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**

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

_{1P }= (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,

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

Fc

This comment has been removed by the author.

DeleteSorry 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.

DeleteSo 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.

DeleteWhat 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.

DeleteFor 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.

(y)

Delete