Soderberg Curve MCQ Quiz in मल्याळम - Objective Question with Answer for Soderberg Curve - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Soderberg Curve ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Soderberg Curve MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Soderberg Curve MCQ Objective Questions

Top Soderberg Curve MCQ Objective Questions

Soderberg Curve Question 1:

A machine component satisfies the following inequality for safe operations where\(% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX! \frac{{{\sigma _{mean}}}}{{360}} + \frac{{{\sigma _a}}}{{210}} \leqslant 1% MathType!End!2!1!\) σmean is the average or mean stress in MPa,  σa is the amplitude in MPa. If the machine member is subjected to stress σ = 120 + k sin (20 t + 0.7).  The ultimate tensile strength and yield strength of the material are 360 MPa and 290 MPa respectively. The endurance limit stress is 210 MPa. For safe operation of the machine component, the maximum value of k (in MPa) is_______. ( Use Goodman Criteria)

  1. 70
  2. 140
  3. 280
  4. 320

Answer (Detailed Solution Below)

Option 2 : 140

Soderberg Curve Question 1 Detailed Solution

Concept:

 Soderberg 

 Criterion

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

 Goodman

 Criterion

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

 Gerber

 Criterion

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

Goodman Criterion:

\(\frac{{{\sigma _m}}}{{{\sigma _{ut}}}} + \frac{{{\sigma _a}}}{{{\sigma _e}}} \le 1\)

Calculation:

σ = 120 + k sin (20 t + 0.7)

σmax = 120 + K (when sin (20t +0.7) = 1) 

σmin = 120 – K (when sin (20t +0.7) = – 1) 

\(\therefore {\sigma _{mean}} = \frac{{{\sigma _{max}} + {\sigma _{min}}}}{2} = \frac{{120 + K + 120 - K}}{2} = 120\;MPa\)

\({\sigma _a} = \frac{{{\sigma _{max}} - {\sigma _{min}}}}{2} = \frac{{120 + K - \left( {120 - K} \right)}}{2} = K\)

So for safe design

\(\frac{{120}}{{360}} + \frac{K}{{210}} \le 1\)

⇒ K ≤ 140

Soderberg Curve Question 2:

In the design of mechanical components for dynamic loading, which expression correctly represents Goodman line for assessing fatigue failure?

Where Kf = Fatigue stress concertation factor

Se = modified endurance limit

Syt = Yield strength

Sut = Ultimate strength in tension

σm = mean stress

σa = alternating stress

  1. \(\frac{K_fn_s\sigma_a}{S_e}+\left(\frac{n_s\sigma_m}{S_{ut}}\right)=1\)
  2. \(\frac{K_f\sigma_a}{S_e}+\frac{\sigma_m}{S_{yt}}=\frac{\sigma_m}{n_s}\)
  3. \(\frac{K_f\sigma_a}{S_e}+\left(\frac{n_s\sigma_m}{S_{ut}}\right)^2=\frac{1}{n_s}\)
  4. \(\frac{K_f\sigma_a}{S_e}+\frac{\sigma_m}{S_{ut}}=\frac{1}{n_s}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{K_f\sigma_a}{S_e}+\frac{\sigma_m}{S_{ut}}=\frac{1}{n_s}\)

Soderberg Curve Question 2 Detailed Solution

Explanation:

When designing mechanical components that are subjected to dynamic loading, it is crucial to assess the risk of fatigue failure. Fatigue failure occurs due to the repeated application of loads, which can cause a material to fail at stress levels lower than its ultimate tensile strength. One of the methods used to assess fatigue failure is the Goodman line, which provides a criterion for safe design based on the mean and alternating stresses experienced by the component.

The Goodman line is an empirical relationship that helps engineers determine whether a component will fail under a given set of loading conditions. It is particularly useful for components subjected to cyclic loading, where both mean stress (\(\sigma_m\)) and alternating stress (\(\sigma_a\)) are present.

The Goodman line can be represented mathematically by an equation that relates the mean stress, alternating stress, and the material properties of the component. The general form of the Goodman equation is:

\(\frac{K_f \sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n_s}\)

Where:

  • K_f = Fatigue stress concentration factor
  • S_e = Modified endurance limit
  • S_{ut} = Ultimate tensile strength
  • \(\sigma_m\) = Mean stress
  • \(\sigma_a\) = Alternating stress
  • n_s = Factor of safety

This equation is used to evaluate the safety of a component under cyclic loading. If the left-hand side of the equation is less than or equal to the right-hand side, the component is considered safe from fatigue failure under the given loading conditions.

Solution:

Given the options in the question, we need to identify which expression correctly represents the Goodman line for assessing fatigue failure.

Option 1: \(\frac{K_f n_s \sigma_a}{S_e} + \frac{n_s \sigma_m}{S_{ut}} = 1\)

This option incorrectly includes the factor of safety \(n_s\) in both terms, which is not consistent with the standard Goodman equation.

Option 2: \(\frac{K_f \sigma_a}{S_e} + \frac{\sigma_m}{S_{yt}} = \frac{\sigma_m}{n_s}\)

This option incorrectly uses the yield strength \(S_{yt}\) instead of the ultimate tensile strength \(S_{ut}\) and also misplaces the factor of safety \(n_s\).

Option 3: \(\frac{K_f \sigma_a}{S_e} + \left(\frac{n_s \sigma_m}{S_{ut}}\right)^2 = \frac{1}{n_s}\)

This option incorrectly squares the term involving the mean stress and factor of safety, which is not consistent with the standard Goodman equation.

Option 4: \(\frac{K_f \sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n_s}\)

This option correctly represents the Goodman line. It includes the fatigue stress concentration factor \(K_f\), modified endurance limit \(S_e\), ultimate tensile strength \(S_{ut}\), mean stress \(\sigma_m\), alternating stress \(\sigma_a\), and factor of safety \(n_s\) in the correct form.

Therefore, the correct answer is option 4:

\(\frac{K_f \sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n_s}\)

This equation is used to assess the safety of a mechanical component under cyclic loading conditions. It ensures that the combined effect of mean and alternating stresses does not exceed the material's capacity to withstand fatigue, considering the factor of safety.

Soderberg Curve Question 3:

A simply supported beam has a concentrated load at the centre which fluctuates from a value P to 4P. The span of the beam is 50 cm and its cross section is circular with a diameter of 60 mm. Assume that the yield and endurance strength of material are 390 and 260 MPa respectively. Calculate the maximum permissible value of P (in kN) for a factor of safety of 1.4.

Answer (Detailed Solution Below) 9.8 - 10.1

Soderberg Curve Question 3 Detailed Solution

Concept:

Mean or average stress:

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

Reversed stress component or alternating or variable stress:

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

Max bending Moment for simply supported Beam at centre: \(M = \frac{{PL}}{4}\)

Calculation:

MD1 sneha images Q2a

\(\begin{array}{l} Z = \frac{\pi }{{32}}{d^3} = \frac{\pi }{{32}}{\left( {60} \right)^3} = 21205.75m{m^3}\\ {P_m} = \frac{{{P_{max}} + {P_{min}}}}{2} = \frac{{4P + P}}{2} = 2.5P\\ {P_a} = \frac{{{P_{max}} - {P_{min}}}}{2} = \frac{{4P - P}}{2} = 1.5P \end{array}\)

Bending Moment:

\(\begin{array}{l} {M_m} = \frac{{{P_m} \times L}}{4}\\ {M_a} = \frac{{{P_a} \times L}}{4} \end{array}\)

Bending Stress:

\(\begin{array}{l} {\sigma _m} = \frac{{{M_m}}}{Z} = \frac{{{P_m} \times L}}{{4Z}} = \frac{{2.5P \times 500}}{{4 \times 21205.75}} = 14.74 \times {10^{ - 3}}P\;MPa\\ {\sigma _a} = \frac{{{M_a}}}{Z} = \frac{{{P_a} \times L}}{{4Z}} = \frac{{1.5P \times 500}}{{4 \times 21205.75}} = 8.84 \times {10^{ - 3}}P\;MPa \end{array}\)

Soderberg Equation:

\(\begin{array}{l} \frac{1}{{F.S}} = \frac{{{\sigma _m}}}{{{\sigma _y}}} + \frac{{{\sigma _a}}}{{{\sigma _e}}}\\ \frac{1}{{1.4}} = P\left( {\frac{{14.74 \times {{10}^{ - 3}}}}{{390}} + \frac{{8.84 \times {{10}^{ - 3}}}}{{260}}} \right) = 71.79 \times {10^{ - 6}}P \end{array}\)

P = 9949 N = 9.9 kN

Soderberg Curve Question 4:

Match List I with List II and select the correct answer using the code given below the lists :

List I      

List II

A. Von - Mises yield criterion

1. Fatigue design

B. Soderberg’s law

2. Failure in machine elements made of ductile materials

C. Sommerfield number

3. Design of journal bearings

D. Buckingham equation

4. Dynamic tooth load in gears in mesh

  1. A - 4, B - 1, C - 3, D - 2
  2. A - 2, B - 1, C - 3, D - 4
  3. A - 4, B - 3, C - 1, D - 2
  4. A - 2, B - 3, C - 1, D - 4

Answer (Detailed Solution Below)

Option 2 : A - 2, B - 1, C - 3, D - 4

Soderberg Curve Question 4 Detailed Solution

Explanation:

Von - mises yield criterion: This criterion is based on the determination of the distortion energy in a given material. It is used to find failure stress in the ductile materials.

Soderberg’s Law: It is used for the component subjected to fluctuating load for fatigue design.

Sommer field Number: It is a dimensionless quantity used in the design of hydrodynamic journal bearings. It is very important in lubrication analysis because it contains all design variables.

Buckingham equation: This equation gives the durability of gears and the dynamic loads to which they are subjected.

Soderberg Curve Question 5:

When designing a shaft for fluctuating loads, the primary consideration to prevent fatigue failure is:

  1. Ultimate tensile strength
  2. Endurance limit
  3. Rigidity
  4. Yield point

Answer (Detailed Solution Below)

Option 2 : Endurance limit

Soderberg Curve Question 5 Detailed Solution

Explanation:

Endurance Limit:

  • The endurance limit (also known as the fatigue limit) is the maximum stress amplitude below which a material can endure an essentially infinite number of load cycles without failing.
  • For many materials, particularly ferrous alloys (like steel), there is a well-defined endurance limit.
  • If the stress amplitude is kept below this limit, the material theoretically will not fail due to fatigue, regardless of the number of cycles.

Importance of Endurance Limit in Shaft Design:

When designing a shaft for fluctuating loads, the endurance limit is crucial because:

  • Preventing Fatigue Failure: By ensuring that the fluctuating stresses in the shaft are below the endurance limit, designers can prevent fatigue failure, which is vital for the safe and reliable operation of machinery.
  • Long-Term Performance: Components designed with the endurance limit in mind can withstand a large number of cycles, ensuring long-term performance and durability.
  • Safety: Fatigue failure can be sudden and catastrophic. Designing below the endurance limit enhances the safety of mechanical systems.

In summary, the endurance limit is the primary consideration when designing a shaft for fluctuating loads because it directly relates to the material's ability to resist fatigue failure over an infinite number of cycles.

Soderberg Curve Question 6:

if the estimated actual endurance strength of the material is S'n' the ultimate tensile strength is Su' the mean stress is σm and the alternating stress is σa, the equation for the Goodman line is

  1. \(\rm \frac{\sigma_a}{S'_n}\times \frac{S_u}{\sigma_m}\) = 1
  2. \(\rm \frac{\sigma_a}{S'_n}\times \frac{\sigma_m}{S_u}\) = 1
  3. \(\rm \frac{\sigma_a}{S'_n}+ \frac{\sigma_m}{S_u}\) = 1
  4. \(\rm \frac{\sigma_a}{S'_n}+ \frac{S_u}{\sigma_m}\) = 1

Answer (Detailed Solution Below)

Option 3 : \(\rm \frac{\sigma_a}{S'_n}+ \frac{\sigma_m}{S_u}\) = 1

Soderberg Curve Question 6 Detailed Solution

The Goodman line equation is used in mechanical engineering to predict the failure of materials under varying loads. It combines the effects of mean stress and alternating stress to determine the endurance limit of the material. The equation accounts for the ultimate tensile strength (Su') and the estimated actual endurance strength (S'n') of the material. By plotting the mean stress and alternating stress on a graph, the Goodman line helps in assessing whether the material will withstand the applied cyclic loads without failure.

Soderberg Curve Question 7:

A machine component is subjected to a flexural stress which fluctuates between +300 MN/m2 and -150 MN/m2 . The yield strength and endurance strength are 0.55 and 0.5 times the ultimate strength, respectively; and, factor of safety = 2. The ultimate strength according to Soderberg relation is ______.

  1. 1008 MN/m2 
  2. 1056 MN/m2 
  3. 925 MN/m2 
  4. 1173 MN/m2 

Answer (Detailed Solution Below)

Option 4 : 1173 MN/m2 

Soderberg Curve Question 7 Detailed Solution

Concept:

Soderberg Line - 

  • When a component is subjected to fluctuating stresses, there is mean stress (Sm) and amplitude stress (Sa) play an important role. To design these components certain principles/methods are used based on the Endurance Limit of the material (Se).
  • The main methods for designing objects for fluctuating loads are:
    1. Soderberg Line Method
    2. Goodman Line Method
    3. Gerber Line Method
  • When a straight line joins Endurance Limit on the y-axis and Yield Strength (Syt) on the x-axis, then it becomes Soderberg Line.
  • The y-axis represents amplitude stress and the x-axis represents mean stress.
  • Soderberg Line is the most Conservative Method for designing objects on the basis of fluctuating stresses. 
  • Soderberg Line is suitable for ductile materials.

The equation of the Soderberg Line is given by:

\(\frac{S_a}{S_e} + \frac{S_m}{S_{yt}} = \frac{1}{FOS}\)

F1 R.Y 14.12.19 Pallavi D10

Mean Stress, Sm\(\frac{\sigma_{max} + \sigma_{min}}{2}\)   

Amplitude Stress, Sa\(\frac{\sigma_{max} - \sigma_{min}}{2}\)

Calculation:

Given:

\(\sigma _{max} = +300 \) MN/m2\(\sigma _{min} = -150 \) MN/m2, Syt = 0.55Su, Se = 0.5Su, FOS = 2

\(S_m = \frac{300-150}{2} = 75\) MN/m2

\(S_a = \frac{300+150}{2} = 225\) MN/m2

\(\frac{225}{0.5S_u} + \frac{75}{0.55S_u} = \frac{1}{2}\)

\(\frac{450}{S_u} + \frac{136.36}{S_u} = \frac{1}{2}\)

\(586.36 \times 2 = S_u\)

Su = 1172.72  MN/m \(\simeq\) 1173  MN/m2

Soderberg Curve Question 8:

A steel rod is subjected to a reversed axial load of 200 kN. The ultimate tensile strength and yield strength of the material are 1080 MPa and 900 MPa respectively. Use the following and select the true statements.

Axial loading = 0.7, machined surface = 0.8

Size Factor = 0.83, for stress concentration = 1

  1. The stress ratio = - 1
  2. For factor of safety 2, diameter using soderberg’s equation is 45.045 mm
  3. Variable load = 400 kN
  4. None of these

Answer (Detailed Solution Below)

Option :

Soderberg Curve Question 8 Detailed Solution

Concept:

For reversed loading |ωmax| = |ωmin|

\({\sigma _m} = mean\;stress = \frac{{{ω _{max}} + {ω _{min}}}}{2}\)

\({\sigma _v} = variable\;stress = \frac{{{ω _{max}} - {ω _{min}}}}{2}\)

Soderberg’s formula,

\(\frac{1}{{FOS}} = \frac{{{\sigma _m}}}{{{\sigma _y}}} + \frac{{{\sigma _y} \times {k_f}}}{{{\sigma _{ea}} \times {k_{surface}} \times {k_{sz}}}}\)

Calculation:

ωmax = +200 kN, ωmin = -200 kN

ωm = 0,

Variable load (ωv) = (ωmax - ωmin)/2 = (200 - (- 200))/2 = 200 kN

Now,

\({\rm{Stress\;ratio\;}} = \frac{{{\sigma _{min}}}}{{{\sigma _{max}}}} = \frac{{{ω _{min}}}}{{{ω _{max}}}} = - 1\)

σea = 1080 × 0.5 Ka

σea = 378 MPa

Now,

\(\frac{1}{{FOS}} = 0 + \frac{{200 \times {{10}^3} \times 1}}{{{d^2}\frac{\pi }{4} \times 378 \times 0.8 \times 0.83}}\)

d = 45.045 mm

Soderberg Curve Question 9:

Forged steel link with a uniform diameter of 30 mm at the centre is subjected to an axial force that varies 40 kN in compression to 160 kN in tension. Ultimate tensile strength, yield strength and corrected endurance strength are 600 MPa, 420 MPa and 240 MPa respectively. Then which of the following statements are true?

  1. Factor of safety according to Soderberg criterion is 1.26
  2. Factor of safety according to Goodman’s criterion is 1.37
  3. Factor of safety according to Soderberg criterion is 1.37
  4. Factor of safety according to Goodman’s criterion is 1.26

Answer (Detailed Solution Below)

Option :

Soderberg Curve Question 9 Detailed Solution

Concept:

For Soderberg criterion of design yield strength is considered.

For Goodman’s criterion of design Ultimate tensile strength is considered.

Calculation:

Given:

D = 30 mm, P = 40 KN (compression), P = 160 KN (Tension)

Syt = 420 MPa, Sut = 600 MPa, Sen = 240 MPa

Now,

\({\sigma _m} = \;\frac{{\left( { - 40 + 160} \right) \times {{10}^3}}}{{2A}}\;MPa\; = \frac{{60 \times {{10}^3}}}{A}MPa\)

\({\sigma _v} = \;\frac{{\left( {160 - \left( { - 40} \right)} \right) \times {{10}^3}}}{{2A}}\;MPa = \;\frac{{100 \times {{10}^3}}}{A}MPa\)

Here,

\(A = \frac{{\pi {D^2}}}{4} = \;\frac{{\pi {{30}^2}}}{4}\;m{m^2} = 706.858\;m{m^2}\)

Now,

According to Soderberg criterion

\(\frac{{{\sigma _m}}}{{{S_{yt}}}} + \;\frac{{{\sigma _v}}}{{{S_{en}}}} = \;\frac{1}{N}\;\)     ---(1)

Substituting all the values in (1), we get

∴ N = 1.26 (option a)

Now,

According to Goodmann’s criterion

\(\frac{{{\sigma _m}}}{{{S_{ut}}}} + \frac{{{\sigma _v}}}{{{S_{en}}}} = \frac{1}{N}\)   ---(2)

Substituting all the values in (2), we get

∴ N = 1.37 (option b)

Soderberg Curve Question 10:

Which one of the following failure theories is the most conservative design approach against fatigue failure?

  1. Modified Goodman line
  2. Yield line
  3. Gerber line
  4. Soderberge line

Answer (Detailed Solution Below)

Option 4 : Soderberge line

Soderberg Curve Question 10 Detailed Solution

Explanation:

Soderberg line:
Soderberg line for ductile materials gives upper limit for any combination of mean and alternating stress. Soderberg line is the most conservative fatigue failure criterion. Following diagram depicts the same.

F1 R.Y 14.12.19 Pallavi D10

Considering following notations

σa = limiting safe stress amplitude

Se = endurance limit of the component

σm = limiting safe mean stress

Sut = ultimate tensile strength

Syt = Yield strength

N = Factor of safety

Soderberg line:

The line joining Syt (yield strength of the material) on the mean stress axis and Se (endurance limit of the component) on stress amplitude axis is called as Soderberg line. This line is used when yielding defines failure (Ductile materials).

The equation for the Soderberg line:

\(\frac{{{{\rm{\sigma }}_{\rm{m}}}}}{{{{\rm{S}}_{{\rm{yt}}}}}} + \frac{{{{\rm{\sigma }}_{\rm{a}}}}}{{{{\rm{S}}_{\rm{e}}}}} = \frac{1}{N}\)

Additional Information

Goodman line:

Line joining Se on stress amplitude axis and Sut on mean stress axis is known as Goodman line. The triangular region below this line is considered a safe region. 
The equation for Goodman line:

\(\frac{{{\sigma _m}}}{{{S_{ut}}}} + \frac{{{\sigma _a}}}{{{S_e}}} = \frac{1}{N}\)

Gerber Line: 

Line joining Se on stress amplitude axis and Sut on mean stress axis is joined by a parabolic curve. 
The equation for Gerber line:

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