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1. Structures, loads and
stresses

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
This course is concerned with structures:
A structure is a solid object or assembly. A
structure connects components, carr...
F22, 2002

Wright Flyer, 1903
Burj al Arab

Qutab
• Distance between any two points does not
change when a force is applied on it.
An upward reaction force at support

Thus, an unbalanced moment results. a moment at
But equilibrium requires a force and
...
Bar

Resisting Force

As force increases, elongation increases till the
equilibrium is restored again.
This implies that t...
Stress, ζ = P/A
Strain, ε = δ/L
Area A
P

Robert Hooke in 1678 showed

l
E is the elastic modulus, or
simply the Elasticit...


Dimensions of stress: F/Area = F/L2
Units of stress = N/m2 = (Pa)scal,
same as that of pressure.
A very small unit.
Sta...


Strain δ/L is dimensionless, hence NO UNITS.



E = Stress / strain, and therefore, has
dimensions of stress, i.e., F/L2.
Units of E are, accordingly, Pa(scal).
Material
Aluminium 2024-T3
Aluminium 6061-T6
Aluminium 7075-T6
Concrete
Copper
Glass fibre
Cast iron
Steel, High strength
...
The external forces acting on a
structure result in strains.
The strains so produced result in stresses
within the materia...
Stresses due to various loads

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Bar
F
L

δ
Tension in the belt
P

P

P

P
Stress:
Force Intensity
20 mm

5 mm

ζ = F/A =

300N

20×5×10− 6 m2
= 3 MPa

300 N
Uniform stress is an approximation.
Valid only in simple loadings.
Away from ends.
105

Steel

Concrete
Soil

N

Bearing strengths
Steel >> concrete >> soil
Required areas
Steel << concrete << soil
105 N

Permissible compressive
stress in steel is about 400
MPa.
Steel

Concrete
Soil

So the area of steel
required is 10...
105 N

Permissible compressive
stress concrete is about 60
MPa.
Steel

Concrete
Soil

So the area of concrete
required is ...
Permissible bearing strength
105 N of soils varies widely. For
good cohesive soil, it could be
between 100 to 400 kPa, if ...
Shear Stresses
Bearing (Compressive)

Shear
Bearing load

Shear load
Blanking force = shear strength
Shear area = perimeter

shear area

sheet thickness
Compression
Shear

Compression

Shear
P

P

P
Shear

Shear
area
Shear stresses on
the back face of the
shaft
The stresses result in a moment that
balances the twisting moment
Compression near top
Extension near bottom
Bending of Beams
Net tensile force is zero!
• Forces that tend to reduce the size of a structural
member produce compressive strains which, in turn,
produce compressi...
• A twisting moment applied to a shaft
produces shear strains. These shear strains
give rise to shear stresses which resul...
Tensile members and trusses

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
10 kN 10 kN

10 kN
10 kN

10 kN

10 kN
10 kN

10 kN 10 kN
A member on which
external forces act
only at two distinct
points (and there is no
external torque acting
on it) is termed...
The forces acting on a two-force member are equal, opposed
and collinear.
Two-force
members

60o

30o
F1cos30o – F2cos60o = 0
500 N

F1sin30o + F2sin60o – 500 = 0
F1 = 125 N; F2 = 433 N
Simply supported:

No moment. Only
reaction force.

• Cannot translate,
• Can rotate.
Load

Pinned
Support:
Reaction could
be inclined. (1
DoF)

Roller
Support:
Normal
reaction only.
(2 DoF)
FRICTION
must act
RH2X4 – 600X0.5 = 0
or RH2 = 75 N

RH2

4m

Or,

N

RV1
600 N
RH1
1m

y
x
z
While in the pinned support,
the member is restrained
from translating, in the
clamped support, the
member cannot even rot...
• Cannot translate,
• Cannot rotate.
No DoF at all.
Also called built-in support
P
Fy
Fx

Mz

L
y

Statically determinate
x

z
Resisting moment

Tension will build up faster than the
moments due to bending, and
therefore, can treat the joint as pinn...
Two-force members
3m

θ

cos = 4/5,
sin = 3/5
Method of Joints
Symmetry:
FGB

FGC
FGF

FGA
15 kN

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

FCB

FCG

FCF
- FCB – FCGcosθ + FCFcosθ = 0
FCB = –2FCGc...
FFC
∑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

FF...
B

D

C
20 kN
B
B
A two-force member

D
D

C
20 kN

FBD,D
F2
B

F1

D
20 kN

C

D

20 kN

∑Fy = 0
• Co-planar
• Concurrent
Notation for stresses

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
First index: normal to the
plane on which acting.

z
x

Second index: direction of

y

the stress component
itself
Stress

Intensity of
Force
The stress vector t depends upon the location as
well as the direction of the surface.
z
x

y

The sign of a stress
component depends on
the direction of normal
and the direction of force:
If both have same si...
ζyy is negative
ηyx is positive
ηyz is negative

z
x

y
ζxx is positive
ηxy is negative
ηxz is negative

z
x

y
Face
Area
Force
xy(direction of (assume
outward
unit depth) compone compone
nt
nt
normal)
x
δy•1
σxx• δy
τxy• δy
−x
δy•1
−...
Thin-walled cylinders are used extensively in
industry and homes because they are very
efficient structures.
• Oil storage...




Cylindrical and spherical pressure vessels
are commonly used for storing gas and
liquids under pressure.
A thin cyli...




In thin cylinders, it can be assumed that
the variation of stress within the metal
is negligible, and that the mean
...




The internal pressure, p tends to increase
the diameter of the cylinder and this
produces a hoop or circumferential ...
C o n sid e r th e h a lf cylin d e r sh o w n . F o rce d u e to in te rn a l p re ssu re , p is b a la n ce d b y th e
f...
T h e in te rn a l p re s s u re , P a ls o p ro d u c e s a te n s ile s tre s s in
lo n g itu d in a l d ire c tio n a s...
1. Since hoop stress is twice longitudinal
stress, the cylinder would fail by tearing along a
line parallel to the axis, r...
Take section of a pressurized cylinder

And the
upper half

FBD of the lower half
σθθ

p

σθθ
FT
FR
p

We can show by symmetry arguments that:
(a) Both shear should be inwards or outwards
(b) Shear should be ZERO
FT

FBD of the ‘contents’

p

Net forced on the curved surface = p×2r×δl

Equilibrium: FT = ζ 2δl t = p×2r×δl
This gives:
...
Forces on the rim
Pressure on
ζ
the back cap

Axial stresses are lone-half
of hoop stresses
Forces on the rim
Pressure on
the ‘content’

p

Maximum stress in a spherical vessel is
one half that of a cylindrical ves...
Shaped structures
Arch

Keystone

All stones are
subjected to
compressive forces
only.
Towers
Load bearing cables

Deck of bridge

Cables support the bridge through tension.
Towers carry compression,
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...
Tension in the cable at the lowest point is;
To = 2.96×108 N
Max tension = 3.23×108 N
Each cable consists of 27,572
strand...
z
x

y
Take section of a pressurized cylinder

And the
upper half

FBD of the lower half
σθθ

p

σθθ
FT
FR
p

We can show by symmetry arguments that:
(a) Both shear should be inwards or outwards
(b) Shear should be ZERO
p
2. Deformations, strains
and material properties

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
P/A
L
δ/L
Stress = E×Strain
P
Deformation depends on
loading, material and
geometry
Strain depends on stress
AND material. NOT on
geometry

Stress
depen...
Load

Equilibrium

Stress

Macro

Geometry

Micro

Strain
Micro

Geometry

Material
Property
Deformation
Macro
Cross-section: 6 cm2

Section Tension, N Stress, kPa

Strain

Length, m Elongation, m

AB
BC
CD

250
500
800

416.7 4.16×1...
W = ρAxg = T
ζ = T/A = ρxg

72 m

ε = ζ/E = ρxg/E
dδ = εdx = ρgxdx/E
T

dx
x
W(x)
or
For a Nylon wire: density, ρ ~ 0.8X103 kg/m3, and E
~ 400 MPa. We get δ ~ 52 mm
For steel: density is 7.6X103 kg/m3, an...
RA,y
RA,x

A

B

RB,y
Member Force
kN

Lengt
h
m

Area
m2

AC

28.3

1.41

1.77×10−4

BC

− 20

1

1.77×10−4

RA,y = 20 kN...
A

A

y
x
45o
B

E

C
D

C1

45o
B

E

C
D

45o
G

F

C1

X-displacement of C ~ shortening of BC =−0.54 mm
y-displacement ...
Statically-indeterminate
structures

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
PP

P

R1

R2

R3

Reaction at middle support (and hence, at all supports
depends on the bending of plank.
1. Consideration of static equilibrium
and determination of loads
2. Consideration of relations between loads
and deformat...
Indeterminate because x is not known!
1.3 m

2.6 m

F2

F1

150 kN

1m
R1

Taking moments about
the pivot point,
2R1 + 2R2 – P = 0

P

Indeterminate because 4
unknown forces and only three
equa...
R2 - F - R1 = 0; R1L – Fx = 0
Geom. Comp. h + δ = 2(h - δ )
1
2
F
h

R1 = kδ1
R2 = kδ2
L

x
P
P = R1+ R2
R2
R1

R1 = (E1A1/L1)δ1
R2 = (E2A2/L2)δ2

Geom. Comp.
δ1 = δ2
(a) Tendon being stressed during
casting. Tension in tendon, no
stress in concrete.

(b) After casting, the force is
relea...
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 o...
2.42 mm
δs

δc

δs + δc = 2.42 mm
Lateral strains:
Poisson ratio, ν

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
σxx

εxx = σxx/E
εyy = - ν εxx
ν is Poisson ratio

σxx

Another material property
Let us consider εxx.
σxx produces an εxx = σxx /E
σyy produces an εyy = σyy /E,
which through Poisson ratio
gives εxx = -ν...
Shear stresses do not cause any normal strain
Therefore,
εxx = ζxx/E – νζyy/E - ν ζzz/E
= [ζxx – ν(ζyy + σzz)]/E
Similarly...
F

ζyy = −F/A, ζzz = 0
What is ζxx and εyy

y

x

Geometric compatibility:
εxx = 0
εxx = [ζxx – ν(ζyy + ζzz)]/E

0 = [ζxx ...
σyy

Steel: εx = 0.6×10−4
εy = 0.3×10−4
Find σxx and σyy :
Plug in:
εxx = [σxx – ν(σyy + σzz)]/E
εyy = [σyy – ν(σzz + σxx)...
Shear strains and stresses

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
Apply shear stresses to a block:

Shear strain γ is π/2 − θ
Shear strain is also seen as: θ1 + θ2

θ2
θ

θ1

Since angles ...
A square blocks 0.2 mm × 0.2 mm deforms under
shear ystress
Coordinates after
C
D
deformation (in mm) are:

θ2

θ1
A

A(0,...
Shear strain γ is related to shear
stress τ by

γxy = τxy/G,
where G is shear modulus
It can be shown that γxy does not
de...
Material
Aluminium

G, GPa
25

Steel

80

Glass

26-32

Soft Rubber

0.003- 0.001
8,000 N

4,000 N

Shear stress τ =
4,000 N/ (0.1 m)(0.12 m)
= 3.33×105 Pa
Shear strain γ = τ/G
3.33×105 Pa/1 MPa
= 0.33

W...
Consider the rubber block on the left:

8,000 N

γ =0.33

Wall

And therefore,
The vertical deflection of load
= 0.33×0.10...
We have so far introduced three elastic
properties of materials.
Material

ν

E, GPa

G, GPa

70

25

0.33

Steel

200

80...
Thermal strains and stresses

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
On heating, there is linear expansion:

There is no thermal shear strain
Material
Steel
Aluminium

α (×10-6/oC)
~ 10
~ 20
Putting Hooke law, Poisson effect and thermal
strains all together,
εxx = [σxx – ν(σyy + σzz)]/E + αΔT
εyy = [σyy – ν(σzz ...
Aluminium rod, rigid supports.
Temperature raised by ΔT.
What are the stresses?
εxx = 0 = [σxx/E + αΔT]
σxx = −αEΔT

x
Tank is flush when empty.
Find end forces when pressure is p
Due to p: σzz = pr/ 2t, σθθ = pr/ t

z

If end forces F, axia...
Determining stress-strain
relations

Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013


A material property.

Tensile Test Machine, UTM
σ (= F/Ao)

Ductile

Brittle
ε (=∆L/Lo)

Ductile Failure
cup-and-cone

Necking

Brittle Failure
Y

σ (= F/A0)

Yield stress, σY

0.02% Permanent set

ε (= ΔL/L0)
Y1

Ultimate stress

σ (= F/A0)

Y

B
ε (= ΔL/L0)
σ

σ

σ

(a) Rigid

ε

ε
(b) Perfectly elastic

σ

ε
(c) Elastic-Plastic

σ
Increase in
yield strength

ε
(d) Perfectly pl...
= 9.82×10-6 m4
Let us check on the stresses:

Quite safe
Φ 10 cm

Φ 5 cm

F

Φ 2 cm
Φ 6 cm

1m

F

0.6 m

150 N.m
−250 N.m

150 N.m
Φ 10 cm

Φ 5 cm

F

Φ 2 cm
Φ 6 cm

1m

F

0.6 m

150 N.m
Φ 10 cm

Φ 5 cm

F

Φ 2 cm
Φ 6 cm

1m

F

0.6 m

150 N.m

Angle θ2 which represents the counter-clockwise movement of
the ...
By geometry: γθz= rdΦ/dz
Therefore, τθz = GrdΦ/dz

r varies from R1 to R2, and
θ varies from 0 to 2π
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0

Weight
0

0.2

0.4

0.6

0.8

1
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0

0.2

0.4

0.6

0.8

Weight
Stiffness
1
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0

Strength to weight

Stiffness
Weight
0

0.2

0.4

0.6

0.8

1
M1

Mo

M2

TMD

Equilibrium Condition: - M1 + Mo – M2 = 0

Geometric Condition: Φ1 +Φ2 = 0
Shear flow on horizontal surfaces
is same as on the vertical surfaces

q1 = q2
Relating q to twisting moment T

h

o

qds

dT at O = qds×h
= q×2×Grey area

q = T/2A
T 100 Nm
R 20 mm
R 16 mm

This gives τ = (49 N/m)/0.004 m = 12.25 MPa
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
Strength of Materials
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Strength of Materials

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Strength of Material complete stress strain torsion etc etc

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Strength of Materials

  1. 1. 1. Structures, loads and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  2. 2. This course is concerned with structures: A structure is a solid object or assembly. A structure connects components, carries loads, provides form and integrity.
  3. 3. F22, 2002 Wright Flyer, 1903
  4. 4. Burj al Arab Qutab
  5. 5. • Distance between any two points does not change when a force is applied on it.
  6. 6. An upward reaction force at support Thus, an unbalanced moment results. a moment at But equilibrium requires a force and the support. Where does it come from?
  7. 7. Bar Resisting Force 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.
  8. 8. Stress, ζ = P/A Strain, ε = δ/L Area A P Robert Hooke in 1678 showed l E is the elastic modulus, or simply the Elasticity δ P P
  9. 9.  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
  10. 10.  Strain δ/L is dimensionless, hence NO UNITS.
  11. 11.   E = Stress / strain, and therefore, has dimensions of stress, i.e., F/L2. Units of E are, accordingly, Pa(scal).
  12. 12. Material Aluminium 2024-T3 Aluminium 6061-T6 Aluminium 7075-T6 Concrete Copper Glass fibre Cast iron Steel, High strength Steel, Structural Titanium Wood Value of E in GPa 70 70 70 20 – 35 100 65 100 200 200 100 10-15
  13. 13. 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. The constant of proportionality is termed as the modulus of elasticity.
  14. 14. Stresses due to various loads Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  15. 15. Bar F L δ
  16. 16. Tension in the belt
  17. 17. P P P P
  18. 18. Stress: Force Intensity 20 mm 5 mm ζ = F/A = 300N 20×5×10− 6 m2 = 3 MPa 300 N
  19. 19. Uniform stress is an approximation. Valid only in simple loadings. Away from ends.
  20. 20. 105 Steel Concrete Soil N Bearing strengths Steel >> concrete >> soil Required areas Steel << concrete << soil
  21. 21. 105 N Permissible compressive stress in steel is about 400 MPa. Steel Concrete Soil So the area of steel required is 105 N/400 MPa, or 2.5 10−4, or about 16 mm 16 mm
  22. 22. 105 N Permissible compressive stress concrete is about 60 MPa. Steel Concrete Soil So the area of concrete required is 105 N/ 60 MPa, or 1.67 10−3, or about 41 mm 41 mm
  23. 23. Permissible bearing strength 105 N of soils varies widely. For good cohesive soil, it could be between 100 to 400 kPa, if it is above the water table. Steel Concrete Soil So the area of the footing required is 105 N/ 200 kPa (say), or 0.5 m2, or about 710 mm 710 mm
  24. 24. Shear Stresses
  25. 25. Bearing (Compressive) Shear
  26. 26. Bearing load Shear load
  27. 27. Blanking force = shear strength Shear area = perimeter shear area sheet thickness
  28. 28. Compression Shear Compression Shear
  29. 29. P P P Shear Shear area
  30. 30. Shear stresses on the back face of the shaft The stresses result in a moment that balances the twisting moment
  31. 31. Compression near top Extension near bottom
  32. 32. Bending of Beams Net tensile force is zero!
  33. 33. • Forces that tend to reduce the size of a structural member produce compressive strains which, in turn, produce compressive stresses.  Forces that tend to distort the shape of a member produce shear strains which in turn produce shear stresses.
  34. 34. • 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.  A bending moment produces both tensile and compressive strains and stresses. These give rise to a resisting moment which balances the bending moment.
  35. 35. Tensile members and trusses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  36. 36. 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN
  37. 37. 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?
  38. 38. The forces acting on a two-force member are equal, opposed and collinear.
  39. 39. Two-force members 60o 30o F1cos30o – F2cos60o = 0 500 N F1sin30o + F2sin60o – 500 = 0 F1 = 125 N; F2 = 433 N
  40. 40. Simply supported: No moment. Only reaction force. • Cannot translate, • Can rotate.
  41. 41. Load Pinned Support: Reaction could be inclined. (1 DoF) Roller Support: Normal reaction only. (2 DoF)
  42. 42. FRICTION must act
  43. 43. RH2X4 – 600X0.5 = 0 or RH2 = 75 N RH2 4m Or, N RV1 600 N RH1 1m y x z
  44. 44. While in the pinned support, the member is restrained from translating, in the clamped support, the member cannot even rotate.
  45. 45. • Cannot translate, • Cannot rotate. No DoF at all. Also called built-in support
  46. 46. P Fy Fx Mz L y Statically determinate x z
  47. 47. Resisting moment Tension will build up faster than the moments due to bending, and therefore, can treat the joint as pinned
  48. 48. Two-force members
  49. 49. 3m θ cos = 4/5, sin = 3/5
  50. 50. Method of Joints Symmetry: FGB FGC FGF FGA 15 kN 2FGC sinθ – 15 kN = 0 FGC = (15 X 5)/(2 X 3) = 12.5 kN
  51. 51. Method of Joints -FCGsinθ - FCFsin θ = 0 FCF = - FCG = -12.5 kN FCB FCG FCF - FCB – FCGcosθ + FCFcosθ = 0 FCB = –2FCGcosθ = 10 kN
  52. 52. FFC ∑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 FFG RB,V
  53. 53. B D C 20 kN
  54. 54. B B A two-force member D D C 20 kN FBD,D
  55. 55. F2 B F1 D 20 kN C D 20 kN ∑Fy = 0
  56. 56. • Co-planar • Concurrent
  57. 57. Notation for stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  58. 58. First index: normal to the plane on which acting. z x Second index: direction of y the stress component itself
  59. 59. Stress Intensity of Force
  60. 60. The stress vector t depends upon the location as well as the direction of the surface.
  61. 61. z x y 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.
  62. 62. ζyy is negative ηyx is positive ηyz is negative z x y
  63. 63. ζxx is positive ηxy is negative ηxz is negative z x y
  64. 64. Face Area Force xy(direction of (assume outward unit depth) compone compone nt nt normal) 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 Consider the moment balance about the mid-point:
  65. 65. 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.
  66. 66.   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.
  67. 67.   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, .
  68. 68.   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.
  69. 69. C o n sid e r th e h a lf cylin d e r sh o w n . F o rce d u e to in te rn a l p re ssu re , p is b a la n ce d b y th e fo rce d u e to h o o p stre ss,  h . i.e . h o o p stre ss x a re a = p re ssu re x p ro je cte d a re a h x 2 L t = P x d L h = (P d ) / 2 t W h e re : d is th e in te rn a l d ia m e te r o f cylin d e r; t is th e th ickn e ss o f w a ll o f cylin d e r.
  70. 70. T h e in te rn a l p re s s u re , P a ls o p ro d u c e s a te n s ile s tre s s in lo n g itu d in a l d ire c tio n a s s h o w n a b o v e .  d F o rc e b y P a c tin g o n a n a re a lo n g itu d in a l s tre s s ,   d t  L  d 4 L is b a la n c e d b y a c tin g o v e r a n a p p ro x im a te a re a , (m e a n d ia m e te r s h o u ld s tric tly b e u s e d ). T h a t is : x d t  P x  L 4 2  P d 4t 2
  71. 71. 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. 
  72. 72. Take section of a pressurized cylinder And the upper half FBD of the lower half σθθ p σθθ
  73. 73. FT FR p We can show by symmetry arguments that: (a) Both shear should be inwards or outwards (b) Shear should be ZERO
  74. 74. FT FBD of the ‘contents’ p Net forced on the curved surface = p×2r×δl Equilibrium: FT = ζ 2δl t = p×2r×δl This gives: Hoop stress
  75. 75. Forces on the rim Pressure on ζ the back cap Axial stresses are lone-half of hoop stresses
  76. 76. Forces on the rim Pressure on the ‘content’ p Maximum stress in a spherical vessel is one half that of a cylindrical vessel of same radius and thickness
  77. 77. Shaped structures
  78. 78. Arch Keystone All stones are subjected to compressive forces only.
  79. 79. Towers Load bearing cables Deck of bridge Cables support the bridge through tension. Towers carry compression,
  80. 80. 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.
  81. 81. Tension in the cable at the lowest point is; To = 2.96×108 N Max 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
  82. 82. z x y
  83. 83. Take section of a pressurized cylinder And the upper half FBD of the lower half σθθ p σθθ
  84. 84. FT FR p We can show by symmetry arguments that: (a) Both shear should be inwards or outwards (b) Shear should be ZERO
  85. 85. p
  86. 86. 2. Deformations, strains and material properties Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  87. 87. P/A L δ/L Stress = E×Strain P
  88. 88. Deformation depends on loading, material and geometry Strain depends on stress AND material. NOT on geometry Stress depends on loading and geometry
  89. 89. Load Equilibrium Stress Macro Geometry Micro Strain Micro Geometry Material Property Deformation Macro
  90. 90. Cross-section: 6 cm2 Section Tension, N Stress, kPa Strain Length, m Elongation, m AB BC CD 250 500 800 416.7 4.16×10- 3 833.3 8.35×10- 3 1333.3 13.33×10- 3 1.5 2.0 1.5 6.24×10- 3 16.70×10- 3 20.02×10- 3 DE EF 550 300 916.7 9.20×10- 3 500.0 5.00×10- 3 1.5 2.0 13.78×10- 3 10.00×10- 3
  91. 91. W = ρAxg = T ζ = T/A = ρxg 72 m ε = ζ/E = ρxg/E dδ = εdx = ρgxdx/E T dx x W(x)
  92. 92. or For a Nylon wire: density, ρ ~ 0.8X103 kg/m3, and E ~ 400 MPa. We get δ ~ 52 mm For steel: density is 7.6X103 kg/m3, and E is 200 GPa. We get δ ~ 1 mm
  93. 93. RA,y RA,x A B RB,y Member Force kN Lengt h m Area m2 AC 28.3 1.41 1.77×10−4 BC − 20 1 1.77×10−4 RA,y = 20 kN RA,x = − 20 kN RB,y = 20 kN C 20kN Stress MPa TAC = 28.8 kN TBC = − 20 kN Strain Elongation m 7.6×10−4 1.07×10−3 −113.2 −5.4×10−4 −0.54×10−3 160.1
  94. 94. A A y x 45o B E C D C1 45o B E C D 45o G F C1 X-displacement of C ~ shortening of BC =−0.54 mm y-displacement of C ~ EF + FC1 = CD/cos45o + FG(=EC) ~ 1.25 mm
  95. 95. Statically-indeterminate structures Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  96. 96. PP P R1 R2 R3 Reaction at middle support (and hence, at all supports depends on the bending of plank.
  97. 97. 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
  98. 98. Indeterminate because x is not known! 1.3 m 2.6 m F2 F1 150 kN 1m
  99. 99. R1 Taking moments about the pivot point, 2R1 + 2R2 – P = 0 P Indeterminate because 4 unknown forces and only three equations to determine them. R2 Geom. Comp. δ1 = δ2 R1L1/E1A1 = R2L2/E2A2
  100. 100. R2 - F - R1 = 0; R1L – Fx = 0 Geom. Comp. h + δ = 2(h - δ ) 1 2 F h R1 = kδ1 R2 = kδ2 L x
  101. 101. P P = R1+ R2 R2 R1 R1 = (E1A1/L1)δ1 R2 = (E2A2/L2)δ2 Geom. Comp. δ1 = δ2
  102. 102. (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..
  103. 103. 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
  104. 104. 2.42 mm δs δc δs + δc = 2.42 mm
  105. 105. Lateral strains: Poisson ratio, ν Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  106. 106. σxx εxx = σxx/E εyy = - ν εxx ν is Poisson ratio σxx Another material property
  107. 107. Let us consider εxx. σxx produces an εxx = σxx /E σyy produces an εyy = σyy /E, which through Poisson ratio gives εxx = -νεyy = - νσyy/E. Similarly for σzz .
  108. 108. Shear stresses do not cause any normal strain Therefore, εxx = ζxx/E – νζyy/E - ν ζzz/E = [ζxx – ν(ζyy + σzz)]/E Similarly for εyy and εzz
  109. 109. F ζyy = −F/A, ζzz = 0 What is ζxx and εyy y x Geometric compatibility: εxx = 0 εxx = [ζxx – ν(ζyy + ζzz)]/E 0 = [ζxx – ν(ζyy + 0)]/E, → ζxx =νζyy = − ν F/A εyy = [ζyy – ν(ζxx + ζzz)]/E = [−F/A + ν F/A]/E = −(1− ν)F/AE
  110. 110. σyy Steel: εx = 0.6×10−4 εy = 0.3×10−4 Find σxx and σyy : Plug in: εxx = [σxx – ν(σyy + σzz)]/E εyy = [σyy – ν(σzz + σxx)]/E E = 200 GPa, ν = 0.3 σxx
  111. 111. Shear strains and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  112. 112. Apply shear stresses to a block: Shear strain γ is π/2 − θ Shear strain is also seen as: θ1 + θ2 θ2 θ θ1 Since angles are measured positive counter-clockwise, the angle θ2 above is a negative angle. In general terms, then, γ = θ1 − θ2 with θs measured positive when counter-clockwise
  113. 113. A square blocks 0.2 mm × 0.2 mm deforms under shear ystress Coordinates after C D deformation (in mm) are: θ2 θ1 A A(0,0), B(0.194, 0.013), and D(−0.012, 0.196). B x θ1 = 0.013/0.2 = 0. 065 θ2 = 0.012/0.2 = 0. 06 γxy = 0.65 − 0.60 = 0.05 radians
  114. 114. Shear strain γ is related to shear stress τ by γxy = τxy/G, where G is shear modulus It can be shown that γxy does not depend on other components of stress.
  115. 115. Material Aluminium G, GPa 25 Steel 80 Glass 26-32 Soft Rubber 0.003- 0.001
  116. 116. 8,000 N 4,000 N Shear stress τ = 4,000 N/ (0.1 m)(0.12 m) = 3.33×105 Pa Shear strain γ = τ/G 3.33×105 Pa/1 MPa = 0.33 Wall Wall Rubber blocks 10 cm × 10 cm with 12 cm height
  117. 117. Consider the rubber block on the left: 8,000 N γ =0.33 Wall And therefore, The vertical deflection of load = 0.33×0.10 m = 33 mm Wall Rubber blocks 10 cm × 10 cm with 12 cm height
  118. 118. We have so far introduced three elastic properties of materials. Material ν E, GPa G, GPa 70 25 0.33 Steel 200 80 0.27 Glass 50-80 26-32 0.21-.27 0.00080.004 0.0030.001 0.50 Aluminium Soft Rubber
  119. 119. Thermal strains and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  120. 120. On heating, there is linear expansion: There is no thermal shear strain Material Steel Aluminium α (×10-6/oC) ~ 10 ~ 20
  121. 121. Putting Hooke law, Poisson effect and thermal strains all together, εxx = [σxx – ν(σyy + σzz)]/E + αΔT εyy = [σyy – ν(σzz + σxx)]/E+ αΔT εzz = [σzz – ν(σxx + σyy)]/E+ αΔT γxy = τxy/G, γyz = τyz/G, and γzx = τzx/G
  122. 122. Aluminium rod, rigid supports. Temperature raised by ΔT. What are the stresses? εxx = 0 = [σxx/E + αΔT] σxx = −αEΔT x
  123. 123. Tank is flush when empty. Find end forces when pressure is p Due to p: σzz = pr/ 2t, σθθ = pr/ t z If end forces F, axial stress due to it is F/2πrt εzz = [(pr/ 2t − F/2πrt) −νpr/t ]/E Equate it to 0 and determine F p
  124. 124. Determining stress-strain relations Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
  125. 125.  A material property. Tensile Test Machine, UTM
  126. 126. σ (= F/Ao) Ductile Brittle ε (=∆L/Lo) Ductile Failure cup-and-cone Necking Brittle Failure
  127. 127. Y σ (= F/A0) Yield stress, σY 0.02% Permanent set ε (= ΔL/L0)
  128. 128. Y1 Ultimate stress σ (= F/A0) Y B ε (= ΔL/L0)
  129. 129. σ σ σ (a) Rigid ε ε (b) Perfectly elastic σ ε (c) Elastic-Plastic σ Increase in yield strength ε (d) Perfectly plastic ε (e) Elastic- Plastic (strain hardening)
  130. 130. = 9.82×10-6 m4
  131. 131. Let us check on the stresses: Quite safe
  132. 132. Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m −250 N.m 150 N.m
  133. 133. Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m
  134. 134. Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m 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-clockwise Rotation 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.
  135. 135. By geometry: γθz= rdΦ/dz Therefore, τθz = GrdΦ/dz r varies from R1 to R2, and θ varies from 0 to 2π
  136. 136. 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Weight 0 0.2 0.4 0.6 0.8 1
  137. 137. 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 Weight Stiffness 1
  138. 138. 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Strength to weight Stiffness Weight 0 0.2 0.4 0.6 0.8 1
  139. 139. M1 Mo M2 TMD Equilibrium Condition: - M1 + Mo – M2 = 0 Geometric Condition: Φ1 +Φ2 = 0
  140. 140. Shear flow on horizontal surfaces is same as on the vertical surfaces q1 = q2
  141. 141. Relating q to twisting moment T h o qds dT at O = qds×h = q×2×Grey area q = T/2A
  142. 142. T 100 Nm R 20 mm R 16 mm This gives τ = (49 N/m)/0.004 m = 12.25 MPa

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