Belt tension using bending and shear stresses

[kaminho]

Well I know I need to plot Mohr's circle for this, but what are the things i need to do before reaching that stage.here is the question:
A 250mm diameter pulley is mounted on a 50mm diameter shaft at a distance of 200mm from a ball bearing. If the belt tension ratio is 3 to 1 determine the maximum belt tension if the stress in the shaft is not to exceed 150MPa.
Thanks

 

[Skrambles]

Draw a free body diagram and balance your forces.

 

[mawais15]

you need to incorporate the fatigue factor as well .

 

[kaminho]

you need to incorporate the fatigue factor as well .

well at this stage I'm ignoring fatigue. i have to do it by finding the shear stress and bending stress and the mohr's circle eventually. but what i need is how can i find those two stresses using bending theory and torsion theory. I know for one that the torque in the pulley is (3p-p).(r) or 2p.D/2 = pD (D is diameter). but I am stuck here.

 

[mawais15]

you can use Euler's theory for finding those stresses

 

[rock.freak667]

well at this stage I'm ignoring fatigue. i have to do it by finding the shear stress and bending stress and the mohr's circle eventually. but what i need is how can i find those two stresses using bending theory and torsion theory. I know for one that the torque in the pulley is (3p-p).(r) or 2p.D/2 = pD (D is diameter). but I am stuck here.

Hello there kaminho,

You know that both torsion and bending is taking place.

You rightfully got the torque at T=pD.

These equations should be in your relevant equations:

\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}

\frac{T}{J} = \frac{\tau}{r} = \frac{G \theta}{L}

\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \frac{1}{2}\sqrt{(\sigma_x - \sigma_y)^2+4\tau_{xy}^2}

Since you have T, you can compute τ using the above formula. Similarly, you will need to get the bending moment M.

What would be the resultant force acting on the end of the shaft due to the tensions?

 

[kaminho]

Hello there kaminho,

You know that both torsion and bending is taking place.

You rightfully got the torque at T=pD.

These equations should be in your relevant equations:

\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}

\frac{T}{J} = \frac{\tau}{r} = \frac{G \theta}{L}

\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \frac{1}{2}\sqrt{(\sigma_x - \sigma_y)^2+4\tau_{xy}^2}

Since you have T, you can compute τ using the above formula. Similarly, you will need to get the bending moment M.

What would be the resultant force acting on the end of the shaft due to the tensions?

Thank you,
I think one of the things I'm unsure about is whether that 150MPa (in the question) is bending stress or shear stress. assuming its shear, I used the equation:
\frac{T}{J} = \frac{\tau}{r} = \frac{G \theta}{L}
I first worked out J=\pid^4/32 = \pi0.05^4/32 = 6.136 x 10^-7 m^4
so
\frac{0.25P}{6.136x10^-7} = \frac{\150x10^6}{0.025} which gives P=14.73 kN<br /> <br /> if I&#039;m right upto here, then I think Bending moment will be:<br /> BM= (4P)(0.2) = (4 x 14.73 x 10^3)(0.2) = 11.78 kNm<br /> <br /> now i have not used the first equation you mentioned and so I think I&#039;m missing something. also I&#039;m not sure about the answer to the last part of your qoute. Is it 4P ?

 

[kaminho]

Hello there kaminho,

You know that both torsion and bending is taking place.

You rightfully got the torque at T=pD.

These equations should be in your relevant equations:

\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}

\frac{T}{J} = \frac{\tau}{r} = \frac{G \theta}{L}

\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \frac{1}{2}\sqrt{(\sigma_x - \sigma_y)^2+4\tau_{xy}^2}

Since you have T, you can compute τ using the above formula. Similarly, you will need to get the bending moment M.

What would be the resultant force acting on the end of the shaft due to the tensions?

I think i made an awful mistake in my previous post. As \tau is needed, it can not be the 150MPa in the question. Let me ask it this way. In equation:
\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}
what are the unknowns? because i guess M (bending moment) is (4P)(0.2) and I (second moment of area) for the shaft can be calculated using (pi)(d^4)/64 where d=50mm. now is \sigma =150MPa or is it to be found? and is y=d/2=25 mm (because i think its the distance to neutral axis of the shaft). but if these are correct then P can be easily found which i sense its not right. or should i be getting a \sigma value in terms of P (tension) ?
and to find \tau using torsion theory, i guess i had the J value right (in my previous post) therefore i will get a value for \tau in terms of P (tension). Am I in the right direction? Thank you.

 

[nvn]

kaminho: The unknown is P; and y = 25 mm. Notice sigma_1 is the last equation in post 6. I think sigma_1 = 150 MPa. And sigma in the bending stress equation, above, is sigma_x in the last equation of post 6. Also, sigma_y = 0 in the last equation of post 6. Try again.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 200 mm, not 200mm. And 150 MPa, not 150MPa. See the international standard for writing units[/color] (ISO 31-0[/color]).

 

[rock.freak667]

kaminho: The unknown is P; and y = 25 mm. Notice sigma_1 is the last equation in post 6. I think sigma_1 = 150 MPa. And sigma in the bending stress equation, above, is sigma_x in the last equation of post 6. Also, sigma_y = 0 in the last equation of post 6.

In addition to this, the \tau you get from the torsion theory is the \tau_{xy} in the last equation.

 

[kaminho]

In addition to this, the \tau you get from the torsion theory is the \tau_{xy} in the last equation.

I get:
I for shaft = (pi)(d^4)/64 = (pi) (0.05^4)/64 = 3.07 x 10^-7 m^4
\sigma= My/I = (4P x 0.2)(0.025) / (3.07 x 10^-7) = 65147P Pa
(so this is my sigma x in last equation)

for torsion equation:
J= (pi)(d^4)/32 = 6.136 x 10^-7 m^4 (i used shaft's diameter i.e. 50 mm)
T/J = \tau/r so 0.25P/(6.136 x 10^-7) = \tau/ 0.025 so \tau= 10190P Pa (here for torque i used pulley's diameter of 250 mm. i think the torque in the pulley and the shaft is the same though)

so will the last equation be looking like:
150 x 10^6 = 65147P/2 ± (0.5)sqrt((65147P)^2+ 4(10190P)^2 ? i took sigma y = 0
however it does not seem to have a solution?

 

[nvn]

Yes, that is algebra. Go ahead and solve it. For the +/- sign, use the plus sign; ignore the minus sign.

 

[rock.freak667]

so will the last equation be looking like:
150 x 10^6 = 65147P/2 ± (0.5)sqrt((65147P)^2+ 4(10190P)^2 ? i took sigma y = 0
however it does not seem to have a solution?

you would set σ₁=σ_des=150 MPa

such that

150(10⁶) = 65147P/2 + (0.5)√[((65147P)²+ 4(10190P)²]

you would get a solution since the √ will give you something like 150(10⁶)= 65147P/2 + 0.5(<some number>*P) in which you can then solve for P.

 

[kaminho]

Thank you rock.freak667 and nvn.
I'm getting P=2.25 kN...
cheers.