Maximum Bore Diameter for Axial Tensile Load question

[MMCS]

A 25mm diameter solid circular section bar made from given material tested is to be
bored axially to produce a cylinder of uniform thickness. If this cylinder is then subjected
to an axial tensile load of 75kN, what is the maximum diameter of the bore possible if the
stress in the cylinder is not to exceed an allowable stress based on a factor of safety of
1.33 of the yield stress

Yield Stress: 320MPa
E: 205GPa
B = bore radius
Calculated allowed stress σ/1.33 = 241MPA
I use σ = F/A
A=F/σ


∏*( r - B)² = 75000/241MPa

∏*( 0.0125 - B)² = 75000/241MPa

Is this the right up to now? i have continued to work this out and don't get the correct answer of 5.59 x 10^-6

 

[rock.freak667]

Your equation on the left side should be

π(r²-B²)

 

[MMCS]

Ok so i get (1.563*10^-4-B²) = 9.906*10^-5

B²=(9.906*10^-5)-1.563*10^-4B²= -5.719*10^-5

It doesn't seem right have i went wrong?

 

[MMCS]

I have just changed some of that, put it in wrong

 

[SteamKing]

What are your units?

 

[MMCS]

i converted radius into metres before using them in the equation

 

[haruspex]

(1.563*10^-4-B²) = 9.906*10^-5

B²=(9.906*10^-5)-1.563*10^-4

Try that step again, being more careful with the signs.

 

[MMCS]

Paying attention to the signs i get 5.724*10^-5 = b^2
once i square root that my answer becomes way off from the correct answer of 5.59 x 10-6

 

[Chestermiller]

Paying attention to the signs i get 5.724*10^-5 = b^2
once i square root that my answer becomes way off from the correct answer of 5.59 x 10-6

That's because 5.59 x 10^-6 is not the correct answer. You have done the problem correctly now, and the value of b is 0.00756 m. The diameter 2b is 0.01513 m, or 15.3 mm. Congratulate yourself. Where did you get the 5.59 x 10^-6 from?