Strength of Materials

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, carries loads,
provides form and integrity.

• 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
the support. Where does it come from?

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.

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



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



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
Steel, Structural
Titanium
Wood

Value of E in GPa
70
70
70
20 – 35
100
65
100
200
200
100
10-15

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.

Stresses due to various loads


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 105 N/400 MPa,
or 2.5 10−4,
or about 16 mm 16 mm

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,

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,

Blanking force = shear strength
Shear area = perimeter

shear area

sheet thickness

Compression
Shear

Compression

Shear

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 compressive stresses.
 Forces that tend to distort the shape of a

member produce shear strains which in turn
produce shear stresses.

• 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.

Tensile members and trusses


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 as a
two-force member.

The forces acting on a two-force member are equal
and opposed. But is it enough?

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)

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 rotate.

• 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 pinned

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 = –2FCGcosθ = 10 kN

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

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

Notation for stresses


First index: normal to the
plane on which acting.

z
x

Second index: direction of

y

the stress component
itself

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 sign
then
the
stress
component is positive, if
the two have different
signs, then the stress
component is negative.

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
−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:

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.





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.





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, .





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.

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.

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

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.


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:

Hoop stress

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 vessel of
same radius and thickness

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
400 kN/m.

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

2. Deformations, strains
and material properties


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
depends
on
loading and geometry

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 4.16×10- 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

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, and E is 200
GPa. We get δ ~ 1 mm

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

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

Statically-indeterminate
structures


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 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

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
equations to determine them.

R2

Geom. Comp.
δ1 = δ2
R1L1/E1A1 = R2L2/E2A2

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
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..

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

2.42 mm
δs

δc

δs + δc = 2.42 mm

Lateral strains:
Poisson ratio, ν


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 = -νεyy = - νσyy/E.
Similarly for σzz .

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

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

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

Shear strains and stresses


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

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

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.

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

Wall

Wall
Rubber blocks
10 cm × 10 cm

with 12 cm height

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

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

Thermal strains and stresses


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 + σxx)]/E+ αΔT
εzz = [σzz – ν(σxx + σyy)]/E+ αΔT

γxy = τxy/G, γyz = τyz/G, and γzx = τzx/G

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, axial stress due to it
is F/2πrt

εzz = [(pr/ 2t − F/2πrt) −νpr/t ]/E
Equate it to 0 and determine F

p

Determining stress-strain
relations




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 plastic

ε
(e) Elastic- Plastic
(strain hardening)

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 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.

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

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