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(adsbygoogle = window.adsbygoogle || []).push({});

see http://www.utem.edu.my/fkm/index.php?option=com_docman&task=doc_download&gid=276&Itemid=69 [Broken] 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

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# Mohrs Circle Problem

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