This document provides an overview of shear loading and shear flow distribution in beams. It discusses shear in open section beams, closed section beams, and the twist that can occur in closed section beams when loads are not applied through the shear center. Key points covered include: - Shear flow distribution in open section beams is determined using equations involving shear center location and section properties. - For closed sections, the section must first be cut to form an open section, then equations are used to determine the unknown shear flow at the cut and resulting basic open section shear flow. - Twisting can occur in closed sections under shear loads not through the shear center, and an equation is provided to calculate the rate of twist. -

1. Aero Structures-Shear of beams
By
Dr. Mahdi Damghani
2017-2018
1

2. Suggested Readings
Chapter 16
of
Aircraft Structural Analysis
2

3. Topics
• Familiarisation with the following concepts;
3
• Shear of open section beams
• Shear centre of open section beams
• Shear of closed section beams
• Shear centre of closed section beams
• Twist of shear loaded closed section beams

4. Overview
4

5. Introduction
• In this lecture we address thin-walled beams loaded
in shear typical sections in the wing
• Topics discussed here do not apply to thick and solid
sections as they are not typical cross sections used in
aircraft structures
5

6. Why shear force in the wing section?
6

7. Why shear loading?
7
Centre spar
Front spar
Rear spar
Aerodynamic
centre
For symmetric loading
can be assumed clamped
Lift
Pitching
moment

8. Stress system on element of a beam
8
We isolate this
element of the beam
as of the next slide
Direction of positive
shears
Direction of positive
shears

9. Stress system on element of a beam
9
s is distance measured around the cross
section from some convenient origin
An element of size 𝛿𝑠 × 𝛿𝑧 × 𝑡 is in
equilibrium by a system of stresses
t is assumed to be constant along 𝛿𝑠
𝜏 𝑧𝑠 = 𝜏 𝑠𝑧 = 𝜏
𝑞 = 𝜏𝑡
shear flow which is positive in the direction
of increasing 𝑠

10. Stress system on element of a beam
10
Equilibrium of forces in 𝑧 direction
Equilibrium of forces in s direction

11. Shear of open section beams
• Shear forces Sx and Sy
are applied at the point
that does not create
torsion in the section,
i.e. shear centre
• In the absence of hoop
stresses 𝜎𝑠 (assumed)
11

12. Shear of open section beams
12
If we take the origin of 𝑠 at the
open edge of the cross section
then 𝑞 = 0 when 𝑠 = 0
yx SzM 

13. Shear of open section beams
• For a section having either Cx or Cy as an axis of
symmetry Ixy=0 and hence the relationship simplifies
to;
13
0
0
0
0

14. Example
• Determine the shear flow distribution in the thin-
walled Z-section shown in the figure due to a shear
load 𝑆 𝑦 applied through the shear centre of the
section.
14

15. Solution
• Since the load is applied at shear centre that means
no torsion takes place in the section
• Therefore, shear stress and shear flow as the result
of torsion does not exist
• There is no axis of symmetry, hence we use the
following equation for shear flow determination
15

16. Solution
16
0xS

17. Solution
17
Integration must be performed for the
whole length of the cross section, i.e.
path 12, 23 and 34
1
2
s
h
x 
2
h
y 
2
0 1
h
s 

18. Solution
1818
1
2
s
h
x 
2
h
y 
2
0 1
h
s 
Integration must be performed for the
whole length of the cross section, i.e.
path 12, 23 and 34
0x
2
2
s
h
y 
hs  20
Obtained by 𝑆1 = 0.5ℎ in
𝑞12 (shear flow at point 2)

19. Solution
• Positive q means that the
shear flows in direction of
s
• You may use the
following analogy to have
a physical understanding
of this phenomenon;
• Emptying a bucket of
water at points of zero
shear flow (brown spot)
causes some of the water
to go to the left and some
to the right. Each path
continues flowing until it
gets drained, i.e. reach
another zero point (red
spot)
19

20. Shear centre
20
• Shear load
applied at shear
centre does not
create torsion in
the section
• You could see
the wooden
support as
fuselage and
the channel
beam as wing

21. Shear centre
• For cruciform or angle sections below, the shear centre
lies at the intersection of the sides
• Students are required to go through section 16.2.1
of Ref [1] (student centred activity) for obtaining
the location of shear centre
21

22. Example
• Obtain shear flow distribution for a cantilevered beam with the C
channel section with constant thickness 𝑡. Assume the value of
shear force acting on the section is 𝑆 𝑦. The section can be
treated as thin wall section.
22
AB
C D
b
h

23. Solution
23
h
• The location of centroid from the
mid line of BC;
 bhbd /2 
0xS
AB
C D
b
h x
y
d
h/2

24. Solution
24
AB
C D
b
h x
y
d
h/2
 
212
2
2
12
1
23
2
3
bthth
I
h
btthI
xx
xx






















 btd
b
tbIyy
2
3
212
1
2
0xyI

25. Solution
25
AB
C D
b
h x
y
d
h/2





s
xx
y
s
yyxx
yyy
s yds
I
tS
tyds
II
IS
q
00
2
h
y 
s1
bs  10
2
2
s
h
y 
s2
s3
2
h
y 
bs  30
12
0
123
0
123
6
2
1212 11
s
bhh
S
q
ds
h
bhh
S
yds
bhh
S
q
y
DC
s
y
s
y
DC








 
b
bhh
S
q
y
C

 2
6
hs  20

26. Solution
26
AB
C D
b
h x
y
d
h/2





s
xx
y
s
yyxx
yyy
s yds
I
tS
tyds
II
IS
q
00
2
h
y 
s1
bs  10
s2
s3
2
h
y 
bs  30
b
bhh
S
dss
h
bhh
S
qyds
bhh
S
q
y
s
y
C
s
y
CB
















2
0
2223
0
223
6
2
12
12
2
2
2
2
s
h
y 
hs  20

27. Solution
27
b
bhh
Sshs
bhh
S
b
bhh
S
dss
h
bhh
S
qyds
bhh
S
q
yy
y
s
y
C
s
y
CB



























2
2
22
23
2
0
2223
0
223
6
22
12
6
2
12
12
2
2
AB
C D
b
h x
y
d
h/2
2
h
y 
s1
bs  10
s2
s3
2
h
y 
bs  30
2
2
s
h
y 
hs  20
b
bhh
S
ds
h
bhh
S
qyds
bhh
S
q
y
s
y
B
s
y
BA
















2
0
323
0
323
6
2
12
12
3
3
b
bhh
S
s
bhh
hS
q
yy
BA




 2323
66

28. Solution
• Shear flow distribution becomes;
28
x
y
12
6
s
bhh
S
q
y
DC


b
bhh
S
s
bhh
hS
q
yy
BA




 2323
66
b
bhh
Sshs
bhh
S
q
yy
CB










 2
2
22
23
6
22
12

29. Shear of closed section beams
• Similar to open section beams with two differences:
• The shear loads may be applied through points in the cross
section other than the shear centre so that torsional as well
as shear effects are included;
• It is generally not possible to choose an origin for 𝑠 at which
the value of shear flow is known.
29

30. Shear of closed section beams
30
Just the same as
open section beams
Unknown value of
shear flow at origin
(we choose where
origin is)

31. Shear of closed section beams
31
Basic shear flow (that
of an open section)
Unknown shear flow at
an assumed origin
Shear flow of a
closed section beam

32. Shear of closed section beams
• Now the question arises as how do we find the
unknown shear flow at an assumed origin (𝑞 𝑠,0) and
open section flow (𝑞 𝑏)?
32
Convert the closed
section to an open
section
• Cut the section at a
convenient point
• At the cut 𝑠 = 0 and
𝑞 = 0
• Obtain 𝑞 𝑏 as you did for
an open section
Equate torsion created
by external shear force
with that generated by
internal shear flow
𝑞 𝑠,0 is obtained

33. Shear of closed section beams
• For open section we had;
• Value of shear flow at the cut
(𝑠 = 0) can be found by;
• We also know;
33
A is the area
enclosed by midline
of the beam section

34. Shear of closed section beams
• Note that if the moment centre chosen coincide with
point of action of shear forces 𝑆 𝑥 and 𝑆 𝑦 then the
equation can be simplified as;
34
00

35. Twist of shear loaded closed section
beams
• If shear load is not applied to shear centre, twist of
the section takes place
• For detailed information of extracting the twist formula
refer to section 16.3.1 of Ref [1]
• The final twist equation is;
35
Rate of
twist

36. Example/Tutorial
• Determine the shear flow distribution for a boxed
beam under 𝑆 𝑦 shear loading passing through the
shear centre of the cross section.
36
Sy
t
a
b

37. Solution
• The shear load passes through the shear centre, so there
is no torsion in the section;
• The section is a closed section, hence we need to make it
an open section first;
• For this, we cut the section at an arbitrary point;
• It is best to cut it at a point that passes through the neutral
axis as normally shear flow is maximum at that point.
37
cut

38. Solution
• Let’s deal with basic shear flow first (open section
shear flow);
38
cut
y
x0xS SSy  0xyI
  33
2
12
1
tabbaIxx 
  33
2
12
1
tbaabIyy 

s
xx
y
b yds
I
tS
q
0

39. Solution
39
y
x

s
xx
y
b yds
I
tS
q
0
1
23
4 5
6
xx
y
b
as
xx
y
s
xx
y
b
I
tSa
q
I
tSs
dss
I
tS
q
2
2,
5.02@
2
1
0
1112,
125.0
2
1
1

  

s1
s2
   



bs
xx
y
xx
y
xx
y
b
s
xx
y
b
aas
I
tS
I
tSa
I
tSas
qads
I
tS
q
2
2
3@2
2
2
2
2,
0
223,
125.05.0
125.05.0
5.0
 2
3, 125.05.0 aab
I
tS
q
xx
y
b 
01,
b
q
Shear flow at
the cut is zero
1sy 
ay 5.0

40. Solution
40
y
x
1
23
4 5
6
 
   
 aba
I
tS
q
aab
I
tS
ass
I
tS
qdsas
I
tS
q
xx
y
b
as
xx
y
xx
y
b
s
xx
y
b
5.0125.0
125.05.0
2
5.0
2
4,
4@
2
3
2
3
3,
0
3334,
3
3
 




s1
s2
s3
   
 abaas
I
tS
aba
I
tS
as
I
tS
qads
I
tS
q
xx
y
xx
y
xx
y
b
s
xx
y
b
5.0125.05.0
5.0125.05.0
5.0
2
4
2
4
4,
0
445,
4


 
s4
 2
5,
5@
125.0
4
a
I
tS
q
xx
y
b
bs

  
35.0 say 
ay 5.0

41.  
   
 
0
125.05.05.0
125.0
2
5.0
6,
5.06@2
5
2
5
2
5
2
5
5,
0
5556,
5
5

 




b
as
xx
y
xx
y
xx
y
b
s
xx
y
b
q
aass
I
tS
a
I
tS
ass
I
tS
qdsas
I
tS
q
Solution
41
y
x
1
23
4 5
6
s1
s2
s3
s4
s5
asy 5.05 

42. Solution
• Now that we calculated basic shear flow we need to
find 𝑞 𝑠,0
• This is achieved by balancing internal moment and
external moment about any point B
• For convenience we will take point B as the point of
action of shear load
• By doing this the moment resulting from shear load
disappears
42

43. Solution
43
Sy
B
0 OO    tbtaA 
t
a
b
A
dspq
qAqdspq b
ssb
2
20 0,0,

 
1
23
4 5
6
xx
y
b
I
tSs
q
2
2
1
12,   2
223, 125.05.0 aas
I
tS
q
xx
y
b 
   2
3
2
334, 125.05.0
2
aab
I
tS
ass
I
tS
q
xx
y
xx
y
b 
 abaas
I
tS
q
xx
y
b 5.0125.05.0 2
445, 
 2
5
2
556, 125.05.05.0 aass
I
tS
q
xx
y
b 

44. Solution
44
Sy
B
t
a
b

























5
43
21
0
56,56
0
45,45
0
34,34
0
23,23
0
12,12
0,
2
1
2
s
b
s
b
s
b
s
b
s
b
b
s
dsqp
dsqpdsqp
dsqpdsqp
AA
dspq
q
1
23
4 5
6
ba
I
tS
ds
I
tSs
bdsqp
xx
y
xx
y
as
b
3
1
2
1
5.0
00
12,12 0104.0
2
5.0
1









 
   baba
I
tS
dsaas
I
tS
adsqp
xx
y
xx
y
bs
b
322
2
2
2
00
23,23 0625.0125.0125.05.05.0
2






 
 
 223
3
2
3
2
3
00
34,34
25.01041.0
125.05.05.05.05.0
3
baba
I
tS
dsaabass
I
tS
bdsqp
xx
y
xx
y
as
b







 

45. Solution
45
Sy
B
t
a
b
1
23
4 5
6
 
 baba
I
tS
dsabaas
I
tS
adsqp
xx
y
b
xx
y
s
b
322
0
4
2
4
0
45,45
0625.0125.0
5.0125.05.05.0
4

 
 
ba
I
tS
dsaass
I
tS
bdsqp
xx
y
a
xx
y
s
b
3
5
5.0
0
2
5
2
5
0
56,56
0104.0
125.05.05.05.0
5

 

46. Solution
• Now the total shear flow will become (note that 𝑞 𝑠,0 > 0);
• So far, we solved this problem parametically, i.e. in terms of section
variables, and now this could be used for optimisation process
46

























5
43
21
0
56,56
0
45,45
0
34,34
0
23,23
0
12,12
0,
2
1
2
s
b
s
b
s
b
s
b
s
b
b
s
dsqp
dsqpdsqp
dsqpdsqp
AA
dspq
q
  tbta
baba
I
tS
A
dspq
q
xx
yb
s


 
2
5.01562.0
2
223
0,

47. Solution
• I now embed this into excel that enables me to find
shear flow for any box section of any dimension;
• For illustration purposes I use the following values;
47

48. Solution
48
qs,o is constant
Basic shear
flow

49. Solution
• A tidier illustrative shear flow distribution is;
49
+
-
- +
- +

50. Important note
• Determination of shear flow distribution in multi-cellular
section thin walled beams as stipulated in chapters 18 and
22 of Ref. [1] is fulfilled through successful
implementation of student centered coursework released at
the start of the semester;
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51. Tutorial 1
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52. Tutorial 2
52

53. Tutorial 3
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