A marble column of cross-sectional area 1.5 m2 supports a mass of
30,000 kg.

a.What is the stress within the column?


b.What is the strain?


c.By how much is the column shortened if it is 10.5 m high? 
Take the Young's Modulus for marble as 50 ´ 109 Nm-2.

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Request for Question Clarification by hedgie-ga on 02 Apr 2003 22:23 PST

Kindly explain value of E

E = 50 ´ 109 Nm-2.

   One would expect  E to be about 50 G Pa

that is E= 50 E9 Pa = 50,000,000,000 N m-2

is character '  a typo?

This web page might be of use to you:
http://newt.phys.unsw.edu.au/~epe/1250.tutorials/1250.tutorial6.html

a) stress is force/area.  force is 30,000 kg * 9.8 N/kg = 294000 N.
So stress is 196000 N/m^2.

b) strain is stress/YM, so is .00000392  (unitless).

c) the column is shortened by its height * strain, so by .041 mm.

hedgie has made at least three mistakes in his answer (besides the one
he already corrected).  First the problem states that the
cross-sectional area is 1.5 square meters, but hedgie uses the area of
a circle with radius 1.5, which is 7.07.  Second he made an error in
part (a) where he evaluates 30 E3 * 10 as 3 E4, rather than 3 E5. 
Third, he used 1.5 m rather than 10.5 m for the height of the column.

The correct answers were already present in a comment before the
error-laden 'official' answer was posted.  The posted answer is an
embarassment. A retraction is called for.

Racecar

 I have indeed made numerical errors.
 Also I did not review the comments
 before answering, so I did not notice your solution.

 
 I have asked for withdrawal of the answer.
 Because of the weekend, it may take some time.
  Thanks for pointing out the mistakes.