Design of Machine Elements Test 1

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Design of Machine Elements Test 1

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Question 1
1 / -0

A shaft rotating at constant speed is subjected to an axial tensile stress of 20 kN/mm² and bending stress of 30 kN/mm². The stress ratio is

A
- 0.2

B
- 0.3

C
0.35

D
0.25

Solution

Explanation:
Given:
\({\sigma _{axial}} = 20\;{\rm{kN}}/{\rm{m}}{{\rm{m}}^2}\;,\;\;{\sigma _{bending}} = 30\;{\rm{kN}}/{\rm{m}}{{\rm{m}}^2}\)
Now,
σ_max = σ_axial + σ_bending = 20 + 30 = 50 kN/mm²
σ_min = σ_axial - σ_bending = 20 – 30 = -10 kN/mm²
\({\rm{Stress\;ratio}} = \frac{{{\sigma _{min}}}}{{{\sigma _{max}}}} = \frac{{ - 10}}{{50}}\)
∴ Stress ratio = - 0.2
Question 2
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A steel component has an endurance limit of 450 MPa. The theoretical stress concentration factor and notch sensitivity of the material is 2.51 and 0.8 respectively. The corrected endurance limit in MPa is __________

A
203-204

B
2034-2041

C
2037-2046

D
2039-2042

Solution

Concept:
The notch sensitivity factor is given by:
\(q = \frac{{{K_f} - 1}}{{{K_t} - 1}} \Rightarrow {K_f} = 1 + q\left( {{K_t} - 1} \right)\;where,\;{K_f} = fatigue\;stress\;conc.\;factor\)
The modifying factor to account for stress concentration \({K_d} = \frac{1}{{{K_f}}}\)
The corrected endurance limit is: \({S_e}' = {S_e}{K_d} = \frac{{{S_e}}}{{{K_f}}}\)
Calculation:
\({K_f} = 1 + q\left( {{K_t} - 1} \right) = 1 + 0.8\left( {2.51 - 1} \right) = 2.208\)
\({S_e}' = \frac{{{S_e}}}{{{K_f}}} = \frac{{450}}{{2.208}} = 203.8\;MPa\)

Question 3

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For a particular component undergoing fatigue testing, the following data is obtained:

Corrected endurance limit = 400 MPa

Maximum stress = 100 MPa, Minimum stress = 0

As per the Gerber’s relation, the ultimate tensile strength of the material is:

53.5 MPa

46.9 MPa

38.4 MPa

40.2 MPa

Solution

Concept:

Soderberg

Criterion

\(\frac{{{σ _m}}}{{{σ _{yt}}}} + \frac{{{σ _a}}}{{{σ _e}}} = \frac{1}{{FS}}\)

Goodman

Criterion

\(\frac{{{σ _m}}}{{{σ _{ut}}}} + \frac{{{σ _a}}}{{{σ _e}}} = \frac{1}{{FS}}\)

Gerber

Criterion

\({\left( {\frac{{{σ _m}}}{{{σ _{ut}}}}} \right)^2}FS + \frac{{{σ _a}}}{{{σ _e}}} = \frac{1}{{FS}}\)

The Gerber’s relation for fatigue loading is: (Take FS = 1)

\(\frac{{{S_a}}}{{{S_e}}} + {\left( {\frac{{{S_m}}}{{{S_{ut}}}}} \right)^2} = 1\)

\({σ _a} = \frac{{{σ _{max}} - {σ _{min}}}}{2}\)

\({σ _m} = \frac{{{σ _{max}}+ {σ _{min}}}}{2}\)

Calculation:

Given: σ_max = 100 MPa, σmin = 0, σe = 400 MPa

Amplitude stress:

\({σ _a} = \frac{{100}}{2} = 50\;MPa\)

Mean stress:

\({σ _m} = \frac{{100}}{2} = 50\;MPa\)

Putting the above values:

\(\frac{{50}}{{400}} + {\left( {\frac{{50}}{{{S_{ut}}}}} \right)^2} = 1 \Rightarrow {S_{ut}} = 53.452\;MPa\)

Question 4
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A machine component is subjected to two-dimensional stresses. The tensile stress in y-direction varies from 20 to 60 N/mm² while in x-direction varies from 40 to 90 N/mm². The endurance limit is 270 N/mm². Calculate the mean and amplitude stress (in N/mm²)

A
σ_m = 61.32, σ_a = 25.23

B
σ_m = 56.79, σ_a = 22.91

C
σ_m = 56.79, σ_a = 25.23

D
σ_m = 61.32, σ_a = 22.91

Solution

Explanation:
Given:
(σ_y)_min = 20 N/mm², (σ_y)_max = 60 N/mm², (σ_x)_min = 40 N/mm², (σ_x)_max = 90 N/mm²
Mean and amplitude stress in x-direction
\({\sigma _{xm}} = \frac{{{{\left( {{\sigma _x}} \right)}_{max}} + {{\left( {{\sigma _x}} \right)}_{min}}}}{2} = \frac{{90 + 40}}{2} = 65\;N/m{m^2}\)
\({\sigma _{xa}} = \frac{{{{\left( {{\sigma _x}} \right)}_{max}} - {{\left( {{\sigma _x}} \right)}_{min}}}}{2} = \frac{{90 - 40}}{2} = 25\;N/m{m^2}\)
Similarly in y-direction,
\({\sigma _{ym}} = \frac{{{{\left( {{\sigma _y}} \right)}_{max}} + {{\left( {{\sigma _y}} \right)}_{min}}}}{2} = \frac{{60 + 20}}{2} = 40\;N/m{m^2}\)
\({\sigma _{ya}} = \frac{{{{\left( {{\sigma _y}} \right)}_{max}} - {{\left( {{\sigma _y}} \right)}_{min}}}}{2} = \frac{{60 - 20}}{2} = 20\;N/m{m^2}\)
Now,
Mean stress,
\({\sigma _m} = \sqrt {\sigma _{xm}^2 - {\sigma _{xm}}\;{\sigma _{ym}} + \sigma _{ym}^2} \)
\({\sigma _m} = \sqrt {{{65}^2} - \left( {65 \times 40} \right) + {{40}^2}} \)
∴ σ_m = 56.789 N/mm²
Now,
\({\rm{Amplitude\;stress\;}}\left( {{\sigma _a}} \right) = \sqrt {\sigma _{xa}^2 - {\sigma _{xa}} \cdot {\sigma _{ya}} + \sigma _{ya}^2} \)
\({\sigma _a} = \sqrt {{{25}^2} - \left( {25 \times 20} \right) + {{20}^2}} \)
∴ σ_a = 22.912 N/mm²