Question 6-8: Compute the radius of gyration for the cross section of a me...

Objective             Compute the radius of gyration with respect to axes X and Y.

Given                    A rectangular cross section; thickness = t = 2.5 mm; width = w = 30 mm.

Analysis               Use Equation (6–6). r = \sqrt{I/A} .

Results                 r_{x} with respect to the x-axis. First, let’s compute the moment of inertia with respect to the x-axis. From Appendix A–1, we find the general form of the equation for the moment of inertia of a rectangle to be I = bh^{3}/12. Note that b is the dimension parallel to the axis of interest and h is the dimension perpendicular to the axis of interest. In this case, b = t = 2.5 mm and h = w = 30.0 mm. Then,

 I_{X} =   tw³/12 = (2.5 mm)(75.0 mm)³/12 = 5625  mm^{4}  

Also, A = tw = (2.5 mm)(30.0 mm) = 75.0 mm²

Then the radius of gyration is

r_{X} = \sqrt{I_{X}/A}= \sqrt{(5625  mm^{4})/(75.0  mm^{2})} = 8.66 mm

r_{y} with respect to the y-axis. Similarly, let’s first compute the moment of inertia with respect to the y-axis. In this case, b = w = 30.0 mm and h = t = 2.5 mm. Then,

 I_{Y} =   tw³/12 = (30.0 mm)(2.5 mm)³/12 = 39.06 mm^{4}  

Also, A = tw = (2.5 mm)(30.0 mm) = 75.0 mm²

Then the radius of gyration is

r_{Y} = \sqrt{I_{Y}/A}= \sqrt{(39.06  mm^{4})/(75.0  mm^{2})} = 0.722 mm

Comment               It is obvious that the choice of the axis about which to compute the radius of gyration is critical to the result. Here  I_{X} =  is much greater than  I_{Y} =  , and therefore,  r_{X} =  than  r_{Y} =  is much greater . You should gain an appreciation of the importance of the selection of the axis from this example. You will see the effect of this observation in Chapter 11 on column analysis and design. The result is that the meterstick would buckle about its y-axis, not the x-axis. It will always buckle about the axis with the smallest radius of gyration.

Alternate Solution    Where a formula for computing the radius of gyration is available, it makes the computation much simpler. For the rectangular section of this problem, Appendix A–1 gives

r_{X} = h/ \sqrt{12} = 0.2887h

and

r_{Y} = b/ \sqrt{12} = 0.2887b

These formulas for r_{X} and r_{Y} are easily derived from the basic definition here:

r_{X} = \sqrt{I/A} = \sqrt{(bh³/12)/(bh)} = \sqrt{h² /12} = h/ \sqrt{12}

r_{Y} =  \sqrt{I/A} = \sqrt{(bh³/12)/(bh)} = \sqrt{b² /12} = b/ \sqrt{12}

In this case, h = w = 30.0 mm and b = t = 2.5 mm. Then,

r_{X} = 0.2887 w= (0.2887)(30.0 mm) = 8.66 mm

and

r_{Y} = 0.2887 t= (0.2887)(2.5 mm) = 0.722 mm

These values match those found using Equation (6–6).