Calculating Maximum Bending Moment and Radius of Curvature for a Round Beam

[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

 

[haruspex]

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.

 

[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.

 

[haruspex]

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.