Structural Design and Inspection level 3 at UWE (final recap)

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

2. Example
• The structure is clamped at
point D and simply
supported at point A. It
carries a uniform distributed
load q = 10kN/m as shown.
Assuming that flexural
stiffness for beam AB and
column BD constant and
equal to EI = 1000.0Nm2,
where E is the Young's
modulus and I is the
second moment of area.
Find reaction at supports.
2

3. Example continued
• We are going to obtain reaction at supports using two
different approaches;
• Slope deflection
• Principle of stationary values of complementary energy
• We will also confirm principle of virtual work for this
structure
• We are going to obtain vertical deflection at the mid-span
of bay AB using;
• Principle of stationary values of complementary energy
• Unit load method
• Abaqus
3

4. Slope-Deflection Method
• The beam we considered
so far did not have any
external loading from A to B
4
• 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


5. Solution with slope-deflection
5
  F
ABBABAAB Mvv
LL
EI
M 




3
2
2

  F
BABABABA Mvv
LL
EI
M 




3
2
2

   



 0
2
12
3
2
2 ABMAB
BABA
AB
AB
qL
vv
LL
EI
M 
  0
12
2
2 2
 AB
BA
AB
qL
L
EI

 
12
2
2 2
AB
AB
AB
BA
qL
L
EI
M  
   





 03
2
2 DF
BDDBDB
BD
BD Mvv
LL
EI
M 
   F
BDB
BD
BD M
L
EI
M  2
2
   





 03
2
2 DF
DBDBBD
BD
DB Mvv
LL
EI
M 
   F
DBB
BD
DB M
L
EI
M  
2

6. Solution with slope-deflection
6






 43
22
2
4
1
3
2
2
bbL
Lb
L
q
M BD
BD
BD
F
BD












mNqw
mL
mb
a
BD
/000,10
2
1
0





432
3
bL
L
qb
M BD
BD
F
DB
NmM F
BD 67.2291
4
1
2
3
2
2
4
4
10000





NmM F
DB 67.1041
4
1
3
2
4
10000






7. Solution with slope-deflection
7
   0
12
2
2 2
AB
BA
AB
qL
L
EI

   F
BDB
BD
BD M
L
EI
M 2
2
   F
DBB
BD
DB M
L
EI
M 
2
  

 0
12
10000
2
1
10002
BA 
   033.83322000 BA  208.05.0  BA 
  
12
2
2 2
AB
AB
AB
BA
qL
L
EI
M    
12
10000
2
1
2000
ABBAM 
  33.83322000  ABBAM 
   67.22912
2
2000
BBDM 
67.22912000  BBDM 
 0BDBA MM
   67.1041
2
2000
BDBM  67.10411000  BDBM 
 067.2291200033.83320004000 BAB  24.033.0  AB 

8. Solution with slope-deflection
8
208.05.0  BA 
24.033.0  AB 
  17.067.1 A
  33.83322000  ABBAM 
67.22912000  BBDM 
67.10411000  BDBM 
2060.0
1018.0


B
A


0ABM

NmMBA 93.1860
NmMDB 67.1247
NmMBD 67.1879
0ABM
AxF
DyF
DBM
DxF
   05.0115.0120 qqMFM DBAxD
NFAx 16.9376
 0xF
 010 qFF Dyy NFDy 10000
NNNFDx 84.6231000016.9376 

9. Bending moment diagram
9
2
5.0)( qxxM AB 
0
mN.5000
mN.5000
mN.84.623
     22
15.05.05.0)(  xqxqxFqxM AxBD
AxF
M
x
mN.67.1247
M

10. Solution with complementary energy
10
v
L
dx
dP
xdM
xIxE
xM







)(
)()(
)(
P


11. Solution with complementary energy
11
P
v
L
dx
dP
xdM
xIxE
xM







)(
)()(
)(
10;
2
)(
2
 x
qx
xM
P M
P
M
x
x10;5.05.0)( 2
 xqxqPxxM
0
)(

dP
xdM
x
dP
xdM

)(

12. Solution with complementary energy
12
P
x
  21;5.015.0)(  xxqqPxxM
x
dP
xdM

)(
M






 0
)(
)()(
)(
AH
L
dx
dP
xdM
xIxE
xM
     
    
































0
5.05.0
5.05.00
21
2
1,
1
0,
2
1
0,
2
x
xBD
x
xBD
x
xAB
AH
dxxxqqPx
dxxqxqPxdx
qx
EI












 0
2
5.0
32
5.0
34
5.0
2
5.0
3
2
1
23231
0
423
x
q
x
q
x
q
x
P
x
q
x
q
x
P
























 0
333
8
3
8
125.025.0
3
qP
qPqq
P
NqP 2.1015601562.1 

13. Solution with complementary energy
13
AxF
DyF
DBM
DxF
   05.0115.0120 qqMFM DBAxD
NNqFF AxDx 2.156100002.10156 
 010 qFF Dyy
NFDy 10000
NFP Ax 2.10156
  NmFqM AxDB 4.3122 

14. Confirmation of principle of virtual
work
14
A B
D
C
AxF
x
x
2B2A
C

M

15. Confirmation of principle of virtual
work
• We assumed that the
structure is rigid so no
internal work is considered.
• Summation of work of
external forces must be zero
• Since our structure is
indeterminate by one degree,
we also can make use of
equation of equilibrium in
addition to work equation, i.e.
15
A B
D
C
AxF
x
x
2B2A
C

M
  0M
  0eW

16. Confirmation of principle of virtual
work
16
A B
D
C
AxF
x
x
2B2A
C

M
 0eW
  02  MqydxqydxF
BDAB
Ax
  02
2
1
1
0
 MdxqxdxqxFAx
 05.15.02  MqqFAx
qMFqMF AxAx 2222 
 0DM  05.15.02 qqMFAx
qMFAx 22  Both came up with
the same equation

17. Deflection
• Now we desire to obtain vertical deflection of mid-
span of AB using two different approaches;
17

18. Vertical Displacement at mid span of
AB using complementary energy method
• I need to solve the structure first by considering a
virtual load Q at mid-span of AB
18
Q Q
We solved this
one before

19. Vertical Displacement at mid span of
AB using complementary energy method
19
Q
HA
L
dx
dP
xdM
xIxE
xM
,
)(
)()(
)(







A B
D
C
P
0
)(

dP
xdM AB
x
PxQxMBD  5.0)(
x
dP
xdMBD

)(
    2
0
32
2
0
, 33.025.05.0
1
0 PxQxdxxPxQ
EI
HA  
QP 375.0

20. Vertical Displacement at mid span of
AB using complementary energy method
20
Q
QFAx 375.0
ABv
L
dx
dQ
xdM
xIxE
xM
mid,
)(
)()(
)(







M
x
5.00;
2
)(
2
 x
qx
xM
0
)(

dQ
xdM
M
x
Q
  15.0;5.0
2
)(
2
 xxQ
qx
xM
x
dQ
xdM
 5.0
)(
QFAx 375.0
QFAx 375.0

21. Vertical Displacement at mid span of
AB using complementary energy method
21
ABv
L
dx
dQ
xdM
xIxE
xM
mid,
)(
)()(
)(







Q
M
x
  10;5.05.05.0375.0)( 2
 xqxqQxQFxM Ax
5.0375.0
)(
 x
dQ
xdM
x
M
Q
    21;5.015.05.0375.0)(  xxqqQxQFxM Ax
5.0375.0
)(
 x
dQ
xdM
QFAx 375.0
QFAx 375.0

22. Vertical Displacement at mid span of
AB using complementary energy method
22






 ABv
L
dx
dQ
xdM
xIxE
xM
mid,
)(
)()(
)(
      
    
      
































 











0
2
1,
1
0,
2
1
5.0,
2
5.0
0,
2
mid,
5.0375.05.05.05.0375.0
5.0375.05.05.05.0375.0
5.05.05.00
2
1 Q
x
xBD
Ax
x
xBD
Ax
x
xAB
x
xAB
ABv
dxxxqqQxQF
dxxqxqQxQF
dxxxQqxdx
qx
EI
  
   
    
















 0
2
1
222
1
33
1
0
321
0
423
1
5.0
334
mid,
5.05.05.05.05.05.0125.0125.0
167.05.05.05.05.0047.0094.0125.0
5.033.0083.0125.0
1 Q
AxAx
AxAxABv
xqqxQxxFqxxF
qxqxQxxFqxqxxF
xQqxqx
EI

23. Vertical Displacement at mid span of
AB using complementary energy method
23
   q
EI
qFqFq
EI
AxAxABv 11.0
1
125.0125.01925.0125.0044.0
1
mid, 
  mABv 1.11000011.0
1000
1
mid, 
  
   
    
















 0
2
1
222
1
33
1
0
321
0
423
1
5.0
334
mid,
5.05.05.05.05.05.0125.0125.0
167.05.05.05.05.0047.0094.0125.0
5.033.0083.0125.0
1 Q
AxAx
AxAxABv
xqqxQxxFqxxF
qxqxQxxFqxqxxF
xQqxqx
EI
 
   
    
    




















223
223
34
mid,
5.0125.0125.0125.01125.0125.0
5.0225.0225.0225.02125.0125.0
167.05.05.05.0047.0094.0125.0
5.0083.05.0125.0083.0125.0
1
qqFqF
qqFqF
qqFqqF
qqqq
EI
AxAx
AxAx
AxAx
ABv

24. Vertical Displacement at mid span of
AB using unit load method
• Student task
24

25. FEA hand calculation
• See uploaded excel file for frame analysis using hand
calculation
25

26. FEA hand calculation
26

27. Abaqus Validation
27
U2,mid span=-1.12