It is used to connect two rods under the action of tensile load.

Applications:

1) Joint between the bars in root trusses

2) Joint between links of the suspension bridge

3) Joint between links of a bicycle chain

Q.1 A Knuckle joint is required to transmit reversible load of $25 kN$. Select suitable material for various parts. Decide failure modes of the component, select permissible stresses for material giving reason and design a Knuckle joint.

I) Selection of material:-

As joint is subjected to reversible stresses, then stresses induced will be repeated and joint may fail due to fatigue. Hence the selection of material is carried out by considering low carbon steel which will withstand stresses developed in the above case and cost of material will be within range.

Hence select C-30 as material for all parts [Hint: we can select in between C-15 to C-45]

II) Properties of material

From PSG 1.9

$\sigma_u=55\ kgt/mm^2=550\ N/mm^2$, PSG-1.9

$\sigma_y=30\ kgt/mm^2=300\ N/mm^2$ PSG-1.9

III) Selection of FOS:-

As joint is subjected to reversible load and to account for fatigue failure, let us select higher FOS based on $\sigma_y$ value.

Let us select FOS as 4.

IV) Permissible stresses:-

(a) From maximum principle stress theory-

$\sigma_t=\dfrac {\sigma_y}n=\dfrac {300}4=75\ N/mm^2$

(b) From shear stress theory

$\tau=\dfrac {0.5\sigma_y}{FOS}=37.5\ N/mm^2=37\ N/mm^2$

(c) Imperial relation

$\sigma_{br}=0.5\sigma_t=37.5\ N/mm^2=38\ N/mm^2$

V) Design of rod-

(a) Tensile failure of rod-

$\sigma_t=\dfrac P{\dfrac \pi4d^2}$

$75\times\dfrac\pi4 d^2=25\times10^3$

$d=20.60mm\approx22mm$

VI) Design of knuckle pin-

(a) Shear failure of pin-

$\tau=\dfrac P{2\left(\dfrac\pi4\right)d_p^2}$

$37=\dfrac{25\times10^3}{2\times\dfrac\pi4\times d_p^2}$

$d_p=20.72\ mm$

Diameter of pin=diameter of rod

$d_p=22\ mm$

(b) Bending of pin -

$M_{max}=\dfrac P2\left[\dfrac {t_1}3+\dfrac l2\right]-\dfrac P2\left[\dfrac l4\right]$

$M_{max}=\dfrac {25\times10^3}2\left[\dfrac {17}3+\dfrac {28}2\right]-\dfrac {25\times10^3}2\left[\dfrac {28}4\right]$

$M_{max}=158.33\times10^3\ mm^4$

$\sigma_{bending}=\dfrac MI.y=\dfrac M{\dfrac Iy}=\dfrac {158.33\times 10^3}{\dfrac {\dfrac \pi{64}d_p^4}{\dfrac {d_p}2}}=\dfrac {158.33\times 10^3}{\dfrac \pi{32}d_p^3}$

$\sigma_{bending}=\dfrac {158.33\times10^3}{\dfrac\pi{32}(22)^3}=151.4\ N/mm^2\gt\gt\gt\sigma_t=75\ N/mm^2$

Therefore, pin fails in bending. Hence, increases $d_p$

Therefore, take $d_p\approx28mm$

$\therefore \sigma_{bending}=\dfrac{158.33\times10^3}{\dfrac\pi{32}(28)^3}=73.4\ N/mm^2\lt75\ N/mm^2$

Therefore pin is safe in bending.

VII) Design of single eye end:-

(a) Tensile failure across knuckle pin hole-

$d_1=2d_p$.................Imperial Relation

$d_1=2(28)=56mm$

$\sigma_t=\dfrac P{(d_1-d_p)l}$

$\sigma_t=\dfrac{25\times10^3}{(56-28)28}$

$\sigma_t=31.88N/mm^2\lt75N/mm^2$

Design is safe in tension.

(b) Bending of a pin in the knuckle pin hole of single eye end.

$\sigma_{br}=\dfrac P{d_p\times L}$

$\sigma_{br}=\dfrac {25\times 10^3}{28\times28}$

$\sigma_{br}=31.88\ N/mm^2\lt38\ N/mm^2$

Therefore design is safe in bearing stress.

(c) Shear failure of single eye end beyond knuckle pin hole (double Shear).

$\tau=\dfrac P{2\left(\dfrac {d_1-d_p}2\right)l}$

$\tau=\dfrac {25\times10^3}{2\left(\dfrac {58-28}2\right)28}$

$\tau=31.88\lt37$

Therefore design of single eye end is safe.

VIII) Design of double eye end:-

(a) Tensile failure of double eye end at knuckle pin hole.

$\sigma_t=\dfrac{P}{(d_1-d_p)t_1\times2}$

$\sigma_t=\dfrac{25\times10^3}{(56-28)17\times2}$

$\sigma_t=26.26\ N/mm^2\lt75\ N/mm^2$

Therefore design of double eye end is safe in tension.

(b) Bearing failure of pin in knuckle pin hole of double eye end-

$\sigma_{br}=\dfrac{P}{d_p\times t_1\times2}$

$\sigma_{br}=\dfrac{25\times 10^3}{28\times17\times2}$

$\sigma_{br}=26.26\ N/mm^2\lt38\ N/mm^2$

Therefore Double eye end is safe in nearing failure.

(c) Shear failure of double eye end beyond knuckle pin hole:-

$\tau=\dfrac {P}{\left(\dfrac {d_1-d_p}2\right)\times t_1\times 4}$

$\tau=\dfrac {25\times10^3}{\left(\dfrac {56-28}2\right)\times 17\times 4}$

$\tau=26.26\ N/mm^2\lt37\ N/mm^2$

Therefore double eye end is safe in shear.

Therefore design of double eye end is safe.

(d) Thickness, $t_2$ of forked end $=t_1-2mm$...................Imperial relation

$=17-2\ mm=15\ mm$

IX) Miscellaneous dimensions:-

(a) Pin head diameter $d_3=1.5d$, PSG-7.139

$d_3=1.5\times22=33\ mm$

(b) Thickness of pin head, $t_3=0.5d$, PSG-pg 7.139

$t_3=0.5(22)=11mm$

(c) Collar thickness, $t_4=0.5d=0.5(22)=11mm$