Terminology of leaf spring:-

$n_e \ or \ n_f =$ Number of the extract full length leave

$n_g = graduated \ leave$

n = total number of leaves

$n = n_e + n_g$

b = Width of the each leave

t = thichkness of the each leaves

$2L_1$ = overfall length of the spring

2L = effective length of the spring

2L = $2L_1 - l$

where, $l =$ width of the band/ distance between U-bold i.e. ineffective length

2P = force acting at the centre of the spring

P = force acting of the spring

$\sigma_{be}$ = bending stress inducing the full length leaves

$\sigma_{bg}$ = bending stress induce in the graduated leave

Q.1. The leaf spring as the 12 number of leaf the spring supprots are 1.1 m apart and central bond is 90 mm wde A centre load is to be taken 5.5 KN with a permissible stress of 300 $N/mm^2$. Determine

1) Thickness and width of the steel spring leave

2) Deflection of the spring

Take the ratio of total depth to width of the spring as 3,

Solution:

$n = 12$

$2L_1 = 1.1 m = 1100 mm$

$l = 90 mm$

$2 P = 5.5 KN$

$P = \frac{5.5}{2} = 2750 N$

$\sigma_b = 300 N/mm^2$

Effective length

2L = $2L_1 - l$

$L = \frac{1010}{2} = 505 mm$

$b = 4t$

Assuming all leaves are pre-stress

(PSG 7.104)

$\sigma_b = \frac{6PL}{nbt^2}$

$t = 8.33 mm$

$b = 4\times t$

$b = 33.33 \ mm$

$Y = \delta = \frac{12PL^3}{Ebt^3(3n_e + 2n_9)}$

Assume $n_e$ on $n_F = 2$

$n = n_e + n_g$

$12 = 2+n_g$

$n_g = 10$

$E = 2.1\times 10^5 N/mm^2$

$Y = \delta = \frac{12\times2750\times(505)^3}{2.1\times10^5\times33.33\times8.33^3(3\times2 + 2\times10)}$

$Y = \delta = 40.40 mm$

Note:

1) The approx relation between the radius of the curvature and camber (Y) of the spring is given by $R = \frac{(L_1)^2}{2y}$

The exact relation is given by

$y(2R +y) = (L_1)^2$

2) length of the leaf spring given as

length of $1^{st}$ = $\frac{Effective length}{n - 1} + Ineffective \ length$

length of $2^{nd}$ = $\frac{Effective length}{n - 1}\times 2 + Ineffective \ length$

length of $3^{rd}$ = $\frac{Effective length}{n - 1}\times 3 + Ineffective \ length$

length of $(n-1)^{th}$ = $\frac{Effective length}{n - 1}\times (n-1) + Ineffective \ length$

length of master leaf $= 24 + \pi (d+t)\times 2$

where, $d =$ inside diameter of eye

$t =$ thickness of leave