1. Aug 29, 2015

### Scott Metcalfe

Hi all,

I have hit what seems to be a mental mind / stumbling block on the question below. I am looking for some guidance wherever possible.

1. The problem statement, all variables and given/known data

c) The column exerts a lateral force on the base plate of 50kN due to wind loading. If Poisson's ratio = 0.34 determine the change in the base plate dimensions under this two dimensional loading. Assume a value of Young's Modulus of E = 2 x 1011 N/m2 for steel.

Info relating to the question -
Base plate = 300 x 300 x 15mm
Force in the X direction = 300kN
Force in the Y Direction = 50kN
E = 2 x 10^11
V = 0.34

Here is my attempt at a solution -

Direct Stress

σX = F/A
(300 x 10^3) / (0.3 x 0.3) = 3.33 x 10^6

σY = F/A
(50 x 10^3) / (0.3 x 0.015) = 1.11 x 10^7

Direct Strain

X = (1/E)(σx -Vσy)

(1 / (2 x 10^11))(3.33x10^6 -0.34(1.11x10^7)) = 2.22 x 10^-6

Y = (1/E)(σy -Vσx)
(1 / (2 x 10^11))(1.11x10^7 -0.34(3.33x10^6)) = 4.98 x 10^-5

I am pretty confident with the above, its the dimensional changes i am not so.... here's my attempt -

Dimension Change

ΔL = Direct Strain(x) x X
(2.22 x 10^-6) x 0.3 = 6.66 x 10^-7

ΔL = Direct Strain(y) x Y

(4.98 x 10^-5) x 0.015 = 7.47 x 10^-7

Scott

2. Aug 29, 2015

### OldEngr63

You really need a figure to make your problem clear.