EK Physics Lecture 5 Question 16- Tensile Stress

[qtsjoe]

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A single steel column is to support a mass of 1.5 x 10^8 kg . If the yield strength for steel is 2.5 x 10^8 N/m^2 and safety regulations require the column to withstand five times the weight it presently holds , what should be the approximate cross sectional area of the base of this column.

A).6 m^2
B)3 ''
C)6 ''
D)30 ''

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I think we treat the yield strength as Young's Modulus- 2.5x 10^-8 set that equal to Stress/ Strain or ( (F/A) / (change in dimension/ original dimesion) )

It asks to withstand five times the weight it presently holds ( 1.5 x 10^8) so lets say weight is "X"...then you need to multiply by 10 to get Newtons or Force.

My question is how do we factor strain in when treating this problem, if I carry out the calculation without factoring straight and just solve for A, I get an answer way off.

Thanks in advanced.

 

[MedPR]

I got 30.. Is that right?

2.5*10^8=(5)(10)(1.5*10^8)/A

A=(7.5*10^9)/(2.5*10^8)
A=3*10^1
A=30

 

[chiddler]

I got 30.. Is that right?

2.5*10^8=(5)(10)(1.5*10^8)/A

A=(7.5*10^9)/(2.5*10^8)
A=3*10^1
A=30

The formula is Y = FL/A(deltaL), why didn't you use L and deltaL?

 

[MedPR]

I thought stress = force/area.

So the most stress it can handle = yield strength for a force that is 5x the mass to be supported (5mg) divided by the area of the thing handling the stress.

 

[dmf2682]

I got 30.. Is that right?

2.5*10^8=(5)(10)(1.5*10^8)/A

A=(7.5*10^9)/(2.5*10^8)
A=3*10^1
A=30

You got it buddy.

The problem has nothing to do with deflection or strain. A straightforward stress problem.

 

[chiddler]

I thought stress = force/area.

So the most stress it can handle = yield strength for a force that is 5x the mass to be supported (5mg) divided by the area of the thing handling the stress.

oh. yield strength ≠ young's modulus.