Two round tie rod subjected to tensile loading are connected by a coupling device which is known as turnbuckle.

Q.1 Design turnbuckle to connect to tie rods. Tie rods are subjected to a maximum pull of $80\ kN$. Take axial adjustment $75\ mm$.

Answer: -

I) Material selection:-

(a) For tie rod

Tie rods are subjected to tensile force and torsional moment. Therefore considering strength as a criteria for material selection, from PSG 1.9, select C-30

II) Properties of material:-

(a) For tie rod

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

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

III) Selection of FOS

As tie rod is subjected to tensile and torsional moment it and to account for stress concentration in threads, let us select $FOS=n=4$ based on $\sigma_y$ value.

IV) Permissible stresses

(a) For maximum principal (normal) stress theory -

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

(b) Maximum shear stress theory -

$\tau=\dfrac {0.5\sigma_y}n=\dfrac {0.5\times300}4=37.5\ N/mm^2$

$\tau\approx37\ N/mm^2$

V) Selection of material:-

For coupler:- select cast iron as a material for the coupler. It will reduce the number of machining operations, also the cost of cast iron material is less as compared to mild steel. Cast iron is where resistant material which will occur in threads. From PSG-1.4, GCI-20

VI) Properties of material:-

$\sigma_u=200N/mm^2$

VI) Selection of FOS:-

As cast iron is brittle material, and to consider the effect of high of torsional shear stress and stress concentration in threads, select $FOS=n=5$, based $\sigma_u$ value.

VIII) Permissible stress:- From max. Principal stress theory

$\sigma_t=\tau=\dfrac{\sigma_u}n=\dfrac{200}5=40\ N/mm^2$

IX) Design procedure:-

As tie rods are subjected to tensile stress and torsional shear stress, to compensate the effect of both, increase design load (it will be greater than 30% of given value)

$\therefore P_d=1.3(P)=1.3(80)=104\ kN$

(A) Design of tie rod:-

(a) Tensile failure of tie rod-

$\sigma_t=\dfrac {P_d}{\dfrac\pi4d_C^2}$

$75=\dfrac{104\times10^3}{\dfrac\pi4d_c^2}$

$d_c=42.01\ mm\approx42\ mm$

PSG-5.42, As standard bolt size is not available,

$\therefore$ use $d_c=0.84d$ ......................Imperial relation.

$\therefore \dfrac {42}{0.84}d$

$\therefore d=50mm$............outer diameter

(B) Design of coupler:-

(a) $D_1=d+[6\text{ to }10\ mm]$.............. Imperical formula

$D_1=d+6=50+6=56\ mm$

(b) Consider tensile failure of coupler at section A-A

$\sigma_t=\dfrac{P_d}{\dfrac\pi4[D_2^2-D_1^2]}$

$D_2=80.2\ mm=81\ mm$

(c) Consider tensile failure of coupler at plane B-B

$\sigma_t=\dfrac{P_d}{\dfrac\pi4[D^2-d^2]}$

$\therefore 40=\dfrac{104\times10^3}{\dfrac\pi4[D_2^2-50^2]}$

$\therefore D=76.22\ mm\approx77\ mm$

(d) Shear failure of coupler-

$\tau=\dfrac{P_d}{\pi dl}$

$40=\dfrac{104\times10^3}{\pi\times50\times l}$

$\therefore l=16.55\ mm=17\ mm$

(e) Total length:-

$l+adjustment+clearance+l$

$=17+75+20+17=129\ mm$