Strength of Materials
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Transcript of Strength of Materials
1. Structures, loads and stresses
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Structures
This course is concerned with structures:
A structure is a solid object or assembly. A structure connects components, carries loads, provides form and integrity.
Structures
Wright Flyer, 1903
F22, 2002
Bridges
Trusses
Tall Structures
Burj al Arab Qutab
Roofs
Concept of a Rigid Body
• Distance between any two points does not change when a force is applied on it.
● ●
Resistance Forces and moments
Resistance Forces and moments
An upward reaction force at support
But an unbalanced moment results.Thus, equilibrium requires a force and a moment at the support. Where does it come from?
Load and deflection
Bar
As force increases, elongation increases till the equilibrium is restored again.
This implies that there is a force resisting the deformation, and that force increases with deformation.
Resisting Force
Hooke’s law
δP
Area A
lP
P
Stress, σ = P/AStrain, ε = δ/L
E is the elastic modulus, or simply the Elasticity
Robert Hooke in 1678 showed
Units of stress
Dimensions of stress: F/Area = F/L2
Units of stress = N/m2 = (Pa)scal,same as that of pressure.A very small unit.Standard atmospheric pressure = 1.03×105 Pa
MPa and GPa (106 Pa and 109 Pa, respectively) are commonly used
Units of strain
Strain δ/L is dimensionless, hence NO UNITS.
Units of elasticity
E = Stress / strain, and therefore, has dimensions of stress, i.e., F/L2.
Units of E are, accordingly, Pa(scal).
Elasticity
Material Value of E in GPaAluminium 2024-T3 70Aluminium 6061-T6 70Aluminium 7075-T6 70Concrete 20 – 35Copper 100Glass fibre 65Cast iron 100Steel, High strength 200Steel, Structural 200Titanium 100Wood 10-15
To summarize:
The constant of proportionality is termed as the modulus of elasticity.
The external forces acting on a structure result in strains.
The strains so produced result in stresses within the material of the members.
The stresses, for the most part, are proportional to the strains.
Stresses due to various loads
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
A bar as a linear spring
Bar
F
δL
which gives
Spring constant is defined as force required to produce a unit deflection. i.e., . Its units are N/m
Structures in Tension
Tension in the belt
Stress Distribution
P P PP
Tensile Stress in a Bar
Stress:
Force Intensity
σ = F/A = 300N
20×5×10− 6 m2
= 3 MPa
20 mm5 mm
300 N
Stress...
Uniform stress is an approximation.Valid only in simple loadings.Away from ends.
Non-Uniform Stresses...
Non-Uniform Stresses
Compression loads
Structures Under Compression
Application: Foundations
Concrete
Steel
Soil
105 N Bearing strengthsSteel >> concrete >> soil
Required areasSteel << concrete << soil
Application: Foundations
Concrete
Steel
Soil
105 NPermissible compressive stress in steel is about 400 MPa.
So the area of steel required is 105 N/400 MPa, or 2.5 × 10−4, or about 16 mm×16 mm
Application: Foundations
Concrete
Steel
Soil
105 NPermissible compressive stress concrete is about 60 MPa.
So the area of concrete required is 105 N/ 60 MPa, or 1.67 × 10−3, or about 41 mm×41 mm
Application: Foundations
Concrete
Steel
Soil
105 NPermissible bearing strength of soils varies widely. For good cohesive soil, it could be between 100 to 400 kPa, if it is above the water table.
So the area of the footing required is 105 N/ 200 kPa (say), or 0.5 m2, or about 710 mm×710 mm
Shear Members
Shear Stresses
Shear Action: Rivets
Shear Stresses in a Pin
Shear
Bearing (Compressive)
Rod & Collar
Shear loadBearing load
Blanking force
Blanking force = shear strength × shear area
Shear area = perimeter × sheet thickness
Compression in Riveted Joints
Shear
Compression
Compression
Shear
Belt driven pulley
Weldment and shear loads
P P
P
Shear
Shear area
Twisting of shaft
The stresses result in a moment that balances the twisting moment
Shear stresses on the back face of the shaft
Bending of Beams
Compression near topExtension near bottom
Net tensile force is zero!
Bending of Beams
To summarize…
Forces that tend to distort the shape of a member produce shear strains which in turn produce shear stresses.
• Forces that tend to reduce the size of a structural member produce compressive strains which, in turn, produce compressive stresses.
To summarize…
A bending moment produces both tensile and compressive strains and stresses. These give rise to a resisting moment which balances the bending moment.
• A twisting moment applied to a shaft produces shear strains. These shear strains give rise to shear stresses which result in a moment that balances the external twisting moment.
Tensile members and trusses
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Segmented Axial Bar
10 kN10 kN10 kN
10 kN
10 kN10 kN
10 kN10 kN10 kN
Two-Force Members
A member on which external forces act only at two distinct points (and there is no external torque acting on it) is termed as a two-force member. The forces acting on a two-force member
are equal and opposed. But is it enough?
Two-Force Members
Two-Force Members
The forces acting on a two-force member are equal, opposed and collinear.
Two-Bar Assembly
30o60o
500 N
F1cos30o – F2cos60o = 0
F1sin30o + F2sin60o – 500 = 0
F1 = 125 N; F2 = 433 N
Two-force members
Restraints or SupportsSimply supported:
No moment. Only reaction force.
• Cannot translate,
• Can rotate.
Simply Supported
Simply Supported...
Roller Support: Normal reaction only. (2 DoF)
Pinned Support: Reaction could be inclined. (1 DoF)
Load
Step Ladder
FRICTION
must act
Leaning Ladder
1 m
4 m
600 N
RV1
RH1
RH2
Or, N
zx
y
RH2X4 – 600X0.5 = 0
or RH2 = 75 N
Clamped Support
While in the pinned support, the member is restrained from translating, in the clamped support, the member cannot even rotate.
Clamped Support...
x
xx
• Cannot translate,• Cannot rotate.
No DoF at all.
Also called built-in support
P
Fx
Mz
Fy
zx
y Statically determinate
Clamped Support...
L
Frames and trusses
Frames and trusses
Resisting moment Tension will build up faster than the moments due to bending, and therefore, can treat the joint as pinned
A truss
Two-force members
Truss
3m
θ
cos𝜃 = 4/5,sin𝜃 = 3/5
Method of Joints
FGFFGA
FGBFGC
15 kN
Symmetry:
2FGC sinθ – 15 kN = 0
FGC = (15 X 5)/(2 X 3)
= 12.5 kN
Method of Joints
-FCGsinθ - FCFsin θ = 0
FCF = - FCG = -12.5 kN
FCG
FCB
FCF
- FCB – FCGcosθ + FCFcosθ = 0
FCB = –2FCGcosθ = 10 kN
Method of Joints
FFG
FFCRB,V
∑FV = RB,V +(3/5)FFC = 0
or, RB,V = - (3/5)FFC = + 7.5 kN
∑FH = -(4/5)FFC - FFG = 0
or, FFG = - (4 /5)FFC = 10 kN
An Example: Pinned Truss
B
C D
20 kN
An Example: Pinned Truss...
B
D
FBD,D
B
C D
20 kN
A two-force member
An Example: Pinned Truss...
20 kN
DF1
F2
B
C D
20 kN
∑Fy = 0
Another Example of Two-Force Members
A Three-Force Member
• Co-planar• Concurrent
Notation for stresses
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Double-index Notation
y
z
x
First index: normal to the plane on which acting.
Second index: direction of the stress component itself
Stress at a Point
StressIntensity of
Force
Stress at a Point
The stress vector t depends upon the location as well as the direction of the surface.
�̂�
�̂�
Stresses: Sign Convention
y
z
x
The sign of a stress component depends on the direction of normal and the direction of force: If both have same sign then the stress component is positive, if the two have different signs, then the stress component is negative.
Stress: Sign Convention...
y
z
x
σyy is negative
τyx is positive
τyz is negative
Stress: Sign Convention...
y
z
x
σxx is positive
τxy is negative
τxz is negative
Equivalence of shear on adjacent planes
Consider the moment balance about the mid-point:
Face
(direction of outward
normal)
Area
(assume unit
depth)
Forcex-
component
y-compon
ent
x δy•1 σxx• δy τxy• δy− x δy•1 −σxx•
δy−τxy•δy
y δx•1 σyy• δx τyx• δx− y δx•1 −σyy•
δx−τyx•δx
or, =
Stresses in thin cylinder
Thin-walled cylinders are used extensively in industry and homes because they are very efficient structures.• Oil storage tanks are cylindrical• So are oxygen bottles, cooking gas cylinders• Deodorant bottles are pressurized cylinders.• So are beer cans.
2.2 THIN-WALLED PRESSURE VESSELS
Cylindrical and spherical pressure vessels are commonly used for storing gas and liquids under pressure.
A thin cylinder is normally defined as one in which the thickness of the metal is less than 1/20 of the diameter of the cylinder.
THIN-WALLED PRESSURE VESSELS CONTD
In thin cylinders, it can be assumed that the variation of stress within the metal is negligible, and that the mean diameter, Dm is approximately equal to
the internal diameter, D. At mid-length, the walls are subjected
to hoop or circumferential stress, and a longitudinal stress, .
Hoop and Longitudinal Stress
2.2.1 Hoop stress in thin cylindrical shell
Hoop stress in thin cylindrical shell Contd.
The internal pressure, p tends to increase the diameter of the cylinder and this produces a hoop or circumferential stress (tensile).
If the stress becomes excessive, failure in the form of a longitudinal burst would occur.
Hoop stress in thin cylindrical shell Concluded
Consider the half cylinder shown. Force due to internal pressure, p is balanced by the
force due to hoop stress, h.
i.e. hoop stress x area = pressure x projected area
h x 2 L t = P x d L
h = (P d) / 2 t
Where: d is the internal diameter of cylinder; t is the thickness of wall of cylinder.
2.2.2. Longitudinal stress in thin cylindrical shell
Longitudinal stress in thin cylindrical shell Contd.
The internal pressure, P also produces a tensile stress in
longitudinal direction as shown above.
Force by P acting on an area d2
4 is balanced by
longitudinal stress, L acting over an approximate area,
dt (mean diameter should strictly be used). That is:
L
L
x d t Pxd
Pd
t
2
4
4
Note
1. Since hoop stress is twice longitudinal stress, the cylinder would fail by tearing along a line parallel to the axis, rather than on a section perpendicular to the axis.
The equation for hoop stress is therefore used to determine the cylinder thickness.
Allowance is made for this by dividing the thickness obtained in hoop stress equation by efficiency (i.e. tearing and shearing efficiency) of the joint.
Stresses in thin cylinderTake section of a pressurized cylinder
pσθθ
σθθ
FBD of the lower half
And the upper half
Thin Cylinder
FR
p
FT
FR
FT
We can show by symmetry arguments that:(a) Both shear should be inwards or outwards
(b) Shear should be ZERO
Thin Cylinder
p
FT
Net forced on the curved surface = p×2r×δl
Equilibrium: FT = σ 2δl t = p×2r×δl
This gives: Hoop stress
FBD of the ‘contents’
Thin Cylinder
Axial stresses are lone-half of hoop stresses
Forces on the rimPressure on the back capσ
Thin spherical shell
p
Forces on the rimPressure on the ‘content’
𝜎
Maximum stress in a spherical vessel is one half that of a cylindrical vessel of same radius and thickness
That is why gas storage tanks are spherical
Shaped structures
Shaped structures
Arch
Keystone
All stones are subjected to compressive forces only.
Suspension bridge
Deck of bridge
Towers
Load bearing cables
Cables support the bridge through tension.Towers carry compression,
Golden Gate Bridge
The main span of the Golden Gate suspension bridge is 1.287 km long. The sag in the cables is 140 m. The design loading is 400 kN/m.
Golden Gate Bridge
Tension in the cable at the lowest point is;To = 2.96×108 NMax tension = 3.23×108 N
Each cable consists of 27,572 strands of 4.88-mm diameter wires bundled parallel.Cross-sectional area of the
cable = 27,572×[π×0.004882/4] = 0.516 m2
So stress = 625.5 MPa
Recap: Sign Convention of stress
y
z
x
Recap: Equivalence of shear on adjacent planes
Recap: Stresses in thin cylinder
Take section of a pressurized cylinder
pσθθ
σθθ
FBD of the lower half
And the upper half
Recap: Thin Cylinder
FR
p
FT
FR
FT
We can show by symmetry arguments that:(a) Both shear should be inwards or outwards
(b) Shear should be ZERO
Recap: Thin Cylinder
Recap: thin spherical shell
p
𝜎
That is why gas storage tanks are spherical
2. Deformations, strains and material properties
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Hooke Law
δP
Stress = E×Strain
P
L
/A
/L
The Fundamental Strategy of Deformable-Body Mechanics
Deformation depends on loading, material and geometry
Strain depends on stress AND material. NOT on geometry
Stress depends on loading and geometry
The Fundamental Strategy of Deformable-Body Mechanics
Load StressEquilibrium
Geometry
Strain
Material Property
DeformationGeometry
Macro Micro
Micro
Macro
Tug of War
Cross-section: 6 cm2
Stress, kPa
416.7833.3
1333.3
916.7500.0
Strain
4.16×10- 3
8.35×10- 3
13.33×10- 3
9.20×10- 3
5.00×10- 3
Length, m
1.52.01.5
1.52.0
Elongation, m
6.24×10- 3
16.70×10- 3
20.02×10- 3
13.78×10- 3
10.00×10- 3
Section
ABBCCD
DEEF
Tension, N
250500800
550300
Measuring the height of Kutub
72 m
W = ρAxg = T
σ = T/A = ρxg
ε = σ/E = ρxg/E
dδ = εdx = ρgxdx/E
x
dx
W(x)
T
Measuring the height of Kutub
For steel: density is 7.6X103 kg/m3, and E is 200 GPa. We get δ ~ 1 mm
or
For a Nylon wire: density, ρ ~ 0.8X103 kg/m3, and E ~ 400 MPa. We get δ ~ 52 mm
Deflection in a truss
C
20kN
A
B
RA,x
RA,y
RB,y
RA,y = 20 kNRA,x = − 20 kNRB,y = 20 kN
TAC = 28.8 kNTBC = − 20 kN
Member
Force
kN
Lengthm
Aream2
StressMPa
Strain Elongationm
AC 28.3 1.41 1.77×10−4 160.1 7.6×10−4 1.07×10−3BC − 20 1 1.77×10−4 −113.2 −5.4×10−4 −0.54×10−3
X-displacement of C ~ shortening of BC =−0.54 mmy-displacement of C ~ EF + FC1 = CD/cos45o + FG(=EC)~ 1.25 mm
Deflection in a truss
C1
E
D
A
B45o
CED
C1
F G45o
A
B45o
C
x
y
Statically-indeterminate structures
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Statically indeterminate problems
P
P
R1 R2R3
Reaction at middle support (and hence, at all supports depends on the bending of plank.
P
Statically indeterminate problems
1. Consideration of static equilibrium and determination of loads
2.Consideration of relations between loads and deformations, (first converting loads to stresses, then transforming stresses to strain using the properties of the material, and then converting strains to deformations),
3. Considerations of the conditions of geometric compatibility
A simple example
1.3 m
150 kN1 m
F1F2
2.6 m
Indeterminate because x is not known!
Statically indeterminate structure
P
R1
R2
2R1 + 2R2 – P = 0
Geom. Comp.
δ1 = δ2
R1L1/E1A1 = R2L2/E2A2
Taking moments about the pivot point,
Indeterminate because 4 unknown forces and only three equations to determine them.
Statically indeterminate structure
R1 = kδ1
R2 = kδ2
x
R2 - F - R1 = 0; R1L – Fx = 0
h + δ1 = 2(h - δ2)
Geom. Comp.
h
L
h
F
P
Statically Indeterminate Structure
P = R1+ R2
R1 = (E1A1/L1)δ1
R2 = (E2A2/L2)δ2
Geom. Comp.
δ1 = δ2
R1
R2
Pre-stressing of concrete
(a) Tendon being stressed during casting. Tension in tendon, no stress in concrete.
(b) After casting, the force is released and the structure shrinks.
(c) FBD of tendon. The concrete does not let the tendon shrink as much as it would on its own. This results in residual tension in the tendon.
(d) FBD of concrete. The residual force in the tendon is trying to compress the concrete..
Pre-stressing: A simple example
A concrete beam of cross-sectional area 5 cm×5 cm and length 2 m be cast with a 10 mm dia mild steel rod under a tension of 20 kN. The external tension in steel released after the concrete is set. What is the residual compressive stress in the concrete?
Calculate the extension of steel under the tension of 20 kN
T = 20 kN →σ = 255 Mpa →ε = 1.21×10- 3 →δ = 2.42 mm
Pre-stressing: A simple example
2.42 mm
δs
δs + δc = 2.42 mm
δc
Lateral strains:Poisson ratio, ν
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Stress-strain relationship
σxx
σxx
εxx = σxx/E
εyy = - ν εxxν is Poisson ratio
Another material property
Stress-strain relationship
Let us consider εxx.
σxx produces an εxx = σxx /Eσyy produces an εyy =
σyy /E, which through Poisson ratio gives εxx = -νεyy = - νσyy/E.
Similarly for σzz .
Generalized Hooke’s Law
Shear stresses do not cause any normal strain
Therefore,
εxx = σxx/E – νσyy/E - ν σzz/E
= [σxx – ν(σyy + σzz)]/E
Similarly for εyy and εzz
An example
F σyy = −F/A, σzz = 0
What is σxx and εyy
Geometric compatibility:εxx = 0x
y
εxx = [σxx – ν(σyy + σzz)]/E
0 = [σxx – ν(σyy + 0)]/E, → σxx =νσyy = − ν F/A
εyy = [σyy – ν(σxx + σzz)]/E = [−F/A + ν F/A]/E = −(1− ν)F/AE
Another Example
σxx
σyySteel: εx =
0.6×10−4 εy = 0.3×10−4Find σxx and σyy :
E = 200 GPa, ν = 0.3
Plug in:εxx = [σxx – ν(σyy + σzz)]/E εyy = [σyy – ν(σzz + σxx)]/E
Shear strains and stresses
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Shear Strain
θθ2 θ1
Shear strain γ is π/2 − θShear strain is also seen as: θ1 + θ2Since angles are measured positive counter-clockwise, the angle θ2 above is a negative angle. In general terms, then, γ = θ1 − θ2with θs measured positive when counter-clockwise
Apply shear stresses to a block:
Shear Strain
x
y
DC
B
A
Coordinates after deformation (in mm) are:
A(0,0), B(0.194, 0.013), and D(−0.012, 0.196).
θ2 θ1 θ1 = 0.013/0.2 = 0. 065θ2 = 0.012/0.2 = 0. 06γxy = 0.65 − 0.60 = 0.05 radians
A square blocks 0.2 mm × 0.2 mm deforms under shear stress
Shear Stress
γxy = τxy/G,
where G is shear modulus
It can be shown that γxy does not depend on other components of stress.
Shear strain γ is related to shear stress τ by
Shear Modulus
Material G, GPaAluminium 25
Steel 80
Glass 26-32
Soft Rubber 0.003- 0.001
Vibration Isolator
Wall
8,000 N
WallRubber blocks
10 cm × 10 cmwith 12 cm height
Shear stress τ =4,000 N/ (0.1 m)(0.12 m) = 3.33×105 PaShear strain γ = τ/G3.33×105 Pa/1 MPa= 0.33
4,000 N
Vibration Isolator
Wall
8,000 N
WallRubber blocks
10 cm × 10 cmwith 12 cm heightAnd therefore,The vertical deflection of load = 0.33×0.10 m = 33 mm
γ =0.33Consider the rubber block on the left:
Elastic Properties
We have so far introduced three elastic properties of materials.
Material E, GPa G, GPa νAluminium 70 25 0.33
Steel 200 80 0.27
Glass 50-80 26-32 0.21-.27
Soft Rubber 0.0008-0.004
0.003- 0.001
0.50
Thermal strains and stresses
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
On heating, there is linear expansion:
Thermal Strains
There is no thermal shear strainMaterial α (×10-6/oC)
Steel ~ 10
Aluminium ~ 20
Generalized Hooke’s Law
εxx = [σxx – ν(σyy + σzz)]/E + αΔTεyy = [σyy – ν(σzz + σxx)]/E+ αΔTεzz = [σzz – ν(σxx + σyy)]/E+ αΔTγxy = τxy/G, γyz = τyz/G, and γzx = τzx/G
Putting Hooke law, Poisson effect and thermal strains all together,
An example
Aluminium rod, rigid supports.
Temperature raised by ΔT.What are the stresses?
εxx = 0 = [σxx/E + αΔT]x
σxx = −αEΔT
Another example
p
Tank is flush when empty.Find end forces when pressure is p
zDue to p: σzz = pr/ 2t, σθθ = pr/ t If end forces F, axial stress due to it is F/2πrtεzz = [(pr/ 2t − F/2πrt) −νpr/t ]/E
Equate it to 0 and determine F
Determining stress-strain relations
Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Stress-Strain Relationship A material property.
Tensile Test Machine, UTM
Extensometer
Stress-Strain Curve: Elasticity
Failure Modes
Ductile Failurecup-and-cone Brittle Failure
σ (=
F/A
o)
ε (=∆L/Lo)
Ductile
Brittle
Necking
Plastic Deformation, Yield Strength
ε (= ΔL/L0)
σ (
= F
/A0)
0.02% Permanent set
YYield stress, σY
Strain Hardening
ε (= ΔL/L0)
σ (
= F
/A0)
YY1
B
Ultimate stress
Stress-Strain in Brittle Materials
Idealized Stress-Strain Curves
ε
σ
(a) Rigidε
σ
(b) Perfectly elasticε
σ
(c) Elastic-Plastic
ε
σ
(d) Perfectly plasticε
σ
(e) Elastic- Plastic (strain hardening)
Increase in yield strength
Power shaft
= 9.82×10-6 m4= 𝐼 𝑧𝑧=𝜋 𝑅4
2
= (− 4.5 kNm)(1.5m)/(80 Gpa)(9.82×10−6m4)=0.009 rad = (− 2.25 kNm)(3m)/(80 Gpa)(9.82×10−10m4) =0.009 rad = += 0.018 rad
Power shaft
τ θ z , max=𝑅𝑇𝐼 𝑧𝑧
Let us check on the stresses: (4.5 kNm)/(9.82×10-6m4)=22.9 MPaQuite safe
1 m 0.6 m
Φ 10 cm
Φ 6 cm
150 N.m
Φ 5 cm
Φ 2 cm
Another example
F
F
150 N.m
−250 N.m
𝜏max, first shaft = 𝜏max, second shaft
1 m 0.6 m
Φ 10 cm
Φ 6 cm
150 N.m
Φ 5 cm
Φ 2 cm
Another example
F
F
1 m 0.6 m
Φ 10 cm
Φ 6 cm
150 N.m
Φ 5 cm
Φ 2 cm
Another example
F
F
𝜑 𝑓𝑖𝑟𝑠𝑡 h𝑠 𝑎𝑓𝑡=0.0051rad
𝜑 𝑠𝑒𝑐𝑜𝑛𝑑 h𝑠 𝑎𝑓𝑡=0.12 rad
Angle θ2 which represents the counter-clockwise movement of the smaller gear due to gearing alone is 10/6 of θ1 or 0.0085 rad counter-clockwiseRotation of the right end of second shaft wrt stationary wall is, therefore, 0.0085 rad + 0.12 rad = 0.1285 rad or 7.36 degree.
Hollow Shaft
By geometry: γθz= rdΦ/dzTherefore, τθz = GrdΦ/dz
r varies from R1 to R2, and θ varies from 0 to 2π
Hollow shaft vs. solid shaft
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.20.40.60.8
11.21.41.61.8
2
𝑅𝑖 /𝑅𝑜
Weight
Hollow shaft vs. solid shaft
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.20.40.60.8
11.21.41.61.8
2
𝑅𝑖 /𝑅𝑜
Weight
Stiffness
Hollow shaft vs. solid shaft
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.20.40.60.8
11.21.41.61.8
2
𝑅𝑖 /𝑅𝑜
Weight
Stiffness
Strength to weight
A statically indeterminate case
MoM1 M2
Geometric Condition: Φ1 +Φ2 = 0TMD
Equilibrium Condition: - M1 + Mo – M2 = 0
A Composite Shaft
𝑇=∫𝐴
❑
G 𝑟 2 𝑑𝜑𝑑𝑧
𝑑𝐴=¿ 𝑑𝜑𝑑𝑧 ∫𝐴
❑
𝐺𝑟 2𝑑𝐴¿
𝛾𝜃 𝑧(𝑟 )=𝑟𝑑𝜑𝑑𝑧
𝑇=𝑑𝜑𝑑𝑧 (𝐺1 𝐼 1+𝐺2 𝐼 2 )
Thin-walled shaftShear flow
q1 = q2
Shear flow on horizontal surfaces is same as on the vertical surfaces
Thin-walled shaft
Relating q to twisting moment T
q = T/2AdT at O = qds×h
qds
oh
= q×2×Grey area
An Example
R 20 mm
R 16 mm
T 100 Nm
= 13.4 MPa at R =20 mm, and10.8 MPa at R =16 mm
𝑞=𝑇
2𝐴=
100 Nm2 ×1.02×10 −3 m 2
=49 k N / m
This gives τ = (49 N/m)/0.004 m = 12.25 MPa