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. 4.20

A curved link of the mechanism made from a round steel bar is shown in Fig. 4.67. The material of the link is plain carbon steel 30C8 \left(S_{y t}=400 N / mm ^{2}\right)  and the factor of safety is 3.5. Determine the dimensions of the link.

Step-by-Step

Verified Solution

\text { Given } \quad P=1 kN \quad S_{y t}=400 N / mm ^{2} .

(fs) = 3.5.

Step I Calculation of permissible tensile stress

\sigma_{\max }=\frac{S_{y t}}{(f s)}=\frac{400}{3.5}=114.29 N / mm ^{2} .

Step II Calculation of eccentricity (e)
At the section XX,

R = 4D

R_{i}=4 D-0.5 D=3.5 D .

R_{o}=4 D+0.5 D=4.5 D .

From Eq. (4.60),

R_{N}=\frac{\left(\sqrt{R_{o}}+\sqrt{R_{i}}\right)^{2}}{4}             (4.60).

R_{N}=\frac{\left(\sqrt{R_{o}}+\sqrt{R_{i}}\right)^{2}}{4} .

=\frac{(\sqrt{4.5 D}+\sqrt{3.5 D})^{2}}{4}=3.9843 D .

e=R-R_{N}=4 D-3.9843 D=0.0157 D .

Step III Calculation of bending stress

h_{i}=R_{N}-R_{i}=3.9843 D-3.5 D=0.4843 D .

A=\frac{\pi}{4} D^{2}=\left(0.7854 D^{2}\right) mm ^{2} .

M_{b}=1000 \times 4 D=(4000 D) N – mm .

From Eq. (4.56), the bending stress at the inner fibre is given by,

\sigma_{b i}=\frac{M_{b} h_{i}}{A e R_{i}}             (4.56).

\sigma_{b i}=\frac{M_{b} h_{i}}{A e R_{i}}=\frac{(4000 D)(0.4843 D)}{\left(0.7854 D^{2}\right)(0.0157 D)(3.5 D)}.

=\left(\frac{44886.51}{D^{2}}\right) N / mm ^{2}             (i).

Step IV Calculation of direct tensile stress

\sigma_{t}=\frac{P}{A}=\frac{1000}{\left(0.7854 D^{2}\right)}=\left(\frac{1273.24}{D^{2}}\right) N / mm ^{2}            (ii).

Step V Calculation of dimensions of link Superimposing the bending and direct tensile stresses and equating the resultant stress to permissible stress, we have

\sigma_{b i}+\sigma_{t}=\sigma_{\max } .

\left(\frac{44886.51}{D^{2}}\right)+\left(\frac{1273.24}{D^{2}}\right)=114.29 .

D = 20.10 mm.