A polished steel bar is subjected to axial tensile force that varies from zero to Pmax. It has a groove 2 mm deep and having a radius of 3 mm. The theoretical stress concentration factor and notch sensitivity factor at the groove are 1.8 and 0.95 respectively. The outer diameter of the bar is 30 mm

Question

A polished steel bar is subjected to axial tensile force that varies from zero to [latex] P_{\max } [/latex] It has a groove 2 mm deep and having a radius of 3 mm. The theoretical stress concentration factor and notch sensitivity factor at the groove are 1.8 and 0.95 respectively. The outer diameter of the bar is 30 mm. The ultimate tensile strength of the bar is 1250 MPa. The endurance limit in reversed bending is 600 MPa. Find the maximum force that the bar can carry for [latex] 10^5 [/latex] cycles with 90% reliability.

Accepted Answer

[latex] ext { Given } P=0 ext { to } P_{\max } S_{u t}=1250 MPa [/latex].

[latex] S_{e}^{\prime}=600 MPa \quad K_{t}=1.8 \quad q=0.95 \quad R=90 \%[/latex].

[latex] N=10^{5} ext { cycles } [/latex].

Step I Endurance limit stress for bar

[latex] S_{e}^{\prime}=600 MPa =600 N / mm ^{2} [/latex].

[latex] ext { From Fig. } 5.24 ext { (polished surface), } K_{a}=1 [/latex].

[latex] ext { For } 30 mm ext { diameter wire, } K_{b}=0.85 [/latex].

[latex] ext { For } 90 \% ext { reliability, } K_{c}=0.897 [/latex].

From Eq. (5.12),

[latex] K_{f}=1+q\left(K_{t}-1 ight) [/latex] (5.12).

[latex] K_{f}=1+q\left(K_{t}-1 ight)=1+0.95(1.8-1)=1.76 [/latex].

[latex] K_{d}=\frac{1}{K_{f}}=\frac{1}{1.76}=0.568 [/latex].

[latex] S_{e}=K_{a} K_{b} K_{c} K_{d} S_{e}^{\prime} [/latex].

= 1.0(0.85)(0.897)(0.568)(600) = 259.84 N/mm².

Step II Construction of S–N diagram

[latex] 0.9 S_{u t}=0.9(1250)=1125 N / mm ^{2} [/latex].

[latex] \log _{10}\left(0.9 S_{u t} ight)=\log _{10}(1125)=3.0512 [/latex].

[latex] \log _{10}\left(S_{e} ight)=\log _{10}(259.84)=2.4147 [/latex].

[latex] \log _{10}\left(10^{5} ight)=5 [/latex].

The S–N curve for this problem is shown in Fig. 5.49.
Step III Fatigue strength at [latex] 10^5[/latex] cycles
From Fig. 5.49,

[latex] \overline{A E}=\frac{\overline{A D} imes \overline{E F}}{\overline{D B}}=\frac{(3.0512-2.4147)(5-3)}{(6-3)} [/latex].

[latex] =0.4243 [/latex].

[latex] \log _{10} S_{f}=3.0512-\overline{A E}=2.6269 [/latex].

[latex] S_{f}=423.55 N / mm ^{2} [/latex].

Therefore, at [latex] 10^5[/latex] cycles the fatigue strength is 423.55 N/mm².

Step IV Construction of modified Goodman diagram
Supose,

[latex] P_{\max .}=P \quad ext { and } \quad P_{\min .}=0 [/latex].

[latex] P_{m}=\frac{1}{2}\left[P_{\max }+P_{\min } ight]=\frac{1}{2} P [/latex].

[latex] P_{a}=\frac{1}{2}\left[P_{\max .}-P_{\min } ight]=\frac{1}{2} P [/latex].

[latex] an heta=\frac{\sigma_{a}}{\sigma_{m}}=\frac{P_{a}}{P_{m}}=1 [/latex].

[latex] heta=45^{\circ} [/latex].

The modified Goodman diagram for this example is shown in Fig. 5.50.
Step V Permissible stress amplitude
Refer to Fig. 5.50. The coordinates of point X are determined by solving the following two equations simultaneously.
(i) Equation of line AB

[latex] \frac{S_{a}}{423.55}+\frac{S_{m}}{1250}=1 [/latex] (a).

(ii) Equation of line OX

[latex] \frac{S_{a}}{S_{m}}= an heta=1 [/latex] (b).

Solving the two equations,

[latex] S_{a}=S_{m}=316.36 N / mm ^{2} [/latex].

Step VI Maximum force on bar

[latex] ext { Since } \quad S_{a}=\left(\frac{P_{a}}{ ext { area }} ight)=\left(\frac{P / 2}{ ext { area }} ight) [/latex].

The minimum cross-section of the bar is shown in Fig. 5.51.

[latex] S_{a}=\frac{P / 2}{ ext { area }} ext { or } 316.36=\frac{P / 2}{\frac{\pi}{4}(30-4)^{2}} [/latex].

P = 335 929.5 N or 336 kN.

Question 5.16: A polished steel bar is subjected to axial tensile force tha...

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