5/26/2009
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
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5/26/2009
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
2
5/26/2009
Aircraft Structural Analysis
Fracture Mechanics Linear Elastic Fracture Elasto-Plastic Fracture Damage Tolerance Fatigue Delamination
Mechanics of ThinWalled Structures Beams Plates Shells Stringers and Panels Multi-cell Torque Boxes Tapered Structures Buckling and Instability Vibration
Mechanics of Composite Materials Stress-Strain Relations of FiberReinforced Composites Laminate Design and Architecture Hygrothermal Effects Durability and Failure of Composites
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Course Overview: Torsion Bending Idealized Beams
T.E. Tay Professor Room EA-7-17 6516-2887 6516 2887
[email protected]
Multi-cell Sections & Tapered Beams Circular & Rectangular Plates Instability of Columns & Plates
S.L. S L Toh Associate Professor
Energy Methods in Instability
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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5/26/2009
Torsion • Non-Circular Shafts • Thin-Walled Open Sections • Thin-Walled Closed Sections • Warping of Unrestrained Sections
B di Bending • Unsymmetric Bars • Thin-Walled Open Sections • Thin-Walled Closed Sections • The Shear Center
Idealized dea ed Beams ea s – Bending e d g & Torsion oso Multi-cell Sections – Bending & Torsion Tapered Beams & Beams with Varying Moments of Area
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Books: Mechanics of Materials, A.C. Ugural, McGraw-Hill, 1991 Ch 6.8, 6 8 6.9: 6 9: Torsion, Torsion Torsion of Thin-Walled 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, 1996 Ch 4, 4 4.7, 4 7 4.8: 4 8: Torsion, Torsion Torsion of Noncircular Sections Ch 6, 6.6, 6.8-6.10, 6.12: Bending, Unsymmetric Bending, Shear Flow, Shear Center
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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5/26/2009
Books (advanced): Mechanics of Elastic Structures, J.T. Oden and E.A. Ripperger, McGraw-Hill, 1981 Aircraft Structures for Engineering Students, T.H.G. Megson, Edward Arnold, 1990 (2nd Edition) Analysis of Aircraft Structures, B.K. Donaldson, McGraw-Hill, 1993 Aircraft Structures & Systems, R. Wilkinson, Addison-Wesley Longman, 1996
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
5
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.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
1
The shear strain is defined:
γ = Δx
r
Δφ
rΔ φ Δx
And the shear stress is
τ = Gγ = lim Gr Δx→ 0
= Gr
dφ dx
Δφ Δx
= Gr θ
where G is the shear modulus.
θ=
dφ dx
is the “rate of twist”, and is a constant.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The twisting moment (i.e. torque) is
T =
∫∫ r (τ dA ) A
dF
=G
τ
r ΔA
dφ dx
= GJ
∫∫ r
2
dA
A
dφ dx
T = GJ θ
(1)
And the shear stress
τ = Gr θ =
Tr J
(2)
where A is the area of the cross-section, and J is the polar second moment of area.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
2
Torsion of Non-Circular Sections Bars of Non-Circular Sections subjected to torsion
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
3
Torsion of Non-Circular Sections Bars of Non-Circular Sections subjected to torsion
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Consider a cross-section of arbitrary shape: A is an arbitrary point.
It is displaced to A’ under torque. z
Note: There is no change in shape. Hence
There is a point O, that has no displacement when the torque is applied, called the centre of twist.
γ yz = 0 τ yz = 0
A’
Let’s place the origin of our coordinate system at O.
O
A
y
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
4
z
In general, we write the displacements:
v = −θ xz w = θ xy
Horizontal displacement of A.
(3a) (3b)
− v = φz
where θ is the rate of twist. We further assume that the axial displacements are given in this form:
u = θψ ( y , z )
φ O
Vertical displacement of A.
A’
w = φy A
y
(3c)
where ψ ( y , z ) is the warping function, yet to be determined.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Axial displacements depend on the warping function
ψ ( y, z )
z Our objective is to determine u, v and w, the complete displacement field.
y
x
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
5
Start with the strain-displacement (compatibility) relations:
∂w ∂u ∂v , εy = , εz = , ∂z ∂x ∂y ∂u ∂v ∂v ∂w ∂u ∂w + = + , γ yz = + , γ xz = ∂y ∂x ∂z ∂y ∂z ∂x
εx = γ xy
Substituting Eqns (3) into above,
ε x = ε y = ε z = γ yz = 0
(4a)
⎛ ∂ψ ⎞ − z ⎟⎟ γ xy = θ ⎜⎜ ⎝ ∂y ⎠
(4b)
⎛ ∂ψ ⎞ + y⎟ ∂ z ⎝ ⎠
(4c)
γ xz = θ ⎜
γ yz = 0 implies cross-section does not distort in its own plane (i.e. all points are simply rotated as in rigid-body rotation).
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The relevant force equilibrium equation
∂σ x ∂τ xy ∂τ xz + + =0 ∂x ∂y ∂z with
τ xy = Gγ xy , τ xz = Gγ xz , σ x = 0 and Eqns (4) yields
∂ 2ψ ∂ 2ψ + =0 ∂y 2 ∂z 2 or
∇ 2ψ = 0
(5)
This is the Laplace Equation. Clearly, the warping function ψ satisfies both compatibility and equilibrium. In order to obtain a solution, we need to specify boundary conditions.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
6
Consider a small element on the boundary, which is usually traction free. Force equilibrium in the axial direction provides the equation of the boundary condition.
dx
τ xy
dz
τ xy dzdx − τ xz dydx = 0 z
(6)
y
dy
τ xz
∴τ xy dz = τ xz dy
x 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 Φ which is twice differentiable, and has the property:
∂Φ ∂z ∂Φ τ xz = − ∂y
τ xy =
(7a) (7b)
Substituting (7) into (4b) and (4c), differentiating (4b) with respect to z, and (4c) with respect to y,
−
⎞ ∂ ⎛ ∂Φ ⎞ ∂ ⎛ ∂ψ −θ z ⎟ ⎜ ⎟ = G ⎜θ ∂z ⎝ ∂z ⎠ ∂z ⎝ ∂y ⎠
(8a)
∂ ⎛ ∂Φ ⎞ ∂ ⎛ ∂ψ ⎞ +θ y ⎟ ⎜ ⎟ = G ⎜θ ∂y ⎝ ∂y ⎠ ∂y ⎝ ∂z ⎠
(8b)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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Ludwig Prandtl Father of Modern Fluid Mechanics (1875-1953)
Subtracting (8b) from (8a) results in
∂ 2Φ ∂ 2Φ + = −2Gθ ∂y 2 ∂z 2
(9)
or
The solution to this equation satisfies both compatibility and equilibrium. In order to obtain a solution, we need to specify boundary conditions.
∇ 2 Φ = −2Gθ
For the boundary condition, substituting (7) into (6), we obtain
∂Φ ∂Φ dz − dy = 0 ∂z ∂y or dΦ = 0
(10)
This means that at the boundary, Φ is a constant. Since the value of Φ can be arbitrary, we shall select Φ = 0 for convenience.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
8
z The torque about the x-axis due to the shear stresses is given by
τ xz dA
T = ∫∫ (τ xz y − τ xy z )dA
(11)
A
y
τ xy dA
Area dA
z y
O
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Introducing Eqn (7) , and noting that dA = dydz,
⎛ ∂Φ ∂Φ ⎞ T = − ∫∫ ⎜⎜ y+ z ⎟dydz ∂z ⎟⎠ ⎝ ∂y ∂Φ ∂Φ = − ∫∫ ydydz − ∫∫ zdydz ∂y ∂z
(12)
⎛ B ∂Φ ⎞ ⎛ D ∂Φ ⎞ z dz ⎟⎟dy − ∫ ⎜⎜ ∫ y dy ⎟⎟dz = − ∫ ⎜⎜ ∫ ⎝ A ∂z ⎠ ⎝ C ∂y ⎠
(
)
(Integration by parts)
(
)
= − ∫ (Φ B z B − Φ A z A ) − ∫ Φ dz dy − ∫ (Φ D y D − Φ C yC ) − ∫ Φ dy dz (13)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
9
z
Φ A Φ B ΦC Φ D are values at boundary points, which we have chosen to set to zero. Hence the equation reduces to
B
T = 2 ∫∫ Φdydz
(14)
D
C
y
O
A T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If we retain the expression
T = GJθ
(15)
where now we define the symbol J as the torsional constant, then the formula for J is
J=
2 Gθ
∫∫ Φ dydz
(16)
The product GJ is known as the torsional stiffness. Only for the very special case of the circular cross-section is J equal to the second polar moment of area.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
10
The Membrane (Soap Bubble) Analogy Good for visualizing Φ . Equilibrium equation for small deflection w of a flat membrane subjected to internal pressure p closely resembles the torsion equation. ∂2w ∂2w p + 2 =− 2 ∂y ∂z TM
(17)
Pressure p Tension TM w z (into plane)
y
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Comparing Eqns (9) and (17),
∂ 2Φ ∂ 2Φ + = −2Gθ ∂y 2 ∂z 2
(9)
∂2w ∂2w p + =− ∂y 2 ∂z 2 TM
(17)
p is analogous to 2Gθ TM w is analogous to Φ Furthermore, according to Eqn (14), the volume under the membrane is proportional to the torsion T.
T = 2 ∫∫ Φdydz
(14)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
11
The slope of the membrane is proportional to the shear stresses.
∂Φ ∂w ⇔ ∂z ∂z ∂Φ ∂w ⇔− τ xz = − ∂y ∂y
τ xy =
Local slope
∂w ∂y
w z (into plane)
y Similarly for
∂w ∂z
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is hollow,
dΦ = 0 at both inner and outer boundaries. z y
Φ = constant
∂w =0 ∂y
Φ=0
z (into plane)
y
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
12
Torsion of Thin-Walled Open Sections Consider a narrow rectangular section: z
z y x Profile of w or Φ in x-z plane
∂Φ = 0 over much of y ∂y
∂Φ = 0 ∂z only at the centerline of the section (at z = 0).
x
y Profile of w or Φ in x-y plane
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
∂Φ = 0 , Eqn (9) reduces to Since ∂y
∂ 2Φ = −2Gθ ∂z 2
∂ 2Φ ∂ 2Φ + = −2Gθ ∂y 2 ∂z 2
(9)
(18)
Integrating twice with respect to z yields
Φ = −G θz 2 + C1z + C 2
(19)
Applying the boundary condition that Φ = 0 at z = + t/2 and – t/2, we obtain C 1 = 0 and C 2 = G θ t 2 / 4 .
⎛ t2 ⎞ ⎟ ∴ Φ = − G θ ⎜⎜ z 2 − 4 ⎟⎠ ⎝
(20)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
13
The stresses are therefore
∂Φ = − 2G θz ∂z ∂Φ =− =0 ∂y
τ xy =
(21a)
τ xz
(21b)
Eqn (21a) shows that τxy varies linearly with z and is zero at the centerline of the section (at z =0).
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
From Eqn (16),
J =
2 Gθ
= −∫
∫∫ Φ dydz
b/2
−b / 2
∴J =
∫
t/2
−t / 2
⎛ 2 t2 ⎞ ⎜⎜ 2 z − ⎟⎟ dzdy 2 ⎠ ⎝
bt 3 3
The torsional constant for a thin rectangular section.
(22)
2T z J
(23)
The shear stress
τ xy = − 2 G θ z = −
has maximum values at z = + t/2 and – t/2, i.e.
τ xy MAX = ±
Tt J
(24)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
14
Linear distribution of shear stress
τ
xy
across thickness:
z y
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Now, consider the warping displacements: From Eqns (4),
⎛ ∂ψ ⎞ τ xy = G θ ⎜⎜ − z ⎟⎟ ⎝ ∂y ⎠
(25a)
From Eqn (21a), τxy = - 2Gθ, so Eqn (25a) becomes
− 2z =
∂ψ −z ∂y
⎛ ∂ψ ⎞ + y⎟ ⎝ ∂z ⎠
τ xz = G θ ⎜
(25b)
From Eqn (21b), τxz = 0, so Eqn (25b) becomes
∂ψ = −y ∂z
∂ψ = −z ∂y
Upon integration,
ψ = − yz + C 1 ( y )
Upon integration,
ψ = − yz + C 2 ( z ) Clearly, C1(y) = C2(z) = 0 and
ψ = − yz
(26)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
15
The warping displacement is therefore
u = − θ yz
(27)
A hyperbolic-paraboloid surface
z y x
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
General Open Sections In general,
s=0
J = z y
Bt 3 3
s=B
Linear distribution of shear stress τ xy across thickness
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
16
b2 t2 For a section consisting of n elements,
J =
b1
1 n ∑ bi ti3 3 i =1
t1
b4
t3
t4
b3
(28)
Maximum shear stress is
τ iMAX =
Ti tiMAX Ji
t5
(29)
where the proportion of torque T acting on element i is
⎛J ⎞ Ti = ⎜ i ⎟ T ⎝ J ⎠
b5
(30)
b6
J =
t6
b1t13 b 2 t 23 b 3 t 33 b 4 t 43 b 5 t 53 b 6 t 63 + + + + + 3 3 3 3 3 3
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Torsion of Thin-Walled Closed Sections Consider a thin-walled closed section: z y
Local slope
z (into plane)
∂w almost constant ∂y
y
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
17
Closed Sections
z y
Constant distribution of shear stress τ xy across thickness
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Taking force equilibrium in the axial direction,
τ 1t1 = τ 2t2
Δx
(31)
z
τ 1 t1 Δ x 1
t1 t2
τ 2t2 Δ x
y x
2
Convenient to define shear flow
q =τ t
(32)
∴ q1 = q 2 = constant
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
18
The total moment then equals the torque,
T = ∫ rqds = q ∫ rds
(33)
O r
qds s=0
Moment about an arbitrary point O is therefore qrds, where r is the perpendicular distance.
Force due to shear flow on a small element of wall is qds.
Choose an arbitrary point on the wall, at s = 0.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as
1 ω ( s1 ) = 2
s = s1
∫ rds
(34)
s =0
O
qds
r s=0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
19
As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as
1 ω ( s1 ) = 2
s = s1
∫ rds
s =0
O
s=0 The total area enclosed by the centerline of the wall is
Ω =
1 2
∫
and the torque, from Eqn (33) is
rds
(35)
T = 2Ω q
(36)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The section does not distort under torsion. Points on the wall experience a rotation φ = θ x about the center of twist.
η θx Center of twist O
η is the tangential displacement component in the s direction.
η = rθ x
(37)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
20
Consider a small element of the wall
Δx
z
y
Δs
x
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Δx
z
Δs Δη
y x
Displacement in the s direction is Δ η
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
21
Δx
z
y
Δs
x
Δu Displacement in the x direction is Δ u
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Δx
z
Δs
x
Therefore, the shear strain in the wall is
∂u ∂η + ∂s ∂x ∂u = + rθ ∂s
y
γ xs =
(38)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
22
Since γ xs = τ xs / G it follows that
∂u τ xs = − rθ ∂s G
(39)
Multiplying both sides by ds and integrating around the total periphery S of the tube, q ds (40) ∫ du = G ∫ t − θ ∫ rds 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 Ω from Eqn (35),
ds − 2Ωθ t q ds ∴θ = 2Ω G ∫ t
0=
q G
∫
(41)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Substituting for q from Eqn (36),
θ=
T 4Ω 2G
∫
ds t
(42)
Since T = GJθ , the torsional constant J for a closed section is
J =
4Ω 2 ds ∫t
(43)
If the thickness t is constant,
J =
4Ω 2 t S
(44)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
23
Hybrid Sections
J=
4Ω 2 1 n + bi ti3 ds 3 ∑ i =1 ∫t
(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
R t
Closed tube
t
Open section
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
24
Open section
Closed tube
4Ω 2 ds ∫t 4π 2 R 3t = = 2πR 3t 2π
bt 3 3 2πRt 3 = 3
J Closed = Torsional constant
J Open =
J Closed ⎛R⎞ = 3⎜ ⎟ J Open ⎝t ⎠
2
= 1200 For a given angle of twist, the closed section resists 1200 times the torsion of the open section.
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
Open section
Closed tube
Maximum shear stress
Tt J 3T = 2πRt 2
τ Open MAX =
T 2Ω t T = 2πR 2 t
τ Closed =
τ Open ⎛R⎞ = 3⎜ ⎟ τ Closed ⎝ t ⎠ = 60 For a given torque, the shear stress in the open section is 60 times as high as in the closed tube.
If the allowable shear stresses are equal, the closed tube is 60 times as strong as the open section.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
25
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. 15 2
2
19
1 2 1 Dimensions in mm 20 T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Total area enclosed:
∫
Ω = 20 × 19 = 380 mm 2
ds 20 19 20 19 = + + + = 58 .5 t 1 1 2 2
The torsional constant is
4Ω 2 1 3 + bt ds 3 ∫t 2 4(380 ) 1 3 = + (15 )(2 ) = 9.913 × 10 3 mm 4 58 .5 3
J =
Since T = GJθ , the rate of twist is
θ=
6000 9 .913 × 10 3 80 × 10 3
(
)(
)
= 7 .57 × 10 − 6 rad/mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
26
The contributions to the torsional constant are
4 (380 ) = 9 .873 × 10 3 mm 4 58 .5 1 3 = (15 )(2 ) = 40 mm 4 3 2
J Closed Part = J Fin
}
Compare !
The proportions of torque carried are
9 .873 × 6000 = 5976 Nmm 9 .913 40 = × 6000 = 24 .2 Nmm 9913
TClosed Part = TFin
The maximum stresses are
τ Closed Part MAX = τ Fin MAX =
TClosed Part 2 Ω t MIN
=
5976 = 7 .86 MPa 2 (380 )(1)
TFin t (24 .2 )(2 ) = = 1 .21 MPa J Fin 40
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? 15 2
2
A 19
1 2 1 Dimensions in mm 20 T. E. Tay, Department of Mechanical Engineering, National University of Singapore
27
The torsional constant is
J =
[
]
1 (19 )(1)3 + (20 )(1)3 + (19 )(2 )3 + (35 )(2 )3 = 157 mm 4 3
and
(6000 )(2 ) = 76 .4 MPa Tt MAX = J 157 T 6000 θ= = = 4 .78 × 10 − 4 rad/mm GJ 80 × 10 3 (157 )
τ MAX =
(
)
Note the increase in stress and rate of twist.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Warping of Unrestrained Thin-Walled Sections
Recall from Eqn (39) that
∂u τ xs = − rθ ∂s G Multiplying both sides by ds and integrating from s = 0 to some other point on the wall,
u − u0 =
1 τ xs ds − θ ∫ rds G∫
(46)
where u0 is the displacement of the point s = 0 in the x-direction.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
28
Recall the sectorial swept area (Eqn (34)), defined as ω ( s ) =
1 2
1 (47) ∴ u − u 0 = ∫ τ xs ds − 2θω ( s ) G where r is measured from an arbitrary point O (called the pole).
∫ r ds
O r s=0 The equation, valid for both open and closed sections, can be rewritten as 1
G∫ u = u − u0 u=
τ xs ds − 2θω( s)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
For closed sections, since τ xs = q / t , we have
u=
q ds − 2θω G∫ t
(48)
Here, u is a function of s, and is zero when the integration is completed around the closed tube.
O
u =0 s=0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
29
For open sections, the contribution of the shearing strain γxs to the warping displacement u is negligible because the walls are very thin, so that
∫τ
xs
ds ≈ 0
(49)
Hence the equation reduces to
u = −2θω
(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. tb
Rate of twist:
1
2
T ds 2 ∫ 4Ω G t 2 3 4 1 T ⎛ ds ds ds ds ⎞ = 2 2 ⎜⎜ ∫ + ∫ + ∫ + ∫ ⎟⎟ 4b h G ⎝ 1 tb 2 th 3 tb 4 t h ⎠ T ⎛ 2h 2b ⎞ = 2 2 ⎜⎜ + ⎟⎟ 4b h G ⎝ th tb ⎠
θ= th
T th
tb
3
4
h
=
T ⎛h b⎞ ⎜ + ⎟ 2b 2 h 2G ⎜⎝ t h tb ⎟⎠
b
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
30
Select the pole at O the center of section, and starting point at I. 1
2
For closed sections, the relative warping displacement is
q ds − 2θω G∫ t
u= O
1
I
∴ u1 =
q ds − 2θω I 1 G ∫I t h
We find that q = 3
4
T T and = 2Ω 2bh
⎛ b ⎞⎛ h ⎞ ⎝ 2 ⎠⎝ 2 ⎠
Also, 2ωI 1 = ⎜ ⎟⎜ ⎟ = Therefore, u1 =
bh 4
T ⎛h b⎞ ⎜ − ⎟ 8bhG ⎜⎝ th tb ⎟⎠
1
ds
∫t I
h
=
h 2th
Interestingly, there will be no warping if
h b = t h tb
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
bh bh + 4 2 3bh = 4
2ωI 2 =
Now, 1
2
2
and O
3
∫ I
I
4
1
=
u2 = =
2
ds ds ds = + t ∫I th ∫1 tb h b + 2th tb
⎛ h b ⎞ 3bh T ⎛ h b⎞ T ⎜⎜ + ⎟⎟ − 2 2 ⎜⎜ + ⎟⎟ 2bhG ⎝ 2t h tb ⎠ 2b h G ⎝ t h tb ⎠ 4 T ⎛b h⎞ ⎜ + ⎟ 8bhG ⎜⎝ tb t h ⎟⎠
∴ u 2 = −u1 Similarly, u3 = u1 , u4 = −u1 T. E. Tay, Department of Mechanical Engineering, National University of Singapore
31
Due to symmetry, we know that the axial displacement at I is zero. Since u1 = u1 −u I , we have u1 = u1.
2 1
I 3 4
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 ? 1
2
Torsional constant: J =
O
I
Rate of twist: θ =
(
2 3 hth + btb3 3
)
T 3T = GJ 2G hth3 + btb3
(
)
For an open section, u = −2θω 3
4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
32
1
2
O
I
3
From I to 1, u1 = −
bh 3T 3bhT ⋅ =− 4 2G hth3 + btb3 8G hth3 + btb3
From I to 2, u2 = −
9bhT = 3u1 8G hth3 + btb3
(
(
)
)
)
From I to 3, u3 = 5u1
5bh ⎞ ⎛ ⎜ 2ωI 3 = ⎟ 4 ⎠ ⎝
From I to 4, u4 = 7u1
7bh ⎞ ⎛ ⎜ 2ωI 4 = ⎟ 4 ⎠ ⎝
From I to I’, uI ' = 8u1
8bh ⎞ ⎛ ⎜ 2ωI 3 = ⎟ 4 ⎠ ⎝
4
(
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
2 1 Longitudinal slit
I 3 4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
33
Example 5 Let us return to the closed section. What will happen if we select another point O’ as pole? 2
1
From I to 1, 2ωI 1 = 1
and
∫ I
O’
I
u1 =
3
4
bh 2
ds h = t 2th
⎛ h b ⎞ bh T ⎛ h ⎞ T ⎜⎜ ⎟⎟ − 2 2 ⎜⎜ + ⎟⎟ 2bhG ⎝ 2t h ⎠ 2b h G ⎝ t h tb ⎠ 2
=−
T ⎛b⎞ ⎜ ⎟ 4bhG ⎜⎝ tb ⎟⎠
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
From I to 2, 2ωI 2 = bh 2
∫
and 2
I
1
ds h b = + t 2th tb
⎛h b⎞ T ⎛ h b⎞ T ⎜ ⎜ + ⎟bh + ⎟− 2bhG ⎜⎝ 2t h tb ⎟⎠ 2b 2 h 2G ⎜⎝ t h tb ⎟⎠ T ⎛h⎞ ⎜ ⎟ =− 4bhG ⎜⎝ t h ⎟⎠
u2 = O’
I
3
4
Similarly, we can show that
u3 = −u2 , u4 = −u1
Try it yourself !
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
34
For illustration purposes, let
T h b = 1.0 = 30, = 10 and 2bhG th tb
Then u1 = −2.5, u2 = −7.5, u3 = 7.5, u4 = 2.5 Present Example 5:
From Example 3: 2
2
1
1
I
I 3
3 4
4
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 O’I axis.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
35
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
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
1
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
x-y-z are the centroidal axes Let the loads Vz and Vy act through the shear centre E. Hence pure bending.
z
Vz
y
Centroid O
Vy
x
Shear Centre
E
t
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
2
x-y-z are the centroidal axes
Mz z
The shear loads give rise to the applied moments (positive right hand screw rule)
My
y
Centroid O
x
t
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Contribution of Vz to bending moment
Hence
Vz
dM
y
Vz =
= V z dx ∂M y ∂x
(51a)
dx z
+ dMy E x
y (into page)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
3
Contribution of Vy to bending moment Hence
Vy
dM z + V y dx = 0 ∴ Vy = −
∂M z ∂x
(51b)
dx y
E
+ dMz
x z (out of page)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let σx denote the normal stress due to bending. Since there is no nett axial load,
∫σ z
Area dA = t ds
σx
x
dA = 0
(52a)
A
dx y x
ds
t
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
4
Consider a small element of the wall (in positive z)
The moment about the y-axis contributes to the normal stress σx due to bending. Consider applied moment about the y-axis
Area dA = t ds
z
σx
dx
y
ds
My
x z
t
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Contribution of axial force dF due to σx to the bending moment Area dA = t ds
M y = ∫ σ x z dA
(52b)
A
(Tensile on positive z) σx dx z
z
+ dMy
x y (into page)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
5
The moment about the z-axis also contributes to the normal stress σx due to bending.
Consider a small element of the wall (in positive y)
z
Consider applied moment about the z-axis
σx dx
ds
y
Mz
x
t y T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Contribution of axial force dF due to σx to the bending moment Area dA = t ds
M z = − ∫ σ x y dA
(52c)
A
(Compressive on positive y) σx dx y
y
+ dMz
x z (out of page)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
6
Plane sections remain plane assumption implies that the axial strain εx (and axial stress σx ) 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. z
λ
y
x
Cross-section
Side view along the neutral axis T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let σ x = C1 + C 2 y + C 3 z and substituting this into Eqns (52) and solving for the constants C1, C2, C3, we have C1 = 0 and
⎛ M z I yy + M y I yz ⎞ ⎛ M I + M z I yz ⎟ y + ⎜ y zz 2 ⎟ ⎜ I I − I2 yy zz yz ⎝ I yy I zz − I yz ⎠ ⎝
σ x = − ⎜⎜
⎞ ⎟z ⎟ ⎠
(53)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
7
Eqn (53) can be rewritten in the form
σx = −
M z y M yz + I zz I yy
⎛ M I ⎜ M z + y yz ⎜ I yy Mz = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝ M I ⎛ ⎜⎜ M y + z yz I zz My = ⎝ 2 ⎛ ⎞ I ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝
⎞ ⎟ ⎟ ⎠
⎞ ⎟⎟ ⎠
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The neutral axis can be located by setting σx = 0 in Eqn (53), and find, for angle λ,
λ
z y
tan λ =
z M z I yy + M y I yz = y M y I zz + M z I yz
(54)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
8
General distribution of normal bending stress σ x
Compression z
Neutral axis
y Tension
x
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is symmetric about either the y- and zaxes, Iyz = 0 and Eqn (53) reduces to
σx = −
M zy M yz + I zz I yy
(55)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
9
A word about sign convention for moments:
σx = − z
M zy M yz + I zz I yy Applied Moment My Positive into page
x y (into page)
This causes tensile (+) stresses for positive z
y Applied Moment Mz Positive out of page x z (out of page)
This causes compressive (-) stresses for positive y
Note that Equation (53) for normal bending stresses is valid for all general sections (whether solid or thin-walled, open or closed sections).
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
10
Example 6 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.) Q
Q
z 20 1000
y
1500
80
20
60 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.
Q
Mz = 0.
1000
1500
0.6 Q
0.4 Q My
Free-body diagram: 1000
Positive into page
0.6 Q T. E. Tay, Department of Mechanical Engineering, National University of Singapore
11
First, locate the centroid C, as the equations are related to centroidal axes. Q
The total cross-sectional area: = (20)(100) + (60)(20) = 3200 mm2
z
Define the points A, B and D.
A 20
D 15 C
Therefore,
y
60 0 ⎞ 1 1⎛ ⎜ ydA 20 y dy 100 = + ∫− 20ydy ⎟⎟⎠ = 5 mm A∫ A ⎜⎝ ∫0
y=
80
measured from point B,
5
and
B
1 zdA = −15 mm A∫
z=
60
20
measured from point D.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Q
Section properties (second moments of area):
z
( 20 )(100 ) 3 + ( 2000 )(15 ) 2 12 ( 60 )( 20 ) 3 + + (1200 )( 25 ) 2 12 = 2.907 × 10 6 mm 4
I yy =
20 15 C
y 80
5
I zz = 1 .627 × 10 6 mm 4 I yz = ( 2000 )( − 15 )( − 15 ) + (1200 )( 25 )( 25 )
20
60
= 1 .2 × 10 6 mm 4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
12
Neutral axis:
tan λ =
z Compression A (-25,35)
= 36.4° C Tension
B (-5,-65)
y
∴
M y I yz + M z I yy M y I zz + M z I yz I yz I zz
=
1 .2 1 . 627
since Mz = 0.
λ = 36 . 4 °
Points A and B are furthest from the neutral axis, and therefore experience the greatest stresses. From Eqn (53), we find
(σ x )A (σ x )B
= − 0 . 0158 Q MPa = 0 . 0182 Q MPa
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.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
13
Force equilibrium in x-direction: Shear stress distribution due to bending
z
∂σ x ∂q ⎞ ⎛ ⎞ ⎛ dx ⎟ tds − qdx + ⎜ q + ds ⎟ dx = 0 − σ x tds + ⎜ σ x + ∂x ∂s ⎠ ⎝ ⎠ ⎝ ∂q ∂σ x = −t ∴ (56) ∂s ∂x
σx
dx q y q+
σx +
ds
∂q ds x∂ s
∂σ x dx ∂x
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,
∂ σ x ⎛⎜ V y I yy − V z I yz = ⎜ I I −I2 ∂x yz ⎝ yy zz
⎞ ⎛ V I + V y I yz ⎟ y + ⎜ z zz ⎟ ⎜ I I −I2 yz ⎠ ⎝ yy zz
⎞ ⎟z ⎟ ⎠
(57)
where the relations in Eqns (51) have been used. Substituting Eqn (57) into Eqn (56) and integrating with respect to dA (= t ds), we have
τ xs = where
(V y I yy − V z I yz )Q z + (V z I zz − V y I yz )Q y q =− t I yy I zz − I yz2 t
(
)
Q z = ∫ y dA = Ay
(59a)
Q y = ∫ z dA = Az
(59b)
A
(58)
( y and z are coordinates of the centroid)
A
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
14
Eqn (58) can be rewritten in the form
⎛ V yQz VzQ y q = −⎜ + ⎜ I I yy ⎝ zz
⎞ ⎟ ⎟ ⎠
⎛ V I ⎞ ⎜ V y − z yz ⎟ ⎜ I yy ⎟⎠ Vy = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝ V I ⎞ ⎛ ⎜⎜ V z − y yz ⎟⎟ I zz ⎠ Vz = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
If the section is symmetric about either the y- and zaxes, Iyz = 0 and Eqn (58) reduces to
⎛ V yQz VzQ y + q = −⎜ ⎜ I I yy ⎝ zz
⎞ ⎟ ⎟ ⎠
(60)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
15
Bending of open sections
General distribution of bending shear flow q for open sections
s
z
q is a function of s and may even change direction, depending upon loading q=0
y x
q=0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
∂σ ∂q = −t x We already know that ∂s ∂x . 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.
Bending of closed sections
z
Vz
y
Centroid O
σx
x
Vy qm
∂q qm + ds ∂s Select an arbitrary point m on the wall for s = 0
E
m s
σx +
∂σ x dx ∂x
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
16
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. h/2 z
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
Vz O
y t
2
h/2
th 3 th 3 ⎛ ht ⎞⎛ h ⎞ = I yy = 2 ⎜ ⎟⎜ ⎟ + 12 3 ⎝ 2 ⎠⎝ 2 ⎠ 3
th 3 ⎛ t ⎞⎛ h ⎞ I zz = 2 ⎜ ⎟⎜ ⎟ = 12 ⎝ 3 ⎠⎝ 2 ⎠ 3 ht h h ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ht ⎞⎛ h ⎞⎛ h ⎞ th I yz = ⎜ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ − ⎟⎜ − ⎟ = 8 ⎝ 2 ⎠⎝ 2 ⎠⎝ 4 ⎠ ⎝ 2 ⎠⎝ 2 ⎠⎝ 4 ⎠
h/2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Hence
⎛ V z I yz ⎞ ⎜⎜ − ⎟ I zz ⎟⎠ Vy = ⎝ = − 0 . 86 V z ⎛ I yz2 ⎞ ⎜1 − ⎟ ⎜ I yy I zz ⎟⎠ ⎝
Vz =
Vz ⎛ I yz2 ⎜1 − ⎜ I yy I zz ⎝
⎞ ⎟ ⎟ ⎠
= 2 . 28 V z
The shear flow due to bending: s ⎞ ⎛ Vy s Vz q = −⎜ ytds ztds ⎟ + ∫ ⎟ ⎜I ∫ I yy 0 ⎠ ⎝ zz 0 s ⎞ ⎛ 0 . 86 s 2 . 28 ytds ztds ⎟⎟ = V z ⎜⎜ 3 − 3 ∫ ∫ th / 3 0 ⎠ ⎝ th / 12 0 s s ⎛ ⎞ V = 3z ⎜⎜ 10 . 3 ∫ yds − 6 . 85 ∫ zds ⎟⎟ h ⎝ 0 0 ⎠
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
17
For the bottom flange, define a coordinate s1 such that z = - h/2 and y = - h/2 + s1, where 0 ≤ s1 ≤ h / 2
q 12 = h/2
=
z
h/2
O
h/2
t
1
s1 s1 ⎛ ⎞ ⎜ 10 . 3 ⎛⎜ s1 − h ⎞⎟ ds 1 − 6 . 85 ⎛⎜ − h ⎞⎟ ds 1 ⎟ ∫0 ⎝ 2 ⎠ ∫0 ⎝ 2 ⎠ ⎟ ⎜ ⎝ ⎠
(
Vz 5 . 15 s12 − 1 . 72 hs 1 + c h3
)
Boundary condition: At s1 = 0, q12 = 0; hence c = 0
y
s1
Vz h3
∴ q 12 =
(
Vz 5 . 15 s12 − 1 . 72 hs 1 h3
)
2 h/2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
q 12 =
)
At 2, s1 = h/2, q12 = 0.43 Vz/h.
z
O
(
Vz 5 . 15 s12 − 1 . 72 hs 1 h3
y
Shear flow distribution is parabolic, with a change in sign at s1 = 0.334h. For values of s1 < 0.334h, q12 is negative and therefore in opposite direction to s1.
s1 1
2
The shear flow distribution on the top flange is similar.
0.334h
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
18
Along the vertical web 2-3, z = s2 – h/2, y = 0, and 0 ≤ s2 ≤ h
3
z
q 23 =
O
y
s2 1
2
Vz h3
s2
∫ (3 . 43 h − 6 . 85 s )ds 2
2
+ q2
0
where q2 is the shear flow at point 2, i.e. q2 = 0.43 Vz/h.
∴ q 23 =
(
Vz 0 . 43 h 2 + 3 . 43 hs 2 − 3 . 43 s 22 h3
)
This shear flow distribution is also parabolic, with a maximum at s2 = 0.5h.
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. T. E. Tay, Department of Mechanical Engineering, National University of Singapore
19
Example 8 Determine the position of the shear center of the section. b
Assume the shear center E exists and is located distance ye from vertical web, as shown.
z
h/2
tf O
Vz
y
Apply a shear load Vz through the shear center. E
ye
h/2
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.
Section properties:
tw
2 2 twh3 t h 3 t f bh ⎛h⎞ + 2bt f ⎜ ⎟ = w + 12 12 2 ⎝2⎠ =0 because section is symmetric.
I yy = I yz
tf
Izz is not relevant.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
b
Vy = 0
∴q = −
Vz tz ds I yy ∫0
At the top flange, z = h/2, y = b – s1, where s1 defined.
z tf O
h/2
Vz = Vz , s
s1
h/2
Hence
Vz
y ye
E
∴ q s1 = −
tw
=− tf
Vz tf I yy Vz t f h 2 I yy
s1
h
∫ 2 ds
1
0
s1
At s1 = 0, q s1 = 0. (free edge).
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
20
Take moments about the point A, b
V z y e = h ∫ q s1 ds1
s1
b
∫ q s1 ds1
0
⎡ s2 ⎤ =− Vz ⎢ 1 ⎥ 2 I yy ⎣ 2 ⎦ 0
z
h/2
tf O
=−
Vz
y
E
ye
h/2
b
h 2t f
b
0
∴ ye = −
tw A
=−
tf
b 2 h 2t f 4 I yy
Vz
b 2 h 2t f 4 I yy
(ht
3b 2 t f w
+ 6bt f
)
Therefore, the shear center is to the left of the section. 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.
Assume the shear center E exists and is located distance ye from vertical web, as shown.
z r
θ
Vz
O
y ye
t
The section is symmetric, the positions of the centroid and the shear center must lie somewhere on the horizontal axis of symmetry.
Apply a shear load Vz through the shear center. E Section properties:
I yy =
πr 3 t
I yz = 0
2 because section is symmetric.
Izz is not relevant.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
21
s
Define coordinate s. s
z θ
r
q=−
Vz
O
y
E
ye
θ
=− =−
t
Vz tz ds I yy ∫0
Vz t 2 r cos θ dθ I yy ∫0
where
z = r cosθ , ds = rdθ.
2
V z tr sin θ I yy
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Take moments about the point O, S
V z y e = r ∫ q ds 0
π
= r 2 ∫ q dθ
s
0
z r
θ
=−
Vz
O
y ye
E
=−
∴ ye = − t
V z tr 4 [− cos θ ]π0 I yy 4r
π
Vz
4r
π
= −1.27 r The shear center is to the left of the section.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
22
Example 10 Determine the position of the shear center of the closed section and the distribution of shear flow. Assume the shear center E exists and is located distance ye from vertical web, as shown.
150 t1
Apply a shear load Vz = 1 × 106 N through the shear center.
t1 Vz 250
E
ye
Section properties:
I yy = 25 .78 × 10 6 mm 4 I yz = 0
because section is symmetric.
t1 = 3 mm , t 2 = 6 mm .
t2
Izz is not relevant.
t1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Define coordinates and points as follows. s4
A s1
z
D t1
O
t1
y s3 t2
B
t1
s2
C
From A to B,
s1 = − z + 125
From B to C,
z = – 125
From C to D,
s3 = z + 125
From D to A,
z = 125
Let the shear flow at A be qm, an unknown quantity to be determined.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
23
q AB = q m − qm A s1
z
s4 D t1
t1
∫ zt ds 1
1
(10 6 )(3) (125 − s1 ) ds1 25 .78 × 10 6 ∫ = q m − 0.116 125 s1 − 0.5 s12 = qm −
(
)
At B, qAB = qm
O
y s3
t1
q BC = q m −
Vz I yy
∫ zt ds 1
2
= q m − 0.116 ∫ (− 125 ) ds 2
t2 B
Vz I yy
= q m + 14 .5 s 2
C
s2
At C, qBC = 2175 + qm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
qCD = q m − qm A s1
z
s4 D t1
t1
∫ zt ds 2
3
(10 6 )( 6) (s3 − 125 ) ds3 25 .78 × 10 6 ∫ = q m − 0.233 0.5 s32 − 125 s3 + 2175
= qm −
(
)
At D, qCD = qm + 2175
O
y s3
t1
s2
q DA = q m −
Vz I yy
∫ zt ds 1
4
= q m − 0.116 ∫ (125 ) ds 4
t2 B
Vz I yy
C
= q m − 14 .5 s4 + 2175
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
24
Rate of twist = 0, since load acts through shear center.
1 ⎛ q ⎞ dφ = ds ⎟ = 0 ⎜ dx 2Ω G ⎝ ∫ t ⎠ q q q q ∴ ∫ AB ds1 + ∫ BC ds 2 + ∫ CD ds 3 + ∫ DA ds 4 = 0 t t t t AB 1 BC 1 AB 2 AB 1 which simplifies to
675 q m + 598125 = 0 ∴ q m = −886 .1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
qm = - 886.1 A
q = 1288.9 D
Therefore,
q AB = −886 .1 − 14 .5 s1 + 0.058 s12 q BC = −886 .1 + 14 .5 s 2 qCD = 1289 − 0.116 s32 + 29 s3 q DA = 1289 − 14 .5 s 4
B qm = - 886.1
C q = 1288.9
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
25
Take moments about point O to obtain shear center.
A
150
D
C
D
250 ∫ q BC ds 2 + 150 ∫ qCD ds 3 = V z y e B
C
150
V z y e = 250 ∫ (− 886 .1 + 14 .5 s 2 ) ds 2
Vz O ye
250
E
0
250
+ 150
∫ (1288 .9 − 0.116 s
2 3
)
+ 29 s3 ds 3
0
B
C
= 75525 + 93646250 ∴ y e = 93 .7 mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Combined Bending and Torsion
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
26
P General Case: Superposition of Bending and Torsion
=
+ Pure Bending through shear center
Pure Torsion T = Pd
P d
E
E
T
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Pure Torsion of Open Sections
T
Linear distribution of shear stress τ xy across thickness
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
27
Pure Torsion of Closed Sections
T
Constant distribution of shear stress τ xs across the thickness Shear flow q is constant anywhere in the section, but shear stress τ xs is NOT constant because of varying wall thickness T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Pure Bending of Open Sections
P
E Distribution of shear flow q in walls due to bending such that nett contribution of q must be due to P, applied force. Bending stress σ x exists in walls
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
28
P
Pure Bending of Closed Sections
E
Distribution of shear flow q in walls is similar to that of an open section, except for a constant q0 superimposed. Bending stress σ x exists in walls
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
29
Idealized Beams
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The section is made up of concentrated stringers of area An, joined by thin sheets of thickness tn. Let the loads Vz and Vy act through the shear centre E. Hence pure bending.
z
y Centroid O
Vz x
tn
E
Shear Centre
An
Vy zn yn
The nth stringer is located at positions yn and zn away from the origin at the centroid.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
3
Taking moments as before (see Eqns (51)), we have
Vz =
∂M y
∂x M y = Vz x
(61a)
∂M z ∂x M z = −V y x
(61b)
and
Vy = −
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let σn denote the normal stress due to bending in the nth stringer. Since there is no nett axial load, N
∑σ n =1
σn
z
An = 0
(62a)
where N is the total number of stringers. An
y Centroid O
n
x
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
4
The moment about the y-axis contributes to the normal stress σn due to bending.
My = =
n =1 N
∑σ n =1
σn
z
N
∑M n
yn
An z n (62b)
An
y Centroid O
zn
x
Myn
yn
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Contribution of axial force dF due to σn to the bending moment Area An
M yn = σ n An zn (Tensile on positive z)
σn dx z
zn
+ Myn
x y (into page)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
5
The moment about the z-axis contributes to the normal stress σn due to bending.
Mz =
N
∑M n =1
zn
N
= − ∑ σ n An y n n =1
σn
z
(62c) An
y Centroid O
zn
x
Mzn yn
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Contribution of axial force dF due to σn to the bending moment Area An
M zn = −σ n An yn (Compressive on positive y)
σn dx y
yn
dMzn
x z (out of page)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
6
Invoke again the assumption that plane sections remain plane, which implies that the axial strain ε (and axial stress σ) is a linear function of y and z. Let σ n = C1 + C 2 y n + C 3 z n and substituting this into Eqns (62) and solving for the constants C1, C2, C3, we have C1 = 0 and
⎛ M z I yy + M y I yz ⎞ ⎛ M I + M z I yz ⎟ y n + ⎜ y zz 2 ⎟ ⎜ I I − I2 yy zz yz ⎝ I yy I zz − I yz ⎠ ⎝
σ n = − ⎜⎜
⎞ ⎟ zn ⎟ ⎠
(63)
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
I yy = ∑ An z n2
(64a)
n =1 N
I zz = ∑ An y n2
(64b)
n =1 N
I yz = ∑ An y n z n
(64c)
n =1
They are easier to evaluate than integrals!
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
7
Eqn (63) can, of course, be rewritten in terms of equivalent moments:
σn = −
M z yn M y zn + I zz I yy
⎛ M I ⎜ M z + y yz ⎜ I yy Mz = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝ M I ⎛ ⎜⎜ M y + z yz I zz My = ⎝ 2 ⎛ ⎞ I ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝
⎞ ⎟ ⎟ ⎠
⎞ ⎟⎟ ⎠
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
General distribution of normal bending stress σ n
z Tension y Neutral axis
x
Compression
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
8
Now, consider the shear flows in the sheets or webs.
z
Select a small length dx of a typical nth stringer with two adjacent nth and (n + 1)th sheets.
Anσ n qn
y dx x
qn+1
∂σ ⎛ ⎞ An ⎜σ n + n dx⎟ ∂x ⎠ ⎝
Force equilibrium in x-direction:
∂σ ⎞ ⎛ −σn An + ⎜σn + n dx⎟ An − qndx+ qn+1dx = 0 ∂x ⎠ ⎝ ∂σ ∴ qn+1 = − n An + qn (65) ∂x T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Substituting Eqn (63) into Eqn (65) and using Eqns (61),
⎛ (V y I yy − Vz I yz )yn + (Vz I zz − V y I yz )zn ⎞ ⎟ An + qn qn +1 = −⎜ 2 ⎜ ⎟ − I I I yy zz yz ⎝ ⎠
(
)
(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.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
9
Eqn (66) can be rewritten in the form
⎛ V y yn Vz zn q n +1 = − ⎜ + ⎜ I zz I yy ⎝
⎞ ⎟⎟ An + q n ⎠
⎛ V z I yz ⎞ ⎜⎜ V y − ⎟ I yy ⎟⎠ ⎝ Vy = ⎛ I yz2 ⎞ ⎜⎜ 1 − ⎟ I yy I zz ⎟⎠ ⎝ V y I yz ⎞ ⎛ ⎜ Vz − ⎟ I zz ⎠ Vz = ⎝ ⎛ I yz2 ⎞ ⎜⎜ 1 − ⎟ I yy I zz ⎟⎠ ⎝
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.
q5 + q0 q0 q4 + q0
Replace the missing sheet with the unknown shear flow. 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 + q0
q3 + q0 q2 + q0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
10
Example 11 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. 2
1 z
48 kN
2000 4
I yy = ∑ Ai zi2 = 4(300)(2000)2 = 4.8×109 mm4
y
i =1
2000 3
4 2000 Dimensions in mm T. E. Tay, Department of Mechanical Engineering, National University of Singapore
q12 = − 2
6 N/mm
1
2000
q 23
− 48000 ( 300 )( 2000 ) − 6 = − 12 N/mm 4 . 8 × 10 9 V = − z A3 z 3 + q 23 I yy =
y 2000 6 N/mm 3
− 48000 ( 300 )( 2000 ) = − 6 N/mm 4 . 8 × 10 9 V = − z A2 z 2 + q12 I yy =
z 12 N/mm
Vz A1 z1 + 0 I yy
4
q 34
=
− 48000 ( 300 )( 2000 ) − 12 = − 6 N/mm 4 . 8 × 10 9
2000 Dimensions in mm
Note that we can check vertical and horizontal force equilibrium and show that the resultants are 48 kN and zero, respectively.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
11
Example 12 Find the shear flows in the webs of the unsymmetrical closed beam section. 300 mm2 A
B 100 mm
2
z V z y
D 100 mm2
200 mm
Vy
100 mm
C 300 mm2
Vy = 4 kN Vz = 10 kN
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
4
I yy = ∑ Ai z i2 = 2(300 )(50 ) 2 + 2 (100 )(50 ) 2 = 2 .0 × 10 6 mm 4 i =1 4
I zz = ∑ Ai yi2 = 2(300 )(100 ) 2 + 2(100 )(100 ) 2 = 8.0 × 10 6 mm 4 i =1 4
I yz = ∑ Ai y i z i = (300 )( − 100 )(50 ) + (300 )(100 )( − 50 ) i =1
+ (100 )(50 )(100 ) + (100 )( −50 )( − 100 ) = −2 .0 × 10 6 mm 4 ⎛ − 2.0 ⎞ 4 00 0 − 10 0 00 ⎜ ⎟ ⎝ 2 .0 ⎠ = 1 .86 67 × 1 0 4 Vy = 2 ( − 2 .0 ) 1− ( 2 .0 )(8 .0 ) ⎛ − 2 .0 ⎞ 10 0 00 − 4 00 0 ⎜ ⎟ ⎝ 8 .0 ⎠ = 1.4 66 7 × 10 4 Vz = ( − 2.0 ) 2 1− ( 2 .0)(8 .0 ) T. E. Tay, Department of Mechanical Engineering, National University of Singapore
12
⎛ V y yn Vz zn ⎞ q n +1 = − ⎜ + ⎟ An + q n ⎜ I zz I yy ⎟⎠ ⎝ ⎛ 1.8667 y n 1.4667 z n ⎞ An + q n = −⎜ + 2 2.0 × 10 2 ⎟⎠ ⎝ 8.0 × 10 = ( − 2.33 y n − 7.33 z n ) × 10 − 3 An + q n
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let qAD be the unknown shear flow, i.e. qAD = q0. Consider stringer D: B
A
q0
D(-100,-50) 100 mm2
C 60 + q0
q DC = (− 2 .33 ( − 100 ) − 7 .33 ( − 50 ) ) × 10 − 3 (100 ) + q 0 = 60 + q 0 T. E. Tay, Department of Mechanical Engineering, National University of Singapore
13
Consider stringer C: B
A
q0
100 + q0
D
C(100,-50) 60 + q0
300 mm2
q CB = (− 2 .33 (100 ) − 7 . 33 ( − 50 ) ) × 10 − 3 (300 ) + q DC = 100 + q 0 T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Consider stringer B: 40 + q0 A
100 mm2 B(100,50)
q0
100 + q0
D
C 60 + q0
q BA = (− 2 .33 (100 ) − 7 . 33 (50 ) )× 10 − 3 (100 ) + q CB = 40 + q 0 T. E. Tay, Department of Mechanical Engineering, National University of Singapore
14
Let us now take moments about, say O:
q AD (100 )(100 ) + q BA ( 200 )( 50 ) + q CB (100 )( 100 ) + q DC ( 200 )( 50 ) = 0 4 q 0 = − 200 q 0 = − 50 N/mm
40 + q0
B
A
q0
100 + q0
O
C
D
60 + q0
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
Example 13
2
I yy = ∑ Ai z i2 = 2 Ar 2
Find the shear center.
i =1
s
q=− z
r
Vz
θ O
y ye
V z z n An V =− z I yy 2r
Taking moments about O, E
πr
V z y e = r ∫ q ds 0
π
= r 2 ∫ q dθ Area of each stringer = A mm2
0
= πr 2 q π rV z =− 2 ∴ ye = −
πr 2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
15
Example 14 Find the shear center. 120 B (80,40)
C (-40,40) 120 mm2 45
40 F
z O
60
y 60 mm2
D (-40,-50) 120 mm2
All dimensions in mm
60 mm2
40
A (80,-20)
First, remember to locate the centroid!
y=
1 4 2(60)(120) ∑ Ai yi = 360 = 40 A i =1
z=
1 4 (−60)(60) + (120)(−90) Ai zi = = −40 ∑ A i =1 360
measured from C
measured from C
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Section properties: 4
I yy = ∑ Ai z i2 = (60 + 120 )( 40 ) 2 + (60 )( − 20 ) 2 + (120 )( −50 ) 2 = 0.612 × 10 6 mm 4 i =1 4
I zz = ∑ Ai yi2 = 2(60 )(80 ) 2 + 2(120 )( − 40 ) 2 = 1.152 × 10 6 mm 4 i =1 4
I yz = ∑ Ai y i z i = ( 60 )(80 )( − 20 ) + (60 )(80 )( 40 ) + (120 )( 40 )( − 40 ) i =1
+ (120 )( − 40 )( −50 ) = 0.144 × 10 6 mm 4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
16
Let us first find the z-coordinate position of shear center E. Vy E
C ze
z Assume an arbitrary shear force Vy applied to the shear center.
y
O
F
B
A Hence
D
Vy I zz
= 0.894 × 10−6 Vy
Vz = −0.210 ×10−6 Vy I yy
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
qBC
C
B
z qCD
O
F
qAB
y A
D
⎛ Vy y A Vz z A ⎞ + q AB = − ⎜ ⎟ AA + 0 ⎜ I zz I yy ⎟⎠ ⎝ = − ( 0.894(80) − 0.210(−20) ) 60 × 10−6 Vy = −4.543 × 10−3Vy ⎛ Vy yB Vz zB ⎞ qBC = − ⎜ + ⎟ AB + q AB ⎜ I zz I yy ⎟⎠ ⎝ = − ( 0.894(80) − 0.210(40) ) 60 ×10−6 Vy − 4.543 × 10−3Vy = −8.33 × 10−3Vy qCD = −5.68 × 10−3Vy
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
17
Vy qBC
C ze qCD
B
z qAB
y
O
F
E
5
A
D Taking moments about a convenient point F,
−V y ( z e + 5) = [π (45) 2 ( − 5.68) + ( − 4.543)(60)(120) + ( − 8.330)(120)(45)] × 10 − 3 V y = − 113.83V y ∴ z e = 108.8 mm
measured from O.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Let us now find the y-coordinate position of shear center E. Vz E
C
B
z O
F
Assume an arbitrary shear force Vz applied to the shear center.
y ye A
D
Now,
Vy I zz
= −0.21×10 −6 V y
Vz = 1.68 × 10 −6 V y I yy
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
18
Vz E
C
B
q AB = 3.024 ×10 −3Vz
z qCD
y
O
F
qAB
qBC = 0 qCD = 7.056 ×10 −3Vz
ye A
40 D Taking moments about F,
V z ( y e + 40 ) = [π ( 45 ) 2 ( 7 . 056 ) + ( 3 .024 )( 60 )(120 )] × 10 − 3 V z = 66 . 7V z ∴ y e = 26 .7 mm measured from O.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 15 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. 10 kN 2
3 z
A1 = A8 = 200 mm2 A2 = A7 = 250 mm2 A3 = A6 = 400 mm2 A4 = A5 = 100 mm2 1
4 200 100
O
y 5
60 6
120
8
7 240
240
All dimensions in mm T. E. Tay, Department of Mechanical Engineering, National University of Singapore
19
Section properties:
(
8
I yy = ∑ Ai zi2 = 2 ( 200 )(30 ) 2 + ( 250 )(100 ) 2 + ( 400 )(100 ) 2 + (100 )(50 ) 2
)
i =1
= 13 .86 × 10 6 mm 4 I yz = 0
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.
⎛V z ⎞ q34 = −⎜ z 3 ⎟ A3 + q23 ⎜ I ⎟ ⎝ yy ⎠ (10 ×103 )(100) =− (400) + q23 13.86 ×106 = −28.9 + q23 10 kN 3
q23 – 28.9 z
q23
2
A1 = A8 = 200 mm2 A2 = A7 = 250 mm2 A3 = A6 = 400 mm2 A4 = A5 = 100 mm2 1
4 200 100
O
y 5
60 6
120
8
7 240
240
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
20
⎛V z ⎞ q45 = −⎜ z 4 ⎟ A4 + q34 ⎜ I ⎟ ⎝ yy ⎠ (10 ×103 )(50) =− (100) − 28.9 + q23 13.86 ×106 = −32.5 + q23
Due to symmetry,
q56 = q34 = −28.9 + q23 q67 = q23
10 kN
q23 – 28.9 z 200 100
O
3
q23
2
A1 = A8 = 200 mm2 A2 = A7 = 250 mm2 A3 = A6 = 400 mm2 A4 = A5 = 100 mm2 1
4 q23 – 32.5 y 5
60 6
q23 – 28.9
q23 240
120
8
7 240
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
(10 × 103 )(−100) (250) + q23 13.86 × 106 = 18.1 + q23
q78 = −
(10 × 103 )(−30) (200) + 18.1 + q23 q81 = − 13.86 × 106 = 22.4 + q23
Due to symmetry,
q12 = q 78 = 18 .1 + q 23 A1 = A8 = 200 mm2 A2 = A7 = 250 mm2 A3 = A6 = 400 mm2 A4 = A5 = 100 mm2
10 kN q23 – 28.9 z 200 100
O
3
q23
2
q23 + 18.1 1
4 q23 – 32.5 y 5
60 6
q23 – 28.9 120
8
7 q23 240
q23 + 22.4
q23 + 18.1 240
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
21
Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.
10 kN
z 200 100
4 q45 O
q23
3
q34
2
q12 1
y
5
A
60
6
8
7
120
q81
240
240
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.
(50 q 45 )(120 ) + (120 q34 + 240 q 23 + 240 q12 )(100 ) + ( 70 q12 )( 240 ) + (30 q81 )( 480 ) = 0 70q12
10 kN 120q34
3
z 200 100
4 50q45 O
240q23
5
A
60
6 120
30q81 1
240q12
50q34 y
2
8
7 240
240
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
22
Substituting for q23 and solving, we obtain
q23 = −5.4 N/mm q34 = −34.3 N/mm q45 = −37.9 N/mm q56 = −34.3 N/mm q67 = −5.4 N/mm q78 = 12.7 N/mm q81 = 17.0 N/mm
10 kN 34.3
3
5.4
2
q12 = 12.7 N/mm 12.7 1
4 37.9
17.0 5
6
8
7 12.7
34.3
5.4
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?
10 kN 34.3
3
5.4
2
12.7 1
4 37.9
17.0 5 34.3
6
8
7 12.7 5.4
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
23
Example 15 Determine the position of the shear center. (Let the thickness of each web = 1 mm.)
2
3 z
A1 = A8 = 200 mm2 A2 = A7 = 250 mm2 A3 = A6 = 400 mm2 A4 = A5 = 100 mm2 1
4 200 100
O
y 5
60 6
8
7
120
240
240 All dimensions in mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.
(50q45 )(120) + (120q34 + 240q23 + 240q12 )(100) + (70q12 )(240) + (30q81 )(480) = (10 ×103 ) ye 70q12 120q34
3
z 200 100
4 50q45 O
240q23
50q34 y
5
A 6
120
E
2
30q81 1
240q12 10 kN
ye
60 8
7 240
240
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
24
Substituting for the unknown shear flow,
6000(q23 − 32.5) + 12000q23 + 24000q23 + 24000q23 + 16800q23 + 304080 − 346800 + 434400 + 14400q23 + 322560 = (10 × 103 ) ye ∴ 97.2q23 + 519.24 = ye The additional equation is supplied by the rate of twist equation.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Rate of twist = 0, since load acts through shear center.
dφ 1 ⎛ q ⎞ ds ⎟ = 0 = ⎜ dx 2ΩG ⎝ ∫ t ⎠ q q q q q ∴ 2 ∫ 12 ds12 + 2 ∫ 23 ds 23 + 2 ∫ 34 ds34 + ∫ 45 ds 45 + ∫ 81 ds81 = 0 t t t t t 1− 2 12 2 − 3 23 3− 4 34 4 − 5 45 8 −1 81 3
2
z
1
4 O
y 5
A 6
7
8
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
25
Rate of twist = 0, since load acts through shear center.
2 ∫ (18 .1 + q23 ) ds12 + 2 ∫ q 23 ds 23 + 2 ∫ ( q 23 − 28 .9) ds34 1− 2
+
∫ (q
2 −3
23
− 32 .5) ds 45 +
4 −5
∫ (22.4 + q
3− 4
23
) ds81 = 0
8 −1
2(18 .1 + q 23 )( 250 ) + 2 q 23 ( 240 ) + 2( q 23 − 28 .9)(130 ) + ( q 23 − 32 .5)(100 ) + ( 22 .4 + q 23 )(60 ) = 0 1400 q 23 = 370 ∴ q 23 = 0.264 Substituting this into the moment equation, we have
ye = 544.9 mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
26
5/26/2009
Multicell Thin-walled Sections The 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 Hence, th the rate t off twist t i t off cell ll j becomes b
1 ⎛ dφ ⎞ ⎜ ⎟ = Ωj dx 2 G ⎝ ⎠j where Ωj is the area enclosed by cell j.
⎛ ⎞ ⎜ q ds − q ds − q ds ⎟ i ∫ k ∫ ⎜ j s∫ t ⎟ t t s ji s jk ⎝ j ⎠
(1)
j 1
,
2
,…
qi
qj
qk
…n
k i
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
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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 dφ . Thus dx
dφ ⎛ dφ ⎞ ⎛ dφ ⎞ ⎛ dφ ⎞ ⎛ dφ ⎞ = ⎜ ⎟ = ⎜ ⎟ = ... = ⎜ ⎟ = ... = ⎜ ⎟ dx ⎝ dx ⎠1 ⎝ dx ⎠ 2 ⎝ dx ⎠ j ⎝ dx ⎠ n
(2)
Therefore, for a general case where cell j is bounded by m cells,
dφ 1 ⎡ ds m ⎛ ds ⎞⎤ ⎢ q j ∫ − ∑ ⎜ qr ∫ ⎟ ⎥ = ⎜ dx 2GΩ j ⎢ s j t r =1 t ⎟ ⎝ s jr ⎠⎥⎦ ⎣
(3)
For n cells, there are n such equations.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
2
1
5/26/2009
However, the number of unknowns involved is (n+1), i.e. q1, q2, q3, …, qj, …, qn and dφ . The additional equation required is the moment equilibrium equation, i.e. dx n
T = 2∑ qi Ω i
(4)
i =1
where qi is the shear flow in the
ith
cell.
The foregoing procedure can also be used for multicell thin-walled tubes loaded by ttransverse a sve se forces, o ces, in w which c case tthee moments o e ts induced duced by the t e transverse t a sve se forces o ces should be taken into account in the moment equilibrium equation, and the shear flow expressions for bending should be used.
3
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Example 1: The section shown has constant thickness t throughout. It is subjected to a constant torque T. Determine the rate of twist. 2a
a q1
q2
1
a
2
Consider cell 1:
Consider cell 2:
4a a dφ 1 ⎡ ds1 ds ⎤ = − q2 ∫ 12 ⎥ q 2 ⎢ 1∫ dx 2Ga ⎣ 0 t t ⎦ 0 1 (4q1 − q2 ) = 2tGa
dφ 1 = dx 4Ga 2
(1)
=
a ⎡ 6 a ds d 2 d ⎤ ds − q1 ∫ 12 ⎥ ⎢ q2 ∫ t t ⎦ 0 0 ⎣
1 (6q2 − q1 ) 4tGa
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
(2)
4
2
5/26/2009
Moment equilibrium: 2
T = 2∑ qi Ω i i =1
= 2q1Ω1 + 2q2 Ω 2 = 2a 2 (q1 + 2q2 )
(3)
From (1) and (2) we have:
q1 =
Substituting these into (3),
16 dφ aGt 23 dx
T=
andd
q2 =
dφ 23 T = dx 104 Gta 3
∴
18 dφ aGt 23 dx
104 3 dφ a Gt 23 dx
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
5
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. G (N/mm2)
Thickness (mm)
Cell Area (mm2)
Wall
Length (mm)
12o
1650
1.22
24200
AI = 258000
12i
508
2 03 2.03
27600
AII = 355000 AIII = 161000
13, 24
775
1.22
24200
34
380
1.63
27600
35, 46
508
0.92
20700
56
254
0.92
20700
Note: The superscript symbols o and i are used to distinguish between outer and inner walls connecting the same points. 11300Nm 3
1
III
II
I 2
5
4
6
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
6
3
5/26/2009
∗ Let Gref = 27.6 GPa and t =
G ×t Gref
Hence
24.2 × 1.22 = 1.07 27.6 ∗ t13∗ = t24 = 1.07
t12∗ o =
∫
ds12o
∫
ds12i
12 o
12
∗ ∗ ∗ t35 = t46 = t56 = 0.69
i
∗ 12 o
t
t12∗ i
= =
1650 = 1542 1.07
508 = 250 2.03
ds13 ds = ∫ ∗24 = 725 ∗ t t 13 13 24 24
∫
ds34 = 233 t∗ 34 34
∫
ds35 ds = ∫ ∗46 = 736 ∗ t t 46 46 35 35
∫
ds56 = 368 t∗ 56 56
∫
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
For cell I,
⎛ ds12o ds i ds i 1 dφ ⎜ qI = + qI ∫ ∗12 − qII ∫ ∗12 dx 2Gref (258000 ) ⎜⎝ 12∫o t12∗ o i t i i t i 12 12 12 12 =
7
⎞ ⎟ ⎟ ⎠
1 (qI (1542 + 250) − 250qII ) 2Gref (258000 )
For cell II,
dφ 1 = (− 250qI + (250 + 725 + 233 + 725)qII − 233qIII ) dx 2Gref (355000 )
For cell III,
dφ 1 = (− 233qII + (736 + 736 + 233 + 368)qIII ) dx 2Gref (161000 )
3
1
II
I
qI
2
qII
5
III 4
qIII
6
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
8
4
5/26/2009
From the moment equilibrium equation
2(258000q I + 355000q II + 161000q III ) = 11.3 ×10 6 These equations are solved to give qI = 7.1 N/mm q II = 8.9 N/mm qIII = 4.2 N/mm
8.9 1
7.1
4.2
3
1.8
5
4.7 2
8.9
4
4.2 6
4.2
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
9
The Method of Successive Approximations for Pure Torque Consider a two-cells section free to warp and subjected to torque T.
I
II qI
qII
If the cells were separated, we see that ⎛⎜ dφ ⎞⎟ ≠ ⎛⎜ dφ ⎞⎟ and this violates the constant ⎝ dx ⎠ I ⎝ dx ⎠ II rate of twist condition.
II
I qI
qII
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
10
5
5/26/2009
⎛ dφ ⎞ ⎛ dφ ⎞ ⎟ = G⎜ ⎟ = 1, then ⎝ dx ⎠ I ⎝ dx ⎠ II
If we arbitrarily let G⎜
For cell I,
For cell II,
q ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = II ⎜ ∫ ⎟ = 1 ⎝ dx ⎠ II 2Ω II ⎝ t ⎠ II qII δ II =1 2Ω II 2Ω
q ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = I ⎜ ∫ ⎟ = 1 ⎝ dx ⎠ I 2Ω I ⎝ t ⎠ I qI δ I =1 2Ω I 2Ω
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
11
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.
I
qII qI
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
12
6
5/26/2009
The contribution of qII to the rate of twist is represented by an equivalent shear flow qI′ over the perimeter of cell I.
⎛ ds ⎞ ⎛ ds ⎞ q′I ⎜ ∫ ⎟ = qII ⎜ ∫ ⎟ ⎝ t ⎠I ⎝ t ⎠ I , II
I
qII
q′I
I
13
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Or,
q′I = qII
where
δ I , II δI
δ I , II = ∫
ds for the wall common to cells I and II t
The “correction carry-over factor” is then defined as
C II , I =
δ I , II δI
∴ q′I = qII C II , I
q′I
I qI
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
14
7
5/26/2009
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.
(qI )Final = qI + q′I + q′I′ + q′I′′ + K where
q′I′ = q′II C II , I q′I′′ = q′II′ C II , I
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
15
Similarly for cell II,
C I , II =
δ I , II δ II
qI
II
qII
∴ q′II = qI C I , II
q′II
II qII
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
16
8
5/26/2009
Example 3: Let us solve the problem of the previous example using the method of successive approximations.
⎛ ds ⎞ ⎟ ⎝ t ⎠ ds12o
δI = ⎜ ∫ =
∫
12
3
2
+
∫ 12
i
ds12i t12∗ i
δ II = 250 + 725 + 233 + 725
5
= 1933
III
II
I
t12∗ o
= 1542 + 250 = 1792
11300Nm 1
o
4
δ III = 736 + 233 + 736 + 368
6
= 2073
δ I , II = 250 δ II , III = 233 17
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The carry-over factors are:
δ I , II 250 = = 0.129 δ II 1933 δ 250 = I , II = = 0.140 δ I 1792
C I , II = C II , I
233 = 0.112 2073 233 = = 0.121 1933
The initial assumed values of shear flows:
2
qI = qII = qIII =
=
2Ω II
=
2Ω III
=
δ II
δ III
4
6
2(258000) = 287.9 1792
2Ω I
δI
5
III
II
I
C II , III = C II , III
3
1
2(355000) = 367.3 1933
2(161000 ) = 155.3 2073
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
18
9
5/26/2009
Cell I
Cell II
CI,II = 0.129
CII,I = 0.140
Cell III CII,III = 0.112
CIII,II = 0.121
A Assumed dq
287 9 287.9
367 3 367.3
155 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 5 345.5
435 9 435.9
199 1 199.1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
19
We still need the torque equilibrium equation to arrive at the correct solution. From the torque equilibrium equation,
T = 2qI Ω I + 2qII Ω II + 2qIII Ω III
= 2(345.5)(258000) + 2(435.9 )(355000) + 2(199.1)(161000) = 5.52 ×108
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
Cell I
Cell II
Cell III
Final q
345.5
435.9
199.1
Scaled q
7.07
8.92
4.1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
20
10
5/26/2009
Bending Shear Stresses in Multi-cell Beams Consider a three-cells section carrying a shear load through the shear centre (no twist):
VZ
I
II
III
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
21
If each cell were “cut”, the open-section shear flow would be given by
qb = =
(I
yz
Qz − I zz Q y ) Vz I yy I zz − I yz2
(I ∫ yt ds − I ∫ zt ds ) V yz
zz
I yy I zz − I
z
2 yz
Distribution of qb:
I
II
III
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
22
11
5/26/2009
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:
qb
qI
qII
qIII
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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:
qII qI
qIII
qb
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
24
12
5/26/2009
Since the twist is zero, 1 ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = ⎜∫q ⎟ = 0 t ⎠ II ⎝ dx ⎠ II 2Ω II ⎝
⎛ ds ⎞ ⎜∫q ⎟ = 0 t ⎠ II ⎝
or
From the figure, taking positive torque anti-clockwise, ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ =0 ⎜ ∫ qb ⎟ + q II ⎜ ∫ ⎟ − q I ⎜ ∫ ⎟ − q III ⎜ ∫ ⎟ t ⎠ II ⎝ ⎝ t ⎠ II ⎝ t ⎠ I , II ⎝ t ⎠ II , III
The other two cells have similar equations, i.e.: ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ =0 ⎜ ∫ qb ⎟ + qI ⎜ ∫ ⎟ − qII ⎜ ∫ ⎟ − qIII ⎜ ∫ ⎟ t ⎠I ⎝ ⎝ t ⎠I ⎝ t ⎠ II , I ⎝ t ⎠ I , III
and
ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ − qI ⎜ ∫ ⎟ =0 ⎜ ∫ qb ⎟ + q III ⎜ ∫ ⎟ − qII ⎜ ∫ ⎟ t ⎠ III ⎝ ⎝ t ⎠ III ⎝ t ⎠ II , III ⎝ t ⎠ I , III
These three equations can be solved to obtain qI , qII and qIII The above equations are also applicable to idealized multicell beams.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
25
The Method of Successive Approximations for Bending Shear Flows Since the twist is zero,
If shear force acts through shear centre
1 ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = ⎜∫q ⎟ = 0 t ⎠ II ⎝ dx ⎠ II 2Ω II ⎝
or
d ⎞ ⎛ ds ⎜∫q ⎟ = 0 t ⎠ II ⎝
Hence, from the figure, taking positive torque anti-clockwise, for cell II, ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ =0 ⎜ ∫ qb ⎟ + q II ⎜ ∫ ⎟ − q I ⎜ ∫ ⎟ − q III ⎜ ∫ ⎟ t ⎠ II ⎝ ⎝ t ⎠ II ⎝ t ⎠ I , II ⎝ t ⎠ II , III
qb
ds ⎞ ⎛ ⎜ ∫ qb ⎟ + q II δ II − q I δ I , II − q III δ II , III = 0 t ⎠ II ⎝
Rearranging,
qII = −
ds ⎞ ⎛ ⎜ ∫ qb ⎟ t ⎠ II ⎝
δ II
δ δ + q I I , II + qIII II , III δ II δ II
qI
qII
qIII
= q′II + C I , II q I + C III , II qIII T. E. Tay, Department of Mechanical Engineering, National University of Singapore
26
13
5/26/2009
To begin, we first neglect the contribution of qI and qIII, so that
q II = q′II = −
ds ⎞ ⎛ ⎜ ∫ qb ⎟ t ⎠ II ⎝
δ II
q I = q′I q III = q′III
Subsequently, q II = q′II + C I , II q′I + C III , II q′III = q′II + q′II′
etc.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
27
Example 4. Determine the shear flow distribution in the single symmetrical three-cell wing section below when it carries a shear load of 100 kN applied through its shear centre and hence find the distance of the shear centre from the spar web 34. Assume all direct stresses are resisted by the stringers while the skin is effective only in shear. The shear modulus G is constant throughout. Stringer areas: B1 = B6 = 2500 mm2, B2 = B5 = 3800 mm2, B3 = B4 = 3200 mm2 C ll areas: AI = 265000 mm2, AII = 580000 mm2, AIII = 410000 mm2 Cell Wall
12, 56
23, 45
34o
16
25
34i
Length (mm)
1025
1275
2200
330
450
400
Thickness (mm)
1.25
1.65
2.25
1.65
2.65
2.65
100kN
2
3
1
E
I
4
ξs 1270mm
II
III
5
6 1020mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
28
14
5/26/2009
The carry-over factors:
δ I = 1128.7 δ II = 1870.0 δ III = 2013.6
δ I , II 150.9 = = 0.081 1870 δ II δ 150.9 = I , II = = 0.134 δ I 1128.7
C I , II = C II , I
δ I , II = 150.9
C II , III = 0.086
δ II , III = 173.6
C III , II = 0.093
29
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
For an open idealized section symmetric about the horizontal axis,
qbi = −
∴
Vz Ai zi I yy
100 ×103 (3200)(− 200) = 80.6 N/mm 7.94 ×108 0 = 110.1 N/mm
qb 43 = − qb 52
3
qb 61 = 52.0 N/mm Also,
0
0
2
80.6
0
4
52.0
110.1
0
(80.6)(400) = −12166.0 N/mm ⎛ ds ⎞ −⎜∫q ⎟ = − t ⎠I 2.65 ⎝ (110.1)(460) − 12166.0 = −6945.7 N/mm ⎛ ds ⎞ −⎜∫q ⎟ = − t ⎠ II 2.65 ⎝
0 1
0
5
0
6
⎛ ds ⎞ − ⎜ ∫ q ⎟ = 6218.9 N/mm t ⎠ III ⎝
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
30
15
5/26/2009
4.4
3
11.4
The solution proceeds in the tabulated form shown. Taking moments about the centre of spar web 34,
2.7
2
4
54.7
III
II
I
1
103.0
73.6
5
4.4
2.7
6
100 ×103 ξ s = 2Ω I qI + 2Ω II qII + 2Ω III q III + (110.1)(460)(1270) + (52)(330)(2290)
∴
ξ s = 946.8 mm Cell I
Cell II
Cell III
1128.7
1870.0
2013.6
-12166.0
-6945.7
6218.9
δ −∫
qb ds t
CI,II=0.08 ⎛ q ds ⎞ q′ = − ⎜ ∫ b ⎟ δ ⎝ t ⎠ q′′
CII,I=0.134
-10.78 10.78
CII,III=0.086
CIII,II=0.093
-3.71 3.71
(0.134)(-3.71) =-0.50
3.09
(0.08)(-10.78)=-0.86
(0.093)(3.09)=0.29
(0.086)(-3.71)=-0.32
q′′′
(0.134)(-0.86)=-0.115
(0.08)(-0.50)=-0.04
(0.093)(0.32)=-0.03
(0.086)(0.29)=0.02
q′′′′
(0.134)(-0.04)=-0.005
(0.08)(-0.115)=-0.01
(0.093)(0.02)=0
(0.086)(-0.03)=0
Final q
-11.4
-4.36
2.7
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
31
Tapered Beams Beams in aerospace vehicle structures are often tapered in order to achieve maximum structural efficiency with minimum weight. The simple beam shown consists of two stringers of equal areas, joined by a vertical web that does not carry bending stresses. x x0
b
α1
h1 h0
α2
h2
Vz T. E. Tay, Department of Mechanical Engineering, National University of Singapore
32
16
5/26/2009
Consider forces acting on a section of the beam at a distance b from the free end. The shear force in the web is Vw = qh
(1)
The total shear force in the stringers is equal to the sum of the vertical forces in the stringers:
P tanα1
V f = P (tan α1 + tan α 2 ) ⎛h h ⎞ = P⎜ 1 + 2 ⎟ ⎝x x⎠ Ph = x
P q
VW (2)
h
P tanα2 P
33
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Now, the presence of P is caused by the bending moment due to the external load Vz , so that th t Ph = VZ b ∴ Vf =
Vz b x
(3) from (2)
Vz
P
(4) h
P b
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
34
17
5/26/2009
Also, considering vertical forces,
Vz = V f + Vw
(5)
∴ Vw = Vz (1 − b x ) = Vz x0 x
from (3)
(6)
From similar triangles, x0 x = h0 h
(7)
so that
Vw = Vz h0 h
(8)
and
V f = Vz (h − h0 ) h
(9)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
35
Hence the shear flow in the web at a section of depth h is q=− =−
Vw I yy
∑Az
i i
Vw
(Ah 2) 2
*Note that we have used Vw instead of Vz
Ah 2
= − Vw h
(10)
h0 h2
(11)
∴ q = −Vz
in terms of applied shear load Vz and geometry.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
36
18
5/26/2009
Let us now seek a more general analysis of the bending and twisting of tapered idealized beams. Consider the section of an idealized tapered multicell tube subjected to shear loads Vy and Vz , acting through an arbitrary point. The reference axis passes through the centroid of the section. Pn is the force in the nth stringer, and is in general composed p of three components p in the x,, y and z directions. Let these components p be denoted Pxn, Pyn and Pzn, respectively. Pn z
Vz
x
y
Vy
37
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
The axial component is given by
Pxn = σ xn An
where σxn is calculated by the flexure formula and An is the cross-sectional area of the nth stringer.
Pxn
We see that
x
Pxn Δx = Pzn Δz n Δz ∴ Pzn = n Pxn Δx Pyn Pzn
=
z
Pzn
(13) Pyn
Δzn
Δyn Δz n
Δy Δy ∴ Pyn = n Pzn = n Pxn Δz n Δx
Δzn Δx
L t us isolate i l t stringer ti id forces f i the th yz Let n andd consider in and xz planes.
Also,
z
Pzn (12)
Δy n (14)
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
y 38
19
5/26/2009
Now consider moment equilibrium about some convenient point C. Let the horizontal and vertical distances from stringer n to the point C be ξn and ηn, respectively. Also let the corresponding distances of the point of application of the shear loads Vy and Vz to the point C be ξ0 and η0. The total number of stringers is N and the total number of sheets is M. Pzn Pyn Vz
ηn
Vy
η0
C
ξn
ξ0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
39
Taking moments about C, N
N
n =1
n =1
M
∴ Vzξ 0 − V yη 0 = ∑ Pznξ n − ∑ Pynη n + ∑ ∫ ri qi
Due to forces in stringers
i =1 i
ds ti
(15)
Due to shear flow in webs
Vertical and horizontal equilibrium of forces yield N
V y = ∑ Pyn + V yw
(16)
n =1
N
and
Vz = ∑ Pzn + Vzw
(17)
n =1
Here Vyw and Vzw are resultants of the shear forces carried by the webs. webs These values should be used in the calculation of shear flows in the webs. The solution procedure is similar to that for multicell boxes, except that the modified moment equilibrium equation (i.e. Eqn 15) should be used.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
40
20
5/26/2009
Example 1: A tapered two-cell tube with symmetrical cross-sections 1.2m apart has the dimensions shown. The tube supports load which produce a bending moment My =-1.65 kNm and a shear force Vz =10 kN in the plane of the internal spar web at the larger cross-section. The shear modulus G is constant throughout. Determine the forces in the stringers and the shear flow distribution at this cross-section. Stringer areas: A1 = A3 = A4 = A6 = 600 mm2 , A2 = A5 = 900 mm2 Vz =10 kN 150mm
100mm
0.8mm
2
3
0.8mm
I 1.0mm
1.0mm
4
II
5
0.8mm
80 mm 180mm
1.0mm
0.8mm
200mm
1
6
400mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
41
At the larger cross-section, I yy = 4(600)(90) + 2(900)(90) = 34 ×106 mm 4 2
2
The direct stresses are given by σ xn = M y zn
Hence
Pxn =
e.g.
Px1 = −
I yy
M y zn I yy
An
(1600)(90)(600)(103 ) = −2620 N 34 ×10 6
The forces in the other stringers are similarly calculated. Note that Δx = 1200mm. e.g. For stringer 1, as x increases, z decreases; ∴ Δz = −50 mm
and
Δz = −0.042 Δx
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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5/26/2009
ξn and ηn are measured from the midpoint of the internal web 2-5. Stringer
Pxn
Δzn/Δx
Pzn
Δ y n / Δx
Pyn
Pznξn
Pynηn
1 2 3 4 5 6
-2620 -3930 -2620 2620 3930 2620
-0.042 -0.042 -0.042 0.042 0 042 0.042 0.042
110 165 110 110 165 110
0 0.083 0.125 0.125 0 083 0.083 0
0 -326 -328 328 326 0
44000 0 -22000 -22000 0 44000
0 29340 29520 29520 29340 0
Note:
Pzn =
Δz n Pxn Δx
6
Therefore,
∑P ξ n =1
zn n
∑P η
yn n
∑P
zn
6
∑P n =1
Hence, the shear forces carried by the webs are V yw = V y − ∑ Pyn = 0
= 117720 Nmm
n =1 6
6
n =1
= 44000 Nmm
Δy n Pxn Δx
6
6
n =1
Pyn =
and
yn
Vzw = Vz − ∑ Pzn = 10000 − 770 = 9230 N
= 770 N
n =1
=0
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
43
Now let shear flows q12 and q23 be unknowns. 1
2
3
II
I
4
5
6
Force equilibrium of stringers: Stringer 2,
Stringer 1,
V z A q25 + q23 = − zw 2 2 + q12 I yy = −22 + q12
V zA q12 = − zw 1 1 + q61 I yy = −14.7 + q61
Stringer 3, q34 = −
Vzw z3 A3 + q23 I yy
= −14.7 + q23
Also, q45 = q23 and q56 = q12
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
44
22
5/26/2009
Rate of twist, Consider cell I:
Consider cell II:
1 dφ ⎛ 2(200 )q23 180q23 − (14.7 )(180 ) 180q25 ⎞ = + − ⎜ ⎟ 1.0 1.0 ⎠ dx 2G (200 )(180 ) ⎝ 0.8 1 (860q23 − 180q12 + 1314) = 72000G
dφ 1 (− 180q23 + 1360q12 − 1314) = dx 72000G
(1)
(2)
Taking moments about the midpoint of web 2-5, we have (1.47)(180)(400) – (14.7)(180)(200) + 72000q23 + 144000q12 + 44000 + 117720 = 0
(3)
E Equating i (1) andd (2), (2) andd solving l i with i h (3), (3) one obtains b i q23 = -4.4 N/mm
and
q12 = -2.6 N/mm
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
45
Effect of Varying Moments of Area The box beam shown is subjected to a pure bending moment about the y axis. The stringer areas change from section A to section B over a distance d along the beam. The cross-sections of the beam are symmetric about the y axis. z Pi d
y Pi+ΔPi x
Section A
Section B
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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23
5/26/2009
Let us consider equilibrium of forces in the axial direction. ΔPi + (qi +1 − qi )d = 0 ∴ qi +1 = ΔPi d + qi
(18)
Now, the direction of ΔPi shown here produces a positive moment My. Thus ΔPi produces a change in the bending moment from sections A to B. Pi
ΔPi =
ΔM y zi Ai
∴ qi +1 =
qi+1
(19)
I yy
qi
ΔM y zi Ai I yy d
+ qi
(20)
Pi+ΔPi
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
47
Example 2: The single cell box beam is loaded by a transverse force of 1.0 kN. Between points B and C, the beam has constant stringer section as shown in section B-B. Between points B and D, two stringers taper uniformly, with the stringer areas at point A as indicated in section A-A. Calculate the shear flow at section A-A. Area of stringers: A1=100 mm2 A2=300 mm2 1.0 kN
A2
A1
A1
100
A
Section A-A
B D
C
A
A2 A1
B
50 C
A1
50 A1
A1 A1
100
*All dimension in mm.
Section B-B
A1 A1 50
50
A1
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
48
24
5/26/2009
Consider section B-B: 3B
2B
1B
Consider section A-A: 3A
I yy = ∑ An z n2
2A
1A
I yy = ∑ An z n2
= 6(100)(50 )
= 4(100 )(50 ) + 2(300 )(50 )
2
2
= 1.5 × 106 mm 4 4B
5B
= 2.5 × 106 mm 4
6B
4A
Stress in stringers due to bending σb = −
2
5A
6A
Stress in stringers due to bending
M yz
σb = −
I yy
(300)(1000)(50) =
M yz I yy
(500)(1000)(50) = 2.5 ×106 = 10 MPa
1.5 ×106 = 10 MPa
∴ Axial force in each stringer = (100)(10) = 1000 N
∴ Axial force in each corner stringer = 1000 N, and axial force in each of stringers 2A and 5A is = 3000 N
49
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
∴ ΔP 1 = ΔP 3 = ΔP 6 = ΔP 6 = 0 N but ΔP2 = ΔP5 = 2000 N
3 2 1
2000 N
4
A-A 5
2000 N
6
B-B
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
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5/26/2009
Assume q61 unknown. Then q61 = q12. Vertical equilibrium of forces gives 2q61 (100) = 1000 ∴ q61 = 5 N/mm
and
q12 = 5 N/mm
Check for the middle stringer, q12 =
5 N/mm
ΔP2 2000 + q23 = + q23 d 200
5 N/mm
∴ q23 = 5 − 10 = −5 N/mm
5 N/mm
which is correct, because q23 = -q12 due to symmetry.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
Note that if we had used qi +1 = −
Vz zi Ai + qi we would have obtained, I yy
for section B-B,
q61 = −
51
for section A-A,
(1000)(50)(100) + q
1.5 ×10 6 = −3.33 + q12
12
q12 = 8.33 N/mm
q61 = −
(1000)(50)(100) + q
2.5 ×10 6 = −3.33 + q12
12
q12 = 7 N/mm
These results would have been wrong.
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
52
26
5/26/2009
T. E. Tay, Department of Mechanical Engineering, National University of Singapore
53
27
THE GREEK ALPHABET A
a
Bfl
r r
A 8 E C
z? 6
K K A h Mr
z e long, as in scene th as in thin
MU
o
ET
v P P
Z ~,fimI s
T T Yu
*+
Zeta
Nu
Nu 8 0
a
Eta Theta Iota Kappa Lambda
Htl Q 0 1
Alpha Beta Gamma Delta Epsilon
X X ./ P # Q W
Xi Omicron Pi Rho . Sigma Tau Uhilon P~ Chi Psi Omega
b g hard, as in begin'
d e short, as in met
i
k 1 m a X
o shmt, as in lot
P
r se t U
ph ch hard, as in lock PS
o long, as in throne
Except that before a, x or another y it is nasal-ng, as in anchor. Sharp as in this, but flat before or p, as in asbestos, dismal.
ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet - Torsion of Thin-Walled Members 1. A hollow aluminium tube of rectangular cross section as shown in Fig. 1 is subjected to a torque 56.6 kNm along its longitudinal axis. Determine the shear stresses and the rate of twist. Assume G = 28 GPa. 500
250
6 10 12
6 Fig. 1
Ans: τ1 = 18.9 MPa; τ2 = 37.7 MPa; τ3 = 22.6 MPa; θ = 0.00687 rad/m 2. The aluminium shafts (G = 28 GPa) of cross-sections shown in Figs. 2(a) to 2(c) are each subjected to a torque of 2 kNm. Neglecting the effect of stress concentrations, calculate (i) the maximum shear stress and (ii) the angle of twist in a 2-m length. A 3 4 A R100 B R75 6
60o B
3 2 D
2 C
R100
150
4 C
(a)
(b)
B 3
4
All dimensions in mm 2
60o
A (c) Ans:
C 120
Fig. 2 (a) τAC = 10.04 MPa, τAB = 6.70 MPa; 0.31° (using exact dimensions) ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 1
(b) τAB = τBC = 8.05 MPa, τAC = 5.37 MPa; 0.31° (using exact dimensions) (c) τAC = 80.18 MPa, τAB = 40.09 MPa, τBC = 53.46 MPa; 6.84°
3. The thin-walled closed section beam shown in Fig.3 is subjected to a torque of 4500 Nm. The section is constrained to twist about O, the centre of the semi-circular arc 512. For the curved wall 512, the thickness is 2 mm and the shear modulus is 22 GPa. For the plane walls 23, 34 and 45, the thickness is 1.6 mm and the shear modulus is 27.5 GPa. Determine the warping displacements of points 2, 3, 4 and 5.
1.6 mm 5
4
1.6 mm
O
1
100 mm R 50 mm
2.0 mm
2
3 1.6 mm 200 mm
Fig. 3 Ans: u 2 = −u 5 = 0.053 mm, u 3 = −u 4 = 0.187 mm.
4. a) The closed cross-section of a thin-walled tube shown in Fig. 4 is subjected to a torque of 6000 Nmm. Determine the magnitude and location of the maximum shear stress experienced by the tube, and the torsional constant. (All dimensions in mm.) ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 2
b) If a longitudinal slit is cut through the wall of the section at Point A, determine the magnitude and location of the maximum shear stress when a torque of 6000 Nmm is applied. Ans: 2.89 MPa, 77412 mm4, 11.96 MPa.
10 2 3
4
30o
30o
60o
A
60o 5
1
20
6 20
30
Fig. 4
ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 3
ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet - Bending of Thin-Walled Members
1.
A beam 250 mm wide by 300 mm deep by 4.2 m long is used as a simply-supported beam on a span of 4 m. It is subjected to a concentrated load P at the midsection of the span. The plane of the loads makes an angle φ = - 4π/9 with the horizontal y-axis. The beam is made of material with a yield stress 25.0 MPa. If the beam has been designed with a factor of safety 2.50 against initiation of yielding, determine the magnitude of P and orientation of the neutral axis. Ans:
2.
31.4 kN, 14.25o
In Fig. 1 let b = 300 mm, h = 300 mm, t = 25.0 mm, L = 2.50 m, and P = 16.0 kN. Calculate the maximum tensile and compressive stresses in the beam, and determine the orientation of the neural axis. Ans:
89.81 MPa, - 66.83 MPa, 30.71°
60 mm 10 mm
10 mm 10 mm
30 mm t
L
70 mm
P 10 mm
h
70 mm t
b
Fig. 2
Fig. 1
3.
An extruded bar of aluminium alloy has the cross section shown in Fig. 2. A 2.10 m length of this bar is used as a simple beam on a span of 2.00 m. A concentrated load P = 5.0 kN is applied at mid-length of the span and makes an angle of φ = 4.54 rad with the y-axis. Determine the maximum tensile and compressive stresses in the beam. Ans: 80.93 MPa, - 75.3 MPa
1
ME4212 Mechanics of Thin-Walled Structures T.E. Tay, Dept of Mechanical Engineering, NUS
4.
A cantilever beam 1.6 m long has the section shown in Fig. 3. It carries a load P = 2.5 kN in the –z direction at the free end. Find the magnitude and location of the maximum tensile flexural stress. (All dimensions in mm.) Ans: 184.1 MPa, at y = 55, z = 70 z
z
10
A
B
70 y 10 70
y
O
100
10 D
55
C 42
55 Fig. 3
Fig. 4
5. A column has the I-section shown in Fig. 4. A compressive force P parallel to the axis of the column is applied at the corner A. Find the axial stresses in terms of P at the remaining three corners B, C and D. (All dimensions in mm.) I yy = 1.105 × 10 6 mm 4 I zz = 4.94 × 10 4 mm 4 A = 680 mm 2
Ans: 0.005195P, 0.009715P and -0.008135P
6. A uniform thin-walled beam has the open cross-section shown in Fig. 5. The wall thickness is constant. Determine the position of the neutral axis and the maximum direct stress for a bending moment of My = 3.5 kNmm. Take r = 5 mm and t = 0.64 mm. Ans: -51.67°, 100 MPa t
z r y
r
Fig. 5 2
ME4212 Mechanics of Thin-Walled Structures T.E. Tay, Dept of Mechanical Engineering, NUS
7. The cross-section of a beam is shown in Fig. 6. The beam is subjected to a bending moment about the y-axis of 6000 Nmm. Determine the bending stresses at A, B, C and D. All dimensions in mm. (All dimensions in mm.) Ans: 0.206 MPa, -0.411 MPa, 0.411 MPa, -0.206 MPa. z 50 D
C 5
50
O
y 5
50
5 A B 50
Fig. 6 8. The closed two-cell cross-section of a fuselage shown in Fig. 7 is subjected to a bending moment My = 5000 kNm. a) Derive an expression for the position of the centroid in terms of r, tw and tf. b) Hence determine the maximum tensile and compressive bending stresses in the crosssection, if r = 1500 mm, tw = 5 mm and tf = 8 mm. Ans: −
rt f 9 19 ⎛⎜ 10 ⎜⎝ 10πt w + 9t f
⎞ ⎟ from O, 143.7 MPa, - 109.1 MPa. ⎟ ⎠ tw r O
tf
9r/5
Fig. 7 3
ME4212 Mechanics of Thin-Walled Structures T.E. Tay, Dept of Mechanical Engineering, NUS
ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet – The Shear Center 1. Fig.1 shows the cross-section of an open thin-walled beam of uniform thickness t. a) Determine the position of the shear center. b) Derive expressions for the torsional constants of the open cross-section, and the cross section if it were closed at point A. If the allowable shear stress is 30 MPa, a = 25 mm and t = 2 mm, determine the maximum torques that can be applied to the open and closed sections. B
A
a t D
C a
Fig. 1 Ans: (a)
a 4 2
from C. (b) 4 Nm, 75 Nm.
2. Find the shear centers of the following open thin-walled beam sections (Fig.2): z
z t
t
y
y r
Fig. 2 (a)
Fig. 2 (b) z
z 60° y
2c
y h
t
60° t b
Fig. 2 (d)
Fig. 2 (c) Ans: (a) -2r (b) -2r (c) -0.577c (d) 3b2/(h+3b)
ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 1
3. Find the shear center if the thickness of the web tw is small compared with the thickness of the flanges tf (Fig. 3). z c y O tf tw b
Fig. 3
Ans: - b/2
4. A thin-walled beam has the cross-section shown in Fig. 4. The thickness of each flange varies linearly from t1 at the tip to t2 at the junction with the web. The web itself has a constant thickness t3. Calculate the position of the shear center from the web. t2
t1
h t3
d
Fig. 4
Ans:
(2t1 + t 2 )d 2 3d (t1 + t 2 ) + ht 3
5. The thin-walled open tube (Fig. 5) of constant thickness t has a narrow longitudinal slit at the corner 1-5. Calculate and sketch the shear flow distribution due to a vertical shear force V acting through the shear center E, and note the values at points 2, 3 and l 4. Hence show that the distance ye is . 2(1 + a / b ) 4 ye 5
3
h E
1 a
b 2 l
Fig. 5 ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 2
6. Determine the positions of the shear centers of the following thin-walled sections:
t
t 20°
r
20° t
t t
r
50 mm
Fig. 6 (b)
Fig. 6 (a)
t
t
45° 45° r t
r
O
O t
t
Fig. 6 (d)
Fig. 6 (c)
Ans: (a) – 1.47r (b) 37.3 mm (c) – 1.35r (d) – 0.53r
ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 3
ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet – Idealized Beams
1. Find the shear centers of the following idealized thin-walled beam sections (Fig.1) (All dimensions in mm): A E
A
1.5
10
C
40
B
B
1.5
40
20
C
1.5
40
F D
10
70
Fig. 1 (a)
2
1.5
D
1.5
Areas of stringers A, B, C and D are 120 mm2 each, Areas of stringers E and F are 60 mm2 each. Thickness of each sheet is 2 mm.
A
E F
20 Area of each stringer is 25 mm2.
Fig. 1 (b)
B z 60
2 y
E 60
2 D
C 60
80
2
Areas of stringers A, B, C and D are 120 mm2 each, Area of stringer E is 80 mm2.
Fig. 1 (c)
Ans: (a) -21.7 to the left of BC (b) 3.33 to the right of CD (c) 40 to the left of E.
ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 1
2. The idealized symmetrical cross-section of an aircraft fuselage (Fig. 2) is subjected to a bending moment of 100 kNm about the horizontal axis. If all the direct (normal) stresses are carried by the stringers, determine the average stress in each stringer. The stringer areas are: A1 = 640 mm2, A2 = 600 mm2, A3 = 600 mm2, A4 = 600 mm2, A5 = 620 mm2, A6 = 640 mm2, A7 = 640 mm2, A8 = 850 mm2, A9 = 640 mm2. Suppose the bending moment arises from the application of a shear force Vz = -100 kN, determine the shear flows in the webs.
1 2
2
3
3
4
4 5
1200 1140
5 960 768
6
6
565 336
7
7
144 8
8 9
38
All dimensions in mm.
Fig. 2 Ans (selected): σx1 = 35.6 MPa, σx2 = 32.3 MPa, σx6 = -11.0 MPa, σx8 = -27.0 MPa, q21 = -11.82 N/mm, q32 = -31.24 N/mm, q54 = -52.21 N/mm, q76 = -46.01 N/mm.
3. Find the stresses in the stringers and the shear flow in the webs of the 1.2 m-long beam shown in Fig. 3. The area of each stringer is A mm2. 1.4 kN
t t
100 mm
45º
30º t
Fig. 3 Ans: 12246/A, 21286/A, -33522.5/A, 0 N/mm, -10.28 N/mm, 17.81 N/mm.
ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 2
4. The single cell cross-section of a thin-walled tube (Fig. 4) is symmetric about the horizontal axis. The normal stresses are carried by the stringers 1 to 4, while the webs are effective only in carrying shear stresses. Calculate the shear stresses in each web. Cell area = 135000 mm2. Stringer areas: A1 = A4 = 450 mm2; A2 = A3 = 550 mm2. Web
Length (mm)
Thickness (mm)
12, 34 23
500 580
0.8 1.0
41
200
1.2 1.0 kN
2
1 100
3
4
100
500
Fig. 4
Ans: τ12 = 2.16 MPa, τ23 = -1.02 MPa, τ34 = 2.16 MPa, τ41 = 3.32 MPa.
ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 3
ME4212 Aircraft Structures Tutorial Sheet – Multicell Sections
1. The cross-section of an idealized three-cell torque box is shown in Figure 1. The thickness of each web is t and the cross sectional area of each stringer is As. The shear modulus of the web material is G and the Young’s modulus of the stringer material is E. If the section is subjected to a constant torque T and allowed to warp freely, determine the rate of twist and the maximum shear flow. A
a
B
C
2a
a
a
D a
E
a 2a
F
a
a 3a
G
H
Figure 1 Ans:
.
,
0.0867
2. The two cell thin-walled section is fres to warp, and is subjected to a load of 10 kN as shown in Figure 2. Determine the rate of twist. R = 50 mm z t1 = 2 mm t2 = 2 mm t1 G = 20 GPa Vz=10 kN R y
2R
O
t2
t1
t1 2R Figure 2 Ans:
3.86
10 rad/mm ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 1
3. Use the method of successive approximations to find the position of the shear centre of the idealized thin-walled section shown in Figure 3. Each stringer has an area A and each web has a thickness t. Assume all direct bending loads are carried by the stringers and the webs carry only shear loads. 5
4
a
a a
a
3 a
a
1
2
Figure 3 Ans: ye = 0.025a =
to the right of 15
4. The cross-section of a three-cell torque box is shown in Figure 4. It is constructed of seven stringer of area 400 mm2 each and nine webs. Aa vertical force of 13.5 kN is applied to stringer 2. Assume that the webs carry only shear stresses and that the section is free to warp. Determine the shear flow distributions in the webs. Shear modulus of the web material G = 60 GPa. Young’s modulus of the stringer material E = 200 GPa. 13.5 kN
(All dimension in mm) 0.5
0.5
3
2
0.4
1
0.5
0.5
0.4
7
0.5
6 0.5
100
4
150
5 0.5
200
200
Figure 4 Ans: q23=+43.56 N/mm, q34=+34.135 N/mm, q12=+42.59 N/mm, q72=+30.96 N/mm, q63=+20.58 N/mm, q45=+4.14 N/mm.
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 2
ME4212 Aircraft Structures Tutorial Sheet – Tapered Beams & Varying Moments of Area
1. A uniformly tapered thin-walled beam consisting of four sheets and four stringers is shown in Figure 1. The thickness of each sheet is 2 mm and the area of each stringer is 200 mm2. The distance between the larger and smaller corss-sections is 800 mm. A vertical load of 2 kN us applied at stringer 2 in the plane of the smaller cross-section. Calculate the forces in the stringers and shear flow distribution at the larger crosssection. (All dimension in mm)
2 kN
1
2 120 100
3
4 100
180 Figure 1 Ans: q31=4.63 N/mm, q12=-2.31 N/mm, q43=-2.31 N/mm, q24=-9.26 N/mm.
2. Figure 2 shows a uniformly tapered thin-walled beam consisting of four sheets and four stringers. The thickness of each sheet is 2 mm and the cross-sectional area of each stringer is 200 mm2. The distance between the larger and smaller cross-sections is 800 mm. A vertical load of 3 kN is applied to stringer 4 in the plane of the smaller cross-section. Assuming that the sheets carry only shear stresses and the stringers carry only normal stresses, determine the shear flow distributions in the sheets at the larger cross-section.
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 1
120 30
40 1 z
2
O
y
3
100
60
4
40
3 kN
Figure 2 Ans: q31=3.5 N/mm, q12=12.5 N/mm, q43=12.5 N/mm, q24=21.5 N/mm. 3. The tapered thin-walled beam shown in Figure 3 has the cross-section 600 mm apart and is made of six stringers and six webs. The webs have a thickness of 2 mm each. A vertical load of 2 kN is applied to stringer 1 in the plane of the smaller cross section. Assuming that the webs carry only sehar stresses and the stringers carry normal stresses, determine the shear flow distributions in the webs ant the larger crosssection. Shear modulus of the web material G = 40 GPa. The areas of the stringers: A1 = A3 = A4 = A6 = 500 mm2 A2 = A5 = 300 mm2 2 kN
2
2
2
1
z
3 2
2 y
200 100
4
100 5
2
150
100
2
6
150 Figure 3
Ans: q12= q54=3.91 N/mm, q23= q65=2.756 N/mm, q36=0.8333 N/mm, q41=5.83 N/mm.
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 2
4. The cantilever box beam of length 2000 mm shown in Figure 4 consists of six sheets and six stringers. It is built-in at one end and loaded by a vertical force of 800 N at the other end. Stringers 2 and 5 have constant cross-sectional areas throughout the length of the beam. Stringers 1, 3, 4 and 6 have uniformly varying cross-sectional areas. Determine the shear flows in the sheets at a distance 600 mm away from the loaded end. Assume that the sheets carry only shear and the stringers carry only normal stresses. 800 N B
B 600 2000 Figure 4 (All dimension in mm) 1
3
2
800 N
1
3
2
120 4
5
6
120 5
4
120
120
Section B-B
Loaded end
Areas of stringers: A1 = A3 = A4 = A6 = 150 mm2 A2 = A5 = 50 mm2
6
Areas of stringers: A1 = A2 = A3 = A4 = A5 = A6 = 50 mm2
Ans: q41= q63=3.33 N/mm, q12= 0.473 N/mm, q65=-0.473 N/mm.
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 3
5. The box beam shown in Figure 5 is loaded by a vertical force of 800 N. Stringers 1, 3 and 5 have uniformly tapered areas, while stringers 2, 4 and 6 have constant areas along the span of the beam. Determine the shear stresses in the webs at section A-A, a distance of 540 mm from the loaded end. (All dimension in mm)
800 N A
A
540
2000 Figure 5
100 1
3
2
5
1
3
2
90
3 4
800 N
100
90
3
6
4
Section A-A
5
6
Loaded end
Areas of stringers: A1 = A3 = A5 = 120 mm2 A2 = A4 = A6 = 80 mm2
Areas of stringers: A1 = A2 = A3 = 80 mm2 A4 = A5 = A6 = 80 mm2
Ans: q41= q63=4.444 N/mm, q12= 1.111 N/mm, q65=-1.905 N/mm.
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 4
ME4212 Mechanics of Thin-Walled Structures Solutions to Tutorials Torsion of Thin-Walled Sections
z, = 4
Hence
-
226400
tI
=
18.87 MPa
0.0 12
z2 = 226400 = 37.73 MPa
0.006
Z3
=
226400 0.01
=
22.64 MPa
To calculate rate of mist. use
6
=
6.87~10-' radlm.
j-
4 ds 2 G 6 =R t
Using
.r
=
2nt
~ B I I
=
ZA(
=
2000 2 x 4 9 . 7 8 3 ~lo-' x 2 x lo-'
=
10.04 MPa
AH
=
TC,/I
=
2000 2 x 49.783 x lo-' x 3 x 1 o
=
6.696 MPa
Angle of twist,
-~
T L 4n2G
4
= ---
=
0.005376391 rad
tan 30'
= 97/x
x
=
168
97 sin 30' = -3 AB = 194 AB
rB(.
=
r~~
Z~~
=
~ U I=
=
Angle of twist,
2000 = 8.05 MPa 2 x 3 1 . 0 6 8 ~ 1 0 -x~4 x 1 0 - ~ 2000 = 5.36 MPa 2 x 3 1 . 0 6 8 ~lo-; x 6 x 1 0 - ~
4
=
2 x lo-; 2 4 x 28x lo9 x ( 3 1 . 0 6 8 ~ 1 0 - ~ ) ~
+ --4
B
1 (120) (103.92) = 6.235 x lo-; m2 2 T From r = 2Qt 2000 ' , 4 ~ = = 40.09 MPa 2xRx4x10-~ Q
=
-
Z,K
=
2000 2 x R~ 2 x lo-'
rB(,
=
2000 = 53.46 MPa 2 x a x3 10-3
Angle of twist,
'=
=
80.19 MPa
4 =
2 A
120
2000 x 2 4~28xl0~~(6.235~10~)~
C
120 cos 30 103.92 mm
Fr&,Locate A m A =
z
@
0 '
rads 2 80.2' I n 9 0
Mere
I
(I- )
I -- 06+60l =
as173
119 cirtulw ,
- IYY, + 6
c~*r&)6205*55)~ = 5.5X~dowm+
3T as2m s ,ts
%at3 tb* Jop-l3b t 3 = 3 OP
@
d
p uLb c & _-&
o
7r
t-r
w 'dosed
'-
'
ia~t
.a.
-'.
Iy45
-- + x b
N&
12
Tyy= fLz[a 5 +b) .
f4w
8 d
s.tavte8 e ws
% ,, ly
ibt
Lr*=O.
C
o
-tb web
walyci+i
%$H. Tf
lroea d 1 2 ,%,> be m'Geu6bGseie
%--&~3,
I = rcor (608) = r(cosw0ctm + ~ i ~ r tsine) 50
srCoSC3
ds, = rde
sa=z+r
r't
Y€
ME4212 Aircraft Structures Tutorial Solutions – Multicell Sections
Question 1. A
a a
B
C
2a
q1
a
q2
a
D a
E
2a
F
q3
a
a
3a
G
H
Thickness of all webs = t Area of all stringers = As Shear modulus of web = G Determine the rate of twist.
Cell ABED: 1
2 1 2 1 2
4
1
4
Cell BCFE: 1 2 2 1 4 1 4
6 6
2 2
2
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 1
Cell DEFHG: 1 3
2 1
8
6
3
2
Moment equilibrium: Ω
2
2
2
2
2
3
4
3
From 1 4
2
1a
From 2
6
2
4
2a
From 3
2
8
6
3a
4
2a
1a
3a
2a
5a
9 10
23
8 4a
18
10
2
15.8
1.253
9
19.8
1.202
4a 5a 6a
1.11
Substitute into □ 4 , 2
1.11
2 1.253
14.44
3 1.202
0.0692
0.0867
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 2
Question 2. Define the coordinates and points as shown. q1T and q2T are constant shear flows due to torsion, to be superposed with the shear flow due to bending. s1
D
C
q2T
θ
q1T E
O
s3 A
B
s2
cos
sin
27.62 sin
1 2
0.017
1 2
2 50
55.2
2
on AB
0.55
cos on DEA
0.552
on BC
At B,
on CD and at C,
55.2 ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 3
1 2
1 2
0.011
50
Rate of twist for the ith cell: 1 2 Ω
For cell ADE, 2
2
2
2 2
2
2 3
2 2
4 3
2
1
For cell ABCD, 1 4
2
2 4
2
2
3
1
2
4
1
2
Substituting values into □ 1 and □ 2 , we have 2
0.03273
2
0.005
0.01273 0.01833
0.2345 and 0.4604
Solving for q1T and q2T, 115.84 140.708
2.921 25.914
Here, 2
14 3
2 3
1 3
1.81
10 mm
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 4
Consider now moment equilibrium about O: 2
2
0
which reduces to 8
4 3
10
0
On substitution for the values, the equation becomes 20000
7853.98
828729.3
0
Substituting for q1T and q2T into the above, we have 7.72
3.86
10 10 rad/mm
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 5
Question 3. Ω
2 2
0.3927
8
Ω √3 2 2
Ω
√3 4
0.433
1
1
2 4 3
2.571
2
,
,
Carry over factors: 1 4
,
,
4
2
,
,
1 3
2 1 3
4
1 4
,
,
,
,
2
0.389
Let open section: 5
4
0 0
3
1
2 2 2
2
4
2
2 ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 6
2
2
2
0
2
2
2
Cell I
δ
Cell II 4
2.571
1 4
0.389
0.195
2 1 3
1 4 0.167
0
0
0.0487
0.042
0
0.019
0
0
0.014
0
0.005
0.214
Final q
3
0
2 C
Cell III
0.0035 0.1
0 0.181
Check for vertical equilibrium: 0.214
0.386
0.581
0.181
0.1
5
4
0.214
0.181
3
0.581
0.386
0.181
1
2
0.1
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 7
Take moments about midpoint of 15, 0.214
2
0.1 2
2
0.181 2
2 cos 60°
0.581 0.181 2
2
sin 60°
0.025
40
to the right of 15.
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 8
Question 4. 13.5 kN 0.5
0.5
0.5
200 mm
2
200 mm
3
4 z
0.4
0.5
1 I 0.5
II
150 mm
III
0.4
125 mm
5
6
7 0.5
1
y
0.5
But
0 400 75
13.5 2
6
10 mm
13500 75 400 13.5 10 30
3
30
4
30
By symmetry, q56= q34 , q67= q23 6
30 30
7
30 30
Assume unknowns q12, q23 and q34. 1 2 Ω
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 9
For cell I, 1 50 150
2
1 7500 1 7500
125 0.5
2
500
375
875
375
150 0.4 30 11250 _____ i
For cell II, 1
2
150 0.4
150 200 1 30000
200 0.5
2
375
30
375
150 0.4 800
30
1 375 30000
1550
375
_____ ii
For cell III, 1
2
2
150 200 1 30000 1 30000
800
200 0.5
150 0.5
300
30
1475
2250
375
150 0.4 375
30
_____ iii
Taking moments about the point 1, 150 100
13500 100
200 75
2 13500 13500
150 150
150
200 75 750
4500 300
600
150
2
450
22500 54000
150 300
450 300
600
150 500 0
300
300
450
0
13500
750
0
0 _____ iv
Combining (ii) & (iii), 375
1550
375
375
1925
1850
1475 2250
2250
375
0 _____ v
Combining (i) & (ii), 300 75
875
375
11250
375
1550
375
3500
1500
45000
375
1550
3875
3050
375
45000
0 _____ vi
375
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 10
From iv ,
360
4
Substitute into v , 137250 Substitute into vi , From vii ,
4 3425
1350000
9.786
350
18550
0 _____ vii 15875
0 _____ viii
392.143
Substitute into (viii), 1350000 173902.75
18550
15875 9.786
392.143
0
7575270.125 43.56 N/mm 34.135 N/mm 42.59 N/mm 30.96 N/mm 20.58 N/mm 4.14 N/mm
(Check vertical equilibrium)
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 11
ME4212 Aircraft Structures Tutorial Solutions – Tapered Beams & Varying Moments of Area
Question 1. At the larger cross‐section,
4 200 60 2.88
2000 800 60 2.88 10
Bending stresses
200
∆
800 mm
10 mm
33.33
33.33 MPa
6.67 kN
Stringer no.
Δz
Δy
Pz
Py
Px
1
-20
+80
+166.75
-667.0
-6670
2
-20
0
+166.75
0
-6670
3
0
+80
0
+667.0
+6670
4
0
0
0
0
+6670
333.5
Δ Δ Δ 6.67 800
Δ Δ
0
Δ Δ
10
∴ Shear force carried by webs: 2000
333.5
1666.5 N
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 1
1
2 z
6.94
y
120 mm
180 mm 3
4
6.94 13.89
Take moments about 2: 180 120
180 180 120
166.75 120
666.7 180
166.75 180
667 120
0
0 2.315
4.63 N/mm
2.31 N/mm
2.31 N/mm
9.26 N/mm
Check vertical equilibrium: 120
1666.8 N
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 2
Question 2. 4 200 50
At larger cross‐section, 2
10 mm
10 mm
3000 800 50 2.88 10
Bending stresses
2
60 MPa
12000 N
∆
800 mm
Δ Δ
and
Δ Δ
Stringer no.
Δz
Δy
Pz
Py
1
-30
+40
-450
+600
2
-30
-40
-450
-600
3
+10
+40
-150
-600
4
+10
-40
-150
+600
1200 N
0
∴ Shear force carried by webs: 3000 1
1200
1800
2
9 3
4
18
Take moments about 4: 100 120
120 100 600 100
450 120 600 100
150 120 3000 40
16 2
7
3.5 N/mm
12.5 N/mm
12.5 N/mm
21.5 N/mm ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 3
Question 3. At larger cross‐section,
26
2000 600 100 26 10
Bending stresses
10 mm
4.615 MPa
Pzn
z Δ Δ
Pxn
Δz
Δ Δ
Δx x Pzn
z
Δ Δ Δ Δ
Pyn
Δz Δy
y
Stringer no.
Pxn
Δz
Δy
Pzn
Pyn
1
-2307.5
-50
100
192.3
-384.58
2
-1384.5
-50
50
115.375
-115.375
3
-2307.5
-50
0
192.3
0
4
2307.5
50
100
192.3
384.58
5
1384.5
50
50
115.375
115.375
6
2307.5
50
0
192.3
0
ΣPzn=1000
ΣPyn=0
Shear forces carried by webs: 0 1000 N
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 4
1
4
2
3
5
6
1.154 1.154
3.077
1.154
_23
Σ moments about 4 = 0 150 200
150 200
200 300
115.375 200
115.375 150
115.375 150
192.3 300
2
384.58 200 192.3 300 0
8.332
Substitute for q23 & q36 and solving for q12, 2
1.154
2
2 3.077 4
8.332 15.64 3.91 N/mm 2.756 N/mm 0.833 N/mm 5.83 N/mm
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 5
Question 4. At section B‐B:
4 150 60
Forces in streingers:
600 800 60 2.52 10
1714.3 N
571.43 N
1
2
26
10 mm
Top bending stress
2 50 60
11.43 MPa
1714.3 N 571.43 N 3
Vertical equilibrium: 2
120
800 N
1 3 N/mm 3 Δ 4
6
5
1714.3 3.33 600 1714.3 3.33 600
0.473 N/mm
1
2
0.473 N/mm 0.473 N/mm
0.473 N/mm
3
3.33 N/mm
3.33 N/mm
4 0.473 N/mm
5 0.473 N/mm
6
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 6
Question 5. At section A-A: 3 80
120
1
600 mm 2
1
1 90 80 600
+ C 1.21
48
42 mm
42
240
80
48
120
160
10 mm
1.2096
10 mm
6
5
Forces in stringers:
800 540 42 1.2096 10
Top, 1800
1371.44 2
17.143
2057.16 3
15 1800
1200
800 540 48 1.2096 10
Bottom,
1
120
3 42
4
80
1371.44
Vertical equilibrium: 2
90
800 4.444
4
6
5 1800 540
1.111 N/mm
1
4.444
2
1.111 N/mm
3
1.111 N/mm 1371.44 540
4.444 N/mm
4.444
4.444 N/mm
1.905 N/mm 4
1.905 N/mm 5
1.905 N/mm
6
ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 7