A Beginner's Guide to Structural Mechanics/Analysis Transformed Moments of Interia (c) 2007, T. Bartlett Quimby
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Section ITR.2

Equivalent Stiffness

Last Revised: 11/04/2014

To find areas of equivalent stiffness consider axial deflection of a bar in tension.

D = PL/AE

If two bars have the same stiffness then they will have the same deflection under the same force when the bars are the same length.

Consider Bar #1 having an area of A1 and a Modulus of Elasticity, E1, and a Bar #2 having an area of A2 and a Modulus of Elasticity, E2.  For equivalent stiffness then:

D1 = D2

PL/(A1E1) = PL/(A2E2)

A1E1 = A2E2

So... for two areas to be equivalent then :

A1 = A2 (E2/ E1)

The ratio (E2/E1) is referred to as the "modular ratio" as is commonly designated with the symbol "n".

Consequently, to have equivalent areas:

A1 = nA2

which means that the area in material 1, A1, is equivalent to the area in material 2, A2, times n, where n is the ratio of the modulus of elasticity of the original material to the modulus of elasticity of the material that we are transforming the member into.

Flexural stiffness is proportional to EI, so similarly loaded and supported beams made of two different materials are equivalent if:

E1I1 = E2I2

Which leads to the condition for equivalent moments of inertia:

I1 = I2 (E2/E1) = nI2

I is linear with respect to the cross sectional dimension parallel to the reference axis used to compute I so n becomes a linear scale factor for that dimension.

Practical Selection of n

Typically, values of n are computed as the ratio of the stiffer material to the less stiff material, making n to be greater than one.  The result is:

To transform a stiffer material into an equivalent less stiff material:

ATR = n Aoriginal

To transform a less stiff material into an equivalent stiffer material:

ATR = Aoriginal / n

The value of "n" is often (particularly when using composite concrete and steel members) rounded to the nearest half integer (i.e. 9, 8.5, 8, 7.5, etc...).