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

http://www.aeroflight.co.uk/types/usa/wright/flyer/wright_flight.jpg

http://www.aeroflight.co.uk/types/usa/wright/flyer/wright_flight.jpg

Bridges

Trusses

Tall Structures

Burj al Arab Qutab

http://www.sights-and-culture.com/India-Delhi/Delhi-Qutb-Minar-2.jpg

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


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


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


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


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


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


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


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

http://www.sights-and-culture.com/India-Delhi/Delhi-Qutb-Minar-2.jpg

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


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


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


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


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


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


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


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

Education

Strength of Materials