Goodman Curve MCQ Quiz - Objective Question with Answer for Goodman Curve - Download Free PDF

Last updated on Apr 14, 2025

Latest Goodman Curve MCQ Objective Questions

Goodman Curve Question 1:

A machine component is subjected to stress which fluctuates between 350 to - 150 MPa. Assume yield and endurance limit as 0.6 and 0.5 times of ultimate strength respectively. Find the value of ultimate strength according to modified Goodman theory taking the factor of safety 2.5. [up to nearest  integer]

Answer (Detailed Solution Below) 1500

Goodman Curve Question 1 Detailed Solution

Concept:

Modified Goodman line: 

The shaded region represents Modified Goodman's Criteria 

F1  Yerra. 26-05-21 Savita D1

The equation for the Modified Goodman line:

It depends upon the condition below, 

σaσm=σe(σutσyt)σut(σytσe)

If, σaσm>σaσmLineAPwillbechosen

If, σaσm<σaσmLinePBwillbechosen

Where,

σa = limiting safe stress amplitude

σe = endurance limit of the component

σm = limiting safe mean stress

σut = ultimate tensile strength

σyt = Yield strength

FOS = Factor of safety

Calculation:

Given:

σmax = 350 N/mm2, σmin = - 150 N/mm2, σ= 200 N/mm2σe = 0.5 σut and σyt = 0.6 σut

σmean=σm=350+(150)2=100MPa

σamplitude=σa=350(150)2=250 MPa

Now, 

σaσm=0.5σut(σut0.6σut)σut(0.6σut0.5σut)=2

Also,

σaσm=250100=2.5

Since,

σaσm>σaσmLineAPwillbechosen

  • For line AP,

σmsut+σase=1N

100σut+2500.5σut=12.5

⇒ σut = 1500 MPa

Goodman Curve Question 2:

A machine component is subjected to fluctuating stress that varies from 40 to 100 MPa. If corrected endurance limit of the component is 300 MPa and ultimate tensile strength of the material is 700 MPa, then the factor of safety of the designed component according to Goodman equation is given by

  1. 1
  2. 2
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 4 : 5

Goodman Curve Question 2 Detailed Solution

Concept

Soderberg line for ductile materials gives an upper limit for any combination of mean and alternating stress. The 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:

σmSyt+σaSe=1N

Goodman line: 

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

σmSut+σaSe=1N

Gerber Line: 

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

(σmSutN)2+σaSeN=1

Calculation:

Given: σmin= 40 MPa, σmax =  100 MPa, Sut =700 MPa, S= 300 MPa

σm=σmax+σmin2=40+1002=70N/mm2

σa=σmaxσmin2=100402=30N/mm2

Using Goodman's theory,

70700+30300=1N

⇒ N = 5 

Goodman Curve Question 3:

A machine part in the form of cantilever beam is subjected to fluctuating load as shown in the figure. The load varies from 800 N to 1600 N. The modified endurance, yield and ultimate strengths of the material are 200 MPa, 500 MPa and 600 MPa, respectively.

F1 Sumit.C 24-02-21 Savita D29

The factor of safety of the beam using modified Goodman criterion is______ (round off to one decimal place).

Answer (Detailed Solution Below) 1.9 - 2.1

Goodman Curve Question 3 Detailed Solution

Concept:

The modified Goodman line is given by

σmσut+σaσe1N

Where

σm=σmax+σmin2,σa=σmaxσmin2

qImage65abc23b807e1f3abcc02296

For a rectangular beam, the bending stress is given by

σb=6Mbd2

Where b – width, d – depth;

Calculation:

Given b = 12 mm, d = 20 mm; σut = 600 MPa, σe = 200 MPa; 

Load varies from 800 N to 1600 N

⇒ Moment varies from 800 × 0.1 N-m to 1600 × 0.1 N-m

⇒ Mmax = 160 N-m, Mmin = 80 N-m;

Now

σmax=6×1600.012×202=200MPa

σmin=6×800.012×202=100MPa

Now

σm = 150 MPa, σa = 50 MPa;

Substituting in modified goodman line,

150600+50200=1N

⇒ N = 2;

Goodman Curve Question 4:

In a structural member under fatigue loading, the minimum and maximum stresses developed at the critical point are 50 MPa and 150 MPa, respectively. The endurance, yield, and the ultimate strengths of the material are 200 MPa, 300 MPa and 400 MPa, respectively. The factor of safety using modified Goodman criterion is

  1. 3/2
  2. 8/5
  3. 12/7
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Goodman Curve Question 4 Detailed Solution

Concept:

Modified Goodman line: 

The shaded region represents Modified Goodman's Criteria 

F1  Yerra. 26-05-21 Savita D1

The equation for Modified Goodman line:

It depends upon the condition below, 

σaσm=σe(σutσyt)σut(σytσe)

If, σaσm>σaσmLineAPwillbechosen

If, σaσm<σaσmLinePBwillbechosen

Where,

σa = limiting safe stress amplitude

Se = endurance limit of the component

σm = limiting safe mean stress

Sut = ultimate tensile strength

Syt = Yield strength

FOS = Factor of safety

Calculation:

Given:

σmax = 150 N/mm2, σmin = 50 N/mm2, S= 200 N/mm2, Sut = 400 N/mm2 and Syt = 300 N/mm2

σm=σmax+σmin2150+502=100N/mm2

σa=σmaxσmin2=150502=50N/mm2

Now, check the condition,

σaσm=σe(σutσyt)σut(σytσe)=200(400300)400(300200)=0.5

σaσm=50100=0.5

σaσm=σaσm=0.5

⇒ So, any of the among AP and PB can be taken as the boundary for modified Goodman's Criteria

The equation for modified Goodman line for line PB,

σaσyt+σmσyt=1N

50300+100300=1N

∴ FOS = N = 2

Goodman Curve Question 5:

A machine component is subjected to a 2D stress state. The tensile stress in the X-direction varies from 40 MPa to 100 MPa while that in the Y-direction ranges from 10 MPa to 80 MPa. The corrected endurance limit of the component is 270 MPa and the ultimate tensile strength is 660 MPa. As per the Goodman’s theory, the factor of safety considered by the designer is _________

Answer (Detailed Solution Below) 4 - 5

Goodman Curve Question 5 Detailed Solution

Concept:

In cases of 2D stress systems, the mean and amplitude stresses are given by:

σm=σxm2+σym2σxmσymandσa=σxa2+σya2σxaσya

Where,

σxm, σym, σxa, σya are the mean and amplitude stresses in the X and Y directions.

The Goodman’s theory is then:

nσaSe+nσmSut=1

Calculation:

σx,min=40MPa,σx,max=100MPa,σy,min=10MPa,σy,max=80MPa

σxm=100+402=70MPa,σxa=100402=30MPa

σym=80+102=45MPa,σya=80102=35MPa

The mean stress is: σm=σxm2+σym2σxmσym=702+45270x45=61.44MPa

Amplitude stress: σa=σxa2+σya2σxaσya=302+35230x35=32.79MPa

Thus, putting the values in the Goodman equation:

n[32.79270+61.44660]=1n=4.66

Top Goodman Curve MCQ Objective Questions

A machine part in the form of cantilever beam is subjected to fluctuating load as shown in the figure. The load varies from 800 N to 1600 N. The modified endurance, yield and ultimate strengths of the material are 200 MPa, 500 MPa and 600 MPa, respectively.

F1 Sumit.C 24-02-21 Savita D29

The factor of safety of the beam using modified Goodman criterion is______ (round off to one decimal place).

Answer (Detailed Solution Below) 1.9 - 2.1

Goodman Curve Question 6 Detailed Solution

Download Solution PDF

Concept:

The modified Goodman line is given by

σmσut+σaσe1N

Where

σm=σmax+σmin2,σa=σmaxσmin2

qImage65abc23b807e1f3abcc02296

For a rectangular beam, the bending stress is given by

σb=6Mbd2

Where b – width, d – depth;

Calculation:

Given b = 12 mm, d = 20 mm; σut = 600 MPa, σe = 200 MPa; 

Load varies from 800 N to 1600 N

⇒ Moment varies from 800 × 0.1 N-m to 1600 × 0.1 N-m

⇒ Mmax = 160 N-m, Mmin = 80 N-m;

Now

σmax=6×1600.012×202=200MPa

σmin=6×800.012×202=100MPa

Now

σm = 150 MPa, σa = 50 MPa;

Substituting in modified goodman line,

150600+50200=1N

⇒ N = 2;

In a structural member under fatigue loading, the minimum and maximum stresses developed at the critical point are 50 MPa and 150 MPa, respectively. The endurance, yield, and the ultimate strengths of the material are 200 MPa, 300 MPa and 400 MPa, respectively. The factor of safety using modified Goodman criterion is

  1. 3/2
  2. 8/5
  3. 12/7
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Goodman Curve Question 7 Detailed Solution

Download Solution PDF

Concept:

Modified Goodman line: 

The shaded region represents Modified Goodman's Criteria 

F1  Yerra. 26-05-21 Savita D1

The equation for Modified Goodman line:

It depends upon the condition below, 

σaσm=σe(σutσyt)σut(σytσe)

If, σaσm>σaσmLineAPwillbechosen

If, σaσm<σaσmLinePBwillbechosen

Where,

σa = limiting safe stress amplitude

Se = endurance limit of the component

σm = limiting safe mean stress

Sut = ultimate tensile strength

Syt = Yield strength

FOS = Factor of safety

Calculation:

Given:

σmax = 150 N/mm2, σmin = 50 N/mm2, S= 200 N/mm2, Sut = 400 N/mm2 and Syt = 300 N/mm2

σm=σmax+σmin2150+502=100N/mm2

σa=σmaxσmin2=150502=50N/mm2

Now, check the condition,

σaσm=σe(σutσyt)σut(σytσe)=200(400300)400(300200)=0.5

σaσm=50100=0.5

σaσm=σaσm=0.5

⇒ So, any of the among AP and PB can be taken as the boundary for modified Goodman's Criteria

The equation for modified Goodman line for line PB,

σaσyt+σmσyt=1N

50300+100300=1N

∴ FOS = N = 2

A machine component is subjected to fluctuating stress that varies from 40 to 100 MPa. If corrected endurance limit of the component is 300 MPa and ultimate tensile strength of the material is 700 MPa, then the factor of safety of the designed component according to Goodman equation is given by

  1. 1
  2. 2
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 4 : 5

Goodman Curve Question 8 Detailed Solution

Download Solution PDF

Concept

Soderberg line for ductile materials gives an upper limit for any combination of mean and alternating stress. The 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:

σmSyt+σaSe=1N

Goodman line: 

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

σmSut+σaSe=1N

Gerber Line: 

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

(σmSutN)2+σaSeN=1

Calculation:

Given: σmin= 40 MPa, σmax =  100 MPa, Sut =700 MPa, S= 300 MPa

σm=σmax+σmin2=40+1002=70N/mm2

σa=σmaxσmin2=100402=30N/mm2

Using Goodman's theory,

70700+30300=1N

⇒ N = 5 

A thin spherical pressure vessel of 200 mm diameter and 1 mm thickness is subjected to an internal pressure varying from 4 to 8 MPa. Assume that the yield, ultimate, and endurance strength of material are 600, 800 and 400 MPa respectively. The factor of safety as per Goodman’s relation is

  1. 2.0
  2. 1.6
  3. 1.4
  4. 1.2

Answer (Detailed Solution Below)

Option 2 : 1.6

Goodman Curve Question 9 Detailed Solution

Download Solution PDF

Given : d = 200 mm, t = 1 mm, σu = 800 MPa, σe = 400 MPa

Circumferential stress induced in spherical pressure vessel is, 
σ=p×r2t=p×1002×1=50p MPa

Given that, pressure vessel is subject to an internal pressure varying from 4 to 8 MPa.

So, σmin =  50 × 4 = 200 MPa

σmax =  50 × 8 = 400 MPa

Mean stress, σm=σmax+σmin2=200+4002=300MPa

Variable stress, σv=σmaxσmin2=4002002=100MPa

From the Goodman method,

1FOS=σmσu+σvσe=300800+100400=1.6

Goodman Curve Question 10:

A machine part in the form of cantilever beam is subjected to fluctuating load as shown in the figure. The load varies from 800 N to 1600 N. The modified endurance, yield and ultimate strengths of the material are 200 MPa, 500 MPa and 600 MPa, respectively.

F1 Sumit.C 24-02-21 Savita D29

The factor of safety of the beam using modified Goodman criterion is______ (round off to one decimal place).

Answer (Detailed Solution Below) 1.9 - 2.1

Goodman Curve Question 10 Detailed Solution

Concept:

The modified Goodman line is given by

σmσut+σaσe1N

Where

σm=σmax+σmin2,σa=σmaxσmin2

qImage65abc23b807e1f3abcc02296

For a rectangular beam, the bending stress is given by

σb=6Mbd2

Where b – width, d – depth;

Calculation:

Given b = 12 mm, d = 20 mm; σut = 600 MPa, σe = 200 MPa; 

Load varies from 800 N to 1600 N

⇒ Moment varies from 800 × 0.1 N-m to 1600 × 0.1 N-m

⇒ Mmax = 160 N-m, Mmin = 80 N-m;

Now

σmax=6×1600.012×202=200MPa

σmin=6×800.012×202=100MPa

Now

σm = 150 MPa, σa = 50 MPa;

Substituting in modified goodman line,

150600+50200=1N

⇒ N = 2;

Goodman Curve Question 11:

In a structural member under fatigue loading, the minimum and maximum stresses developed at the critical point are 50 MPa and 150 MPa, respectively. The endurance, yield, and the ultimate strengths of the material are 200 MPa, 300 MPa and 400 MPa, respectively. The factor of safety using modified Goodman criterion is

  1. 3/2
  2. 8/5
  3. 12/7
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Goodman Curve Question 11 Detailed Solution

Concept:

Modified Goodman line: 

The shaded region represents Modified Goodman's Criteria 

F1  Yerra. 26-05-21 Savita D1

The equation for Modified Goodman line:

It depends upon the condition below, 

σaσm=σe(σutσyt)σut(σytσe)

If, σaσm>σaσmLineAPwillbechosen

If, σaσm<σaσmLinePBwillbechosen

Where,

σa = limiting safe stress amplitude

Se = endurance limit of the component

σm = limiting safe mean stress

Sut = ultimate tensile strength

Syt = Yield strength

FOS = Factor of safety

Calculation:

Given:

σmax = 150 N/mm2, σmin = 50 N/mm2, S= 200 N/mm2, Sut = 400 N/mm2 and Syt = 300 N/mm2

σm=σmax+σmin2150+502=100N/mm2

σa=σmaxσmin2=150502=50N/mm2

Now, check the condition,

σaσm=σe(σutσyt)σut(σytσe)=200(400300)400(300200)=0.5

σaσm=50100=0.5

σaσm=σaσm=0.5

⇒ So, any of the among AP and PB can be taken as the boundary for modified Goodman's Criteria

The equation for modified Goodman line for line PB,

σaσyt+σmσyt=1N

50300+100300=1N

∴ FOS = N = 2

Goodman Curve Question 12:

A machine component is subjected to fluctuating stress that varies from 40 to 100 MPa. If corrected endurance limit of the component is 300 MPa and ultimate tensile strength of the material is 700 MPa, then the factor of safety of the designed component according to Goodman equation is given by

  1. 1
  2. 2
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 4 : 5

Goodman Curve Question 12 Detailed Solution

Concept

Soderberg line for ductile materials gives an upper limit for any combination of mean and alternating stress. The 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:

σmSyt+σaSe=1N

Goodman line: 

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

σmSut+σaSe=1N

Gerber Line: 

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

(σmSutN)2+σaSeN=1

Calculation:

Given: σmin= 40 MPa, σmax =  100 MPa, Sut =700 MPa, S= 300 MPa

σm=σmax+σmin2=40+1002=70N/mm2

σa=σmaxσmin2=100402=30N/mm2

Using Goodman's theory,

70700+30300=1N

⇒ N = 5 

Goodman Curve Question 13:

 A steel link of diameter 40 mm is subjected to an axial load that varies from 140 kN in tension to 50 kN in compression. Find the factor of safety using Goodman’s criteria, if σut = 610 MPa, σyt = 400 MPa and σe = 240 MPa

Answer (Detailed Solution Below) 2.66 - 2.68

Goodman Curve Question 13 Detailed Solution

Concept:

According to Goodman’s criteria,

σvσe+σmσut=1n(1)

Where

Variable stress,

σv=(PmaxPmin)2×Area(2)

Mean stress,

σm=(Pmax+Pmin)2×Area(3)

σe = Endurance strength, n = factor of safety, σut = Ultimate tensile strength

Calculation:

Given:

Pmax  = 140 kN, Pmin  = 50 kN, σut = 610 MPa, σyt = 400 MPa, σe = 240 MPa, d = 40 mm

By using equation (2) and (3),

Pv=140+502=1902=95kN

Pm=140502=45kN

σm=PmA=45000πd2×4=35.81MPa

σv=PvA=95000×4πd2=75.598MPa

By using equation (1),

1n=75.598240+35.81610

⇒ n = 2.675

Goodman Curve Question 14:

A machine component is subjected to a 2D stress state. The tensile stress in the X-direction varies from 40 MPa to 100 MPa while that in the Y-direction ranges from 10 MPa to 80 MPa. The corrected endurance limit of the component is 270 MPa and the ultimate tensile strength is 660 MPa. As per the Goodman’s theory, the factor of safety considered by the designer is _________

Answer (Detailed Solution Below) 4 - 5

Goodman Curve Question 14 Detailed Solution

Concept:

In cases of 2D stress systems, the mean and amplitude stresses are given by:

σm=σxm2+σym2σxmσymandσa=σxa2+σya2σxaσya

Where,

σxm, σym, σxa, σya are the mean and amplitude stresses in the X and Y directions.

The Goodman’s theory is then:

nσaSe+nσmSut=1

Calculation:

σx,min=40MPa,σx,max=100MPa,σy,min=10MPa,σy,max=80MPa

σxm=100+402=70MPa,σxa=100402=30MPa

σym=80+102=45MPa,σya=80102=35MPa

The mean stress is: σm=σxm2+σym2σxmσym=702+45270x45=61.44MPa

Amplitude stress: σa=σxa2+σya2σxaσya=302+35230x35=32.79MPa

Thus, putting the values in the Goodman equation:

n[32.79270+61.44660]=1n=4.66

Goodman Curve Question 15:

A brittle bar of cross-sectional area 1 mis loaded with sinusoidal variable tensile load from 80 MN to 160 MN, with Yield strength = 240 MPa, Ultimate yield strength = 360 MPa, Endurance limit = 100 MPa.

The factor of safety of bar is____________.

Answer (Detailed Solution Below) 1.3 - 1.4

Goodman Curve Question 15 Detailed Solution

Concept:

When a component is subjected to fluctuating stresses, there is mean stress (σmean) as well as stress amplitude (σamp).

To design those components certain methods are used based on the Endurance limit of the material σe.

Goodman Line:

When a straight-line joins Endurance limit σe on the ordinate and Ultimate strength σut on abscissa it is known as Goodman line.

The abscissa represents σmean and the ordinate represents σamp.

When no method is given, we can choose on the basis of material, here brittle material is given and for that σut will be used.

Goodman Line is given by:

σmeanσut+σampσe=1FOS

Calculation:

Given:

Pmax = 160 MN = 160 × 106 N, Pmin = 80 MN = 80 × 106 N, Area = 1 m2,

Yield strength (σyt) = 240 MPa, Ultimate yield strength (σut) = 360 MPa, Endurance limit (σse) = 100 MPa.

σmax=PmaxArea=160×1061=160×106N/m2=160MPa

σmin=PminArea=80×1061=80×106N/m2=80MPa

Mean stress is given by:

σm=σmax+σmin2

σm=160+802=120 MPa

Stress amplitude is given by:

σa=σmaxσmin2

σa=160802=40 MPa

Goodman Line is given by:

σmeanσut+σampσe=1FOS

120360+40100=1FOS

FOS = 1.36

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