Answer to Question #229117 in Mechanical Engineering for SRABANI

Question

A foot-lever is 1 metre from the center of shaft to the point of application of 600 N load. Find: i) diameter of the shaft, ii) dimensions of the key, iii) dimensions of rectangular arm of the foot lever at 60 mm from the center of shaft assuming width of the arm as 3 times thickness. The allowable tensile stress may be taken as 75 N/mm2 and allowable shear stress as 70 N/mm2.

Answer 1

(i)T=P*L

=600*1000=600*10³Nmm

$600*10^{3}=\frac{π}{16}*\tau*d^{3}$

$600*10^{3}=\frac{π}{16}*70*d^{3}$

600*10³=13.74d³

$d^{3}=\frac{600*10^{3}}{13.74}=43668.12$

d=35.21mm.

$d\approx 35mm$

(ii)the standard dimensions of the key for a 35mm diameter shaft are:

width of key= 12mm

thickness of key= 8mm

length of key is obtained by considering the shearing of the key

T=600*10³=l₁*w*$\tau*\frac{d}{2}$

=l₁*12*70*17.5=14700l₁

l₁=$\frac{600*10^{3}}{14700}=40.82mm$

This is always taken to be equal to the length of the boss l₁=l₂$\approx 41mm$m

(iii)let t=thickness of arm in mm

B=width of arm in mm=3t

Bending moment at 60mm from the center of the shaft; M=600(1000-60)=564*10³Nmm

Section modulus,Z=$\frac{1}{6}*t*B^{2}$

=$\frac{1}{6}*t(3t)^{2}=0.5t^{3}$

tensile stress$(\sigma_t)=75=\frac{M}{Z}$

$75=\frac{564*10^{3}}{0.5t^{3}}=\frac{11.28*10^{5}}{t^{3}}$

$t^{3}=\frac{11.28*10^{5}}{75}=15040$

t=24.68

$t\approx 25mm$

B=3t=3*25=75mm

Width of arm is tapered while thickness is kept constant throughout

Width of the arm on the foot plate side;B₁=$\frac{B}{2}=37.5mm$