Question 1 |
The S-N curve from a fatigue test for steel is shown. Which one of the options gives
the endurance limit?
S_{ut} | |
S_{2} | |
S_{3} | |
S_{4} |
Question 1 Explanation:
For some ferrous (iron base) and titanium alloys,
the S-N curve becomes horizontal at higher N
values; or there is a limiting stress level, called the
fatigue limit (also sometimes the endurance limit),
below which fatigue will not occur. This fatigue
limit represents the largest value of fluctuating
stress that will not cause failure for essentially an infinite number of cycles. For many steels, fatigue
limits range between 35% and 60% of the tensile
strength.
Question 2 |
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).
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).
163840 | |
124589 | |
365247 | |
457812 |
Question 2 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 3 |
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).
68.5 | |
78.2 | |
122.5 | |
100 |
Question 3 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}
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 4 |
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)
0 | |
-0.8 | |
-0.2 | |
-0.5 |
Question 4 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 5 |
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).
\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).
150.25 | |
195.61 | |
240.36 | |
280.48 |
Question 5 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}
There are 5 questions to complete.