Subject: Machine Design -I

Topic: Design against static Loads, Bolted and welded joints under eccentric loading. Power Screw

Difficulty: Medium

Step 1: Selection of material:

The rods are subjected to tensile force and strength is the criterion for the selection of rod material. On the basis of strength, the material of the two rods and the cotter is selected as plain carbon steel of Grade 30C8 (Syt=400MPa)

Step 2: Selection of factor of safety:

The FOS for the rods, spigot end and socket end is assumed as 6, while for cotter, it is 4. Ther are two reasons for assuming a lower FOS for the cotter. They are as follows.

(i). No stress concentration on the cotter. (ii). Low cost of the cotter.

Step 3: Calculation of permissible stresses:

$\sigma_t=\frac{S_yt}{fs}=\frac{400}{6}=66.67N/mm^2$

$\sigma_t=\frac{S_yc}{fs}=\frac{2*400}{6}=133.33N/mm^2$

$\tau=\frac{S_sy}{fs}=\frac{0.5*400}{6}=33.33N/mm^2$

Permissible stresses for the cotter are as follows:

$\sigma_t=\frac{S_yt}{fs}=\frac{400}{4}=100N/mm^2$

$\tau=\frac{S_sy}{fs}=\frac{0.5*400}{4}=50N/mm^2$

Step 4: Calculation of dimensions:

Diameter of rods:

$d=\sqrt\frac{4P}{\pi \sigma_t}$=$\sqrt\frac{4(50*10^3)}{\pi (66.67)}=30.90 or 32mm$

Thickness of cotter:

$t=0.31d=0.31(32)=9.92 or 10mm$

Diameter(d2) of spigot:

$P=[\frac{\pi}{4}d_2^2-d_2t]*\sigma_t$

$50*10^3=[\frac{\pi}{4}d_2^2-d_2*10]*66.67$

$d_2^2-12.73d_2-954.88=0$

$d_2=\frac{12.73 \pm \sqrt(12.73^2-4(-954.88))}{2}$

$d_2=37.91 or 40mm$

Outer diameter (d1) or socket:

$P=[\frac{\pi}{4}(d_1^2-d_2^2)-(d_1-d_2)t]*\sigma_t$

$50*10^3=[\frac{\pi}{4}(d_1^2-40^2)-(d_1-40)10]*66.67$

$or d_1^2-12.73d_1-2045.59=0$

$d_1=\frac{12.73 \pm \sqrt12.73^2-4(-2045.59)}{2}$

$d_1=52.04 or 55mm$

Diameters of spigot collar(d3) and socket collar(d4):

$d_3=1.5d=1.5(32)=48mm$ $d_4=2.4d=2.4(32)=76.8 or 80mm$

Dimensions a and c

$a=c=0.75d=0.75(32)=24mm$

Width of cotter:

$b=\frac{P}{2 \tau t}=\frac{50*10^3}{2*50*10}=50mm$

Check for crushing and shear stresses in spigot end:

$\sigma_c=\frac{P}{td_2}=\frac{50*10^3}{10*40}=125N/mm^2$

$\tau=\frac{P}{2ad_2}=\frac{50*10^3}{2*24*40}=26.04N/mm^2$

Check for crushing and shear stresses in socket end:

$\sigma_c=\frac{P}{(d_4-d_2)t}=\frac{50*10^3}{(80-40)(10)}=125 N/mm^2$

$\tau=\frac{P}{2(d_4-d_2)c}=\frac{50*10^3}{2(80-40)(24)}=26.04 N/mm^2$

$\sigma_c \lt 133.33 N/mm^2 and \tau\lt33.33 N/mm^2$

The stresses induced in the spigot and the socket ends are within limits.

Thickness of spigot collar:

$t_1=0.45d=0.45(32)=14.4 or 15mm$

The taper for the cotter is 1 in 32.