# Fatigue Strength and S-N Diagram

 Question 1
The figure shows the relationship between fatigue strength (S) and fatigue life (N) of a material. The fatigue strength of the material for a life of 1000 cycles is 450 MPa, while its fatigue strength for a life of $10^6$ cycles is 150 MPa. The life of a cylindrical shaft made of this material subjected to an alternating stress of 200 MPa will then be cycles (round off to the nearest integer).
 A 163840 B 124589 C 365247 D 457812
GATE ME 2021 SET-2   Machine Design
Question 1 Explanation: Equation of line $A \bar{B}:$
\begin{aligned} y-y_{1} &=\frac{\left(y_{2}-y_{1}\right)}{\left(x_{2}-x_{1}\right)}\left[x-x_{1}\right] \\ \log _{10} 200-\log _{10} 450 &=\frac{\log _{10} 150-\log _{10} 450}{(6-3)}\left[\log _{10} N-3\right] \\ N &=163840.580 \text { cycles } \end{aligned}
 Question 2
A machine member is subjected to fluctuating stress $\sigma =\sigma _0 \cos (8 \pi t)$. The endurance limit of the material is 350 MPa. If the factor of safety used in the design is 3.5 then the maximum allowable value of $\sigma _0$ is __________ MPa (round off to 2 decimal places).
 A 68.5 B 78.2 C 122.5 D 100
GATE ME 2020 SET-2   Machine Design
Question 2 Explanation:
\begin{aligned} \text{Fluctuating stress, } \quad \sigma&=\sigma_{o} \cos (8 \pi t) \\ \sigma_{\max } &=\sigma_{o} \\ \sigma_{\min } &=-\sigma_{o} \\ \sigma_{\text {man }} &=\frac{\sigma_{\max }+\sigma_{\min }}{2}=0 \\ \sigma_{v} &=\frac{\sigma_{\max }-\sigma_{\min }}{2}=\frac{2 \sigma_{o}}{2}=\sigma_{o} \\ \sigma_{e} &=350 \mathrm{MPa} \\ \mathrm{FOS} &=3.5 \end{aligned}
From strength criteria, $\frac{\sigma_{v}}{\sigma_{e}} \leq \frac{1}{\mathrm{FOS}}$
\begin{aligned} \frac{\sigma_{o}}{350} & \leq \frac{1}{3.5} \\ \sigma_{o} & \leq 100 \mathrm{MPa} \end{aligned}
 Question 3
During a high cycle fatigue test, a metallic specimen is subjected to cyclic loading with a mean stress of +140 MPa, and a minimum stress of -70 MPa. The R-ratio (minimum stress to maximum stress) for this cyclic loading is_______ (round off to one decimal place)
 A 0 B -0.8 C -0.2 D -0.5
GATE ME 2019 SET-1   Machine Design
Question 3 Explanation:
$\begin{array}{l} \sigma_{\min }=-70 \mathrm{MPa} \\ \sigma_{\text {mean }}=140 \mathrm{MPa} \\ \Rightarrow \sigma_{\text {mean }}=\frac{\sigma_{\min }+\sigma_{\max }}{2}=140 \\ \Rightarrow \frac{-70+\sigma_{\max }}{2}=140 \\ \Rightarrow \sigma_{\max }=350 \mathrm{MPa} \\ \text { The R-ratio, } \mathrm{R}=\frac{\sigma_{\min }}{\sigma_{\max }}=-\frac{70}{350}=-0.2 \end{array}$
 Question 4
Fatigue life of a material for a fully reversed loading condition is estimated from
$\sigma_{a}=1100N^{-0.15}$
where $\sigma_{a}$
is the stress amplitude in MPa and N is the failure life in cycles.The maximum allowable stress amplitude (in MPa) for a life of $1\times10^{5}$ cycles under the same loading condition is ________(correct to two decimal places).
 A 150.25 B 195.61 C 240.36 D 280.48
GATE ME 2018 SET-2   Machine Design
Question 4 Explanation:
\begin{aligned} \frac{\sigma_{\max }-\sigma_{\min }}{2} &=1100 \mathrm{N}^{-0.15} \\ \frac{\sigma_{\max }-\left(-\sigma_{\max }\right)}{2} &=1100 \mathrm{N}^{-0.15} \\ \frac{2 \sigma_{\max }}{2} &=1100 \mathrm{N}^{-0.15} \\ \sigma_{\max } &=1100 \mathrm{N}^{-0.15}=1100 \times\left(10^{5}\right)^{-0.15} \\ &=1100 \times(10)^{-0.75}=\frac{1100}{5.62} \\ \sigma_{\max } &=195.61 \mathrm{MPa} \end{aligned}
 Question 5
A machine element has an ultimate strength $(\sigma _{u})$ of $600\, N/mm^{2}$ , and endurance limit $(\sigma _{en})$ of $250\, N/mm^{2}$ , The fatigue curve for the element on log- log plot is shown below. If the element is to be designed for a finite of 10000 cycles, the maximum amplitude of a completely reversed operating stress is _______ $N/mm^{2}$. A 350.25 B 365.23 C 386.19 D 395.48
GATE ME 2017 SET-1   Machine Design
Question 5 Explanation: From two similar triangles,
$(i.e. \Delta A B D \text { and } \Delta E C D)$
\begin{aligned} \frac{A B}{E C}&=\frac{B D}{C D} \\ \frac{\left(\log _{10} 480-\log _{10} 250\right)}{\left(\log _{10} S-\log _{10}-250\right)}&=\frac{6-3}{6-\log _{10} 10000} \\ \frac{2.68124-2.39794}{\log _{10} S-2.39794}&=1.5 \\ S&=386.195 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}
 Question 6
In a structural member under fatigue loading, the minimum and maximum stresses developed at the critical point are 50 MPaand 150 MParespectively. 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
 A $\frac{3}{2}$ B $\frac{8}{5}$ C $\frac{12}{7}$ D 2
GATE ME 2016 SET-2   Machine Design
Question 6 Explanation:
\begin{aligned} \sigma _{max}&=150MPa\\ \sigma _{min}&=50MPa\\ S_e&=200MPa\\ S_{yt}&=300MPa\\ S_{ut}&=400MPa\\ \sigma _a&=\frac{\sigma _{max}-\sigma _{min}}{2}\\ &=\frac{150-50}{2}\\ &=50 MPa\\ \sigma _m&=\frac{\sigma _{max}+\sigma _{min}}{2}\\ &=\frac{150+50}{2}\\ &=100 MPa\\ &\text{Goodman sequation}\\ \frac{\sigma _a}{S_e}+\frac{\sigma _m}{S_{ut}}&=\frac{1}{FS}\\ \frac{50}{200}+\frac{100}{400}&=\frac{1}{FS}\\ FS&=2 \end{aligned}
 Question 7
Which one of the following is the most conservative fatigue failure criterion?
 A Soderberg B Modified Goodman C ASME Elliptic D Gerber
GATE ME 2015 SET-1   Machine Design
Question 7 Explanation:
When a component is subjected to fluctuating stresses there is mean stress $(\sigma _m)$ as well as stress amplitude $(\sigma _a)$. It has been observed that the mean stress component has an effect on fatigue failure when it is present in combination with an alternating component.
When the component is subjected to both component of stress $(\sigma _m)$ and $(\sigma _a)$ the actual failure occurs at different scattered points as shown in figure. There exist a border which decides sole region from unsole region for various combinations of $(\sigma _m)$ and $(\sigma _a)$ and different criterion are proposed to constant the borderline dividing safe zone and failure zone, they include Gerber line, soderberg line, Goodman line, ASME elliptic.

Gerber Line:
A parabolic curve jointing $\sigma _e$ on the ordinate to $\sigma _{ut}$ on the abscissa is called Gerber line. It is written as
$\left (\frac{\sigma _a}{\sigma _e} \right )+\left (\frac{\sigma _m}{\sigma _{ut}} \right )^2=1$

Soderberg Line:
A straight line joining $\sigma _e$ on the ordinate to $\sigma _{yt}$ on the abscissa is called soderberg line. It is written as
$\left (\frac{\sigma _a}{\sigma _e} \right )+\left (\frac{\sigma _m}{\sigma _{yt}} \right )=1$

Goodman Line:
A straight line joining $\sigma _e$ on the ordinate to $\sigma _{ut}$ on the abscissa is called the Goodman line. It is written as
$\left (\frac{\sigma _a}{\sigma _e} \right )+\left (\frac{\sigma _m}{\sigma _{ut}} \right )=1$

ASME Elliptic:
The ASME elliptic is written as
$\left (\frac{\sigma _a}{\sigma _e} \right )^2+\left (\frac{\sigma _m}{\sigma _{yt}} \right )^2=1$

The soderberg line is a more conservative failure criterion and there is no need to consider even yielding in this case.
 Question 8
A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycle is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will be
 A 1071 cycles B 15000 cycles C 281914 cycles D 928643 cycles
GATE ME 2006   Machine Design
Question 8 Explanation: Equation of straight line connecting $\left(3, \log _{10} 490\right)$
and $\left(6, \log _{10} 70\right)$
\begin{aligned} \frac{y-1.8451}{x-6}&=\frac{1.84-2.6902}{6-3} \\ y-1.8451&=-0.2817(x-6)\\ \text{ At }\quad y&=\log _{10} 100=2 \\ \Rightarrow 2-1.8451 &=-0.2817(x-6) \\ x &=5.4501 \\ \log _{10} N &=5.4501 \\ N&=281914 \end{aligned}
 Question 9
The S-N curve for steel becomes asymptotic nearly at
 A $10^{3}cycles$ B $10^{4}cycles$ C $10^{6}cycles$ D $10^{9}cycles$
GATE ME 2004   Machine Design
Question 9 Explanation: Question 10
In terms of theoretical stress concentration factor ($K_{t}$) and fatigue stress concentration factor ($K_{f}$), the notch sensitivity 'q' is expressed as
 A $\frac{K_{f}-1}{K_{t}-1}$ B $\frac{K_{f}-1}{K_{t}+1}$ C $\frac{K_{t}-1}{K_{f}-1}$ D $\frac{K_{f}+1}{K_{t}-1}$
GATE ME 2004   Machine Design
Question 10 Explanation:
We know that
$K_{f}=1+q\left(K_{t}-1\right), \quad q=\frac{K_{f}-1}{K_{t}-1}$
There are 10 questions to complete. 