Design of Hollow Shaft to Replace Solid Shaft for Same Shear Stress and Torsion

[series111]

Homework Statement

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.

Homework Equations

(T) torque/ (J) polar second moment of area and (t) shear stress/ ( r) radius

polar second moment of area = pie (d4-d4)/32

radius = D/2

The Attempt at a Solution

(j)= pie ((0.6)4-d4)/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 don't no what i have done wrong..

 

[PhanthomJay]

Homework Statement

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.

Homework Equations

(T) torque/ (J) polar second moment of area and (t) shear stress/ ( r) radius

polar second moment of area = pie (d4-d4)/32

that's pie(d_o^4 - d_i^4)/32

radius = D/2

please watch your subscripts, r = d_o/2

The Attempt at a Solution

(j)= pie ((0.6)4-d4)/32 = 0.08545 d4

the equation should be pie(d_o^4 - (0.6d_o)^4))/32, but your answer is correct

(r) = 0.6/2 = 0.3

r is d_o/2

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 don't no what i have done wrong..

Yes, it should be larger in diameter; please correct your errors and try again. Be careful with the math, there are a lot of places where you can go wrong..

 

[series111]

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!

 

[PhanthomJay]

Yes, that looks about right.

 

[series111]

thanks for checking