Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 3)
3.
The ratio of strengths of solid to hollow shafts, both having outside diameter D and hollow having inside diameter D/2, in torsion, is
Discussion:
42 comments Page 1 of 5.
Dwarika prasad sahoo said:
9 years ago
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.
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.
UMESH said:
1 decade ago
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.
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.
Salman said:
8 years ago
Shear stress= 16T/πD3(solid).
Shear stress=16T/πD3(1-d4/D4)
=16T/π(D4-d4)/D
=16T/π/(15D4/16)/D
=16*16T/(π*15*D3)
Ratio=16T/πD3:16*16T/15πD3.
= 15/16.
Shear stress=16T/πD3(1-d4/D4)
=16T/π(D4-d4)/D
=16T/π/(15D4/16)/D
=16*16T/(π*15*D3)
Ratio=16T/πD3:16*16T/15πD3.
= 15/16.
(1)
Anil said:
1 decade ago
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.
Torsional Equation:
T/J = s/R = G0/l.
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.
Torsional Equation:
T/J = s/R = G0/l.
Raju Katara said:
9 years ago
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.
Give me the answer.
(1)
Raj said:
1 decade ago
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.
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.
Pradhyumna said:
1 decade ago
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.
So, simply calculate the ratios of polar modulus of solid to hollow shafts.
Polar modulus = Polar moment of inertia/Radius of shaft.
AMIT PANDIT said:
10 years ago
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.
= (15/16)d^4.
So, the ratio of hollow shaft is 15/16.
Inner diameter, moment of inertia = (d/2)^4 = (d^4)/16.
For hollow shaft = d^4-(d^4)/16.
= (15/16)d^4.
So, the ratio of hollow shaft is 15/16.
Avinash said:
1 decade ago
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.
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.
Shatakshi said:
9 years ago
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.
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.
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