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Microsoft Word - ES2F XXXXXXXXXXCoursework_v2.docx ES2F4 – Structural Analysis Indeterminate Structures 18/02/2021 Dr Justin Russell ES2F4 – Structural Analysis Indeterminate Structures Report –...

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Microsoft Word - ES2F XXXXXXXXXXCoursework_v2.docx
ES2F4 – Structural Analysis
Indeterminate Structures
18/02/2021
Dr Justin Russell
ES2F4 – Structural Analysis
Indeterminate Structures Report – Briefing Document 2021
Aim
The aim of this assignment is to demonstrate understanding of the structural behaviour of
statically indeterminate structures. This will assessed by obtaining, by calculation, the
ending moment diagram of an example structure and discussing some key points.
Assessment
This assignment is worth 30% of this module and therefore a good amount of time is expected
to be spent solving the problem and writing the discussion points.
It should be 6 pages, including calculations, annotated solutions and written answers.
Submission
The submission must be on Tabula in PDF format. Deadline of submission is midday Thursday
15th April 2021.
Feedback
Marks and feedback will be provided by Friday 14th May 2021. Individual comments on each
submission will be provided, along with a summary of the cohort performance.
ES2F4 – Structural Analysis
Indeterminate Structures
18/02/2021
Dr Justin Russell
Structure
Figure 1 - Indeterminate Structure
Table 1 – Loading and Section Properties
Student
ID L1(m) L2 (m)
IBeam
(cm4)
IColumn
(cm4)
w
(kN/m) P1 (kN) P2 (kN)
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ES2F4 – Structural Analysis
Indeterminate Structures
18/02/2021
Dr Justin Russell
Questions
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 2nd moment 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:

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

. Apply the Moment Distribution Method to find all element end moments.
You can stop your iterations when the ca
y over moments are less than
1kNm or you have done 8 iterative steps.
(20 marks)

c. 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) 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) 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
eferenced co
ectly.
(30 marks)
ES2F4 – Structural Analysis
Indeterminate Structures
18/02/2021
Dr Justin Russell
Appendix A – Loaded Structure
ES2F4 – Structural Analysis
Indeterminate Structures
18/02/2021
Dr Justin Russell
Appendix B – Blank structures
ES2F4 – Structural Analysis
Indeterminate Structures
18/02/2021
Dr Justin Russell
Answered 2 days After Apr 12, 2021

Solution

Prateek answered on Apr 15 2021
156 Votes
1(a)
Design Data-
W= 5KN/m, P1= 30 KN
L1= 7m
L2= 8m
E= 200 GPa
Ibeam= 4000cm4
Icolumn= 3200cm4
Calculation of Fixed End Moment
Distribution Factor Calculation
    
    
    I (cm4)
    I (mm4)
    L (mm)
    K
    Sum K
    DF
    B
    BA
    4000
    40000000
    7000
    4285.714
     
    0.24
    
    BF
    3200
    32000000
    3000
    8000
    18000
    0.44
    
    BC
    4000
    40000000
    7000
    5714.286
     
    0.32
    C
    CB
    4000
    40000000
    7000
    5714.286
     
    0.22
    
    CG
    3200
    32000000
    3000
    10666.67
    26380.95
    0.40
    
    CD
    8000
    80000000
    8000
    10000
     
    0.38
    D
    DC
    8000
    80000000
    8000
    10000
     
    0.48
    
    DH
    3200
    32000000
    3000
    10666.67
    20666.67
    0.52
    
    DE
    8000
    80000000
    4000
     
     
     
    
    A
    B
    C
    D
    F
    G
    H
    Members
    AB
    BA
    BC
    BF
    CB
    CG
    CD
    DC
    DH
    DE
    FB
    GC
    HD
    DF
     
    0.24
    0.44
    0.32
    0.22
    0.40
    0.38
    0.48
    0.52
    …
    ….
     
     
    FEMS
    20.42
    -20.42
    21.88
    0
    -21.88
    0
    0
    0
    0
    40
    0
    0
    0
     
    -20.42
    -10.21
     
     
     
     
     
     
     
     
     
     
     
    Initial...
SOLUTION.PDF

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