Solving the Mohrs Circle Problem for Aluminium Angle Iron

[Kalus]

The basic problem involves a piece of aluminium angle iron (L section) pinned at the top, with a force acting at the bottom horizontally. This angle iron can be rotated between 0 and 180 degrees. The aim is to find the deflections and then calculate the second moment of area of the beam

see http://www.utem.edu.my/fkm/index.php?option=com_docman&task=doc_download&gid=276&Itemid=69 for a better idea of the experiment

The deflection is read by two dial gauges, U (the deflection in the direction of the force) and V (the deflection normal to the line of F) for masses between 0 and 500g and head angles every 22.5 degrees between 0 and 180.

The U and V values (in mm) i have plotted (y axis) against mass (x axis, in grams) for each of the 9 head angles. I've then read off the gradients dU/dP and dV/dP (where p is the mass) for each experiment. These gradients I've then converted from units of mm/g to m/N.

I've then plotted the 9 (one coordinate point for each head angle) sets of dU/dP, dV/dP on a graph to form a mohrs circle. The center of this circle is 1.3x10^-7 from the origin and 7x10^-8 as a radius.

Then to find the values for the second moment of area Ix (Iy is done in a similar fashion but replacing OC +R with OC-R)

Ix = L^3/ (3E(OC+R))

Where
L = length of specimen =(0.47m)
E= youngs modulus of material= (AL= 69x10^9 Pa)
OC= Distance from origin center of mohrs circle (m/N) =1.3x10^-7
R= Radius of mohrs circle (m/N)= 7x10^-8

This gives an answer of around 2.5x10^-6 (m^4) or 2500000 mm^4

This seems quite small, is it feasible? I wish i had the dimensions of the L shape of the beam so i could calculate Ix and Iy the easy way to verify my answer.

Thanks,
Kalus

 

[Achy47]

Dear Kalus,

Thank you for sharing your experiment and calculations with us. It seems like you have done a thorough job in obtaining your data and analyzing it. However, in order to accurately determine the second moment of area of the beam, we would need more information about the dimensions and material properties of the angle iron.

The second moment of area, also known as the moment of inertia, is a property of a cross-sectional shape that is related to its resistance to bending. It is calculated by integrating the cross-sectional area with respect to the distance from an axis of rotation. Therefore, in order to calculate the second moment of area for your angle iron, we would need to know the dimensions of the angle iron, such as the width and thickness of the flanges and the length of the legs.

Additionally, the material properties of the angle iron, such as the yield strength and modulus of elasticity, would also affect the calculation of the second moment of area. These properties can vary depending on the type and grade of the aluminum used.

Without this information, it is difficult to determine the accuracy of your calculated value of 2.5x10^-6 m^4. It is always good to verify your results by using multiple methods or cross-checking with known values. If possible, try to obtain the dimensions and material properties of the angle iron to ensure the accuracy of your results.

I hope this helps and good luck with your experiment!