4. Topics
• Familiarisation with the following concepts;
4
• Shear of open section beams
• Shear centre of open section beams
• Shear of closed section beams
• Shear centre of closed section beams
• Tw ist of shear loaded closed section beams
6. 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
6
7. Why shear force in the wing section?
(https://www.youtube.com/watch?v=YyeX6ArxCYI)
7
8. Why shear loading?
8
Centre spar
Front spar
Rear spar
Aerodynamic
centre
For symmetric loading
can be assumed clamped
Lift
Pitching
moment
10. Stress system on element of a beam
10
We isolate this
element of the beam
as of the next slide
Direction of positive
shears
Direction of positive
shears
11. Stress system on element of a beam
11
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
𝜏𝑧𝑠 = 𝜏𝑠𝑧 = 𝜏
𝑞 = 𝜏𝑡
shear flow w hich is positive in the direction
of increasing 𝑠
12. Stress system on element of a beam
12
Equilibrium of forces in 𝑧 direction
Equilibrium of forces in s direction
13. 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)
13
14. Shear of open section beams
14
If w e take the origin of 𝑠 at the
open edge of the cross section
then 𝑞 = 0 w hen 𝑠 = 0
y
x S
z
M =
15. 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;
15
0
0
0
0
16. Example
• Determine the shear flow distribution in the thin-
walled Z-section shown in the figure due to a shear
load 𝑆𝑦 applied through the shear centre of the
section.
16
17. 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
17
19. Solution
19
Integration must be performed for the
w hole 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 +
20. Solution
20
20
1
2
s
h
x +
−
=
2
h
y −
=
2
0 1
h
s +
Integration must be performed for the
w hole length of the cross section, i.e.
path 12, 23 and 34
0
=
x
2
2
s
h
y +
−
=
h
s
2
0
Obtained by 𝑆1 = 0.5ℎ in
𝑞12 (shear flow at point 2)
21. 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)
21
22. Finite element simulation of previous
example
22
100
1
1000
y
h mm
t mm
S N
=
=
=
Undeformed
shape
Deformed
shape
Shear
flow
23. Shear centre
23
• 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
24. Shear centre
• For cruciformor 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
24
25. Example
• Obtain shear flowdistribution for a cantilevered beam with the C
channel section with constant thickness 𝑡. Assume the value of
shear force acting on the section is 𝑆𝑦. The section can be
treated as thin wall section.
25
A
B
C D
b
h
26. Solution
26
h
• The location of centroid from the
mid line of BC;
( )
b
h
b
d /
2 +
=
0
=
x
S
A
B
C D
b
h x
y
d
h/2
27. Solution
27
A
B
C D
b
h x
y
d
h/2
( )
2
12
2
2
12
1
2
3
2
3
bth
th
I
h
bt
t
h
I
xx
xx
+
=
=
+
=
−
+
= bt
d
b
tb
Iyy
2
3
2
12
1
2
0
=
xy
I
28. Solution
28
A
B
C D
b
h x
y
d
h/2
−
→
−
=
s
xx
y
s
yy
xx
yy
y
s yds
I
t
S
tyds
I
I
I
S
q
0
0
2
h
y −
=
s1
b
s
1
0
2
2
s
h
y +
−
=
s2
s3
2
h
y +
=
b
s
3
0
1 1
1 1
3 2 3 2
0 0
1
2
12 12
6 6 2
6
6
s s
y y
DC
y
DC
S S h
q yds ds
h bh h bh
S
q s
h bh
− −
= = − =
+ +
=
+
2
6
6
y
C
S
q b
h bh
=
+
h
s
2
0
29. Solution
29
A
B
C D
b
h x
y
d
h/2
−
→
−
=
s
xx
y
s
yy
xx
yy
y
s yds
I
t
S
tyds
I
I
I
S
q
0
0
2
h
y −
=
s1
b
s
1
0
s2
s3
2
h
y +
=
b
s
3
0
2
2
2
3 2
0
2 2
3 2 2
0
12
6
12 6
6 2 6
s
y
CB C
s
y y
S
q yds q
h bh
S S
h
s ds b
h bh h bh
−
= + =
+
−
− + +
+ +
2
2
s
h
y +
−
=
h
s
2
0
30. Solution
30
2
2
2
3 2
0
2 2
3 2 2
0
2
2 2
3 2 2
12
6
12 6
6 2 6
12 6
6 2 2 6
s
y
CB C
s
y y
y y
S
q yds q
h bh
S S
h
s ds b
h bh h bh
S S
hs s
b
h bh h bh
−
= + =
+
−
− + + =
+ +
−
− + +
+ +
A
B
C D
b
h x
y
d
h/2
2
h
y −
=
s1
b
s
1
0
s2
s3
2
h
y +
=
b
s
3
0
2
2
s
h
y +
−
=
h
s
2
0
3
3
3
3 2
0
3
3 2 2
0
12
6
12 6
6 2 6
s
y
BA B
s
y y
S
q yds q
h bh
S S
h
ds b
h bh h bh
−
= + =
+
−
+ +
+ +
3
3 2 2
6 6
6 6
y y
BA
S h S
q s b
h bh h bh
−
= +
+ +
31. Solution
• Shear flow distribution becomes;
31
x
y
1
2
6
6
y
DC
S
q s
h bh
=
+
3
3 2 2
6 6
6 6
y y
BA
S h S
q s b
h bh h bh
−
= +
+ +
2
2 2
3 2 2
12 6
6 2 2 6
y y
CB
S S
hs s
q b
h bh h bh
−
= − + +
+ +
32. Students’ task
• Find the location of shear centre (distance e) for the
C section of previous example.
32
A
B b
x
y
d
h/2
s1
s2
s3
1
1 1
2
0
2
6
6
3
6
s b
y
y
S
S e s hds
h bh
b
e
h b
=
= →
+
=
+
e
Sy
33. 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.
33
34. Shear of closed section beams
34
Just the same as
open section beams
Unknow n value of
shear flow at origin
(w e choose w here
origin is)
35. Shear of closed section beams
35
Basic shear flow (that
of an open section)
Unknow n shear
flow at an assumed
origin
Shear flow of a
closed section beam
36. Shear of closed section beams
• Now the question arises as how do we find the
unknown shear flow at an assumed origin (𝑞𝑠,0) and
open section flow (𝑞𝑏)?
36
Convert the closed
section to an open
section
• Cut the section at a
convenient point
• At the cut 𝑠 = 0 and 𝑞
= 0
• Obtain 𝑞𝑏 as you did for
an open section
Equate torsion created
by external shear force
with that generated by
internal shear flow
𝑞𝑠,0 is obtained
37. 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;
37
A is the area
enclosed by midline
of the beam section
38. Shear of closed section beams
• Note that if the moment centre chosen coincide with
point of action of shear forces 𝑆𝑥 and 𝑆𝑦 then the
equation can be simplified as;
38
0
0
39. 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;
39
Rate of
twist
40. Example/Tutorial
• Determine the shear flow distribution for a boxed
beam under 𝑆𝑦 shear loading passing through the
shear centre of the cross section.
40
Sy
t
a
b
41. 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.
41
cut
42. Solution
• Let’s deal with basic shear flow first (open section
shear flow);
42
cut
y
x
0
=
x
S S
Sy = 0
=
xy
I
( )( )
( )
3
3
1
2 2
12
xx
I ba b t a t
= − − −
( )( )
( )
3
3
1
2 2
12
yy
I ab a t b t
= − − −
−
=
s
xx
y
b yds
I
t
S
q
0
44. Solution
44
y
x
1
2
3
4 5
6
( )
( ) ( )
( )
ab
a
I
t
S
q
a
ab
I
t
S
as
s
I
t
S
q
ds
a
s
I
t
S
q
xx
y
b
a
s
xx
y
xx
y
b
s
xx
y
b
5
.
0
125
.
0
125
.
0
5
.
0
2
5
.
0
2
4
,
4
@
2
3
2
3
3
,
0
3
3
34
,
3
3
+
−
=
⎯
⎯
⎯ →
⎯
+
−
+
−
−
=
+
+
−
−
=
=
→
s1
s2
s3
( ) ( )
( )
ab
a
as
I
t
S
ab
a
I
t
S
as
I
t
S
q
ads
I
t
S
q
xx
y
xx
y
xx
y
b
s
xx
y
b
5
.
0
125
.
0
5
.
0
5
.
0
125
.
0
5
.
0
5
.
0
2
4
2
4
4
,
0
4
45
,
4
−
−
=
+
−
+
=
+
−
−
=
s4
( )
2
5
,
5
@
125
.
0
4
a
I
t
S
q
xx
y
b
b
s
−
=
⎯
⎯
⎯ →
⎯ =
→
3
5
.
0 s
a
y −
=
a
y 5
.
0
−
=
45. ( )
( ) ( )
( )
0
125
.
0
5
.
0
5
.
0
125
.
0
2
5
.
0
6
,
5
.
0
6
@
2
5
2
5
2
5
2
5
5
,
0
5
5
56
,
5
5
=
⎯
⎯
⎯ →
⎯
+
−
−
=
−
−
−
=
+
−
−
=
=
→
b
a
s
xx
y
xx
y
xx
y
b
s
xx
y
b
q
a
as
s
I
t
S
a
I
t
S
as
s
I
t
S
q
ds
a
s
I
t
S
q
Solution
45
y
x
1
2
3
4 5
6
s1
s2
s3
s4
s5
a
s
y 5
.
0
5 −
=
46. Solution
• Now that we calculated basic shear flow we need to
find 𝑞𝑠,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
46
47. Solution
47
Sy
B
0
=
= O
O
( )( ) ab
A
t
b
t
a
A =
−
−
= or
t
a
b
A
ds
pq
q
Aq
ds
pq b
s
s
b
2
2
0 0
,
0
,
−
=
→
+
=
1
2
3
4 5
6
xx
y
b
I
t
S
s
q
2
2
1
12
, −
= ( )
2
2
23
, 125
.
0
5
.
0 a
as
I
t
S
q
xx
y
b +
−
=
( ) ( )
2
3
2
3
34
, 125
.
0
5
.
0
2
a
ab
I
t
S
as
s
I
t
S
q
xx
y
xx
y
b +
−
+
−
−
=
( )
ab
a
as
I
t
S
q
xx
y
b 5
.
0
125
.
0
5
.
0 2
4
45
, −
−
=
( )
2
5
2
5
56
, 125
.
0
5
.
0
5
.
0 a
as
s
I
t
S
q
xx
y
b +
−
−
=
Since it is a thin w all section, I could
ignore thickness effects. This w ill have
negligible impact on the final solution.
49. Solution
49
Sy
B
t
a
b
1
2
3
4 5
6
( )
( )
b
a
b
a
I
t
S
ds
ab
a
as
I
t
S
a
ds
q
p
xx
y
b
xx
y
s
b
3
2
2
0
4
2
4
0
45
,
45
0625
.
0
125
.
0
5
.
0
125
.
0
5
.
0
5
.
0
4
+
−
=
−
−
=
( )
b
a
I
t
S
ds
a
as
s
I
t
S
b
ds
q
p
xx
y
a
xx
y
s
b
3
5
5
.
0
0
2
5
2
5
0
56
,
56
0104
.
0
125
.
0
5
.
0
5
.
0
5
.
0
5
−
=
+
−
−
=
50. Solution
• Now the total shear flow will become (note that 𝑞𝑠,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
50
+
+
+
+
−
=
−
=
5
4
3
2
1
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
ds
q
p
ds
q
p
ds
q
p
ds
q
p
ds
q
p
A
A
ds
pq
q
( )( )
t
b
t
a
b
a
b
a
I
t
S
A
ds
pq
q
xx
y
b
s
−
−
+
+
=
−
=
2
5
.
0
1562
.
0
2
2
2
3
0
,
51. 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;
51
53. Solution
• A tidier illustrative shear flow distribution is;
53
+
-
- +
- +
What does
negative mean?
54. Important note
• Determination of shear flow distributionin multi-cellular
section thin walled beamsas stipulatedin chapters18 and
22 of Ref. [1] is fulfilled throughsuccessful
implementationof student centered coursework releasedat
the start of the semester;
54
