Calculating Dimensional Change in 2-D Loading System with Mechanical Principles

[tone999]

Hey guys I am a first timer here. I need some help with a couple of questions I am quite stumped.

1. A bar in a 2-D loading system is 1 metre long and 20 millimeters square in section before loading. Determine the change in its "x" and "y" dimensions if E=200GPa and Poisson's Ratio is 0.3

The force acting in the "x" direction is 60KN and in the "y" direction 40KN.

ok, i know i have to find the stress in the x and y directions first. By dividing the Force by the Area of the bar. e.g. 60000 Newtons/ mm2.

i can't seem to work out what the area of the bar would be, i dnt understand "20 millimeters square in section" i need the area in mm2 also.

Then i would work out the combined strain in the xx direction and yy direction.

Finally to work out the change in length i multiply the strain by the original length.

Any feed back would be great, thanks

 

[FredGarvin]

1. A bar in a 2-D loading system is 1 metre long and 20 millimeters square in section before loading. Determine the change in its "x" and "y" dimensions if E=200GPa and Poisson's Ratio is 0.3

The force acting in the "x" direction is 60KN and in the "y" direction 40KN.

ok, i know i have to find the stress in the x and y directions first. By dividing the Force by the Area of the bar. e.g. 60000 Newtons/ mm2.

i can't seem to work out what the area of the bar would be, i dnt understand "20 millimeters square in section" i need the area in mm2 also.

You are directly given the area of the cross section. The area is A=20 mm^2 You will need to convert that to m^2. So now you have the combined tensile stress with the bending stress. You also have Poisson's ratio for calculating the transverse strain.

Then i would work out the combined strain in the xx direction and yy direction.

Finally to work out the change in length i multiply the strain by the original length.

Any feed back would be great, thanks

From there on out you seem to have a good grasp of the problem.

 

[tone999]

Hi Fred thanks for the reply :). I think i got a little messed up with my formula but hopefully its ok this time if you could take a look?

To get the stress in the axial direction i divided the force (60,000N) by the area 20mm2 (0.0004m) which gave me 150,000,000 or 150x10 to the power of 6.

Then to get the axial strain i divided the stress 150,000,000 by the Elastic modulus 200x10 to the power of 9 which gave me 0.00075.

Then to get the change in the x direction i multiply the strain in the x direction by the orignal length, 0.00075 x 1 = 0.0075m or 0.75mm ?

Then to get the lateral or transverse strain i multiply poisson's ratio by the axial strain. 0.3 x 0.0075 = 0.000225

Finally to get the change in the y direction i multiply the y strain by the original length 0.000225 x 0.02 = -0.0045mm

Does that sound right?

 

[radou]

To get the stress in the axial direction i divided the force (60,000N) by the area 20mm2 (0.0004m) which gave me 150,000,000 or 150x10 to the power of 6.

Watch out for the units! 1 [mm^2] = 1 \cdot 10^{-6} [m^2].

 

[FredGarvin]

Like was already mentioned, your units conversion is not correct on your area. The original area is A = 20 mm^2, that equates to A = 20x10^{-6} m^2