Elastic modulus:

Now as bars BA is fixed at A, and its elastic modulus; moment of inertia and length is E₁, I₁, L₁ then the moment M′_BA which can cause a rotation θ at end B will be equal to 4 E₁ I₁/L₁. θ from Eq. (3.7).

Similarly, for member BD, M′_BD =   4 E₂ I₂/L₂. θ and so on.

But the sum of all these central moments should be equal to the balancing moment M. In other words the balancing moment M will be distributed in certain ratios to each of these members. Algebraically it could be expressed as

M′_BA + M′_BD + M′_BC + M′_BE = M

Or (4 E₁ I₁/L₁) θ + (4 E₂ I₂/L₂) θ +(4 E₃ I₃/L₃) θ+(4 E₄ I₄/L₄) θ =M

Still, if all the bars are of the same material, E is same for all of them and can be taken as E.

∴          (4E θ)(I₁/l₁ +I₂/l₂ +I₃/l₃ +I₄/l₄) =M

or        4E θ =   M/∑ I/l

Here, the ratio, Moment of inertia/ Length = I/l of any member is called its stiffness ratio and is denoted by K.

∴          4E θ =   M∑ K =        M/K₁+K₂+K₃+K₄

Substituting back in M′_BA etc., we get

 M′ _BA = (4E θ)I1/L1 =M/∑ K (I1/L1) =M (K1/∑ K)

M'_DA = M/∑K (I₂/L₂) =M(K₂/∑K)

M'_CA = M/∑K (I₃/L₃) =M(K₃/∑K)

M'_EA = M/∑K (I₄/L₄) =M(K₄/∑K)

Thus, the balancing moment M has to be allotted to each member in ratio of their respective stiffnesses, and the multiplying factors.K₁/∑ K etc. are called the moment distribution factors or simply distribution factors (DF).

DF = K_i/∑ K

It is exciting to remember that the sum of the distribution factors of all the members meeting at a joint is equal to 1.