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I need help in this assignment problem that requires the use of stiffness matrix and excel to analyze the structure.

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I need help in this assignment problem that requires the use of stiffness matrix and excel to analyze the structure.
Answered Same Day May 06, 2021

Solution

Hemalatha answered on May 09 2021
148 Votes
The problem statement is as follows
SOLUTION
Horizontal displacements of joints C and B and rotations of B, C, D and E are independent displacement components. Also, the horizontal joint displacement of joint at D is same as that of joint C. The horizontal joint displacement of joint at B is same as that of joint E, as well.
Therefore, DOF (degrees of freedom) is six. Figure (a) shows the possible displacement shown as coordinates.
Figure (a)
Moments at the coordinates due to fixed end moments (caused due to externally applied loads) are represented in matrix form as [P’].
In this case [P’] is a null matrix.
Loads acting at the coordinates are shown in a matrix which is denoted by [P].
[P] =
    35
    20
    0
    0
    0
    0
Calculation of Stiffness Matrix
First column is determined by applying unit displacement at coordinate 1 at a time and measuring the forces at all the coordinates as shown figure (b).
First column is determined by applying unit displacement at coordinate 1 at a time and measuring the forces at all the coordinates as shown figure (c).
First column is determined by applying unit displacement at coordinate 1 at a time and measuring the forces at all the coordinates as shown figure (d).
First column is determined by applying unit displacement at coordinate 1 at a time and measuring the forces at all the coordinates as shown figure (e).
First column is determined by applying unit displacement at coordinate 1 at a time and measuring the forces at all the coordinates as shown figure (f).
First column is determined by applying unit displacement at coordinate 1 at a time and measuring the forces at all the coordinates as shown figure (h).
The stiffness coefficients are shown in the following table

    k11
    (12 EI / L3 )BC + (12 EI / L3 )ED
    k21
    (-12 EI / L3 )BC + (-12 EI / L3 )ED
    k31
    -(6 EI / L2 )BC
    k41
    -(6 EI / L2 )DE
    k51
    -(6 EI / L2 )BC
    k61
    -(6 EI / L2 )DE
    k12
     k21
    k22
    (12 EI / L3 )BC + (12 EI / L3 )ED +(12EI / L3 )BC + (12 EI / L3 )ED
    k32
    (6 EI / L2 )BC
    k42
    (6 EI / L2 )DE
    k52
    (6 EI / L2 )BC
    k62
    (6 EI / L2 )DE
    k13
     k31
    k23
     k32
    k33
    4(EI/L)BC + 4(EI/L)CD
    k43
    2(EI/L)CD
    k53
    2(EI/L)BC
    k63
    0
    k14
     k41
    k24
     k42
    k34
     k43
    k44
    4(EI/L)CD +...
SOLUTION.PDF

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