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A 2D non-sway structure is given in Figure 1 (and Appendix A). It consists of a continuous beam ABCDE attached with rigid joints to columns BF and DH. A pin connection is at the top of column CG....

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  1. A 2D non-sway structure is given in Figure 1 (and Appendix A). It consists of a continuous beam ABCDE attached with rigid joints to columns BF and DH. A pin connection is at the top of column CG. Section DE is a cantilever. Beam CDE has twice the 2ndmoment of area,I, of beam ABC. A uniformly distributed load,w,is applied along AB and DE and one span has a point load,P, in the middle. Young’s Modulus,E, is constant for all sections.

    Using the loading and section properties from Table 1 according to your student ID, complete the following:

    1. Determine the Distribution Factors (DF) and Fixed End Moments (FEMs) at each joint as part of the Moment Distribution Method.

      (15 marks)

    2. Apply the Moment Distribution Method to find all element end moments. You can stop your iterations when the carry over moments are less than 1kNm or you have done 8 iterative steps.

      (20 marks)

    3. Draw the bending moment diagram for the entire structure (drawn on the tension side), clearly marking all key points, including the midspan moments for elements AB, BC and CD.

      (15 marks)

  2. 2) For the structure shown in Figure 1, describe what the changes in internal forces and deflections may be if element CG is removed. You do not need to do more calculations, but may wish to make use of sketches.

    (20 marks)

  3. 3) Explain what is meant by the term ‘redundancy’ within a structural engineering context and how this is linked to static indeterminacy. You should give clear examples from real structures to demonstrate you points. Any sources must be referenced correctly.

Answered 2 days After Apr 12, 2021

Solution

Prateek answered on Apr 15 2021
158 Votes
1(a)
Design Data-
W= 10KN/m P1= 25 KN
L1= 6m
L2= 8m
E= 200 GPa
Ibeam= 4000cm4
Icolumn= 3200cm4
Calculation of Fixed End Moment
Calculation of Distribution Facto
    
    
    I (cm4)
    I (mm4)
    L (mm)
    K
    Sum K
    DF
    B
    BA
    4000
    40000000
    6000
    5000
    Â 
19666.67
 
    0.25
    
    BF
    3200
    32000000
    3000
    8000
    
    0.41
    
    BC
    4000
    40000000
    6000
    6666.667
    
    0.34
    C
    CB
    4000
    40000000
    6000
    6666.667
    Â 
27333.33
 
    0.24
    
    CG
    3200
    32000000
    3000
    10666.67
    
    0.39
    
    CD
    8000
    80000000
    8000
    10000
    
    0.37
    D
    DC
    8000
    80000000
    8000
    10000
    Â 
20666.67
 
    0.48
    
    DH
    3200
    32000000
    3000
    10666.67
    
    0.52
    
    DE
    8000
    80000000
    4000
    Â ------
    
    Â ----
1(B) Moment Distribution table
    
    A
    B
    C
    D
    F
    G
    H
    Members
    AB
    BA
    BC
    BF
    CB
    CG
    CD
    DC
    DH
    DE
    FB
    GC
    HD
    DF
    Â 
    0.25
    0.41
    0.34
    0.24
    0.39
    0.37
    0.48
    0.52
    â€¦
    â€¦.
    Â 
    Â 
    FEMS
    30
    -30
    18.75
    0
    -18.75
    0
    0
    0
    0
    80
    0
    0
    0
    Â 
    -30
    -15
    Â 
    Â 
    Â 
    Â 
    Â 
    Â 
    Â 
    Â 
    Â 
    Â 
    Â 
    Initial...
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