Several codes of practice in the world allow us to idealise structures into 2-dimensional frames for simplified analysis. For sub-frames, it is obvious that the force method becomes less handy due to the high number of redundants, and the next best alternative is the displacement method (stiffness method), where we solve for the unknown displacements.

The displacement method, also known as the stiffness method, has become a cornerstone of structural analysis. It offers a powerful and versatile framework for solving equilibrium problems in various structures, from trusses and beams to complex frames and continua.

At its core, the stiffness method focuses on determining the unknown displacements of points within a structure under the action of applied loads. This is achieved by establishing a relationship between the displacements and the internal forces generated within the structure. This relationship is expressed through the stiffness matrix, which encodes the inherent rigidity of the structure and its resistance to deformations.

**Solved Example Using the Stiffness Method**

In this article, we have a typical example where a problem that would have generated a (21 x 21) matrix using the force method, has been solved using a (4 x 4) matrix by the stiffness (displacement) method. Another approach to the solution of this problem is the moment distribution method. But in this case, the stiffness method 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 the nodes 1, 2, 3 and 4 against rotation as shown below.

We will now have to evaluate the basic system for different cases of a unit rotation Z_{i} = 1.0, applied at the fixed nodes.

Analysis of Case 1

Z_{1} = 1.0; Z_{2} = Z_{3 }= Z_{4 }= 0

K_{11} = (4EI/3) + (4EI/4) + (13.6EI/3.825) = 5.889EI

K_{21} = (6.8EI/3.825) = 1.778EI

K_{31} = 0

K_{41} = 0

**Analysis of Case 2**

Z_{2} = 1.0; Z_{2} = Z_{3 }= Z_{4 }= 0

K_{12} = (6.8EI/3.825) = 1.778EI

K_{22} = (4EI/3) + (4EI/4) + (13.6EI/3.825) + (13.6EI/2.8) = 10.7460EI

K_{32} = (6.8EI/2.8) = 2.4286EI

K_{42} = 0

**Analysis of Case 3**

Z_{3} = 1.0; Z_{1} = Z_{2 }= Z_{4 }= 0

K_{13} = 0

K_{23} = (4EI/3) + (4EI/4) + (13.6EI/3.825) + (13.6EI/2.8) = 10.7460EI= 2.4286EI

K_{33} = (4EI/3) + (4EI/4) + (13.6EI/3.325) + (13.6EI/2.8) = 11.2807EI

K_{43} = (6.8EI/3.325) = 2.0451EI

**Analysis of Case 4**

Z_{4} = 1.0; Z_{1} = Z_{2 }= Z_{3 }= 0

K_{14} = 0

K_{24} = 0

K_{34} = (6.8EI/3.325) = 2.045EI

K_{44} = (4EI/3) + (4EI/4) + (13.6EI/3.325) = 6.4236EI

**Stiffness coefficient due to externally applied load;**

K_{1P }= -(q_{1}L_{1}^{2}/12) = -(26.148 × 3.825^{2}) / 12 = -31.880 kNm

K_{2P }= (q_{1}L_{1}^{2}/12) – (q_{2}L_{2}^{2}/12) = [(26.148 × 3.825^{2}) / 12] – [(25.437 × 2.80^{2}) / 12] = 15.261 kNm

K_{3P }= (q_{2}L_{2}^{2}/12) – (q_{3}L_{3}^{2}/12) = [(25.437 × 2.8^{2}) / 12] – [(27.345 × 3.325^{2}) / 12] = -8.5741 kNm

K_{4P }= (q_{3}L_{3}^{2}/12) = (27.345 × 3.325^{2}) / 12 = 25.193 kNm

The appropriate cannonical equation;

K_{11}Z_{1} + K_{12}Z_{2} + K_{13}Z_{3} + K_{14}Z_{4} + K_{1P} = 0

K_{21}Z_{1} + K_{22}Z_{2} + K_{23}Z_{3} + K_{24}Z_{4} + K_{2P} = 0

K_{31}Z_{1} + K_{32}Z_{2} + K_{33}Z_{3} + K_{34}Z_{4} + K_{3P} = 0

K_{41}Z_{1} + K_{42}Z_{2} + K_{43}Z_{3} + K_{44}Z_{4} + K_{4P} = 0

On substituting;

5.889Z_{1} + 1.778Z_{2} + 0Z_{3} + 0Z_{4} = 31.880

1.778Z_{1} + 10.746Z_{2} + 2.4286Z_{3} + 0Z_{4} = -15.261

0Z_{1} + 2.4286Z_{2} + 11.287Z_{3} + 2.045Z_{4} = 8.5741

0Z_{1} + 0Z_{2} + 2.045Z_{3} + 6.4236Z_{4} = -25.193

On solving;

Z_{1} = 6.3102/EI (radians)

Z_{2} = -2.9701/EI (radians)

Z_{3} = 2.2384/EI (radians)

Z_{4} = -4.6346/EI (radians)

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

**M _{i} = M_{1}Z_{1} + M_{2}Z_{2} + M_{3}Z_{3} + M_{4}Z_{4} + P**

**Bottom column support moments**

M_{A} = (6.3102/EI) × (2EI/4) = 3.1551 kNm

M_{B} = (-2.9701/EI) × (2EI/4) = -1.485 kNm

M_{C} = (2.2384/EI) × (2EI/4) = 1.1192 kNm

M_{D} = (-4.6346/EI) × (2EI/4) = -2.3173 kNm

**Beam Support Moments**

M_{1}^{R} = [(6.3102/EI) × (13.6EI/3.825)] – [(2.9701/EI) × (6.8EI/3.825)] – 31.880 = -14.7239 kNm

M_{2}^{L} = [(6.3102/EI) × (6.8EI/3.825)] – [(2.9701/EI) × (13.6EI/3.825)] + 31.880 = 32.537 kNm

M_{2}^{R} = [-(2.9701/EI) × (13.6EI/2.8)] + [(2.23841/EI) × (6.8EI/2.8)] – 16.619 = -25.6090 kNm

M_{3}^{L} = [-(2.9701/EI) × (6.8EI/2.8)] + [(2.2384/EI) × (13.6EI/2.8)] + 16.619 = 20.278 kNm

M_{4}^{R} = [(2.2384/EI) × (13.6EI/3.325)] – [(4.6346/EI) × (6.8EI/3.325)] – 25.193 = -25.5157 kNm

M_{3}^{L} = [(2.2384/EI) × (6.8EI/3.325)] – [(4.6346/EI) × (13.6EI/3.325)] + 25.193 = 10.814 kNm

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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

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

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

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.

This comment has been removed by the author.

(y)

Where did the 3.4 in 3.4EI come from?

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).

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.

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.

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.