Subject: Machine Design -I

Topic: Curved beams and Thin cylinder

Difficulty: High

Step 1: Calculation of permissible tensile stress:

$\tau_max=\frac{S_ut}{fs}=\frac{200}{3}=66.67N/mm^2$

Step 2: Calculation of eccentricity (e):

b_i=3t

h=3t

R_i=2t

R_o=5t

t_i=t

t=0.75t

$R_N=\frac{t_i(b_i-t)+th}{(b_i-t)log_e(\frac{R_i+t_i}{R_i})+tlog_e\frac{R_o}{R_i}}$

$R_N=\frac{t(3t-0.75t)+0.75t(3t)}{(3t-0.75t)log_e(\frac{2t+t}{2t})+0.75tlog_e\frac{5t}{2t}}$

$=2.8134t$

$R=R_i+\frac{0.5th^2+0.5t_i^2(b_i-t)}{th+t_i(b_i-t)}$

$R=2t+\frac{0.5*0.75t*3t^2+0.5t^2(3t-0.75t)}{0.75t*3t+t(3t-0.75t)}=3t$

$e=R-R_N=3t-2.8134t=0.1866t$

Step 3: Calculation of bending stress:

$h_i=R_N-R_i=2.8134t-2t=0.8134t$

$A=3t*t+0.75t*2t=4.5t^2 mm^2$

$M_b=100*10^3(1000+3t) N-mm$

bending stress at the inner fibre is given by,

$\sigma_bi=\frac{M_bh_i}{AeR_i}=\frac{100*10^3(1000+3t)(0.8134t)}{4.5t^2*0.1866t*2t}=\frac{100*10^3(1000+3t)(2.1795)}{4.5t^2} N/mm^2$

Step 4: Calculation of direct tensile stress

$\sigma_t=\frac{P}{A}=\frac{100*10^3}{4.5t^2} N/mm^2$

Step 5: Calculation of dimensions of cross-section

$\sigma_N+\sigma_t=\sigma_max$

$\frac{100*10^3(1000+3t)(2.1795)}{4.5t^2}+\frac{100*10^3}{4.5t^2}=66.67$

$t^3-2512.83t-726500=0$

Adding the two stresses and equating the resultant stress to permissible stress,

t=99.2mm or t=100mm