Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 5.3

A plate made of steel 20C8 $\left(S_{u t}=440 N / mm ^{2}\right)$ in hot rolled and normalised condition is shown in Fig. 5.28. It is subjected to a completely reversed axial load of 30 kN. The notch sensitivity factor q can be taken as 0.8 and the expected reliability is 90%. The size factor is 0.85. The factor of safety is 2. Determine the plate thickness for infinite life.

## Verified Solution

$\text { Given } P=\pm 30 kN \quad S_{u t}=440 N / mm ^{2} \quad(f s)=2$.

$R=90 \% \quad q=0.8 \quad K_{b}=0.85$.

Step I Endurance limit stress for plate

$S_{e}^{\prime}=0.5 S_{u t}=0.5(440)=220 N / mm ^{2}$

$\text { From Fig. } 5.24 \text { (hot rolled steel and } S_{u t}= \left.440 N / mm ^{2}\right)$.

$K_{a}=0.67$.

$K_{b}=0.85$.

For 90% reliability,

$K_{c}=0.897$.

$\frac{d}{w}=\frac{10}{50}=0.2$.

From Fig. 5.2,    $K_{t}=2.51$.

From Eq. (5.12),

$K_{f}=1+q\left(K_{t}-1\right)$                (5.12).

$K_{f}=1+q\left(K_{t}-1\right)=1+0.8(2.51-1)=2.208$.

$K_{d}=\frac{1}{K_{f}}=\frac{1}{2.208}=0.4529$.

$S_{e}=K_{a} K_{b} K_{c} K_{d} S_{e}^{\prime}$.

$=0.67(0.85)(0.897)(0.4529)(220)$.

$=50.9 N / mm ^{2}$.

For axial load, (Eq. 5.30)

$\left(S_{e}\right)_{a}=0.8 S_{e}$            (5.30).

$\left(S_{e}\right)_{a}=0.8 S_{e}=0.8(50.9)=40.72 N / mm ^{2}$.

Step II Permissible stress amplitude

$\sigma_{a}=\frac{\left(S_{e}\right)_{a}}{(f s)}=\frac{40.72}{2}=20.36 N / mm ^{2}$              (a).

Step III Plate thickness

$\sigma_{a}=\frac{P}{(w-d) t}=\frac{(30)\left(10^{3}\right)}{(50-10) t} N / mm ^{2}$            (b).

From (a) and (b),

$20.36=\frac{(30)\left(10^{3}\right)}{(50-10) t}$.

∴            t = 36.84 mm.