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Numericals on Design against fluctuating loads

Q.1 A rotating steel shaft $(S_{ut}=1200 \ MPa)$ shown in fig below must support a load $F = 2\ KN$ with reliability of 99% Determine FOS.

$S_{ut}=1200\ MPa\ ( N/mm^2)$

R = 99

h = ?

$\sum M_B=0$

$R_A(400)-2\times 10^3(270)=0$

$R_A=1350N$

$\sum Fv=0 \uparrow +ve$

$R_A+R_B-2\times 10^3=0$

$R_B=650N$

BM calculation

$M_c=R_A \times 130 =1350 \times 130=175.5\times 10^3 \text{ N.mm}$

$M_D=R_A\times 250=2\times 10^3\times 120=97.5\times 10^3 \text{ N.mm}$

We know that,

$\sigma _b=\frac{M}{I}\cdot y=\frac{M}{Z}$

$where \ Z = \frac{\pi}{32}d^3$

$\sigma _{b_c}=\frac{M_c}{Z_c}=\frac{175.5\times 10^3}{\frac{\pi(20)^3}{32}}=223.45 N/mm^2$

$\sigma {b_d}=\frac{M_D}{Z_D}=\frac{97.5\times 10^3}{\frac{\pi}{32}(15)^3}=294.2 N/mm^2$

As member rotates with uniform speed and carrying constant load, therefore, assume reverse stress

$\sigma _{max}=294.2 N/mm^2$

$\sigma _{min}=-\sigma _{max}=-294.2 N/mm^2$

$\sigma _m=0$

$\sigma _a=\sigma _{max}=294.2 N/mm^2$

Standard endurance limit

$\sigma _{-1}=0.5 \sigma _u$.........Ductile material

$=0.5\times 1200$

$=600 \text{ Mpa}$

From PSG 7.17,

$S_{ut}=1200 N/mm^2$

$=1200 kgt/cm^2$

From graph,$K_a=0.75$

To get $K_b$,

From table,

$K_b=0.85$

To get $K_c,$

From table, for 99% reliability,

$K_c=0.814$

$K_f=1+q(K_t-1)$

From PSG 7.14

To get $K_t$ from graph

$\frac{D}{d}=\frac{20}{15}=1.33$

$\frac{r}{d}=\frac{2}{15}=0.13$

From graph, $K_t=1.45$

From PSG 7.8,

To get q from graph,

$r=2 \text{ mm}$

Assume, annealed condition,

$\therefore q=0.9$

$K_f=1+0.9(1.45-1)=1.405$

Actual endurance limit

$\sigma _{-1}=K_a\cdot K_b\cdot K_L\cdot \frac{\sigma _{-1}}{K_f}$

$=0.75\times 0.85\times 0.814\times \frac{600}{1.405}$

$=221.60 \text{ MPa}$

Use soderbergs equation,

$\frac{1}{n}=\frac{\sigma _m}{\sigma _y}+\frac{\sigma _a}{\sigma _{-1}}$

$\frac{1}{n}=0+\frac{294.2}{221.60}$

$\therefore n=0.7532$

Q.2 A machine component is subjected to a Hexural stress which fluctuates between $300 MN/m^2$ and $-150MN/m^2$. Determine the value of minimum ultimate strength according to (1) Goodman relation (2) Soderberg relation. Take yield strength=0.55ultimate strength, Endurance strength=0.5ultimate strength, FOS=2.

Solution:

$\sigma _{max}=300N/mm^2$

$\sigma _{min}=-150N/mm^2$

$\sigma _y=0.55N/mm^2$

$\sigma _y=0.5N/mm^2$

$FOS=2$

$\sigma _m=\frac{300-150}{2}=75N/mm^2$

$\sigma _a=\frac{300-(-150)}{2}=225N/mm^2$

Goodman criteria,

$\frac{1}{n}=\frac{\sigma _m}{\sigma _u}+\frac{\sigma _a}{\sigma _{-1}}$

$\frac{1}{2}=\frac{7.5}{\sigma _u}+\frac{225}{0.5\sigma _u}$

$\therefore \sigma _u=1050N/mm^2$

Soderberg criteria,

$\frac{1}{n}=\frac{\sigma _m}{\sigma _y}+\frac{\sigma _a}{\sigma _{-1}}$

$\frac{1}{2}=\frac{75}{0.55\sigma _u}+\frac{225}{0.5\sigma _u}$

$\therefore \sigma _u=1172.72N/mm^2$

Q.3 A bar of circular cross-section is subjected to alternating tensile force varying from a minimum of $200 KN$ to a maximum of $500 KN$. It is to be manufactured of a material with an ultimate tensile strength of $900 MPa$ and an endurance limit of $700 MPa$. Determine the diameter of bar using FOS of 3.5 related to ultimate tensile strength and 4 related to endurance limit. Use Goodman straight line as basis of design.

Solution:

$W_{min}=200KN$

$W_{max}=500KN$

$\sigma _u=900MPa$

$\sigma _{-1}=700Mpa$

$\text{FOS related to }\sigma _u=3.5$

$\text{FOS related to }\sigma _{-1}=4$

Find: d=? by using Goodmans criteria

Soln.

$\sigma _{max}=\frac{W_{max}}{A}=\frac{500\times 10^3}{\frac{\pi}{4}d^2}=\frac{636.61\times 10^3}{d^2}$

$\sigma _{min}=\frac{W_{min}}{A}=\frac{200\times 10^3}{\frac{\pi}{4}d^2}=\frac{254.62\times 10^3}{d^2}$

$\sigma _m=\frac{[\frac{636.61\times 10^3}{d^2}+\frac{254.62\times 10^3}{d^2}]}{2}=\frac{445.61\times 10^3}{d^2}$

$\sigma _a=\frac{[\frac{636.61\times 10^3}{d^2}-\frac{254.62\times 10^3}{d^2}]}{2}=\frac{190.9\times 10^3}{d^2}$

$\therefore \frac{\frac{190.9\times 10^3}{d^2}}{(\frac{700}{4})}=1-\frac{\frac{445.61\times 10^3}{d^2}}{(\frac{900}{3.5})}$

$d=53.1mm$

Q.4 Determine the thickness of a $120 mm$ wide plate for safe continuous operation if the plate is to be subjected to a tensile load that has a maximum value of $250 KN$ and a minimum value of $100 KN$. The properties of plate materials are as follows:

Endurance limit stress=225 MPa, Yield point stress=300 MPa, FOS based on yeild pooint may be taken as 1.5.

Solution:

$t=?$

$b=120mm$

$W_{max}=250KN$

$W_{min}=100KN$

$\sigma _{-1}=225MPa$

$\sigma _y=300MPa$

$FOS=1.5$

$\sigma _{max}=\frac{W_{max}}{A}=\frac{250\times 10^3}{b\times t}=\frac{250\times 10^3}{120\times t}=\frac{2083.33}{t}$

$\sigma _{min}=\frac{W_{min}}{A}=\frac{100\times 10^3}{120\times t}=\frac{833.33}{t}$

$\sigma_m=\frac{\frac{2083.33}{t}+\frac{833.33}{t}}{2}=\frac{1(1458.33)}{t}$

$\sigma _a=\frac{\frac{2083.33}{t}-\frac{833.33}{t}}{2}=\frac{625}{t}$

Soderberg equation

$\frac{1}{n}=\frac{\sigma _m}{\sigma _y}+\frac{\sigma _a}{\sigma _{-1}}$

$\frac{1}{1.5}=\frac{\frac{1458.33}{t}}{300}+\frac{\frac{625}{t}}{225}$

$t=11.46mm$

Q.5 .A cantilever bean made of cold drawn carbon steel of circular cross section as shown in fig. is subjected to a load which varies from -F to 3F. Determine the maximum load that this member can withstand for an indefinite life using FOS as 2. The theoretical stress concentration factor is 1.42 and the notch sensitivity is 0.9. Assume the following values. Ultimate stress = 550 $N/mm^2$, Yield stress = 470 $N/mm^2$, size factor = 0.85, surface finish factor = 0.89.

$W_{min}=-F$

$W_{max}=3F$

$FOS=2$

$K_f=2$

$q=0.9$

$\sigma _u=550N/mm^2$

$\sigma _y=470N/mm^2$

$K_b=0.85$

$K_a=0.89$

$K_c=1$.....Assume 50% reliability (Refer table in textbook)

$\sigma _{-1}'=0.5\sigma _u=0.5\times 550=275 MPa$

Actual endurance limit

$\sigma _{-1}=(0.89)(0.85)(1).\frac{275}{1.378}$

$\sigma _{-1}=150.97 MPa$

$M_{max}=3F\times 125=375F$

$M_{min}=-F\times 125=-125F$

$\sigma _{b_{max}}=\frac{M_{max}}{Z}=\frac{375F}{\frac{\pi}{32}d^3}=\frac{375F}{\frac{\pi}{32}(13)^3}=1.738F$

$\sigma _{min}=\frac{M_{min}}{Z}=\frac{-125F}{\frac{\pi}{32}(13)^3}=-0.5795F$

$\sigma _M=\frac{1.738F+(-0.5795F)}{2}=0.57925F$

$\sigma _a=\frac{1.738F-(-0.5795F)}{2}=1.159F$

Soderberg equation,

$\frac{1}{2}=\frac{0.57925F}{470}+\frac{1.159F}{150.97}$

$F=56.11N$

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