Deflection and Modulus of Elasticity

[karmatic]

Homework Statement

Calculate the deflection at the end of a rod whose dimensions are 1m x 0.1m x 0.1m, when a load of 100N is applied. The modulus of elasticity is given as 1 x 10¹¹ (N/m²)

Homework Equations

d = PL/AE where

d = end deflection of bar in metres (in m)
P = the applied load in Newtons (N)
L = length of the bar (in m)
A = cross sectional area of bar (in m²)
E = modulus of elasticity (in N/m²)

The Attempt at a Solution

d = PL/AE
d = 100N x 1m/1^-4(1 x 10¹¹)
d = 100Nm/1^-4 x 1 x 1^-4 x 10¹¹
d = 1^-05

I'm not sure if I've calculated that in the right order, or if I have missed any steps. I'm a little bit lost on how to approach this problem!

 

[PhanthomJay]

Your approach is fine, but your maths is not so good. The Area of the rod is 1 X 10^-2 m² (that is, 0.01 m²). Please redo the maths and don't forget to note the units.

 

[karmatic]

Okay had another try at this one just now..

d=PL/AE
d=(1m*100N)/((1m*〖10〗^(-2))(1*〖10〗^11))
d=100Nm/((1m*1/100)(100000000000))
d=1/1000000
d=1^(-6)

I think that's the right answer, have I missed anything? The cross sectional area got me the first time round, pretty stupid mistake!

edit - I forgot the units for the final answer but I'm not sure what they should be, is it in metres?

 

[PhanthomJay]

Okay had another try at this one just now..

d=PL/AE
d=(1m*100N)/((1m*〖10〗^(-2))(1*〖10〗^11))
d=100Nm/((1m*1/100)(100000000000))
d=1/1000000
d=1^(-6)

I think that's the right answer, have I missed anything? The cross sectional area got me the first time round, pretty stupid mistake!

edit - I forgot the units for the final answer but I'm not sure what they should be, is it in metres?

If you stick with Newton and meter units, your result for the deflection should be in meters (PL/AE has units of N*m/((m^2)(N/m^2)) = N*m/N = m). You are off by a decimal point, the answer should be d = 1 X 10^-7 m.. Watch your scientific notation, and keep track of the decimal point especially when using SI units of measure.