Design of a Cantilever Beam for Fluctuating Bending

AIM: TO DESIGN A CANTILEVER BEAM FOR FLUCTUATING LOADING USING MATLAB/OCTAVE.

GIVEN:

A feed-roll assembly is to be mounted at its ends on support brackets cantilevered from the machine frame as shown in the Figure. The feed rolls experience a total fluctuating load that varies from a minimum of 200 lb. to a maximum of 2200 lb., split equally between the two support brackets.

The load-time function shape is shown in the Figure. The operating environment is room air at a maximum temperature of 120°F. The space available allows a maximum cantilever length of 6 in. Only ten of these parts are required.

OBJECTIVE:

• To design a cantilever bracket to support a fluctuating bending load of 100 lb. to 1100 lb. amplitude for 10⁹ cycles with no failure.
• To calculate the factor of safety.
• To ensure that Its dynamic deflection cannot exceed 0.02 in.
• To plot the S-N diagram.

OVERVIEW:

FATIGUE LOADS-

Any loads that vary with time can potentially cause fatigue failure. The character of these loads may vary substantially from one application to another. In rotating machinery, the loads tend to be consistent in amplitude over time and repeat with some frequency.

ENDURANCE LIMIT –

The fatigue strength S falls steadily and linearly (on log-log coordinates) as a function of N until reaching a knee at about 10⁶ to 10⁷ cycles. This knee defines an endurance limit Se' for the material, which is a stress level below which it can be cycled infinitely without failure

CANTILEVER BENDING TESTS –

If a cantilever beam is oscillated at its tip by a linkage mechanism, any combination of mean and alternating stresses as shown in Figure above.

Correction Factors to the Theoretical Fatigue Strength or Endurance Limit –

1.SIZE EFFECTS

The rotating-beam and static test specimens are small. If the part is larger than that dimension, a strength-reduction size factor needs to be applied to account for the fact that larger parts fail at lower stresses due to the higher probability of a flaw being present in the larger stressed volume.

2. SURFACE EFFECTS

The rotating-beam specimen is polished to a mirror finish to preclude surface imperfections serving as stress raisers. It is usually impractical to provide such an expensive finish on a real part. Rougher finishes will lower the fatigue strength by the introduction of stress concentrations and/or by altering the physical properties of the surface layer. A forged surface is both rough and decarburized and the reduced carbon levels weaken the surface where stresses are often highest. A strength-reduction surface factor C_surf is needed to account for these differences.

3. TEMPERATURE

Fatigue tests are most commonly done at room temperature. The fracture toughness decreases at low temperatures and increases at moderately high temperatures (up to about 350°C).

4. RELIABILITY

Many of the reported strength data are mean values. There is considerable scatter in multiple tests of the same material under the same test conditions. The standard deviations of endurance strengths of steels seldom exceed 8% of their mean values. Note that a 50% reliability has a factor of 1 and the factor reduces as you choose higher reliability. For example, if you wish to have 99.99% probability that your samples meet or exceed the assumed strength, multiply the mean strength value by 0.702.

5. LOADING EFFECTS

Based on discussion of axial and bending fatigue tests, we define a strength-reduction load factor C_load as

For bending: C_load = 1

For axial loading: C_load = 0.70

Lets solve the given problem:

This is a typical design problem. Very few data are given except for the loading on the device, some limitations on size, and the required cycle life. We will have to make some basic assumptions about part geometry, materials, and other factors as we go. Some iteration should be expected.

Assumptions:

The bracket can be clamped between essentially rigid plates bolted at its root. The normal load will be applied at the effective tip of the cantilever beam from a rod attached through a small hole in the beam. Since the bending moment is effectively zero at the beam tip, the stress concentration from this hole can be ignored. Given the small quantity required, machining of stock mill-shapes is the preferred manufacturing method.

• Let the mill stock is purchased thicker than the desired final dimension and machined top and bottom to dimension D, then machined to thickness d over the length l.
• A fillet radius r is provided at the clamp point to reduce fretting fatigue and achieve a lower Kt.
• With suitable control of the r/d and D/d ratios for a stepped flat bar in bending, the geometric stress-concentration factor Kt can be kept under about 1.5.

• Let choose SAE 1040 normalized carbon steel because of infinite life, low cost, and ease of fabrication. The fact that the deflection is of concern is also a good reason to choose a material with a large E. Low- to medium-carbon, ductile steels have the requisite endurance limit knee for the infinite life required in this case and also have low notch sensitivities.

S_ut = 90 Kpsi

S_y = 60 Kpsi

E = 90 Kpsi

• We will assume the trial dimensions to be b = 2 in, d = 1 in, D = 1.125 in, r = 0.5 in, a = 5 in, and l = 6.0 in.

BODY OF CONTENT:

Given that,

F_max = 1100 lb.

F_min = 100 lb.

CODE:

clear all
close all
clc

% Inputs

Sut = 80000;                    % ultimate tensile strength(psi)
Sy = 60000;                     % yield strength(psi)

Fmax = 1100;                    % maximum applied load (lb)
Fmin = 100;                     % minimum applied load (lb)

b = 2;                           % beam width(in)
d = 1;                           % beam depth over length (in)
D = 1.125;                       % beam depth in wall (in)
r = 0.5;                         % fillet radius (in)
a = 5;                           % beam length to load F (in)
l = 6;                           % length constraint (in)

% Calculation of "The mean and alternating components of the load" 
Fm = (Fmax+Fmin)/2;
Fa = (Fmax-Fmin)/2;

printf("\n The mean and alternating components of the load :")
fprintf('\n Fm is : %.2f lb',Fm)
fprintf('\n Fa is : %.2f lb',Fa)

% Calculation of "The reaction forces of the mean and alternating components of the load" 
Ra = Fa;
Rm = Fm;
Rmax = Fmax;

printf("\n\n The reaction forces of the mean and alternating components of the load :")
fprintf('\n Ra is : %.2f lb',Ra)
fprintf('\n Rm is : %.2f lb',Rm)

% Caltulating  "The mean and alternating moments, and the maximum moment(acting at the root of the cantilever beam)"
Ma = Ra*l-Fa*(l-a);
Mm = Rm*l-Fm*(l-a);
Mmax = Rmax*l-Fmax*(l-a);

printf("\n\n The mean and alternating moments, and the maximum moment:")
fprintf('\n Ma is : %.2f lb-in',Ma)
fprintf('\n Mm is : %.2f lb-in',Mm)
fprintf('\n Mmax is : %.2f lb-in',Mmax)

% Caltulating "The cross-section area moment of inertia and the distance to the outer fiber"
I = b*d^3/12;
c = d/2;

% Caltulation of "The nominal bending stresses[psi]"
Siganom = Ma*c/I;                  % nominal bending stress for alternating load
Sigmnom = Mm*c/I;                  % nominal bending stress for mean load
Sigmax = Mmax*c/I;                 % Maximum nominal bending stress

printf("\n \n The nominal bending stresses :")
fprintf('\n Siganom is : %.2f psi',Siganom)
fprintf('\n Sigmnom is : %.2f psi',Sigmnom)
fprintf('\n Sigmax is : %.2f psi',Sigmax)

% To find the geometric stress-concentration factor 
printf("\n \n Geometric stress-concentration factor :")
Kt = 1.012*(r/d)^-0.221



% To find the fatigue stress-concentration factor
printf("\n Neuber’s constant :")
Nu = 0.08                     % Neuber’s constant
printf("\n Fatigue stress-concentration factor :")
q = 1/(1+(Nu/r^(1/2)))        % The notch sensitivity
printf("\n Notch sensitivity :")
Kf = 1+q*(Kt-1)


if Kf*abs(Sigmax)<Sy;
    Kfm = Kf;
elseif   Kf*abs(Sigmax)>Sy;
    Kfm = (Sy-Kf*Siganom)/abs(Sigmnom);
elseif Kf*abs(Sigmax)>(2*Sy);
    Kfm = 0;
end;

% To find the local mean and alternating notch stresses
SigaA = Kf*Siganom;
SigmM = Kfm*Sigmnom;
Siga1 = SigaA;
Sigm1 = SigmM;

printf("\n \n The local mean and alternating notch stresses:")
fprintf('\n SigaA is : %.2f psi',SigaA)
fprintf('\n SigmM is : %.2f psi',SigmM)

% The uncorrected endurance limit 
if Sut<200000;
    Se1 = 0.5*Sut;
else Se1 = 100000;
end;

% The size factor 
printf("\n \n The size factor,load factor,surfacefinish factor,reliability factor,temperature factor:\n")
A95 = 0.05*d*b;
dequiv = (A95/0.0766)^(1/2);
Csize = 0.869*(dequiv)^(-.097)

% The corrected endurance limit 
Cload = 1                               % load factor
Csurf = 0.85                            % surface finish factor
Creliab = 0.753                         % reliability factor(99.9%)
Ctemp = 1                               % temperature factor


Se = Cload*Csize*Csurf*Ctemp*Creliab*Se1;
printf("\n \n The corrected endurance limit:")
fprintf('\n Se is : %.2f psi',Se)

% Factor of safety
printf("\n \n Factor of safety :\n")
Nf = (Se*Sut)/(Siga1*Sut+Sigm1*Se)

% The maximum deflection 
printf("\n \n The maximum deflection :\n")
E=30000*1000;                 % Elastic constant (psi)
ymax = (Fmax/(6*E*I))*[l^3 - 3*a*l^2 -(l-a)^3]


% To plot the "S-N curve"

N1=10^3;
N2=10^6;
N = 10^9;

Sm = 0.9*Sut;         % fatigue strength for N1 cycle
z = log10(N1)-log10(N2);          
p = log10(Sm/Se)/z;
q = 10^(log10(Sm)-3*p);
Sn = q*(N)^p;

% for low cycle fatigue
x = 0:0.001:3;
m = 0;                       % m is used to indicate slope
c = log10(Sm);
y = m*x+c;
plot(x,y,'linewidth',3,'color','g');
hold on

% for high cycle fatigue
x = 3:0.001:6;
m = p;
c = log10(q);
y = m*x+c;
plot(x,y,'linewidth',3,'color','b');

% fatigue limit
x = 6:0.001:9;
m = 0;
c = log10(Se);
y = m*x+c;
plot(x,y,'linewidth',3,'color','r');
xlim([0,9]);
ylim([4,5]);
grid on
title('S-N Diagram')
xlabel('Nt, number of cycles in logrithmic scale')
ylabel('Se, fatigue strength in logrithmic scale')
legend('low cycle fatigue','High cycle fatigue','Fatigue limit')

EXPLANATION OF CODE:

• Clear all, close all, clc command is used to clear all previous commands and plots.
• ‘fprintf’ command is used to print the values of all calculation parameters.
• The mean and alternating components of the load are calculated with conceptual formulas.
• Then the reaction forces of the mean and alternating components of the load are calculated.
• Then the mean and alternating moments, and the maximum moment are calculated with conceptual formulas.
• The nominal bending stresses are calculated.
• For the calculation of the geometric stress-concentration factor the values (1.012 and -0.221) are taken by interpolation of the datas given in the above figure in the overview section.
• ‘if-else’ is used to provide the conditional statement.
• The size factor, load factor ,surface finish factor, reliability factor ,temperature factor are used to find the Actual endurance strength.
• ‘plot’ command is used to plot the S-N curve.
• ‘xlim’ and ‘ylim’ is to limit the axis values of x and y respectively.
• ‘title’ is to entitle the graph.
• ‘xlabel’ and ‘ylabel’ is for labeling the axis.
• ‘legend’ command is to clarify the all lines in the plot.

RESULTS:

PARAMETERS -

The factor of safety is approx '1.9' and the maximum deflection is also in the safe limit.

TO IMPROVE THE FACTOR OF SAFETY:

We will assume the trial dimensions to be b = 2 in, d = 1.2 in, D = 1.4 in, r = 0.5 in, a = 5 in, and l = 6.0 in.

CODE:

clear all
close all
clc

% Inputs

Sut = 80000;                    % ultimate tensile strength(psi)
Sy = 60000;                     % yield strength(psi)

Fmax = 1100;                    % maximum applied load (lb)
Fmin = 100;                     % minimum applied load (lb)

b = 2;                           % beam width(in)
d = 1.2;                           % beam depth over length (in)
D = 1.4;                       % beam depth in wall (in)
r = 0.5;                         % fillet radius (in)
a = 5;                           % beam length to load F (in)
l = 6;                           % length constraint (in)

% Calculation of "The mean and alternating components of the load" 
Fm = (Fmax+Fmin)/2;
Fa = (Fmax-Fmin)/2;

printf("\n The mean and alternating components of the load :")
fprintf('\n Fm is : %.2f lb',Fm)
fprintf('\n Fa is : %.2f lb',Fa)

% Calculation of "The reaction forces of the mean and alternating components of the load" 
Ra = Fa;
Rm = Fm;
Rmax = Fmax;

printf("\n\n The reaction forces of the mean and alternating components of the load :")
fprintf('\n Ra is : %.2f lb',Ra)
fprintf('\n Rm is : %.2f lb',Rm)

% Caltulating  "The mean and alternating moments, and the maximum moment(acting at the root of the cantilever beam)"
Ma = Ra*l-Fa*(l-a);
Mm = Rm*l-Fm*(l-a);
Mmax = Rmax*l-Fmax*(l-a);

printf("\n\n The mean and alternating moments, and the maximum moment:")
fprintf('\n Ma is : %.2f lb-in',Ma)
fprintf('\n Mm is : %.2f lb-in',Mm)
fprintf('\n Mmax is : %.2f lb-in',Mmax)

% Caltulating "The cross-section area moment of inertia and the distance to the outer fiber"
I = b*d^3/12;
c = d/2;

% Caltulation of "The nominal bending stresses[psi]"
Siganom = Ma*c/I;                  % nominal bending stress for alternating load
Sigmnom = Mm*c/I;                  % nominal bending stress for mean load
Sigmax = Mmax*c/I;                 % Maximum nominal bending stress

printf("\n \n The nominal bending stresses :")
fprintf('\n Siganom is : %.2f psi',Siganom)
fprintf('\n Sigmnom is : %.2f psi',Sigmnom)
fprintf('\n Sigmax is : %.2f psi',Sigmax)

% To find the geometric stress-concentration factor 
printf("\n \n Geometric stress-concentration factor :")
Kt = 1.012*(r/d)^-0.221



% To find the fatigue stress-concentration factor
printf("\n Neuber’s constant :")
Nu = 0.08                     % Neuber’s constant
printf("\n Fatigue stress-concentration factor :")
q = 1/(1+(Nu/r^(1/2)))        % The notch sensitivity
printf("\n Notch sensitivity :")
Kf = 1+q*(Kt-1)


if Kf*abs(Sigmax)<Sy;
    Kfm = Kf;
elseif   Kf*abs(Sigmax)>Sy;
    Kfm = (Sy-Kf*Siganom)/abs(Sigmnom);
elseif Kf*abs(Sigmax)>(2*Sy);
    Kfm = 0;
end;

% To find the local mean and alternating notch stresses
SigaA = Kf*Siganom;
SigmM = Kfm*Sigmnom;
Siga1 = SigaA;
Sigm1 = SigmM;

printf("\n \n The local mean and alternating notch stresses:")
fprintf('\n SigaA is : %.2f psi',SigaA)
fprintf('\n SigmM is : %.2f psi',SigmM)

% The uncorrected endurance limit 
if Sut<200000;
    Se1 = 0.5*Sut;
else Se1 = 100000;
end;

% The size factor 
printf("\n \n The size factor,load factor,surfacefinish factor,reliability factor,temperature factor:\n")
A95 = 0.05*d*b;
dequiv = (A95/0.0766)^(1/2);
Csize = 0.869*(dequiv)^(-.097)

% The corrected endurance limit 
Cload = 1                               % load factor
Csurf = 0.85                            % surface finish factor
Creliab = 0.753                         % reliability factor(99.9%)
Ctemp = 1                               % temperature factor


Se = Cload*Csize*Csurf*Ctemp*Creliab*Se1;
printf("\n \n The corrected endurance limit:")
fprintf('\n Se is : %.2f psi',Se)

% Factor of safety
printf("\n \n Factor of safety :\n")
Nf = (Se*Sut)/(Siga1*Sut+Sigm1*Se)

% The maximum deflection 
printf("\n \n The maximum deflection :\n")
E=30000*1000;                 % Elastic constant (psi)
ymax = (Fmax/(6*E*I))*[l^3 - 3*a*l^2 -(l-a)^3]


% To plot the "S-N curve"

N1=10^3;
N2=10^6;
N = 10^9;

Sm = 0.9*Sut;         % fatigue strength for N1 cycle
z = log10(N1)-log10(N2);          
p = log10(Sm/Se)/z;
q = 10^(log10(Sm)-3*p);
Sn = q*(N)^p;

% for low cycle fatigue
x = 0:0.001:3;
m = 0;
c = log10(Sm);
y = m*x+c;
plot(x,y,'linewidth',3,'color','g');
hold on

% for high cycle fatigue
x = 3:0.001:6;
m = p;
c = log10(q);
y = m*x+c;
plot(x,y,'linewidth',3,'color','b');

% fatigue limit
x = 6:0.001:9;
m = 0;
c = log10(Se);
y = m*x+c;
plot(x,y,'linewidth',3,'color','r');
xlim([0,9]);
ylim([4,5]);
grid on
title('S-N Diagram')
xlabel('Nt, number of cycles in logrithmic scale')
ylabel('Se, fatigue strength in logrithmic scale')
legend('low cycle fatigue','High cycle fatigue','Fatigue limit')

IMPROVED RESULTS:

PARAMETERS -

Now The factor of safety is approx '2.62' and Maximum deflection is far less then the limiting criteria.

PLOT -

ERROR:

Error as 'EI undefined' because of '*' was missing in the code between 'E' and 'I'

REFERENCES:

'Machine design book by Robert L Norton' for the Engineering concepts.