# Analysis of Sub-Frames Using Stiffness Method: A solved Example Several codes of practice in the world allow us to idealise structures into 2-dimensional frames for the purpose of simplified analysis. For sub-frames, it is obvious that the force method becomes less handy due to high number of redundants, and the next best alternative is the displacement method, where we solve for the unknown displacements.

In this post, we have a typical example where a problem that would have generated (21 x 21) matrix using force method has been solved using (4 x 4) matrix by displacement method. Another approach to the solution of this problem is the moment distribution method. But in this case, the displacement remains the fastest.

For the frame loaded as shown above, we have to start by drawing the kinematic basic system of the structure. This has been achieved by fixing all the nodes against rotation as shown below.

We will now have to evaluate the basic system for different cases of rotation.

Analysis of Case 1
Z1 = 1.0; Z2 = Z= Z= 0

K11 = (4EI/3) + (4EI/4) + (13.6EI/3.825) = 5.889EI
K21 =  (6.8EI/3.825) = 1.778EI
K31 = 0
K41 = 0

Analysis of Case 2
Z2 = 1.0; Z2 = Z= Z= 0

K12 = (6.8EI/3.825) = 1.778EI
K22 = (4EI/3) + (4EI/4) + (13.6EI/3.825) + (13.6EI/2.8) = 10.7460EI
K32 = (6.8EI/2.8) =  2.4286EI
K42 = 0

Analysis of Case 3
Z3 = 1.0; Z1 = Z= Z= 0

K13 = 0
K23 = (4EI/3) + (4EI/4) + (13.6EI/3.825) + (13.6EI/2.8) = 10.7460EI= 2.4286EI
K33 = (4EI/3) + (4EI/4) + (13.6EI/3.325) + (13.6EI/2.8) = 11.2807EI
K43 = (6.8EI/3.325) = 2.0451EI

Analysis of Case 4
Z4 = 1.0; Z1 = Z= Z= 0

K14 = 0
K24 = 0
K34 =  (6.8EI/3.325) = 2.045EI
K44 = (4EI/3) + (4EI/4) + (13.6EI/3.325) = 6.4236EI

Stiffness coefficient due to externally applied load;

K1P = -(q1L12/12) = -(26.148 × 3.8252) / 12 = -31.880 KNm

K2P = (q1L12/12) – (q2L22/12) = [(26.148 × 3.8252) / 12] – [(25.437 × 2.802) / 12] = 15.261 KNm

K3P = (q2L22/12) – (q3L32/12) = [(25.437 × 2.82) / 12] – [(27.345 × 3.3252) / 12] = -8.5741 KNm

K1P = (q3L32/12) = (27.345 × 3.3252) / 12 = 25.193 KNm

The appropriate cannonical equation;

K11Z1 + K12Z2 + K13Z3 + K14Z4 + K1P = 0
K21Z1 + K22Z2 + K23Z3 + K24Z4 + K2P = 0
K31Z1 + K32Z2 + K33Z3 + K34Z4 + K3P = 0
K41Z1 + K42Z2 + K43Z3 + K44Z4 + K4P = 0

On substituting;

5.889Z1 + 1.778Z2     +    0Z3        + 0Z4         = 31.880
1.778Z1 + 10.746Z2   + 2.4286Z3  + 0Z4         = -15.261
0Z1        + 2.4286Z2   + 11.287Z3  + 2.045Z4  = 8.5741
0Z1        + 0Z2            + 2.045Z3  + 6.4236Z4  = -25.193

On solving;

Now that we have obtained the rotations, we can now substitute and obtain the moments at the critical points;

Mi = M1Z1 + M2Z2 + M3Z3 + M4Z4 + P

Bottom column support moments
MA = (6.3102/EI) × (2EI/4) = 3.1551 KNm
MB = (-2.9701/EI) × (2EI/4) = -1.485 KNm
MC = (2.2384/EI) × (2EI/4) = 1.1192 KNm
MD = (-4.6346/EI) × (2EI/4) = -2.3173 KNm

Beam Support Moments
M1R = [(6.3102/EI) × (13.6EI/3.825)] – [(2.9701/EI) × (6.8EI/3.825)] – 31.880 = -14.7239 KNm

M2L = [(6.3102/EI) × (6.8EI/3.825)] – [(2.9701/EI) × (13.6EI/3.825)] + 31.880 = 32.537 KNm

M2R = [-(2.9701/EI) × (13.6EI/2.8)] + [(2.23841/EI) × (6.8EI/2.8)] – 16.619 = -25.6090 KNm

M3L = [-(2.9701/EI) × (6.8EI/2.8)] + [(2.2384/EI) × (13.6EI/2.8)] + 16.619 = 20.278 KNm

M4R = [(2.2384/EI) × (13.6EI/3.325)] – [(4.6346/EI) × (6.8EI/3.325)] – 25.193 = -25.5157 KNm

M3L = [(2.2384/EI) × (6.8EI/3.325)] – [(4.6346/EI) × (13.6EI/3.325)] + 25.193 = 10.814 KNm

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1. Hello Bro,

The coefficient stiffness 13.6 i'm not seeing where it's derived from? Can you explain how you derived it? thanks

Fc

2. Sorry that I did not make the post very 'down to earth'. The moment from unit rotation at any near end is given by 4EI/L and at any far end is 2EI/L.

So 4 x 3.4EI = 13.6EI/L

3. The beam stiffness was given, I missed that! Just to be clear, the beam stiffness is factored up due to the column stiffness, this is my understand.

What if the upper column had a smaller stiffness compared to the lower column would it be and average factor times the beam stiffness?

Thanks
Frank

4. No, there is nothing like average in handling relative stiffness of Structural members. Note that there is no rule that you must work in terms of column stiffness. You can also work in terms of beam stiffness.

For example, if the upper column has lesser stiffness, we can have something like; 0.75EI for upper column, EI for lower column, and 3.4 EI for beams.

5. mr nobody

Where did the 3.4 in 3.4EI come from?

• Ubani Obinna

It simply means that the flexural rigidity of the beams is 3.4 times higher than that of the columns. It is usually due to a difference in the geometry of the sections.

For instance if the beam is a T-beam with a flange width of 1500mm, and the columns are just 230 x 230mm, you should expect a significant difference in their flexural rigidity (EI).

6. Eng. Michael Amarasekera

If we assume that the bottom column ends are pinned to foundation, how do we solve the problem? In the book “Designed and Detailed (BS 8110 : 1997) authored by J.B. Higgins and B.R. Rogers and published by British Cement Association (BCA) – A multi Storey Building has been designed using sub frame analysis, assuming bottom columns pinned at foundation level. But calculations are not given. It says that “A linear elastic analysis either produced by hand or by computer program is used to obtain the moments and forces”. I such a case will there be a change in moments and forces at beam-column joints at first floor level.
I am a civil engineer in Sri Lanka. Thank You.

7. Michael Amarasekera

How do we proceed if the bottom columns are assumed pinned at foundation level?
Will there be any changes in moments at beam-column joints at floor level.
Thank you.

8. Michael Amarasekera

I found the method.
In such a case, the respective Beam Stiffness have to be taken as 0.75 of the normal stiffness.
The results obtained tally with the results of the reference book.
I solved both examples using SMath Studio programming language which is totally free.
I will be happy to share if anyone needs them.
Thank you so much.