Bending Moment

  • Thread starter Confusedbiomedeng
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  • #1
Confusedbiomedeng

Homework Statement


A round bar .125mm in diameter , is to be used as a beam. IF Youngs modulus For the material E=200x103N/mm2 and the stress due to bending is limited to 17.5N/mm2
A) Maximum allowable bending moment
B) the radius of curvature at the point of Maximum bending moment

Homework Equations


σ/y=m/i=E/R
Ina=πD4/64

The Attempt at a Solution


A) 17.5/(125/2)=M/1.1x107 Bending moment =3080000
Ina=1.1x107

B) 17.5/125/2=7/25

=> R= 7/25/200x103 R=1.4mm
 

Answers and Replies

  • #2
haruspex
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Homework Statement


A round bar .125mm in diameter , is to be used as a beam. IF Youngs modulus For the material E=200x103N/mm2 and the stress due to bending is limited to 17.5N/mm2
A) Maximum allowable bending moment
B) the radius of curvature at the point of Maximum bending moment

Homework Equations


σ/y=m/i=E/R
Ina=πD4/64

The Attempt at a Solution


A) 17.5/(125/2)=M/1.1x107 Bending moment =3080000
Ina=1.1x107

B) 17.5/125/2=7/25

=> R= 7/25/200x103 R=1.4mm
As mentioned in your other thread, there seem to be some issues with units conversion.
Please post your working again, in a bit more detail, specifying units at all times.
 
  • #3
Confusedbiomedeng
so for allowable bending moment i took the stress 17.5 N/mm2 and divided it by 125/2 assuming neutral axis to be down the centre which gave me and answer of 7/25 this was then equal to m over π(1254 )/64which is the equation for moment of inertia, that was an answer of 1.1x107. to get M alone i multiplied both sides by 1.1x107 giving 308x104 N/mm4 since in the question the values are given in terms of mm i left the diameter in mm.

and for B) again I put stress over neutral axis and got 7/25 put that equal to 200x103 over R. To get R alone i divided both sides by 200x103 to get R out as 1.4mm

does this answer your previous question.
 
  • #4
haruspex
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so for allowable bending moment i took the stress 17.5 N/mm2 and divided it by 125/2 assuming neutral axis to be down the centre which gave me and answer of 7/25 this was then equal to m over π(1254 )/64which is the equation for moment of inertia, that was an answer of 1.1x107. to get M alone i multiplied both sides by 1.1x107 giving 308x104 N/mm4 since in the question the values are given in terms of mm i left the diameter in mm.

and for B) again I put stress over neutral axis and got 7/25 put that equal to 200x103 over R. To get R alone i divided both sides by 200x103 to get R out as 1.4mm

does this answer your previous question.
That isn't quite what I asked for, but anyway...
> R= 7/25/200x103 R=1.4mm
I still do not understand this line. Your quoted equation is that σ/y = E/R, but you seem to have done R=(σ/y)/E. Further, I do not understand how 0.28/(200x103) gives 1.4. Shouldn't it be 1.4x10-6? Since you believe you have standardised on mm as the unit of distance, there should be no final units conversion.
 

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