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.

Chapter 4

Q. 4.8

An offset link subjected to a force of 25 kN is shown in Fig. 4.27. It is made of grey cast iron FG300 and the factor of safety is 3. Determine the dimensions of the cross-section of the link.

Step-by-Step

Verified Solution

\text { Given } P=25 kN \quad S_{u t}=300 N / mm ^{2} \quad(f s)=3 .

Step I Calculation of permissible tensile stress for the link

\sigma_{t}=\frac{S_{u t}}{(f s)}=\frac{300}{3}=100 N / mm ^{2}         (a).

Step II Calculation of direct tensile and bending stresses
The cross-section is subjected to direct tensile stress and bending stresses. The stresses are maximum at the top fibre. At the top fibre,

\sigma_{t}=\frac{P}{A}+\frac{M_{b} y}{I} .

=\frac{25 \times 10^{3}}{t(2 t)}+\frac{25 \times 10^{3}(10+t)(t)}{\left[\frac{1}{12} t(2 t)^{3}\right]}             (b).

Step III Calculation of dimensions of cross-section
Equating (a) and (b),

\frac{12500}{t^{2}}+\frac{37500(10+t)}{t^{3}}=100 .

\text { or, } t^{3}-500 t-3750=0 .

Solving the above equation by trial and error method,

t \cong 25.5 mm .