
#1
Dec2609, 02:13 PM

P: 54

1. The problem statement, all variables and given/known data
assuming that the maximum shear stress and torsion are the same in both shafts design a hollow shaft to replace the solid one . the ratio diameters is to be 0.6 for the maximum shear stress i calculated 79.95 x 10 ^6 mn/m2 and for the torsion i calculated 122.78x10^3 in the previous question. 2. Relevant equations (T) torque/ (J) polar second moment of area and (t) shear stress/ ( r) radius polar second moment of area = pie (d4d4)/32 radius = D/2 3. The attempt at a solution (j)= pie ((0.6)4d4)/32 = 0.08545 d4 (r) = 0.6/2 = 0.3 solid shaft = j/r = 152.41x10^6 / 99.25 x 10 ^3 = 1.535 x 10^3values from previous question hollow shaft = j/r = 0.08545/0.3 = 284.83 x 10^ 3 d 3 square root with 1.535 x 10^3/284.83 x 10^3 under it = 175.32 x 10^3 D= 0.6 x D = 0.6 x 175.32 x 10^3 = 105.19 x 10^3 therefore the dimensions are 175mm and 105mm i think these values are wrong as the solid shaft is 198.5mm and i always thought the hollow shaft would be larger in diameter please help as i dont no what i have done wrong.. 



#2
Dec2709, 09:06 AM

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#3
Dec2709, 10:43 AM

P: 54

Had another look today and relised where i have went wrong here is my new answers :
torsion / polar second moment of area = tau/ radius : (t) 122.78 x 10^3/(j) 0.08545 = (tau) 79.95 x 10 ^6/d/2 : (t) 122.78 x 10 ^3/(tau) 79.95 x 10 ^6 = (j) 0.08545 / d/2 D= 3 square root with 122.78 x 10^3/79.95 x 10 ^6 divide by 2 x 0.08545 = 207.90x10 ^3m d= 0.6 x D = 0.6 x 207.90 x 10 ^3 = 124.74 x 10 3m the dimensions for the hollow shaft are D=208mm and d=125mm hopefully i am correct this time!!!!! 



#4
Dec2709, 12:15 PM

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P: 5,963

solid and hollow shafts
Yes, that looks about right.




#5
Dec2709, 12:54 PM

P: 54

thanks for checking



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