examination preparation for UWE aerospace students

1. Structural Analysis Example with
Solutions
By
Dr. Mahdi Damghani
2016-2017
1

2. Objective(s)
• Review of structural analysis methods and
procedures learnt in this module
• Preparation for final exam (familiarity with a typical
question) as this year’s exam will be different from
previous years
2

3. Session plan
Introduction
(5-10 mins)
Slope-Deflection
theory by students
(5-10 mins)
Q&A
(5 mins)
Solving a
representative
example (30 mins)
Q&A
(5-10 mins)
3

4. Asking Questions
• Ask in publicly
• Write your questions in the following link and I will
answer them all at the end of the session:
https://padlet.com/mahdi_damghani/SDI
4

5. Topics covered in this module
• Direct integration method
• Macaulay’s method
• Slope-deflection method
• Principle of virtual work
• Rigid bodies
• Deformable bodies
• Unit load method
• Principle of complementary energy
• Castigliano’s second theorem
• Principle of total potential energy
• Castigliano’s first theorem
• Rayleigh-Ritz
• Finite element analysis of truss structures
• Finite element analysis of beam structures
• Finite element analysis of frame structures
5
Objectives satisfied
by the assignment
Assessment in the
final exam
Assessment in the
final exam with all
details

6. Slope-Deflection Method
• The beam we considered
so far did not have any
external loading from A to B
6
• In the presence of mid-span loading (common engineering
problems) the equations become:
  F
ABBABAAB Mvv
LL
EI
M 




3
2
2
   F
ABBABAAB Svv
LL
EI
S 




26
2

  F
BABABABA Mvv
LL
EI
M 




3
2
2
   F
BABABABA Svv
LL
EI
S 




26
2


7. Fixed End Moment/Shear
• MAB
F, MBA
F are fixed end moments at nodes A and B,
respectively.
• Moments at two ends of beam when beam is clamped at
both ends under external loading (see next slides)
• SAB
F, SBA
F are fixed end shears at nodes A and B,
respectively.
• Shears at two ends of beam when beam is clamped at both
ends under external loading (see next slides)
7

8. Fixed End Moment/Shear
8

9. Fixed End Moment/Shear
9

10. Example
10
PL
2P
CDB
A
P
3EI/L 2EI/L
EI/L

11. Question 1
Is this structure
determinate or
indeterminate?
Degree of
indeterminacy
is 9-3=6
1
1

12. Question 2
(slope-deflection equations?)
12
  F
ABBABAAB Mvv
LL
EI
M 




3
2
2

  F
BABABABA Mvv
LL
EI
M 




3
2
2

 
8
2
8
2
2 0 PL
L
EI
M
PL
L
EI
M DADDAAD
A
  
 
 
8
4
8
2
2 0 PL
L
EI
M
PL
L
EI
M DDAADDA
A
  
 
  DBDDBBD
L
EI
M
L
EI
M B
  6
2
32 0
 

 
  DDBBDDB
L
EI
M
L
EI
M B
  12
2
32 0
 

 
 
4
4
8
2
2
22 0 PL
L
EI
M
PL
L
EI
M DCDDCCD
C
 

 
 
 
4
8
8
2
2
22 0 PL
L
EI
M
PL
L
EI
M DDCCDDC
C
 

 
 

13. Question 3
(Equation of equilibrium at joints?)
1313
PL
D
MDB
MDA
MDC
0 PLMMM DCDADB

14. Question 4
(Rotation of joint D?)
14
0 PLMMM DCDADB
8
4 PL
L
EI
M DDA  DDB
L
EI
M 
12

4
8 PL
L
EI
M DDC  
2
7
192
D
PL
EI
 

15. Question 5
(Reaction at support locations?)
15
2
2 2 7
0.2
8 192 8
AD D
EI PL EI PL PL
M PL
L L EI

 
        
 
2
4 4 7
0.02
8 192 8
DA D
EI PL EI PL PL
M PL
L L EI

 
         
 
6
0.22BD D
EI
M PL
L
   
12
0.43DB D
EI
M PL
L
   
4
0.10
4
CD D
EI PL
M PL
L
    
8
0.54
4
DC D
EI PL
M PL
L
    

16. Question 5 continued
16
A
P
EI/L
MAD=-0.2PL
MDA=-0.02PL
D
0
0.02 0.2 0.5
0.72
D
Ax
Ax
M
R L PL PL PL
R P
 
    
 

RAx

17. Question 5 continued
17
B
MDB=-0.43PLMBD=-0.22PL
D
0
0.22 0.43
0.65
D
By
By
M
R L PL PL
R P
 
  
 

RBy

18. Question 5 continued
18
D
MCD=+0.10PLMDC=-0.54PL
C
0
0.54 0.10
0.56
D
Cy
Cy
M
R L PL PL PL
R P
 
    
 

RCy
2P

19. Question 6
(Shear and bending moment diagram o f AD?)
19
A
P
MAD=-0.2PL
MDA=-0.02PL
D
RAx=-0.72P
M(y)
V(y)
( ) 0.72 ; 0 0.5
0
( ) 0.28 ; 0.5
x
V y P y L
F
V y P L y L
   
  
   

 
( ) 0.2 0.72 ; 0 0.5
0
( ) 0.2 0.72 0.5 ; 0.5
M y PL Py y L
M
M y PL Py P y L L y L
    
  
      



-0.72P
+0.28P


-0.2PL
+0.16PL
-0.02PL