Aircraft Structures Notes 1
-
Upload
wizardwannabe -
Category
Documents
-
view
309 -
download
37
description
Transcript of Aircraft Structures Notes 1
-
5/26/2009
1
ME4212
Aircraft Structures(formerly Mechanics of Thin-Walled
Structures)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Why study thin-walled structures?
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
-
5/26/2009
2
Why study thin-walled structures?
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Why study thin-walled structures?
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
-
5/26/2009
3
Linear Elastic Fracture
Elasto-Plastic Fracture
Damage Tolerance
Fatigue
Delamination
Fracture MechanicsAircraft Structural Analysis
Mechanics of Composite Materials
Stress-Strain Relations of Fiber-Reinforced Composites
Laminate Design and Architecture
Beams
Plates
Shells
Stringers and Panels
Multi-cell Torque Boxes
Mechanics of Thin-Walled Structures
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Laminate Design and Architecture
Hygrothermal Effects
Durability and Failure of Composites
Tapered Structures
Buckling and Instability
Vibration
Torsion
Bending
Course Overview:
T.E. TayProfessorRoom EA-7-17
6516 2887Idealized Beams
Multi-cell Sections & Tapered Beams
Circular & Rectangular Plates
Instability of Columns & Plates
S L Toh
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Instability of Columns & Plates
Energy Methods in Instability
S.L. TohAssociate Professor
-
5/26/2009
4
Torsion Non-Circular Shafts
Thin-Walled Open Sections
Thin-Walled Closed Sections
Warping of Unrestrained Sections
B diBending
Unsymmetric Bars
Thin-Walled Open Sections
Thin-Walled Closed Sections
The Shear Center
Idealized Beams Bending & Torsion
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
dea ed ea s e d g & o s o
Multi-cell Sections Bending & Torsion
Tapered Beams & Beams with Varying Moments of Area
Mechanics of Materials, A.C. Ugural, McGraw-Hill, 1991Ch 6 8 6 9: Torsion Torsion of Thin Walled Shafts
Books:
Ch 6.8, 6.9: Torsion, Torsion of Thin-Walled Shafts
Ch 7: Bending of Beams
Ch 8.6, 8.8: Unsymmetric Bending, Shear Center
Mechanics of Materials, Roy R. Craig Jr., Wiley & Sons, 1996Ch 4 4 7 4 8: Torsion Torsion of Noncircular Sections
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Ch 4, 4.7, 4.8: Torsion, Torsion of Noncircular Sections
Ch 6, 6.6, 6.8-6.10, 6.12: Bending, Unsymmetric Bending, Shear Flow, Shear Center
-
5/26/2009
5
Mechanics of Elastic Structures, J.T. Oden and E.A. Ripperger, McGraw-Hill, 1981
Books (advanced):
Aircraft Structures for Engineering Students, T.H.G. Megson, Edward Arnold, 1990 (2nd Edition)
Analysis of Aircraft Structures, B.K. Donaldson, McGraw-Hill, 1993
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Aircraft Structures & Systems, R. Wilkinson, Addison-Wesley Longman, 1996
-
1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Torsion
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Torsion of Circular Sections
Bars of Circular Sections subjected to torsion
Radial lines on cross-section remains straight. But square elements are distorted.
-
2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
r x
The shear strain is defined:
xr
=
And the shear stress is
GrdxdGr
xGr
G
x
=
=
=
=
0lim
where G is the shear modulus.
dxd = is the rate of twist, and is a constant.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
rA
The twisting moment (i.e. torque) is
( )
GJTdxdGJ
dArdxdG
dArT
A
A
=
=
=
=
2
(1)
where A is the area of the cross-section, and
JTrGr
=
=
And the shear stress
J is the polar second moment of area.
(2)
dF
-
3
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Torsion of Non-Circular Sections
Bars of Non-Circular Sections subjected to torsion
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Bars of Non-Circular Sections subjected to torsion
Torsion of Non-Circular Sections
-
4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Bars of Non-Circular Sections subjected to torsion
Torsion of Non-Circular Sections
Note: There is no change in shape. Hence
0
0
=
=
yz
yz
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Consider a cross-section of arbitrary shape:
A is an arbitrary point.
A
It is displaced to A under torque.
A
There is a point O, that has no displacement when the torque is applied, called the centre of twist.
OLets place the origin of our coordinate system at O.
y
z
-
5
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
A
A
O y
z
yw =
zv =
Horizontal displacement of A.
Vertical displacement of A.
xywxzv
==
In general, we write the displacements:
where is the rate of twist.
(3a)
(3b)
),( zyu =
We further assume that the axial displacements are given in this form:
),( zywhere is the warping function, yet to be determined.
(3c)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Axial displacements depend on the warping function ),( zy
y
z
x
Our objective is to determine u, v and w, the complete displacement field.
-
6
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Start with the strain-displacement (compatibility) relations:
xw
zu
yw
zv
xv
yu
zw
yv
xu
xzyzxy
zyx
+
=
+
=
+
=
=
=
=
,,
,,,
Substituting Eqns (3) into above,
+
=
=
====
yz
zy
xz
xy
yzzyx
0 (4a)
(4c)
(4b)
implies cross-section does not distort in its own plane (i.e. all points are simply rotated as in rigid-body rotation).
0=yz
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The relevant force equilibrium equation
0,,
0
===
=
+
+
xxzxzxyxy
xzxyx
GG
zyx
This is the Laplace Equation. Clearly, the warping function satisfies both compatibility and equilibrium.
In order to obtain a solution, we need to specify boundary conditions.
with
and Eqns (4) yields
0
0
2
2
2
2
2
=
=
+
zy
(5)
or
-
7
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
Consider a small element on the boundary, which is usually traction free.
dz
xz
xy
dy
dx
Force equilibrium in the axial direction provides the equation of the boundary condition.
dydz
dydxdzdx
xzxy
xzxy
=
= 0(6)
In general, it may be necessary to solve the Laplace Equation numerically.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Stress Functions for Torsion
Prandtl assumed a stress function which is twice differentiable, and has the property:
y
z
xz
xy
=
=
(7a)
(7b)
Substituting (7) into (4b) and (4c), differentiating (4b) with respect to z, and (4c) with respect to y,
G zz z z y
G yy y y z
= = +
(8a)
(8b)
-
8
Ludwig PrandtlFather of Modern Fluid Mechanics
(1875-1953)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Subtracting (8b) from (8a) results in
The solution to this equation satisfies both compatibility and equilibrium.
In order to obtain a solution, we need to specify boundary conditions.
G
Gzy
2
2
2
2
2
2
2
=
=
+ (9)
or
0
0
=
=
d
dyy
dzz
or
For the boundary condition, substituting (7) into (6), we obtain
This means that at the boundary, is a constant. Since the value of can be arbitrary, we shall select = 0 for convenience.
(10)
-
9
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
dAxz
dAxyArea dA
z
O y
z
y
( ) =A
xyxz dAzyT
The torque about the x-axis due to the shear stresses is given by
(11)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )( ) ( )( )dzdyyydydzzzdzdyy
ydydzz
z
dydzzz
dydzyy
dydzzz
yy
T
CCDDAABB
D
C
B
A
=
=
=
+
=
Introducing Eqn (7) , and noting that dA = dydz,
(12)
(13)
(Integration by parts)
-
10
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
B
D
O y
z
C
A
B
= dydzT 2
are values at boundary points, which we have chosen to set to zero. Hence the equation reduces to
A C D
(14)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If we retain the expression
GJT = (15)
Only for the very special caseof the circular cross-section is J equal to the second polar moment of area.
= dydzGJ 2
where now we define the symbol J as the torsional constant, then the formula for J is
The product GJ is known as the torsional stiffness.
(16)
-
11
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The Membrane (Soap Bubble) Analogy
yz (into plane)
w
MTTensionPressure p
Good for visualizing .
Equilibrium equation for small deflection w of a flat membrane subjected to internal pressure pclosely resembles the torsion equation.
MTp
zw
yw
=
+
2
2
2
2
(17)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Comparing Eqns (9) and (17),
MTp
zw
yw
=
+
2
2
2
2
(17)Gzy22
2
2
2
=
+ (9)
= dydzT 2
Furthermore, according to Eqn (14), the volume under the membrane is proportional to the torsion T.
(14)
w is analogous to MTp
G2is analogous to
-
12
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The slope of the membrane is proportional to the shear stresses.
yz (into plane)
w
yw
y
zw
z
xz
xy
=
=
yw
Local slope
zw
Similarly for
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is hollow,
yz (into plane)
0=
yw
yz
0=d at both inner and outer boundaries.
= constant
= 0
-
13
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Torsion of Thin-Walled Open Sections
y
z
Profile of w or in x-y plane
0=y
over much of y
y
x
Consider a narrow rectangular section:
Profile of w or in x-z plane
0=z
only at the centerline of the section (at z = 0).
x
z
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Gzy
222
2
2
=
+ (9)
0=y
Since , Eqn (9) reduces to
Gz
222
=
(18)
212 CzCzG ++=
Integrating twice with respect to z yields
(19)
Applying the boundary condition that = 0 at z = + t/2 and t/2, we obtain and .4/0 221 tGCC ==
=
4
22 tzG (20)
-
14
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
0
2
=
=
=
=
y
zGz
xz
xy
The stresses are therefore
(21b)
Eqn (21a) shows that xy varies linearly with z and is zero at the centerline of the section (at z =0).
(21a)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
3
22
2
3
2/
2/
2/
2/
22
btJ
dzdytz
dydzG
J
b
b
t
t
=
=
=
From Eqn (16),
(22)The torsional constant for a thin rectangular section.
zJTzGxy
22 ==
The shear stress
(23)
JTt
MAXxy=
has maximum values at z = + t/2 and t/2, i.e.
(24)
-
15
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
Linear distribution of shear stress across thickness:xy
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Now, consider the warping displacements:
= zy
Gxy
From Eqns (4),
(25a)
+
= yz
Gxz (25b)
From Eqn (21b), xz = 0, so Eqn (25b) becomes
yz
=
yz= (26)
Clearly, C1(y) = C2(z) = 0 and
Upon integration,
)(1 yCyz +=
From Eqn (21a), xy = - 2G, so Eqn (25a) becomes
zy
zy
z
=
=
2
Upon integration,
)(2 zCyz +=
-
16
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
A hyperbolic-paraboloid surface
yzu = (27)
The warping displacement is therefore
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
General Open Sections
yz
Linear distribution of shear stress across thickness
xy
In general,
3
3BtJ =
s = 0
s = B
-
17
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
b1
b2
b3
b4
b5
b6
t1
t2
t3t4
t5
t6
333333
366
355
344
333
322
311 tbtbtbtbtbtbJ +++++=
For a section consisting of n elements,
=
=n
iiitbJ
1
3
31 (28)
Maximum shear stress is
i
iMAXiiMAX J
tT=
where the proportion of torque T acting on element i is
TJJT ii
=
(29)
(30)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Consider a thin-walled closed section:
yz
yz (into plane)
yw
Local slope almost constant
Torsion of Thin-Walled Closed Sections
-
18
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
yz
Constant distribution of shear stress across thickness
xy
Closed Sections
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
x
t1
t2
1
2
xt 11
xt 22
Taking force equilibrium in the axial direction,
2211 tt = (31)
Convenient to define shear flow
tq = (32) constant21 == qq
-
19
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Choose an arbitrary point on the wall, at s = 0.
s = 0qds
Force due to shear flow on a small element of wall is qds.
O
r
Moment about an arbitrary point O is therefore qrds, where r is the perpendicular distance.
== rdsqrqdsT
The total moment then equals the torque,
(33)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
s = 0
O
qdsr
=
=
=1
01 2
1)(ss
s
rdss
As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as
(34)
-
20
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
O
s = 0
=
=
=1
01 2
1)(ss
s
rdss
As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as
= rds21
The total area enclosed by the centerline of the wall is
(35)qT = 2
and the torque, from Eqn (33) is
(36)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
x =The section does not distort under torsion. Points on the wall experience a rotation about the center of twist.
x
is the tangential displacement component in the s direction.
xr =
OCenter of twist
(37)
-
21
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
xConsider a small element of the wall
s
y
z
x
x
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
s
Displacement in the sdirection is
-
22
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
x
su
Displacement in the xdirection is u
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
x
sTherefore, the shear strain in the wall is
rsu
xsu
xs
+
=
+
=
(38)
-
23
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Since it follows that Gxsxs / =
rGs
u xs = (39)
Multiplying both sides by ds and integrating around the total periphery S of the tube,
= rdstds
Gqdu (40)
The integral on the left-hand-side is zero, since u at s = 0 and u at s = S are the same (i.e. same point), so that the difference is zero.
Substituting for from Eqn (35),
=
=
tds
Gqt
dsGq
2
20
(41)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Substituting for q from Eqn (36),
= tds
GT
24 (42)
Since T = GJ , the torsional constant J for a closed section is
=
tdsJ
24(43)
If the thickness t is constant,
StJ
24= (44)
-
24
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
=
+
=n
iiitb
tdsJ 1
32
314
Hybrid Sections
(45)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 1
The round tube shown has an average radius R and a wall thickness t. Compare the torsional strength and rigidity of this tube with that of a similar tube which is slit along its entire length. Assume R/t = 20.
R
t
R
t
Closed tube Open section
-
25
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Torsional constant
tRtRt
dsJ Closed
332
2
22
4
4
==
=
Closed tube Open section
323
3
3
Rt
btJ Open
=
=
For a given angle of twist, the closed section resists 1200 times the torsion of the open section.
1200
32
=
=
tR
JJ
Open
Closed
For a given torque, the open section twists through an angle 1200 times as great as the closed section.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Maximum shear stress
tRT
tT
Closed
22
2
=
=
Closed tube Open section
223RtT
JTt
MAXOpen
=
=
For a given torque, the shear stress in the open section is 60 times as high as in the closed tube.
3
60
Open
Closed
Rt
=
=
If the allowable shear stresses are equal, the closed tube is 60 times as strong as the open section.
-
26
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 2
A torque of T = 6000 Nmm is applied to the section shown. Determine the rate of twist and maximum shear stress. Assume shear modulus G = 80 GPa.
Dimensions in mm
1
2
1
2 215
20
19
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
5.582
192
201
19120
=+++= tds
The torsional constant is
( ) ( )( ) 4332
32
mm10913.921531
5.583804
314
=+=
+
=
bt
tdsJ
Total area enclosed: 2mm3801920 ==
Since T = GJ , the rate of twist is
( )( )rad/mm1057.7
108010913.96000
6
33
=
=
-
27
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The contributions to the torsional constant are
( )
( )( ) 43
432
mm4021531
mm10873.95.58
3804
==
==
Fin
PartClosed
J
J } Compare !The proportions of torque carried are
Nmm2.2460009913
40
Nmm59766000913.9873.9
==
==
Fin
PartClosed
T
T
The maximum stresses are
( )( )( )( ) MPa21.1
4022.24
MPa86.713802
59762
===
==
=
Fin
FinMAXFin
MIN
PartClosedMAXPartClosed
JtTt
T
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is slit longitudinally at A as shown, what are the rate of twist and maximum shear stress?
Dimensions in mm
1
2
1
2 215
20
19
A
-
28
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )( )
( )( ) rad/mm1078.415710806000
MPa4.76157
26000
43
=
==
===
GJT
JTt MAX
MAX
The torsional constant is
( )( ) ( )( ) ( )( ) ( )( )[ ] 43333 mm15723521912011931
=+++=J
Note the increase in stress and rate of twist.
and
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Multiplying both sides by ds and integrating from s = 0 to some other point on the wall,
where u0 is the displacement of the point s = 0 in the x-direction.
Warping of Unrestrained Thin-Walled Sections
Recall from Eqn (39) that
= rdsdsGuu xs 1
0
rGs
u xs =
(46)
-
29
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
s = 0
O
r
dsrs = 21)(Recall the sectorial swept area (Eqn (34)), defined as
The equation, valid for both open and closed sections, can be rewritten as
0
)(21
uuu
sdsG
u xs
=
=
where r is measured from an arbitrary point O (called the pole).
)(210 sdsGuu xs = (47)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
O
s = 00=u
Here, is a function of s, and is zero when the integration is completed around the closed tube.
u
For closed sections, since , we havetqxs /=
2= tds
Gqu (48)
-
30
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
For open sections, the contribution of the shearing strain xs to the warping displacement u is negligible because the walls are very thin, so that
0 dsxs
2=u
Hence the equation reduces to
(49)
(50)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 3
The closed box section shown is subjected to a torque T. Determine the warping displacements.
th
b
h
th
tb
tb
T
12
43
+=
+=
+++=
=
bh
bh
hbhb
tb
th
GhbT
tb
th
GhbT
tds
tds
tds
tds
GhbT
tds
GT
22
22
2
1
1
4
4
3
3
222
2
2
224
4
4
Rate of twist:
-
31
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Select the pole at O the center of section, and starting point at I.
12
43
OI
bhTTq
22=
= and
hI h th
tds
2
1
=We find that
For closed sections, the relative warping displacement is
2= tds
Gqu
1
1
1 2 II ht
dsGqu =
4222 1
bhhbI =
=Also,
=
bh tb
th
bhGTu
81Therefore,
Interestingly, there will be no warping if
bh tb
th=
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
12
43
OI
12
222
8
43
222
uuth
tb
bhGT
bhtb
th
GhbT
tb
th
bhGTu
hb
bhbh
=
+=
+
+=
43
242 2
bh
bhbhI
=
+=Now,
1413 , uuuu ==Similarly,
bh
bI hI
tb
th
tds
tds
tds
+=
+=
2
2
1
12
and
-
32
1
2
4
3
I
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Due to symmetry, we know that the axial displacement at I is zero. Since we have .11 uu =
,11 Iuuu =
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 4
What would the warping displacements be if the section is slit longitudinally along its entire length at point I ?
12
43
OI ( )332
3
bh bthtGT
GJT
+==Rate of twist:
( )3332
bh bthtJ +=Torsional constant:
2=uFor an open section,
-
33
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
12
43
OI
( ) ( )33331 83
23
4 bhbh bthtGbhT
bthtGTbhu
+=
+=From I to 1,
( ) 1332 389 u
bthtGbhTu
bh
=+
=From I to 2,
==
4525 313bhuu IFrom I to 3,
==
4727 414bhuu IFrom I to 4,
==
4828 31'bhuu II From I to I,
1
2
4
3
I
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Longitudinal slit
-
34
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 5
Let us return to the closed section. What will happen if we select another point O as pole?
12
43
O I
22 1
bhI =From I to 1,
=
+
=
b
bhh
tb
bhGT
bhtb
th
GhbT
th
bhGTu
4
2222 221
hI th
tds
2
1
=and
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
12
43
O I
bhI =22From I to 2,
bhI tb
th
tds
+= 22
and
=
+
+=
h
bhbh
th
bhGT
bhtb
th
GhbT
tb
th
bhGTu
4
222 222
1423 , uuuu ==
Similarly, we can show that
Try it yourself !
-
35
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
10,30 ==bh tb
thFor illustration purposes, let 0.1
2=
bhGT
and
5.2,5.7,5.7,5.2 4321 ==== uuuuThen
1
2
4
3
I
1
2
4
3
I
From Example 3: Present Example 5:
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Hence changing the pole does not alter the problem (i.e. the stresses are also unaffected), but merely changes the reference plane from which the axial displacements are measured.
In this example, the new reference plane is rotated about the OI axis.
-
1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Bending
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Bending of Bars
Bars of Non-Circular Sections subjected to pure bending
Bending Moment M
-
2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Pure bending occurs only when the force F acts through the Shear Center E.
Built-in end
When the force F acts through any other point, there is combined bending and twisting.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
t
Let the loads Vz and Vy act through the shear centre E. Hence pure bending.
Vy
Vz
Shear Centre
E
y
z
xCentroid O
x-y-z are the centroidal axes
-
3
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
t
y
z
xCentroid O
x-y-z are the centroidal axes
The shear loads give rise to the applied moments (positive right hand screw rule)
My
Mz
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y (into page)
z
x
Contribution of Vzto bending moment
dx
Vz
E
Hence
xM
V
dxVdM
yz
zy
=
=
(51a)
+ dMy
-
4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
z (out of page)
y
x
dx
Vy
E
Hence
xMV
dxVdM
zy
yz
=
=+ 0
(51b)
Contribution of Vyto bending moment
+ dMz
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
t
y
z
x
dx
ds
Let x denote the normal stress due to bending. Since there is no nett axial load,
0= dAA
x (52a)
xArea dA = t ds
-
5
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
t
y
z
x
Consider a small element of the wall (in positive z)
Consider applied moment about the y-axis
dx
ds My
x
Area dA = t ds
z
The moment about the y-axis contributes to the normal stress xdue to bending.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y (into page)
z
x
+ dMy
dx
dAzMA
xy = (52b)
x
Area dA = t ds
z
Contribution of axial force dF due to x to the bending moment
(Tensile on positive z)
-
6
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
t
y
z
x
Consider a small element of the wall (in positive y)
Consider applied moment about the z-axis
dx ds
Mz
x
y
The moment about the z-axis also contributes to the normal stress xdue to bending.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
z (out of page)
y
x
+ dMz
dx
x
y
Area dA = t dsdAyM
Axz = (52c)
Contribution of axial force dF due to x to the bending moment
(Compressive on positive y)
-
7
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Plane sections remain plane assumption implies that the axial strain x (and axial stress x ) is a linear function of y and z. The cross-section rotates about a neutral axis whose orientation is dependent upon the loading and geometry of the cross-section.
y
z
Cross-sectionx
Side view along the neutral axis
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let zCyCCx 321 ++=
and substituting this into Eqns (52) and solving for the constants C1, C2, C3, we have C1 = 0 and
zIII
IMIMy
IIIIMIM
yzzzyy
yzzzzy
yzzzyy
yzyyyzx
++
+= 22 (53)
-
8
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Eqn (53) can be rewritten in the form
+
=
+
=
+=
zzyy
yz
zz
yzzy
y
zzyy
yz
yy
yzyz
z
yy
y
zz
zx
III
IIM
MM
III
IIM
MM
IzM
IyM
2
2
1
1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The neutral axis can be located by setting x = 0 in Eqn (53), and find, for angle ,
yz
yzzzzy
yzyyyz
IMIMIMIM
yz
++
==tan (54)
-
9
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
Neutral axis Compression
Tension
xGeneral distribution of normal bending stress
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is symmetric about either the y- and z-axes, Iyz = 0 and Eqn (53) reduces to
yy
y
zz
zx I
zMI
yM+= (55)
-
10
A word about sign convention for moments:
Applied Moment MyPositive into page
yy
y
zz
zx I
zMI
yM+=
y (into page)
z
x
z (out of page)
y
xPositive out of page
Applied Moment Mz
This causes tensile (+) stresses for positive z
This causes compressive (-) stresses for positive y
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Note that Equation (53) for normal bending stresses is valid for all general sections(whether solid or thin-walled, open or closed sections).
-
11
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The simply-supported beam has an unsymmetrical cross-section. Locate the neutral axis and determine the largest tensile and compressive stresses in terms of load Q and where they occur. (Assume Q acts through the shear center.)
Example 6
1000 1500
Q
y
z
20
80
20 60
Q
Cross-section
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The maximum bending moment occurs at the point of application of load Q and is given by
My = - (0.6Q)(1000) = -600 Q Nmm.
Mz = 0.
1000 1500
Q
0.6 Q 0.4 Q
1000
0.6 Q
MyFree-body diagram:
Positive into page
-
12
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
First, locate the centroid C, as the equations are related to centroidal axes.
20
80
20 60
QThe total cross-sectional area:
= (20)(100) + (60)(20) = 3200 mm2
Cy
z
AD
B
Define the points A, B and D.
mm510020110
20
60
0
=
+==
dyydyyA
ydAA
y
mm151 == zdAAz
Therefore,
measured from point B,and
measured from point D.
15
5
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
20
80
20 60
Q
Cy
z
15
5
Section properties (second moments of area):
46
23
23
mm10907.2
)25)(1200(12
)20)(60(
)15)(2000(12
)100)(20(
=
++
+=yyI
46 mm10627.1 =zzI
46 mm102.1
)25)(25)(1200()15)(15)(2000(
=
+=yzI
-
13
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Cy
z
Neutral axis:
=
==
++
=
4.36627.1
2.1
tan
zz
yz
yzzzzy
yyzyzy
II
IMIMIMIM
since Mz = 0.Compression
Tension
36.4
A (-25,35)
B (-5,-65)
Points A and B are furthest from the neutral axis, and therefore experience the greatest stresses.
From Eqn (53), we find
( )( ) MPa0182.0
MPa0158.0Q
Q
Bx
Ax
=
=
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
There is also a shear stress distribution due to bending.
We will now discuss shear stresses due to bending in thin-walled members only.
-
14
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
Shear stress distribution due to bending
qdx
dsds
sqq
+
x
dxx
xx
+
xt
sq
dxdssqqqdxtdsdx
xtds
x
xxx
=
=
++
++
0
Force equilibrium in x-direction:
(56)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Recall that Eqn (53) holds for normal bending stresses. Differentiating with respect to x,
where the relations in Eqns (51) have been used.
zIIIIVIV
yIIIIVIV
x yzzzyyyzyzzz
yzzzyy
yzzyyyx
+
+
=
22
(57)
( ) ( )( )tIII
QIVIVQIVIVtq
yzzzyy
yyzyzzzzyzzyyyxs 2
+== (58)
Substituting Eqn (57) into Eqn (56) and integrating with respect to dA (= t ds), we have
zAdAzQ
yAdAyQ
Ay
Az
==
==
where (59a)
(59b)
( and are coordinates of the centroid)
zy
-
15
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Eqn (58) can be rewritten in the form
=
=
+=
zzyy
yz
zz
yzyz
z
zzyy
yz
yy
yzzy
y
yy
yz
zz
zy
IIIIIV
VV
III
IIV
VV
IQV
IQV
q
2
2
1
1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is symmetric about either the y- and z-axes, Iyz = 0 and Eqn (58) reduces to
+=
yy
yz
zz
zy
IQV
IQV
q (60)
-
16
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
q is a function of s and may even change direction, depending upon loading
General distribution of bending shear flow q for open sections
s
Bending of open sections
q = 0
q = 0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Vy
Vz
E
y
z
xCentroid O
Bending of closed sections
Select an arbitrary point m on the wall for s = 0
smx
dxx
xx
+
ds
sqqm
+
qm
We already know that . This gives the same solution for the shear flow distribution as in the open section case. But qm is not zero because the section is closed. However, qm can be found by equating to zero the sum of moments about x-axis, since no torque is applied.
xt
sq x
=
-
17
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 7
Determine the shear flow distribution in the z-section of uniform thickness t, due to a shear load Vz applied through the shear center of the section.
First thing to note is that due to anti-symmetry, the centroid and the shear center are the same point O. This is a special case.h/2
t
h/2
y
z
O
h/2
h/2
8422422
12232
312222
3
33
332
thhhhthhhtI
thhtI
ththhhtI
yz
zz
yy
=
+
=
=
=
=+
=
Vz
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Hence
z
zzyy
yz
zzz
zzyy
yz
zz
yzz
y V
III
VVV
IIIIIV
V 28.21
86.01
22=
==
=
The shear flow due to bending:
=
=
+=
ssz
ss
z
s
yy
zs
zz
y
zdsydshV
ztdsth
ytdsth
V
ztdsIVytds
IV
q
003
03
03
00
85.63.10
3/28.2
12/86.0
-
18
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
h/2
t
h/2
y
z
O
h/2
h/2
For the bottom flange, define a coordinate s1 such that z = - h/2 and y = - h/2 + s1, where 2/0 1 hs
( )chsshV
dshdshshVq
z
ssz
+=
=
1213
01
011312
72.115.5
285.6
23.10
11
1 2
s1
Boundary condition: At s1 = 0, q12 = 0; hence c = 0
( )121312 72.115.5 hsshVq z =
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
O
1 2
s1
( )121312 72.115.5 hsshVq z =
For values of s1 < 0.334h, q12 is negative and therefore in opposite direction to s1.
At 2, s1 = h/2, q12 = 0.43 Vz/h.
Shear flow distribution is parabolic, with a change in sign at s1 = 0.334h.
0.334h
The shear flow distribution on the top flange is similar.
-
19
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( ) 220
2323
2
85.643.3 qdsshhVq
sz +=
where q2 is the shear flow at point 2, i.e. q2 = 0.43 Vz/h.
y
z
O
1 2
s2
3Along the vertical web 2-3, z = s2 h/2, y = 0, and hs 20
This shear flow distribution is also parabolic, with a maximum at s2 = 0.5h.
( )2222323 43.343.343.0 shshhVq z +=
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The Shear Center
There is a point where the resultant force must pass through in order that no twisting moments (torque) are developed. This point is called the shear center.
Why do we wish to find the shear center?
So that we will know, for a given load, how to proportion the stresses (and deformations) due to bending and torsion.
How do we find the shear center ?
We make use of its definition. A general procedure:
1. Find the centroid.
2. Find section properties.
3. Apply Vz or Vy, in turn.
4. The moment due to shear flows is due to application of the shear force through the shear center. The position of the shear center with respect to a convenient point can thus be obtained.
-
20
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 8
Determine the position of the shear center of the section.
First thing to note is that since the section is symmetric, the positions of the centroid and the shear center must lie somewhere on the horizontal axis of symmetry.
Vz
Assume the shear center E exists and is located distance ye from vertical web, as shown. h/2
tw
tf
y
z
O
b
h/2
tfApply a shear load Vz through the shear center.
yeE
02122
212
2323
=
+=
+=
yz
fwf
wyy
I
bhththbthtI
Section properties:
because section is symmetric.
Izz is not relevant.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Vzh/2
tw
tf
y
z
O
b
h/2
tf
yeE
At the top flange, z = h/2, y = b s1, where s1 defined.
Hence 0, == yzz VVV
dstzIVq
s
yy
z =0
s1
.0,0
2
2
1
1
1
1
1
10
==
=
=
s
yy
fz
s
fyy
zs
qs
sI
htV
dshtIVq
At (free edge).
-
21
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Vzh/2
tw
tf
y
z
O
b
h/2
tf
yeE
Therefore, the shear center is to the left of the section.
Take moments about the point A,
zyy
f
b
zyy
f
b
sez
VI
thb
sVIth
dsqhyV
4
2222
0
21
2
10
1
=
=
= s1
( )fwf
yy
fe
bthttb
Ithb
y
63
42
22
+=
=
A
b
s dsq0
11
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 9
Determine the position of the shear center of the semi-circular section.
The section is symmetric, the positions of the centroid and the shear center must lie somewhere on the horizontal axis of symmetry.
Vz
Assume the shear center E exists and is located distance ye from vertical web, as shown.
y
z
OApply a shear load Vz through the shear center.
yeE
02
3
=
=
yz
yy
I
trI
Section properties:
because section is symmetric.
Izz is not relevant.
t
r
-
22
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Vz
y
z
Oye
E
t
r
Define coordinate s.
where.,cos rddsrz ==
sin
cos
2
0
2
0
yy
z
yy
z
s
yy
z
ItrV
drI
tV
dstzIVq
=
=
=
s
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Vz
yeEy
z
O
t
r
s
Take moments about the point O,
[ ]
z
yy
z
S
ez
Vr
ItrV
dqr
dsqryV
4
cos 04
0
2
0
=
=
=
=
The shear center is to the leftof the section.
r
rye
27.1
4
=
=
-
23
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 10
Determine the position of the shear center of the closed section and the distribution of shear flow.
Vz
Assume the shear center E exists and is located distance ye from vertical web, as shown.
t1
150
t1Apply a shear load Vz = 1 106 N through the shear center.
.mm6,mm3
0
mm1078.25
21
46
==
=
=
tt
I
I
yz
yy
Section properties:
because section is symmetric.
Izz is not relevant.t1
250
t2
yeE
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Define coordinates and points as follows.
Let the shear flow at A be qm, an unknown quantity to be determined.
t1 t1
t1
t2
y
z
O
s4
s2
s1
s3
A
B C
D1251 += zsFrom A to B,
1253 += zsFrom C to D,
From B to C, z = 125
From D to A, z = 125
-
24
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )
( )211116
6
11
5.0125116.0
1251078.25
)3)(10(
ssq
dssq
dsztIVqq
m
m
yy
zmAB
=
=
=
At B, qAB = qm
( )2
2
21
5.14
125116.0
sq
dsq
dsztIVqq
m
m
yy
zmBC
+=
=
=
At C, qBC = 2175 + qm
t1 t1
t1
t2
y
z
O
s4
s2
s1
s3
A
B C
Dqm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
qm ( )
( ) 21751255.0233.0125
1078.25)6)(10(
323
336
6
32
+=
=
=
ssq
dssq
dsztIVqq
m
m
yy
zmCD
At D, qCD = qm + 2175
( )21755.14
125116.0
4
4
41
+=
=
=
sq
dsq
dsztIVqq
m
m
yy
zmDA
t1 t1
t1
t2
y
z
O
s4
s2
s1
s3
A
B C
D
-
25
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Rate of twist = 0, since load acts through shear center.
0
02
1
41
32
21
11
=+++
=
=
dst
qdst
qdst
qdst
q
dstq
Gdxd
AB
DA
AB
CD
BC
BC
AB
AB
which simplifies to
1.8860598125675
==+
m
m
qq
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
4
323
2
211
5.14128929116.01289
5.141.886058.05.141.886
sqssq
sqssq
DA
CD
BC
AB
=+=
+=+=
Therefore,A
B C
D
qm = - 886.1
qm = - 886.1 q = 1288.9
q = 1288.9
-
26
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Take moments about point O to obtain shear center.
( )
( )
mm7.939364625075525
29116.09.1288150
5.141.886250
150250
3
250
03
23
2
150
02
32
=+=
++
+=
=+
e
ez
ez
D
CCD
C
BBC
y
dsss
dssyV
yVdsqdsq
Vz
150
250
yeE
O
A
B C
D
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Combined Bending and
Torsion
-
27
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
General Case: Superposition of Bending and Torsion
P
E
P
=Pure Bending through shear center
+Pure Torsion T = Pd
E
d
T
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Pure Torsion of Open Sections
T
Linear distribution of shear stress across thickness
xy
-
28
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Pure Torsion of Closed Sections
T
Shear flow q is constant anywhere in the section, but shear stress is NOT constant because of varying wall thickness
xs
Constant distribution of shear stress across the thickness
xs
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Pure Bending of Open Sections
Bending stress exists in walls
x
E
P
Distribution of shear flow q in walls due to bending such that nett contribution of q must be due to P, applied force.
-
29
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Pure Bending of Closed Sections
E
P
Bending stress exists in walls
x
Distribution of shear flow q in walls is similar to that of an open section, except for a constant q0superimposed.
-
1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Idealized Beams
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
-
2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
-
3
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
xCentroid O tn
An
The section is made up of concentrated stringers of area An, joined by thin sheets of thickness tn.
zn
yn
The nth stringer is located at positions yn and zn away from the origin at the centroid.
Let the loads Vz and Vy act through the shear centre E. Hence pure bending.
VyVz
Shear Centre
E
-
4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Taking moments as before (see Eqns (51)), we have
xVMx
MV
zy
yz
=
=
(61a)
and
xVMx
MV
yz
zy
=
=
(61b)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
xCentroid O
An
Let n denote the normal stress due to bending in the nthstringer. Since there is no nett axial load,
01
==
N
nnn A (62a)
n where N is the total number of stringers.
-
5
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
xCentroid O
An
zn
yn
=
=
=
=
N
nnnn
N
nyny
zA
MM
1
1
(62b)
Myn
n
The moment about the y-axis contributes to the normal stress ndue to bending.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y (into page)
z
x
+ Myn
dx
n
Contribution of axial force dF due to n to the bending moment
(Tensile on positive z)
nnnyn zAM =Area An
zn
-
6
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
xCentroid O
An
zn
yn
=
=
=
=
N
nnnn
N
nznz
yA
MM
1
1
(62c)
The moment about the z-axis contributes to the normal stress ndue to bending.
Mzn
n
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
z (out of page)
y
x
dMzn
dx
Contribution of axial force dF due to n to the bending moment
(Compressive on positive y)
nnnzn yAM =Area An
n
yn
-
7
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let nnn zCyCC 321 ++=
and substituting this into Eqns (62) and solving for the constants C1, C2, C3, we have C1 = 0 and
nyzzzyy
yzzzzyn
yzzzyy
yzyyyzn zIII
IMIMy
IIIIMIM
++
+= 22 (63)
Invoke again the assumption that plane sections remain plane, which implies that the axial strain (and axial stress ) is a linear function of y and z.
Note the similarity with Eqn (53).
The expression for the neutral axis remains the same as Eqn (54).
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
However, the section properties are now
=
=
=
=
=
=
N
nnnnyz
N
nnnzz
N
nnnyy
zyAI
yAI
zAI
1
1
2
1
2 (64a)
(64b)
(64c)
They are easier to evaluate than integrals!
-
8
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Eqn (63) can, of course, be rewritten in terms of equivalent moments:
+
=
+
=
+=
zzyy
yz
zz
yzzy
y
zzyy
yz
yy
yzyz
z
yy
ny
zz
nzn
III
IIM
MM
III
IIM
MM
IzM
IyM
2
2
1
1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
nGeneral distribution of normal bending stress
Neutral axis
Compression
Tension
-
9
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
x
Now, consider the shear flows in the sheets or webs.
qn
nnA
+ dxx
A nnn
nnn
n
nnnn
nnn
qAx
q
dxqdxqAdxx
A
+
=
=+
++
+
+
1
1 0
Force equilibrium in x-direction:
(65)
qn+1
Select a small length dx of a typical nth stringer with two adjacent nthand (n + 1)th sheets.
dx
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Substituting Eqn (63) into Eqn (65) and using Eqns (61),
( ) ( )( ) nnyzzzyy
nyzyzzznyzzyyyn qAIII
zIVIVyIVIVq +
+=+ 21 (66)
This equation relates the shear flow of the (n + 1)th sheet to the shear flow of the adjacent nth sheet.
If the section is open, starting from one free edge, the shear flows in successive sheets may be easily found.
Note that the shear flow does not vary with s within each sheet or web.
-
10
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Eqn (66) can be rewritten in the form
1
2
2
1
1
y n z nn n n
zz yy
z yzy
yyy
yz
yy zz
y yzz
zzz
yz
yy zz
V y V zq A qI I
V IV
IV
II I
V IV
IV
II I
+
= + +
=
=
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is closed, we first analyze the problem as if one of the sheets (pick one arbitrarily) is missing.
The unknown shear flow is then superposed onto the problem and solved by taking moments of the combined shear flows about a convenient point.
q1
q2
q3
q4
q5
Replace the missing sheet with the unknown shear flow.
q0
+ q0
+ q0
+ q0
+ q0
+ q0
-
11
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Determine the shear flow distribution in the idealized channel section produced by a vertical load of 48 kN acting through its shear center. Assume the skin is effective only in resisting shear stresses, while the stringers, each of area 300 mm2, resist all the direct stresses.
Example 11
y
z2000
2000
2000
48 kN
2
43
1
4924
1
2 mm108.4)2000)(300(4 ====i
iiyy zAI
Dimensions in mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
N/mm612)2000)(300(108.4
48000
N/mm126)2000)(300(108.4
48000
N/mm6)2000)(300(108.4
48000
0
9
233334
9
122223
9
1112
=
=
+=
=
=
+=
=
=
+=
qzAIVq
qzAIVq
zAIVq
yy
z
yy
z
yy
z
y
z2000
2000
2000
2
43
1
Dimensions in mm
12 N/mm
6 N/mm
6 N/mm
Note that we can check vertical and horizontal force equilibrium and show that the resultants are 48 kN and zero, respectively.
-
12
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Find the shear flows in the webs of the unsymmetrical closed beam section.Example 12
Vy = 4 kN
Vz = 10 kN
Vz
Vy
A
200 mm
100 mm
CD
B300 mm2 100 mm2
300 mm2 100 mm2
y
z
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
46
4
1
46224
1
2
46224
1
2
mm100.2)100)(50)(100()100)(50)(100(
)50)(100)(300()50)(100)(300(
mm100.8)100)(100(2)100)(300(2
mm100.2)50)(100(2)50)(300(2
=++
+==
=+==
=+==
=
=
=
iiiiyz
iiizz
iiiyy
zyAI
yAI
zAI
42
42
2 .04000 100002.0 1.8667 10
( 2.0)1(2.0)(8 .0)
2 .010000 40008.0 1.4667 10
( 2.0)1(2.0)(8 .0)
y
z
V
V
= =
= =
-
13
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )
1
2 2
3
1.8667 1.46678.0 10 2.0 10
2.33 7.33 10
y n z nn n n
zz yy
n nn n
n n n n
V y V zq A qI I
y z A q
y z A q
+
= + +
= + +
= +
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let qAD be the unknown shear flow, i.e. qAD = q0.
( )0
03
60)100(10)50(33.7)100(33.2
qqqDC
+=+=
A
CD(-100,-50)
B
q0
Consider stringer D:
100 mm2 60 + q0
-
14
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )0
3
100)300(10)50(33.7)100(33.2
qqq DCCB
+=+=
A
C(100,-50)D
B
q0
Consider stringer C:
300 mm2 60 + q0
100 + q0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )0
3
40)100(10)50(33.7)100(33.2
qqq CBBA
+=+=
A
CD
B(100,50)
q0
Consider stringer B:
100 mm2
60 + q0
100 + q0
40 + q0
-
15
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
N/mm502004
0)50)(200()100)(100()50)(200()100)(100(
0
0
==
=+++
qq
qqqq DCCBBAAD
A
CD
B
q0
Let us now take moments about, say O:
60 + q0
100 + q0
40 + q0
O
qAD = - 50 N/mm qDC = 10 N/mm
qCB = 50 N/mm qBA = - 10 N/mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Find the shear center.Example 13
Area of each stringer = A mm2
y
zr Vz
yeE
rV
IAzVq
ArzAI
z
yy
nnz
iiiyy
2
2 22
1
2
==
== =
Taking moments about O,
2
2
20
2
0
ry
rVqr
dqr
dsqryV
e
z
r
ez
=
=
=
=
=
O
s
-
16
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Find the shear center.Example 14
A
B
45
F
120
60
C
D
60 mm2
60 mm2
120 mm2
120 mm2 All dimensions in mm
40360
)90)(120()60)(60(1
40360
)120)(60(21
4
1
4
1
=+
==
===
=
=
iii
iii
zAA
z
yAA
y
First, remember to locate the centroid!
measured from C
measured from C
y
z
O40
40
(-40,-50)
(80,-20)
(80,40)(-40,40)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
46
4
1
46224
1
2
462224
1
2
mm10144.0)50)(40)(120(
)40)(40)(120()40)(80)(60()20)(80)(60(
mm10152.1)40)(120(2)80)(60(2
mm10612.0)50)(120()20)(60()40)(12060(
=+
++==
=+==
=+++==
=
=
=
iiiiyz
iiizz
iiiyy
zyAI
yAI
zAI
Section properties:
-
17
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let us first find the z-coordinate position of shear center E.
A
B
F
C
D
Assume an arbitrary shear force Vy applied to the shear center.
y
z
O
Vy
6
6
0.894 10
0.210 10
yy
zz
zy
yy
VV
IV VI
=
=
Hence
ze
E
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )
( )
6
3
6 3
3
3
0
0.894(80) 0.210( 20) 60 10
4.543 10
0.894(80) 0.210(40) 60 10 4.543 10
8.33 10
5.68 10
y A z AAB A
zz yy
y
y
y B z BBC B AB
zz yy
y y
y
CD y
V y V zq AI I
V
V
V y V zq A qI I
V V
V
q V
= + +
=
=
= + +
=
=
=
A
B
F
C
D
y
z
O
qAB
qBC
qCD
-
18
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
A
B
F
C
D
y
z
O
qBC
qABqCD
Taking moments about a convenient point F,
2 3( 5) [ (45) ( 5.68) ( 4.543)(60)(120) ( 8.330)(120)(45)] 10
113.83
108.8 mm
y e y
y
e
V z V
V
z
+ = + +
=
= measured from O.
5
Vy
ze
E
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let us now find the y-coordinate position of shear center E.
A
B
F
C
D
Assume an arbitrary shear force Vz applied to the shear center.
y
z
O
yyy
z
yzz
y
VIV
VIV
6
6
1068.1
1021.0
=
=
Now,
Vz
ye
E
-
19
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
zCD
BC
zAB
Vq
qVq
3
3
10056.7
010024.3
=
==
A
B
F
C
D
y
z
O
qABqCD
Taking moments about F,
mm7.267.66
10)]120)(60)(024.3()056.7()45([)40( 32
==
+=+
e
z
zez
yV
VyV
measured from O.
40
Vz
ye
E
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The idealized tube is loaded by a vertical shear force of 10 kN acting in the plane of the section. Assuming the direct stresses are carried by the stringers while the webs are effective only in carrying the shear stresses, calculate the distribution of shear flow around the section.
Example 15
100
All dimensions in mm
y
z
O 60200
240120 240
1
23
4
56 7 8
10 kNA1 = A8 = 200 mm2
A2 = A7 = 250 mm2
A3 = A6 = 400 mm2
A4 = A5 = 100 mm2
-
20
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
( )
0mm1086.13
)50)(100()100)(400()100)(250()30)(200(2
46
22228
1
2
==
+++== =
yz
iiiyy
I
zAI
Section properties:
because the section is symmetric about the horizontal axis.
Izz does not enter into calculation because Vy = 0.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Select the shear flow in web 2-3 to be the unknown shear flow, q23.
100 y
z
O 60200
240120 240
1
23
4
56 7 8
10 kNA1 = A8 = 200 mm2
A2 = A7 = 250 mm2
A3 = A6 = 400 mm2
A4 = A5 = 100 mm2
23
236
3
2333
34
9.28
)400(1086.13
)100)(1010(
q
q
qAI
zVqyy
z
+=
+
=
+
=
q23q23 28.9
-
21
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
100 y
z
O 60200
240120 240
1
23
4
56 7 8
10 kNA1 = A8 = 200 mm2
A2 = A7 = 250 mm2
A3 = A6 = 400 mm2
A4 = A5 = 100 mm2
23
236
3
3444
45
5.32
9.28)100(1086.13
)50)(1010(
q
q
qAI
zVqyy
z
+=
+
=
+
=
2367
233456 9.28qq
qqq=
+==
Due to symmetry,
q23q23 28.9
q23 32.5
q23 28.9 q23
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
100 y
z
O 60200
240120 240
1
23
4
56 7 8
10 kN
A1 = A8 = 200 mm2
A2 = A7 = 250 mm2
A3 = A6 = 400 mm2
A4 = A5 = 100 mm2
23
236
3
81
23
236
3
78
4.22
1.18)200(1086.13
)30)(1010(
1.18
)250(1086.13
)100)(1010(
q
qq
q
qq
+=
++
=
+=
+
=
237812 1.18 qqq +==Due to symmetry,
q23q23 28.9
q23 32.5
q23 28.9 q23q23 + 18.1
q23 + 22.4
q23 + 18.1
-
22
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
100 y
z
O 60200
240120 240
1
23
4
56 7 8
10 kN
Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.
q23q34
q45
q12
q81
A
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
100 y
z
O 60200
240120 240
1
23
4
56 7 8
10 kN
0)480)(30()240)(70()100)(240240120()120)(50(
8112
12233445
=+++++
qqqqqq
Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.
50q45
50q34
120q34 240q23240q12
70q12
30q81
A
-
23
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
N/mm7.12N/mm0.17N/mm7.12N/mm4.5N/mm3.34N/mm9.37N/mm3.34
N/mm4.5
12
81
78
67
56
45
34
23
========
qqqqqqqqSubstituting for q23 and solving, we obtain
1
23
4
56 7 8
10 kN
5.434.3
37.9
12.7
17.0
34.35.4
12.7
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Question to ponder:
How do we determine the separate proportions of these shear flows due to bending and torsion?
1
23
4
56 7 8
10 kN
5.434.3
37.9
12.7
17.0
34.35.4
12.7
-
24
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Determine the position of the shear center.
(Let the thickness of each web = 1 mm.)
Example 15
100
All dimensions in mm
y
z
O 60200
240120 240
1
23
4
56 7 8
A1 = A8 = 200 mm2
A2 = A7 = 250 mm2
A3 = A6 = 400 mm2
A4 = A5 = 100 mm2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
100 y
z
O 60200
240120 240
1
23
4
56 7 8
eyqq
qqqq
)1010()480)(30()240)(70(
)100)(240240120()120)(50(3
8112
12233445
=++
+++
Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.
50q45
50q34
120q34 240q23240q12
70q12
30q81
A10 kN
ye
E
-
25
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
e
e
yqy
qqqqqq
=+=+
+++++++
24.5192.97)1010(322560
1440043440034680030408016800240002400012000)5.32(6000
23
32323
23232323
Substituting for the unknown shear flow,
The additional equation is supplied by the rate of twist equation.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y
z
O
1
23
4
56 7 8
A
Rate of twist = 0, since load acts through shear center.
0222
02
1
8118 81
8145
54 45
4534
43 34
3423
32 23
2312
21 12
12 =++++
=
=
dstqds
tqds
tqds
tqds
tq
dstq
Gdxd
-
26
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Rate of twist = 0, since load acts through shear center.
0)4.22()5.32(
)9.28(22)1.18(2
8118
234554
23
3443
232332
231221
23
=+++
+++
dsqdsq
dsqdsqdsq
264.03701400
0)60)(4.22()100)(5.32()130)(9.28(2)240(2)250)(1.18(2
23
23
2323
232323
==
=++++++
qq
qqqqq
Substituting this into the moment equation, we have
mm9.544=ey
-
5/26/2009
1
Multicell Thin-walled SectionsThe multicell tube in pure torsion is a statically indeterminate problem. The condition of consistent twist must be used for each cell of the tube in order to obtain the solution.
Consider the n-celled tube of general shape subjected to a torque T. Clearly, the shear flow in the web between cell i and cell j is qj-qi. Similarly, in the web between cell j and cell k, the shear flow is qj-qk. H th t f t i t f ll j bHence, the rate of twist of cell j becomes
where j is the area enclosed by cell j.
=
j ji jks s skij
jj tdsq
tdsq
tdsq
Gdxd
21
n
j
21 , , qi qjqk
(1)
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore
The first integral is evaluated over the entire cell parameter sj, the second integral over the web between cell i and cell j (i.e. sji), and the third integral over the web between cell j and cell k (i.e. sjk).
1
ki
Although cross-sections warp, they do not distort in their own plane. This means that
the entire cross-section and each cell rotate at the same rate of twist . Thusdxd
Therefore, for a general case where cell j is bounded by m cells,
nj dxd
dxd
dxd
dxd
dxd
==
==
=
=
......21
(2)
= m
jdsqdsqd 1 (3)
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore
For n cells, there are n such equations.
2
=
j jrs r srj
j tq
tq
Gdx 12(3)
-
5/26/2009
2
However, the number of unknowns involved is (n+1), i.e. q1, q2, q3, , qj, , qnand . The additional equation required is the moment equilibrium equation, i.e.
dxd
(4) n
T 2
where qi is the shear flow in the ith cell.
The foregoing procedure can also be used for multicell thin-walled tubes loaded by
transverse forces, in which case the moments induced by the transverse forces
(4)=
=i
iiqT1
2
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore
t a sve se o ces, w c case t e o e ts duced by t e t a sve se o ces
should be taken into account in the moment equilibrium equation, and the shear
flow expressions for bending should be used.
3
Example 1:The section shown has constant thickness t throughout. It is subjected to a constant torque T. Determine the rate of twist.
a 2a
Consider cell 1: Consider cell 2:
4 61 ddd aa
aq1
1 2
q2
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 4
(2)( )21
0
122
4
0
112
42
1
21
qqtGa
tdsq
tdsq
Gadxd aa
=
=
( )12
0
121
6
0
222
64
1
41
qqtGa
tdsq
tdsq
Gadxd aa
=
=
(1)
-
5/26/2009
3
Moment equilibrium:
2211
2
1
22
2
qq
qTi
ii
+=
= =
From (1) and (2) we have:
d
(3)( )212 22 qqa +=
dxdaGtq
2316
1 =
Substituting these into (3),
dxdGtaT 3
23104
=
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore
and
5
dxdaGtq
2318
2 =
3104
23Gta
Tdxd
=
Example 2. Calculate the shear stress distribution in the walls of the three-cell wing
section shown when it is subjected to an anticlockwise torque of 11.3kNm.
Wall Length (mm) Thickness (mm) G (N/mm2) Cell Area (mm2)
12o 1650 1.22 24200 AI = 258000
12i 508 2 03 27600 A = 355000
Note: The superscript symbols o and i are used to distinguish between outer and inner walls connecting the same points.
12i 508 2.03 27600 AII = 355000
13, 24 775 1.22 24200 AIII = 161000
34 380 1.63 27600
35, 46 508 0.92 20700
56 254 0.92 20700
11300Nm
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 6
I II III
6
5
4
3
2
1
-
5/26/2009
4
Let andGPa 6.27=refG tGGt
ref
= Hence
07.122.16.272.24
12==t o
508
154207.1
1650
12 12
12 ==
ds
tds
o o
o
69.0
07.16.27
564635
2413
===
==
ttt
tt
736
233
725
25003.2
508
4635
34 34
34
24 24
24
13 13
13
12 12
12
=
==
==
dsdst
dst
dst
ds
tds
i i
i
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 7
368
736
56 56
56
46 46
46
35 35
35
=
==
tds
tt
For cell I,( )
( ) ( )( )IIIref
IIIIref
qqG
tds
qt
dsq
tds
qGdx
di i
i
i i
i
o o
o
250250154225800021
25800021
12 12
12
12 12
12
12 12
12
+=
+=
For cell II,
For cell III,
( ) ( )( )IIIIIIrefqqq
Gdxd 233725233725250250
35500021
++++=
( ) ( )( )IIIIIrefqq
Gdxd 368233736736233
16100021
++++=
531
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 8
I II III
642qI qII qIII
-
5/26/2009
5
From the moment equilibrium equation
( ) 6103.111610003550002580002 =++ IIIIII qqq
These equations are solved to give
N/mm 2.4N/mm 9.8N/mm 1.7
===
III
II
I
qqq
7.18.9 4.21 3 5
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 9
4.2
4.2
4.7
8.9
1.8
2 46
The Method of Successive Approximations for Pure Torque
Consider a two-cells section free to warp and subjected to torque T.
I II
If the cells were separated, we see that and this violates the constant rate of twist condition.
qIqII
III dxd
dxd
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 10
qI
I
qII
II
-
5/26/2009
6
If we arbitrarily let , then 1=
=
III dxdG
dxdG
For cell I, For cell II,
12
12
=
=
=
II
II
I
I
qt
dsqdxdG
12
12
=
=
=
IIII
IIII
II
II
qt
dsqdxdG
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 11
2 I 2 II
Considering the separated cell I, we see that at the wall shared with cell II, the shear
flow must be corrected for the shear flow due to cell II.
qI
qIII
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 12
-
5/26/2009
7
The contribution of qII to the rate of twist is represented by an equivalent shear flow qI
over the perimeter of cell I.
IIIII
II t
dsqt
dsq,
=
q
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 13
I IIq I Iq
Or,
where for the wall common to cells I and II
I
IIIIII qq
,=
= tds
III ,
The correction carry-over factor is then defined as
t,
IIIIII
I
IIIIII
Cqq
C
,
,,
=
=
q
I Iq
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 14
qI
-
5/26/2009
8
Since the correction carry-over factor is less than unity, if the procedure is repeated,
the correction become successively smaller until they are finally negligible. The
number of iterations depends on the accuracy required.
where
( ) K++++= IIIIFinalI qqqqq
IIIIII
IIIIII
CqqCqq
,
,
=
=
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 15
Similarly for cell II,
IIIIIIC
,, =
IIqI
IIIIII
II
Cqq ,=
IIq
qII
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 16
qIIIIIIq
-
5/26/2009
9
Example 3:Let us solve the problem of the previous example using the method of successive approximations. 1212 +=
=
i i
i
o o
o
tds
tdst
dsI
I II III
6
531
11300Nm 17922501542
12 1212 12
=+=
i io o tt
1933725233725250
=+++=II
368736233736 +++=III
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 17
642 2073=
250, =III
233, =IIIII
The carry-over factors are:
531
1400250
129.01933250
,
,, ===
III
II
IIIIII
C
C
The initial assumed values of shear flows:
I II III
642
121.01933233
112.02073233
140.01792
,
,
,,
==
==
===
IIIII
IIIII
IIII
C
C
C
( ) 9.287179225800022
==
=I
IIq
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 18
( )
( ) 3.1552073
16100022
3.367193335500022
==
=
==
=
III
IIIIII
II
IIII
I
q
q
-
5/26/2009
10
Cell I Cell II Cell III
CI,II = 0.129 CII,I = 0.140 CII,III = 0.112 CIII,II = 0.121
A d 287 9 367 3 155 3Assumed q 287.9 367.3 155.3
q (0.140)(367.3)=51.4 (0.129)(287.9)=37.14 (0.121)(155.3)=18.8 (0.112)(367.3)=41.1
q (0.140)(37.14)=5.2 (0.129)(51.4)=6.63 (0.121)(41.1)=4.97 (0.112)(18.8)=2.10
q 0.93 0.67 0.25 0.56
q 0.09 0.12 0.07 0.03
Final q 345 5 435 9 199 1
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 19
Final q 345.5 435.9 199.1
We still need the torque equilibrium equation to arrive at the correct solution.
From the torque equilibrium equation,
( )( ) ( )( ) ( )( )1610001.19923550009.43522580005.3452222
++=++= IIIIIIIIIIII qqqT
But the actual applied torque is 0.113x108
So the values of q must be scaled down by a factor of 0.113/5.52=0.0205
81052.5 =
Cell I Cell II Cell III
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 20
Final q 345.5 435.9 199.1
Scaled q 7.07 8.92 4.1
-
5/26/2009
11
Bending Shear Stresses in Multi-cell Beams
Consider a three-cells section carrying a shear load through the shear centre
(no twist):
I II III
VZ
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 21
If each cell were cut, the open-section shear flow would be given by
( )
( )2yzzzyy
zyzzzyzb
VdsztIdsytI
IIIVQIQI
q
=
Distribution of qb:
( )2yzzzyy
zzzyz
III
VdsztIdsytI
=
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 22
I II III
-
5/26/2009
12
If we now close the cells, the constant shear flows qI, qII and qIII must be superposed
onto the shear flow qb in the respective cells:
qI qII qIII
qb
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 23
Considering each cell separately, we now have to correct for the shear flows in the
shared walls. Consider, for example, cell II, where qI acts on the left wall, and qIII on the
right wall:
qI
qII
qIII
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 24
qb
-
5/26/2009
13
Since the twist is zero,
From the figure, taking positive torque anti-clockwise,
0+ dsdsdsds
02
1=
=
IIIIII t
dsqdxdG 0=
IItdsqor
The other two cells have similar equations, i.e.:
and
0,,
=
+
IIIIIIII
IIII
IIII
IIb t
qt
qt
qt
q
0,,
=
+
IIII
IIIIII
III
II
b tdsq
tdsq
tdsq
tdsq
0,,
=
+
IIIII
IIIIIII
IIIIII
IIIb t
dsqt
dsqt
dsqt
dsq
T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 25
These three equations can be solved to obtain
The above equations are also applicable to idealized multicell beams.
,,
qI , qII and qIII
The Method of Successive Approximations for Bending Shear Flows
Since the twist is zero,If shear force acts through shear centre
02
1=
=
IIIIII tdsq
dxdG
d
Hence, from the figure, taking positive torque anti-clockwise, for cell II,