Discussion :: Strength of Materials - Section 1 (Q.No.3)
|Maggie said: (Aug 11, 2013)|
|The strength of a solid shaft in torsion is given by,
T(solid shaft) = (pi/16)*(max stres)*(D^3).
T(hollow shaft) = (pi/16)*(max stres)*(D^4-d^4)/D.
On solving we get 15/16.
|Avinash said: (Aug 27, 2013)|
|max stress = Tymax/Ip; where Ip = polar moment of inertia.
ymax = D/2.
max stress(solid) = 16T/piD^3.
max stress(hollow) = 32T(D/2)/pi(D^4-d^4).
Replacing d = D/2.
The ratio we get is 15/16.
|Pradhyumna said: (Oct 6, 2013)|
|Polar modulus is a measure of strength of a shaft in a torsion.
So, simply calculate the ratios of polar modulus of solid to hollow shafts.
Polar modulus = Polar moment of inertia/Radius of shaft.
|Pawan said: (Sep 26, 2014)|
|But ratio obtained 16/15 how?|
|Murli Meena said: (Oct 5, 2014)|
|Dia of solid shaft = D.
DIA OF HOLLOW SHAFT = d/2.
The ratio of solid shaft = 16.
The ratio of hollow shaft = 15.
Total tension in shafts = 15/16.
|Anil said: (Nov 19, 2014)|
|The maximum stress due to torsion in solid shaft is 16T/d3.
The maximum stress due to torsion in solid shaft is 16x16T/15d3.
Therefore ratio maximum stress of solid shaft to hollow shaft is simply: 15/16.
T/J = s/R = G0/l.
|Virendra said: (Nov 25, 2014)|
|Ratio will be 16/15.
Simply if equal outer diameter then solid shaft will be more strong. But if material are same then hollow shaft will be more strong.
|Raj said: (Dec 10, 2014)|
|As @Pradhyumna said earlier that polar moment of inertia(J) decide the strength of the member.
J for solid section = D^4 and for Hallow section = inner D^4-outer D^4.
i.e D^4-(D/2)^4 = 15/16D^4. So the ratio is 15/16.
|Bharat said: (Jun 9, 2015)|
|Zp(solid)/Zp(hollow) = (pi*D^3/16)/(pi*(D^4-(D/2)^4)/16D) = 16/15.|
|Parna Sengupta said: (Jul 6, 2015)|
|For a hollow shaft whose external diameter is d and internal diameter is d/2, then what will be the polar modulus?|
|Umesh said: (Aug 8, 2015)|
|Shaft is said to be more strong if it resist moment and tmax is not the parameter to be used for it.
Tmax is used to compared material property and here we are calculating section strength, so maximum moment of resistant should be compared and answer for it will be 16/15.
|Ajeet said: (Aug 28, 2015)|
|What is diameter of sun and internal and external heat?|
|Amit Pandit said: (Dec 26, 2015)|
|For outer diameter (odd) moment of inertia = d^4.
Inner diameter, moment of inertia = (d/2)^4 = (d^4)/16.
For hollow shaft = d^4-(d^4)/16.
So, the ratio of hollow shaft is 15/16.
|Raju Katara said: (Nov 21, 2016)|
|A hollow steel shaft of 300 mm external diameter and 200 mm internal diameter has to be replaced by a solid steel shaft of 300 mm external diameter and 200 mm internal diameter has to be replaced by a solid steel shaft of?
Give me the answer.
|Dwarika Prasad Sahoo said: (Dec 24, 2016)|
|If Solid and Hollow shaft are of the same material and same weight, then Hollow shaft has more strength.
But when outer diameter is same (weight will not be same) then Solid shaft has more strength, so here ratio of the strength of solid and Hollow shaft should be more than 1 as outer diameter same.
|Shatakshi said: (Jan 26, 2017)|
|It's true, ultimately the ratio of Tmax (Solid) / Tmax (Hollow) = D^4/(D^4-d^4).
Let D=20 mm & d=10mm, then D^4=160000 mm^4 & d^4=10000 mm^4.
Therefore, D^4/(D^4-d^4) = 16/15.
|Raghavendra said: (Apr 26, 2017)|
|Yeah, the answer must be 16/15.|
|Krishnavn said: (May 3, 2017)|
|Here, the answer must be 16/15.|
|Bhagchand Golhani said: (Jun 29, 2017)|
|Maharaj said: (Jul 11, 2017)|
Yes, you are right.
The answer is 16/15.
|Ravikumar said: (Aug 11, 2017)|
|The solid section max shear stress is 16T/pi D3.
Hollow section is 16T/piD^3*[1-[D^4/d^4]].
The ratio of these to is 15/16 once remembering d=D/2.
|Akshay said: (Aug 14, 2017)|
|I also get Answer is 16/15.|
|Salman said: (Nov 27, 2017)|
|Shear stress= 16T/πD3(solid).
|Gangadhar said: (Jan 9, 2018)|
|Simply says (1-(1^4÷2÷4)=15÷16.|
|Nitish said: (Jan 15, 2018)|
|Abdul Sovan Khan said: (Feb 17, 2018)|
|((π/16)*maximum shere stress*(outside dia)^3)/((π/16)*maximum shere stress*((out side dia^4-inside dia^4)/inside dia)=16D^4/15D^4=16/15.|
|Neel said: (Feb 27, 2018)|
|Ration obtain is 16/15.|
|Azzu said: (Mar 8, 2018)|
|Strength proportional to section modulus,
So, I got 16/15.
|Ekon said: (Apr 16, 2018)|
|Answer should be 16/15. If the answer is 15/16 then it is implying that the hollow shaft is stronger than the solid shaft of same outer diameter, which is not correct.|
|Basavaraj said: (Apr 29, 2019)|
|I think we get 16/15.|
|Santosh said: (May 8, 2019)|
|It is 16/15.|
|Prachi said: (Aug 31, 2019)|
|Yes, it is 16/15 because it has asked for the ratio of solid to the hollow shaft.|
|Abidha Karthu said: (Jan 10, 2020)|
|Yes, the ratio of solid to hollow will be 16/15.|
|Deepak Kumar said: (Mar 29, 2020)|
Ds is diameter of solid shaft,
we get 15/16,
|Pratyush said: (May 10, 2020)|
|Here torsion of solid shaft/torsion of the hollow shaft.
|Bikalp said: (May 23, 2020)|
|The Answer should be 16/15.|
|Parth said: (Aug 4, 2020)|
The question is ratio of the solid shaft to the hollow shaft not hollow shaft to solid shaft.
So the answer will be : 16/15.
|Boss said: (May 20, 2021)|
|Therefore the torsion = th/ts.
Hence, torsion (t) =th/ts=15/16.
|Rajvir Singh said: (Jun 8, 2021)|
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