A Beginner's Guide to Structural Mechanics/Analysis

Transformed Moments of Interia

(c) 2007, T. Bartlett Quimby

Section ITR.3

Transformed Moment of Inertia

Last Revised: 11/04/2014

A Moment of Inertia, I, is a section property (i.e. solely depended on cross sectional dimensions) taken about a specific axis.  The general equation for I is:

I = ∫ y²dA

Figure ITR.3.1
General Moment of Inertia

For the case shown in Figure ITR.3.1, the differential area becomes:

dA = b(y) dy

So that the general equation becomes:

I  = ∫ b(y) y²dy

For the special case where b(y) is a constant the equation for moment of inertia becomes:

I  = b ∫ y²dy

For this case, we see that moment of inertia is linear with respect to dimensions that are parallel to the axis about which I is being determined.

The computation of I assumes a homogeneous material.  If a section made of more than one material, each with a different stiffness, E, then the section needs to be transformed into an equivalent section in one material.  In this case, equivalent means that the transformed section would result in a moment of inertia that yields that same defection results in elastic deformation equations as the composite section does by other means.

So, given the derivation above, the key to transforming a section into a homogeneous material is to only change those dimensions that are parallel to the reference axis. 

As previously shown, an area with equivalent stiffness is found as

A₁ = n A₂

For a differential rectangle:

b₁ dy = n b₂ dy

b₁ = n b₂

Where n is the ratio of E₂/E₁.  Following convention, E₂ would be the stiffer material and E₁ the less stiff and the equation above would be transforming a stiff material into a less stiff material, so:

b_{less stiff} = n b_stiffer

To go the other way you get:

b_stiffer = b_{less stiff} / n

Probably the best way to show this is by example.

Example #1:  Composite Wood & Steel

For the section shown in Figure ITR.3.2, the steel and wood are bonded together in such a way that they act as a composite section.  This section is fairly easy to deal with since it is doubly symmetric and will remain so after the transformations.

Use E_s = 29 x 10⁶ psi and E_w = 1.4 x 10⁶ psi.  The steel is the stiffer material.

n = 29/1.4 = 20.7 ... use n = 20.5

There are several routes that we can go here.

First, let's find I_xtr and I_ytr when converting the wood into steel:

First I_xtr,s.  Transform the wood dimension parallel to the "x" axis to an equivalent width:

b_w,s = 3.5" / 20.5 = 0.171 inches

The total width of the transformed-to-steel section is now

b_total,s = 0.375" + 0.171" + 0.375" = 0.921"

The resulting steel section is shown in Figure ITR.3.3a.

Since the section is a rectangle centered on the x axis, we can compute:

I_xtr,s = (0.921") (11.25")³ / 12 = 109.2 in⁴

Next I_ytr,s:  Transform the wood dimension parallel to the "y" axis to an equivalent

Figure ITR.3.3 Equivalent Steel Sections

 width:

b_w,s = 11.25" / 20.5 = 0.549 inches

The transformed section is shown in Figure ITR3.3b.  There are a couple of ways that you could compute I_ytr from this section.  Here is one:

I_ytr,s = (11.25") (4.25")³ / 12 - (11.25" - .549") (3.5")³ / 12 = 33.73 in⁴

These moments of inertia would be used in conjunction with E_s in deflection equations.  Stresses computed on the section would be accurate for the steel, but would needed to be modified by the factor n for the wood.

Now, let's convert the wood to equivalent wood.

For I_xtr,s we have an new width for each plate:

b_s,w = (0.375") (20.5) = 7.69"

for a total transformed section width of

b_total,w = 7.69" + 3.5" + 7.69" = 18.88"

The transformed section is shown in Figure ITR.3.4a.

Figure ITR.3.4
Equivalent Wood Sections

I_xtr,w = (18.88") (11.25")³ / 12 = 2,240 in⁴

To compute I_ytr,w we need to multiply the height dimension of the steel plates by n.

b_s,w = (11.25") (20.5) = 230.6 in

The transformed section is shown in Figure ITR.3.4b.

The result is:

I_ytr,w = (230.6") (4.25")³ /12 - (230.6" - 11.25") (3.5")³ / 12 = 691.5 in⁴

A check on the results should show that EI is approximately the same in both cases:

About the x axis

E_s I_xtr,s = (29x10⁶ psi) (109.2 in⁴) ~ E_w I_xtr,w = (1.4x10⁶ psi) (2240 in⁴)

3.167x10⁹ lb-in² ~ 3.136x10⁹ lb-in²

The difference in the result (about 1%) is the result of the rounding on the value for n and is close enough, in most cases, for what we do as engineers.  A similar result is found if you compare E_s I_ytr,s and E_w I_ytr,w.