Answer:
The maximum deflection is 0.325 mm
Explanation:
we have the following data given:
m = 40 kg
L = 0.5 m
h = 80 mm = 0.08 m
b = 20 mm = 0.02 m
We calculate W:
W = m*g = 40 * 9.8 = 392.4 N
we will calculate the moment of inertia of the area which will be equal to:
I = (b*h^3)/12 = (0.02*0.08^3)/12 = 8.56x10^-7
The Young´s modulus is equal to:
E = (20*270)/(100) + (3.2*(100-20))/100 = 58.6x10^9 Pa
The deflection is equal to:
δ = (W*L^3)/(3*E*I) = (392.4 * (0.5^3))/(3*58.6x10^9*8.56x10^-7) = 0.000325 m = 0.325 mm