A cube with 30-cm-long sides is sitting on the bottom of an aquarium in which the water is one meter deep. (Round your answers t
o the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1000 kg/m3.) (a) Estimate the hydrostatic force on the top of the cube. N (b) Estimate the hydrostatic force on one of the sides of the cube. N
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Answer:
τ (bc. max) =25.37 MPa
Explanation:
From the question, T = 1.3Kn.m;
(G = 77.2 GPa) and from the image of this solid shaft system i attached;
d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m
So ΣT = 0 → Ta + Tc = 1.3Kn.m
So the system is statically indeterminate.
Let's check at the equation that makes it compatible ;
ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0
[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =
ΣT = 0 and T(bc) = T(c)
ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m
Now,
[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0
So, T(c) = 273374 Nmm = 273.374Nm
T(a) = 1300Nm - 273.374Nm = 1026. 63Nm
From the beginning we saw that;
T(bc) = T(c) = 273.374Nm
Now let's find the maximum shear stress in shaft BC;
τ (bc. max) = {τ (bc) x r(bc)} / J(bc)
τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707
= 25.37 MPa
