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```Lecture #3
Torsion of opened cross sections.
Loads on frame due to fuselage bending
SHEAR STRESSES RELATED QUESTIONS
- shear flows due to the shear force, with no torsion;
- shear center;
- torsion of closed contour;
- torsion of opened contour, restrained torsion and
deplanation;
- shear flows in the closed contour under combined action
of bending and torsion;
- twisting angles;
- shear flows in multiple-closed contours.
2
SHEAR CENTER AND TORSION - ILLUSTRATION
3
TORSION IN MECHANICS OF MATERIALS
The measure of resistance to
torsion is a polar moment of
inertia Ir .
Relative twist angle
 
Ir 
Mz
 d
4
32
G  Ir
Maximal shear stress
 m ax 
Mz
Ir
 r m ax
Ir    h  b
3
4
TORSION IN THIN-WALLED CROSS SECTIONS
The polar moment of inertia Ir is calculated as a sum
for rectangular portions of thin-walled cross section.
Relative twist angle
 
Mz
G  Ir
Maximal shear stress
 m ax 
Mz
Ir
Ir 
1
3
  m ax
  bi  
3
i
i
5
TORSION IN THIN-WALLED CROSS SECTION
Torsional moment is 1000 N·m.
Material is steel, G = 77 GPa.
Moment of inertia Ir = 4.14 cm4.
Shear stress max = 241.3 MPa.
6
CALCULATION OF DEPLANATIONS (WARPING)
Since the hypothesis of planar cross section is not
valid, the beam theory is not applicable.
Thus, specific theory developed by Vlasov is used.
Vlasov’s theory is based on two main hypotheses:
1) The cross section keeps its shape and rotates as a
whole around the shear center.
2) There are no shear strains and stresses at the
middle plane (gtz = 0).
7
CALCULATION OF DEPLANATIONS
g tz 
w
t

u
z
0
where w – displacement along z axis (deplanation);
u – displacement along t axis.
w
t

df
dz
 r t
where f – angle of rotation of cross section along z
axis;
r – lever from the shear center to the direction of t
axis at the given point.
8
CALCULATION OF DEPLANATIONS
w  t   w 0     r  t  dt
t
where w0 –displacement at the start point.
If start point is set on the axis of symmetry, we get
w  t     w  t 
where wt – sectorial coordinate (doubled area
w t 
 r  t   dt
t
9
CALCULATION OF DEPLANATIONS
Analytical values:
Max is 0.67 mm
At the corner is 0.51 mm
10
NORMAL STRESSES AT RESCTRICTED TORSION
Normal stresses could be found using the formula
B
w t 
Iw
w t
where Iw – sectorial moment of inertia:
Iw 
w t
2
   dt
t
B – bimoment (kind of scalar force factor):
2
df
B  
2
d z
 E  Iw
11
NORMAL STRESSES AT RESCTRICTED TORSION
The distribution of normal stresses for real structure is
usually quite complex, so it is usually wise to use FEA.
12
COMPARISON OF OPENED AND CLOSED CONTOURS
For a tube of 25 mm diameter and thickness of 2 mm
we get:
For the closed contour we get the polar moment of
inertia of 1.227 cm4, while for opened – 0.021 cm4,
which is 57 times smaller.
If we would increase the diameter, the difference will
be increased dramatically.
Let’s take a thin-walled circle with radius 1 m and
thickness of 2 mm.
For the closed contour we get 628,300 cm4, while for
opened – only 1.67, which is 375 thousands times
smaller.
13
FUSELAGE
If the cross section is subjected to bending with
moment Mx , the specific normal force is equal to
dN z
dt
  z  
M x 
Ix
 R  sin  

Here  is effective thickness of
skin (includes stringers);
Ix is moment of inertia for cross
section:
Ix    R 
3
14
FUSELAGE
The relative bending angle between two frames
equal to
M a
fx 
fx is
x
E  Ix
Here a is a distance between
frames.
Normal stresses have a vertical
projection due to the presence
of bending angle. This projection
tends to compress the frame as
shown at the figure.
15
FUSELAGE
The distributed load on the frame could found as
2
dN z f x  M x  a    R
q  2


 sin  

dt
2
E
 Ix 

By making few transformations,
we get
q  q 0  sin    ;
q0 
M a
2
x
5
  R   E
2
16