Question

Sut = 80 ksi, Sy= 60 ksi, Si = 40 ksi, q=0.85 (notch sensitivity), steel E = 30e6 psi Remember we use Modified-Goodman for all calculations A 1-inch diameter rod that is 8 inches long has a bearing at one end and another bearing at 6 inches (treat bearings as simple supports). A vertical load is applied at the end of the overhanging rod and a torque is transmitted through the length of the rod. (Fmax = 100 lbf Tmax = 50 ft-lbf) Calculate the fatigue factor of safety for these cases. a) Force constant Fmax, Torque varies from Tmax to O b) Torque constant Tmax, Force varies from Fmax to O c) Torque constant Tmax, Force constant at Fmax but shaft is rotating so normal stress is fully reversed

Answer 1

Given data,

$\small \\\\S_{ut}=80 \ ksi \\\\S_{f}=40 \ ksi \\\\S_{y}=60 \ ksi \\\\q= 0.85 \\\\E=30*10^6 \ psi \\\\d= 1 \ inch \\\\L= 8 \ inch \\\\F_{max}=100 \ lbf \\\\T_{max}= 50 \ ft-lbf=600 \ lbf-in$

Fmax 2 in 6 in Ra Rb

So maximum moment is

$\small \\\\M=F_{max}{2} \\\\M=100*2=200 \ lbf-in$

Bending and shear stress

$\small \\\\\sigma_b=\frac{32M}{\pi d^3}=\frac{32*200}{\pi *1^3}=2038.21 \ PSI \\\\\tau_b=\frac{16T}{\pi d^3}=\frac{16*600}{\pi *1^3}=3057 \ PSI$

a)

01 huom+VO + 72 2 01 = 2038 2 + 2038 2 + 30572 01 = 4241 PSI

$\small \\\\\sigma_{2}=\frac{\sigma_b}{2}+ \sqrt{\left (\frac{\sigma_b}{2} \right )^2+\tau^2} \\\\\sigma_{2}=\frac{2038}{2}+ \sqrt{\left (\frac{2038}{2} \right )^2+0^2} \\\\\sigma_2=2038 \ PSI$

$\small \\\\\sigma_m=\frac{\sigma_1+\sigma_2}{2}=\frac{4241.8+2038}{2}=3140 \ PSI \\\\\sigma_a=\frac{\sigma_1-\sigma_2}{2}=\frac{4241.8-2038}{2}=1101 \ PSI \\\\\text{Now from good max theory} \\\\\frac{\sigma_m}{\sigma_{ut}}+\frac{\sigma_a}{\sigma_f}=\frac{1}{f} \\\\\frac{3140}{80*10^3}+\frac{1101}{40*10^3}=\frac{1}{f} \\\\f=14.97$

b)

For Tmax & Fmax

$\small \\\\\sigma_1=4241 \ psi$

For Tmax & Fmax=0

$\small \\\\\sigma_2=\tau=3057 \ psi$

$\small \\\\\sigma_m=\frac{\sigma_1+\sigma_2}{2}=\frac{4241.8+3057}{2}=3649 \ PSI \\\\\sigma_a=\frac{\sigma_1-\sigma_2}{2}=\frac{4241.8-3057}{2}=592.24\ PSI \\\\\text{Now from good max theory} \\\\\frac{\sigma_m}{\sigma_{ut}}+\frac{\sigma_a}{\sigma_f}=\frac{1}{f} \\\\\frac{3649}{80*10^3}+\frac{592.24}{40*10^3}=\frac{1}{f} \\\\f=16.54$

c)Shaft is rotating and normal stress is fully reversed so ,

$\small \\\\\sigma_{1,2}=\frac{\sigma_b}{2}\pm \sqrt{\left (\frac{\sigma_b}{2} \right )^2+\tau^2} \\\\\sigma_{1,2}=\frac{2038}{2}\pm \sqrt{\left (\frac{2038}{2} \right )^2+3057^2} \\\\\sigma_1=4241 \ PSI \\\\\sigma_2=-2203 \ PSI$

$\small \\\\\sigma_m=\frac{\sigma_1+\sigma_2}{2}=\frac{4241.8-2203}{2}=1019 \ PSI \\\\\sigma_a=\frac{\sigma_1-\sigma_2}{2}=\frac{4241.8+2203}{2}=3222\ PSI \\\\\text{Now from good max theory} \\\\\frac{\sigma_m}{\sigma_{ut}}+\frac{\sigma_a}{\sigma_f}=\frac{1}{f} \\\\\frac{1019}{80*10^3}+\frac{3222}{40*10^3}=\frac{1}{f} \\\\f=10.71$

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