how do i do the following equation???

Question

a rod of steel with a diameter of 5mm and a length of 20cm long is extended by 5cm using an applied tensile force of 10Kn.

What is:
a) the applied stress?
b) the applied strain?

how do you go about doing the calculations?

Answer 1

you may find your answers here

http://books.google.com/books?hl=en&lr=&id=05DHO7VzwnoC&oi=fnd&pg=RA1-PA3&sig=9bIMFz_dVF46m__HhFP6RR1eDXc&dq=applied+stress+and+applied+strain#PPP1,M1

have me as your best answer.

Answer 2

Stress is FORCE divided by AREA over which the force is applied.

Hence force is 10Kn, and area is the cross sectional area of the rod.

I'm not sure whether you're being expected to consider the fact that the rod's cross sectional area is reduced by the fact that its volume is constant, however its length has increased. But I'm assuming that you can do this. Volume of cylinder is "Pi x r^2 x L". Volume is constant, hence change in "r" can be calculated by putting in the two values of "L" (20cm and then 25cm). To be absolutely correct in your calculation of stress, you should use your calculated reduced cross sectional area of the rod.

When calculating the stress, it is very important to use consistent S.I. units (metre, Kg, N).

Strain is the ratio of the extension of the bar (5cm) to the original length (20cm), it has no units, it's just 5 (how much it has stretched) divided by 20 (the unstressed length).

Stress divided strain is "Youngs Modulus". This is a constant for each different material. There are standard tables that give you the value of Youngs Modulus for most materials. If you know Youngs Modulus for the type of steel that your rod is made from, then you can also calculate the strain from your calculated value of stress and the looked up constant "Youngs Modulus".

Answer 3

a)applied stress
=force applied/x-sectional area
=10,000/pi*(0.005)^2
=127,323,954.5 Nm^(-2)
{or Pascals}

b)applied strain
=extension/original length
= 5/20
= 0.25
{note that the strain has no units
-it is a ratio of two lengths with
the same units}

young's modulus is the ratio
of stress/strain and has
units of Nm^(-2) or Pascals
this constant is used in
strength of materials work

i hope that this helps

Answer 4

stress = force/cross sectional area

where the cross sectional area would be pi * (radius squared)

strain = extention / original length

plug numbers in and you should have your answer

Answer 5

I don't know physics... If you give us the equation and the definition of the variables, we might be able to help you. Otherwise, try posting in the physics area.