PA, pressure angle (degrees)
PA =

    20

load, type of load applied
load =

bending

o.shape, shape of cross section for output shaft 
o.rotation, rotate value for output shaft 
o.dim, radius dimension for output shaft (in)
size_o = 

       shape: 'circle'
    rotation: 'rotating'
         dim: 1.1423

i.shape, shape of cross section for input shaft 
i.rotation, rotation value for input shaft 
i.dim, radius dimension for input shaft (in)
size_i = 

       shape: 'circle'
    rotation: 'rotating'
         dim: 0.8686

surf, surface condition
surf =

machined

temp, temperature (degrees F)
temp =

    70

reliab, reliability (percent)
reliab =

    50

mat, material used
mat =

steel

Kt, goemetric stress consentration facter - assumed. 
Kt =

     3

Solution

% Tangential Forces
fprintf('---------------------------------------------- \n \n')
fprintf('----- Calculation of Tangential Forces: ------ \n \n')
fprintf('---------------------------------------------- \n \n')
fprintf('Ft_max = T_max / r_g')
Ft_max = T_max / r_g   % lbf
fprintf('Ft_min = T_min / r_g')
Ft_min = T_min / r_g   % lbf


% Total Forces
fprintf('----------------------------------------- \n \n')
fprintf('----- Calculation of Total Forces: ------ \n \n')
fprintf('----------------------------------------- \n \n')
fprintf('F_max = Ft_max / cos(Phi)')
F_max = Ft_max / cos(Phi) % lbf
fprintf('F_min = Ft_min / cos(Phi)')
F_min = Ft_min / cos(Phi) % lbf


% Moments
fprintf('------------------------------------ \n \n')
fprintf('----- Calculation of Moments: ------ \n \n')
fprintf('------------------------------------ \n \n')
fprintf('M_max = F_max * (d_p/4)')
M_max = F_max * (d_p/4) % lbf in
fprintf('M_min = F_min * (d_p/4)')
M_min = F_min * (d_p/4) % lbf in


% Mean/Aplitude Moments
fprintf('-------------------------------------------------- \n \n')
fprintf('----- Calculation of Mean/Aplitude Moments: ------ \n \n')
fprintf('-------------------------------------------------- \n \n')
fprintf('Mm = (1/2) * (M_max + M_min)')
Mm = (1/2) * (M_max + M_min)  % lbf in
fprintf('Ma = (1/2) * (M_max - M_min)')
Ma = (1/2) * (M_max - M_min)  % lbf in


% Mean/Aplitude Torques
fprintf('-------------------------------------------------- \n \n')
fprintf('----- Calculation of Mean/Aplitude Torques: ------ \n \n')
fprintf('-------------------------------------------------- \n \n')
fprintf('Tm = (1/2) * (T_max + T_min)')
Tm = (1/2) * (T_max + T_min)  % lbf in
fprintf('Ta = (1/2) * (T_max - T_min)')
Ta = (1/2) * (T_max - T_min)  % lbf in


fprintf('--------------------------------------------------- \n \n')
fprintf('----- C factores and strength of output shaft ----- \n \n')
fprintf('--------------------------------------------------- \n \n')
% Output Endurance Limit
Se_o = CorrectedStrengthEnglishSh(load, size_o, surf, temp, reliab, mat, S_ut);

fprintf('-------------------------------------------------- \n \n')
fprintf('----- C factores and strength of input shaft ----- \n \n')
fprintf('-------------------------------------------------- \n \n')
% Input Endurance Limit
Se_i = CorrectedStrengthEnglishSh(load, size_i, surf, temp, reliab, mat, S_ut);


fprintf('------------------------------- \n \n')
fprintf('----- Corrected strengths ----- \n \n')
fprintf('------------------------------- \n \n')
fprintf('Se_i, Corrected strength for input shaft (psi)')
Se_i = Se_i.*1000 % psi
fprintf('Se_o, Corrected strength for output shaft (psi)')
Se_o = Se_o.*1000 % psi


% Notch Sensitivity
fprintf('------------------------------- \n \n')
fprintf('----- Notch Sensitivities ----- \n \n')
fprintf('------------------------------- \n \n')

fprintf('r, notch radius - assumed.')
r = 0.01   % in

fprintf('a_b, Neubers constant for bending stress (in).')
a_b = 0.10^2  % in
fprintf('q_b, Notch Sensitivity for bending stress in keyway. \n')
fprintf('q_b = 1 / (1 + sqrt(a_b/r))')
q_b = 1 / (1 + sqrt(a_b/r))

fprintf('a_t, Neubers constant for torsion stress (in).')
a_t = 0.075^2  % in
fprintf('q_t, Notch Sensitivity for torsion stress in keyway. \n')
fprintf('q_t = 1 / (1 + sqrt(a_t/r))')
q_t = 1 / (1 + sqrt(a_t/r))


% Fatigue Stress Consentration Factors
fprintf('------------------------------------------------ \n \n')
fprintf('----- Fatigue Stress Consentration Factors ----- \n \n')
fprintf('------------------------------------------------ \n \n')
fprintf('Kf, Fatigue Stress Consentration Factors for bending stress in keyway. \n')
fprintf('Kf = 1 + q_b*(Kt - 1)')
Kf = 1 + q_b*(Kt - 1)

fprintf('Kfs, Fatigue Stress Consentration Factors for tortion stress in keyway. \n')
fprintf('Kfs = 1 + q_t*(Kt - 1)')
Kfs = 1 + q_t*(Kt - 1)

fprintf('Kfm = Kf \n')
Kfm = Kf;
fprintf('Kfsm = Kfs \n \n')
Kfsm = Kfs;


fprintf('--------------------------------- \n \n')
fprintf('----- Diameter Calculations ----- \n \n')
fprintf('--------------------------------- \n \n')
% Output Shaft Diameter
fprintf('d_output, output diameter (in)')
d_output = (((32.*Nsf)./pi).* ...
    ((sqrt((Kf.*Ma).^2 + (3/4).*(Kfs.*Ta).^2)./Se_o) + ...
    (sqrt((Kfm.*Mm).^2 + (3/4).*(Kfsm.*Tm).^2)./Sut))).^(1/3) % in


fprintf('Tm_in, mean tourque for inner shaft (lbf*in) \n')
fprintf('Tm_in = 0.4*Tm')
Tm_in = 0.4*Tm     % lbf*in
fprintf('Ta_in, alternating tourque for inner shaft (lbf*in) \n')
fprintf('Ta_in = 0.4*Ta')
Ta_in = 0.4*Ta     % lbf*in

% Input Shaft Diameter
fprintf('d_input, input diameter (in)')
d_input = (((32.*Nsf)./pi).* ...
    ((sqrt((Kf.*Ma).^2 + (3/4).*(Kfs.*Ta_in).^2)./Se_i) + ...
    (sqrt((Kfm.*Mm).^2 + (3/4).*(Kfsm.*Tm_in).^2)./Sut))).^(1/3) % in
---------------------------------------------- 
 
----- Calculation of Tangential Forces: ------ 
 
---------------------------------------------- 
 
Ft_max = T_max / r_g
Ft_max =

   160

Ft_min = T_min / r_g
Ft_min =

   -50

----------------------------------------- 
 
----- Calculation of Total Forces: ------ 
 
----------------------------------------- 
 
F_max = Ft_max / cos(Phi)
F_max =

  170.2684

F_min = Ft_min / cos(Phi)
F_min =

  -53.2089

------------------------------------ 
 
----- Calculation of Moments: ------ 
 
------------------------------------ 
 
M_max = F_max * (d_p/4)
M_max =

  170.2684

M_min = F_min * (d_p/4)
M_min =

  -53.2089

-------------------------------------------------- 
 
----- Calculation of Mean/Aplitude Moments: ------ 
 
-------------------------------------------------- 
 
Mm = (1/2) * (M_max + M_min)
Mm =

   58.5298

Ma = (1/2) * (M_max - M_min)
Ma =

  111.7387

-------------------------------------------------- 
 
----- Calculation of Mean/Aplitude Torques: ------ 
 
-------------------------------------------------- 
 
Tm = (1/2) * (T_max + T_min)
Tm =

   275

Ta = (1/2) * (T_max - T_min)
Ta =

   525

--------------------------------------------------- 
 
----- C factores and strength of output shaft ----- 
 
--------------------------------------------------- 
 

C_load =

     1


C_size =

    0.8579


C_surf =

    0.8969


C_temp =

     1


C_reliab =

     1


S_prime =

    32

-------------------------------------------------- 
 
----- C factores and strength of input shaft ----- 
 
-------------------------------------------------- 
 

C_load =

     1


C_size =

    0.8810


C_surf =

    0.8969


C_temp =

     1


C_reliab =

     1


S_prime =

    32

------------------------------- 
 
----- Corrected strengths ----- 
 
------------------------------- 
 
Se_i, Corrected strength for input shaft (psi)
Se_i =

  2.5283e+004

Se_o, Corrected strength for output shaft (psi)
Se_o =

  2.4620e+004

------------------------------- 
 
----- Notch Sensitivities ----- 
 
------------------------------- 
 
r, notch radius - assumed.
r =

    0.0100

a_b, Neubers constant for bending stress (in).
a_b =

    0.0100

q_b, Notch Sensitivity for bending stress in keyway. 
q_b = 1 / (1 + sqrt(a_b/r))
q_b =

    0.5000

a_t, Neubers constant for torsion stress (in).
a_t =

    0.0056

q_t, Notch Sensitivity for torsion stress in keyway. 
q_t = 1 / (1 + sqrt(a_t/r))
q_t =

    0.5714

------------------------------------------------ 
 
----- Fatigue Stress Consentration Factors ----- 
 
------------------------------------------------ 
 
Kf, Fatigue Stress Consentration Factors for bending stress in keyway. 
Kf = 1 + q_b*(Kt - 1)
Kf =

     2

Kfs, Fatigue Stress Consentration Factors for tortion stress in keyway. 
Kfs = 1 + q_t*(Kt - 1)
Kfs =

    2.1429

Kfm = Kf 
Kfsm = Kfs 
 
--------------------------------- 
 
----- Diameter Calculations ----- 
 
--------------------------------- 
 
d_output, output diameter (in)
d_output =

    1.1423

Tm_in, mean tourque for inner shaft (lbf*in) 
Tm_in = 0.4*Tm
Tm_in =

   110

Ta_in, alternating tourque for inner shaft (lbf*in) 
Ta_in = 0.4*Ta
Ta_in =

   210

d_input, input diameter (in)
d_input =

    0.8686