STRENGTH OF MATERIAL
Strength of Materials (Theory)
CONTENTS
Unit Chapters Page No.
1 Stress-Strain and Properties of Metal 2 - 12
2 Shear Force & Bending Moment 13 - 25
3 Complex Stress & Principal Stresses 26 - 39
4 Bending Stress in Beams 40 - 50
5 Shear Stress Distribution in Beams 51 - 55
6 Torsion of Twisting Moment 56 - 62
7 Deflection & Slope 63 - 69
8 Columns and Struts 70 - 72
9 Thin Cylinders and Spheres 73 - 75
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Published at : Ascent Gate Academy
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MECHANICAL ENGINEERING
1 STRESS -STRAIN AND PROPERTIES OF METALS
ASSUMPTIONS MADE IN STRENGTH OF MATERIAL :
a) Material is continous (no cracks & voids)
b) Material is homogenous & isotropic.
c) There are no internal forces in a body prior to loading
d) Principle of superposition is valid
(It states that effect of number of forces is equal to algebric sum of individual forces)
e) Self weight is ignored :
f) St. Venants principle is valid ( It states that except in the region of extreme ends of a bar carrying
direct loading the stress distribution over the cross section is uniform)
P PPP
1----------
2---------- max avg
3----------
= 1.387
max avg
= 1.0.27 =
max avg
P
STRESS
* It is the internal resistance offered by the body against external loading or deformation. Stresses are of
followingtypes :
i) Direct or normal stresses which may be tensile or compressive
= P
A
s = Resisting force against deformation
A=Area on which resistance is acting
P =Applied load
Unit - N/m2 or Pa
N/mm2 or MPa
ii) Shear or tangential stresses - Force acting tangential to surface is called shear force and correspond-
ing stress is called shear stress
Resisting shear force
Area on which shear force is acting
iii) Transverse or bending stress
iv) Torsional or twisting stresses.
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STRENGTH OF MATERIAL
Note : ** For direct stresses, if area under consideration is original area, then it is known as Engineering
stress or nominal stress. But, if area taken is actual area then stress is known as true stress.
Engg. or Nominal stress = P (A = original area)
0
A
0 (Ai = instantaneous area)
True stress = P
A
i
**Stress is a tensor quantity i.e. it has one magnitude and two directions.
Relationship : Volume will be same
A0L0 = AL
A0 = L = 1 + = (1 +
A L0 T
STRAIN : -
It is defined as change in dimension per unit original dimension
change in dimension
original dimension
a) Normal or axial strain or direct strain : It may be tensile or compressive B B’ C C’
b) Shear or tangential strain : It is angular distortion between two
planes at right angle (f). Expressed in radians.
)
)
c) Volumetric strain : Change in volume per unit original volume
change in volume V AD
V original volume V
Unit : Strain is a unitless quantity
Note : * Due to normal stress, there will be changes in dimensions & volume without distortion in shape.
* Due to shear, there will be only distortion, no change in dimension & volume
HOOKE’S LAW
Within elastic limit, Stress a strain
stress
or = const = E
strain
i) Young’s modulus of elasticity (E)
E= t or c
t c
Elastic Constants for different materials : Ediamamond = 1200 GPa
ECI = 100 GPa
Esteel = 200 GPa - for all steels irrespective of carbon E = 10 GPa
EAl = 70 GPa
E = 100 GPa timber
brass EBronze = 80 GPa
Ecu = 120 Gpa
Erubber = 10 GPa
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MECHANICAL ENGINEERING
Note :
* Slope of - e diagram upto proportional limit is E
* More E - More elasticity
* Diamond is more elastic than steel, steel is more elastic than rubber.
* E is constant for a material under all circumstances and its value can be calculated by using slope within
proportional limit
ii) Modulous of rigidity or shear modulous (C, N or G) :-
shear stress
G = shear strain =
* G is also constant. For any material G < E
Significance : More G less deflection more stiffness
iii) Bulk modulous (K)
Normal stress or spherical stress
K= = n
volumetric strain
v
Significance : More K less change in dimension or less compressible
*V is also called dialation. K is called dialation const.
* E > K > G (for any isotropic, homogenous mtl)
LATERALSTRAIN : It is defined as change in lateral dimension per unit to original lateral dimension
d < >< >l d change in diameter
lateral strain = =
<>
d original diameter
<>
d-d
l
P
POISSONS RATIO () or 1
m
= Lateral strain = d/d (for cylindrical rod)
Linear strain l/l
Note : Value of for any material varies from 0 to 0.5
Significance : more more ductility
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STRENGTH OF MATERIAL
Value of for different materials
i) Cork (almosto) = 0
ii) Incompressible fluid = 0.5
iii) Clay = 0.5
iv) Paraffin wax = 0.5
v) Rubber = 0.5
vi) Isotropic matrial = 0.25 to 0.33
vii) Metals > 0.25
viii) Non metals < 0.5
ix) steel = 0.33
x) Concrete = 0.15
Note : There is no normal stress in transverse direction yet there is strain. This is due to poisson effect.
RELATIONSHIP BETWEEN E, K & G
* E = 3K (1 - 2)
* E = 2G (1 +
9 KG
*E=
3K + G
* = 3K - 2G
6K + 2G
Total number of elastic constants for different materials :-
Material Total Elastic constant Independent Elastic constant
i) Isotropic mtl 4. (E,K,G, ) 2 (E, )
9
ii) Orthotropic 12 21
iii) Aleotropic or anisotropic
ELONGATION OF BARS
1) A bar of uniform cross section area
l = P.l P P
< >
AE
< l >< l >
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2) Non-uniform bar
l = P l1 + l2 + l l1A1E1
AE A2E2 3 l2A2E2
11 A3E3
l3A3E3
P
3) Tapering bar of circular criss section whose diameter changes from d1 to d2
P < d1 d 4Pl
2> P
l = d d E
12
4) Elongation due to self weight : ^
a) Bar of uniform section :
l
l = Wl
2 AE
b) Bar of tapering section : (Conical bar)
l = Wl
6 AE
5) Bar of uniform strength :
* For a bar to have constant strength, the stress at any section due to external load & weight of the portion
below it should be constant
A1
l l
eA1 = A2
6) Elongation of composite beam A2
P P P P3
1 2 4 3
1
2
l = l1 + l2 + l3 (algebrically)
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STRENGTH OF MATERIAL
7) Elongation of Compound bar : P
Pl1==P1+l2 P2 1 2l
Load shared by part 1
P = PA1E1
1 A E +A E
11 22 P
Load shared by part 2
P = PA2E2
2 A1E1 + A2E2
Volumetric Strain for a rectangular bar sub to 3 mutually perpendicular
tensile stress
=v V = 1 ( + + z) ( 1 - 2) y
V E x y z
x x
When = = =
x y Z
= ( 1 - 2) z y
vE
Volumetric strain of cylindrical rod sub to axial load :
l d P P
v= +
ld
Volumetric strain of sphere subjected to tensile force
V = 3 = d
V d
d
= strain in dia
d
Volumetric strain of rectangular body subjected to axial force :
V P P< ^ P
>
= t
V btE < L >< b
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STRESS STRAIN DIAGRAMS :
1) For Mild Steel (Ductile Material)
^
C E
B F
D
A
>
e
A - Limit of proportionality
B - Elastic limit point
C - Upper yield point
D - Lower yield point - yielding begins at this point
DE - Strain hardening region (The mtl in this region undergoes change in its atomic & cyrstalline structure
resulting in increased resistance to further drformation. This portion is not used for structural design)
E - Ultimate stress point
E F - Necking region (strain softening)
F - Fracture point
Note : * The magnitude of stress correspending to upper yield poinst depends on cross section area,
shape of specimen & type of equipment used to perform the test. It has no practical significance so lower
yield pt is considered as true characterstic yield stress for M.S.
** Materials that undergo large strains before failure are classified as ductile.Advantage of ductility is
that visible deformation can occur before failure hence remedial action can be taken.
2) For Brittle Material ^
A
B
A = Ultimate Load point
B = Fracturer Point
>
e
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STRENGTH OF MATERIAL
Generalised - curve ^
fail ( Linear elastic )
PL
Perfect ><
rigid ><
Ideal fluid fail
>
( No Viscosity)
Residual strain : Loading beyond elastic limit causes residual strain or permanent set
^^
E.L.
>< P.L.
E.L.
><
><
>
P.L. L = Residual strain or permanent set
> < L> >
Stress Strain Diagram for different materials :
^^ ^
> > >
(a) Ideal plastic (b) Elasto plastic or (c) Rigid strain hardening
linear elastic plastic
^ ^ ^
> Elastic >
necking
(d) Elastic strain hardening (f) Rigid softening
>
(e) Elastic strain softening
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low carton steel ^ Tools steel
Low alloy high
Concrete Al alloy strength steel
C.J. Low corbon
C.I. Rubber steel
wood
> e >
e
Engineering and True Stress Strain Curve
^ Engg. - curve in compression
True - curve in compression
True - curve in tension
Engineering
stress-stain cruve
in section
>
PROOF STRESS :
When a material such as Aluminium which doesnot have an obvious yield point & yet undergoes large
strains after proportional limit, the yield stress is determined by offset method.
A line parallel to initial linear part is drawn, which is offset by some standard amount of strain such as
0.2%. The intersection of the offset point (A) defines the yield stress which is slightly above proportionality
limit and is called proof stress.
^
A.
0.2% offset
e >
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STRENGTH OF MATERIAL
PROPERTIES OF METALS
i) Ductility : It is the property by which material can be stretched
Eg. : M.S., Al, Cu, Mn, Lead, Brass etc.
ii) Brittleness : It is lack of ductility i.e. material cannot be stretched. For brittle materials, fracture point
& ultimate point are same. Materials with strain less than 5% at fracture point are regarded as brittle & those
having strains > 5% at fracture pt are called ductile
iii) Malleability : The property by which mateerial can be uniformly extended in a direction without
reputre.
iv) Toughness : This property enables material to absorb energy without fracture. This property is very
desirable in case of cylic loading or shock loading.
Modulous of toughness : Area under entire stress -strain curve and is the energy absorbed by material
of the specimen / unit vol. upto fracture.
Mod. of toughness = y u f
2
= yield tensile strength
y
= ultimate tensiel strength
u
f = strain at fracture pt
** Material having higher modulation of toughness will be very ductile.
Modulus of resilience : Maximum elastic engery / unit vol. that can be absorbed without attaining plastic
stage. It is area under stress - strain diagram upto dastic limit.
y
Mod. of resilience ( u) = 2E
** Higher u - Higher yield strength
** Higher toughness is desirable for gears, chains, crane hooks, etc, Higher resilience is destrable in
springs
v) Hardness : Reistance to indentation or scratching or surface abrasion.
vi) Fatigue : The behaviour of material under variable loads is referred as fatigue.
vii) Creep :Additional strains over a long peiod of time is called creep.
viii) Relaxation : Awire attached between two rigid supports after sometime stress in wire diminshes &
reaches constant value called relaxation of material. ^
^
- - - -
- - - - -- -
T >
fig : creep in bar under constant load t > Fig : Relaxation
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ix) Tenacity : ultimate strength in tension is called Tenacity.
Note : * Ductile materials are tough & brittle materials are hard.As carbon content increases ductility
decreases but ultimate strength increases.
* Ductile materials are strong in tension, weak in shear, moderate in comprression
* Brittle mtls are : weak in tension, strong in Compression, moderate in shear
* Theoritically, ductile materials are equally strong in tension & in compression but practically due to
buckling this materials are weak in compression & very weak in shear.
Factor of Safety (N)
Used to determine permissible stress. Permissible stress is used to get safe dimensions of a component
under strength critieria
Failure stress or critical or limiting stress
N=
Permissible stress or allowable or working or Design or safe stress
Failure stress For ductile =
FOS per y
per = N
For brittle =
u
per N
Reasons for FOS :
i) Unknown loading conditions
ii) Imperfect workmanship
iii) Unknown environmental conditions
iv) Unrealityof assumptions made in STM equation
v) Effect of true stress
vi) Effect of stress concentration
Margin of Safety = N - 1
Note : Margin of safety is reduced to 0 or less, structure fails.
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STRENGTH OF MATERIAL
2 SHEAR FORCE & BENDING MOMENT
BEAM :
Beam is a structural member that is designed to resist forces acting transverse to the axis.
TYPES OF BEAM
Beam Free Body Diagram
1) Simply supported beam : H
One side hinge, one side roller support
v
2) Cantilever beam : One side free - one fixed
H
MV
<
3) Fixed beam : Both sides fixed
<
<
4) Continous Beam
5) Overhanging beam :
6) Propped cantilever
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TYPES OF LOADING
i) Concentrated / point load W
ii) UniformlyDistributed load / UDL WKN/m
Total load = W. l kN ,
Centroid = l/2 <l>
iii) Uniformlyvarying load (UVL) w/m
a) Triangular load c
Total load = Area of = 1 l x w vvvvv v
2 <l >
< 2l/3 >< >
l/3
b) Trapezoidal load : w1/m w2/m
Total load =Area of trapezoium c
1 vvvv v v v
= 2 l ( w1 + w2) < l>
<x>
l 2w1 + w2
x=
3 w1 + w2
iv) Concentrated moment
v) Couple M KN - mV
Magnitude of couple M = P.d.
P
vi) Load on Link : d
V
P
e. P
PM = P x e
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STRENGTH OF MATERIAL
Shear force and Bending moment (SF and BM) :
* Shear force at the cross section of a beam is defined as the unbalanced vertical force to the right or left
of the section
OR
*Algebric sum of forces on left or right side of section (In this case concentrated moment is ignored)
* Bending moment at section of a beam is defined as algebric sum of moment of forces to the right or left
of the section
Sign convention :
1) Shear force
-------
-------
Positive Negative
RUN - Right side upward is Negtaive
Bendingmoment :
Positive Negative
(sagging) (Hagging)
Shear Force & Bending Moment Diagram (SFD & BMD)
Agraphical representation of variation in S.F. and B.M. along length of beam is called SFD & BMD
respectively.
Use : To locate maximum & minimum point and variation
Relation between Shear Force (F) Bending Moment (M) and Intensity of Loading (W)
dM KN - m kN
F= m kN / m
dx
Note : dF KN
* Slope of Shear Force diagram is W. W= m
dF dx d2M
dx = W = Intensity of loading d dM =
* Slope of Bending Moment Diagram is shear force . W=
dx dx dx2
dM
=F i.e. dM = Fdx
dx
Thus Change in B.M. between two points is the area of SFD between those two points
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Also , dF = W d x.
* Change in SF between any two points = Area of loading diagram between those two points.
* For B.M. to be maximum
dM
dx = 0
or F = 0
Note : At a point of zero Shear Force, Bending Moment is maximum, converse is not true.
Simply supported beams :-
i) Carrying load at mid span :-
Note : * In case of symettrical loading in a simply supported beam the reaction are equal & is equal to
half the load W ^ RB
* In Simply supported beam , moment at supports are zero. c
>< l / 2 >
RA^
< l/2
RB = RA = W/2 W/2
Wl W/2
4
Mc = M max = RA x l / 2 =
WL
4
ii) < l>
RA = Wb RB = W.a < W
a+b a+b a >< b >
RA^
Mc = Wab = Area under SFD between A & C c ^RB
l Wb
lA B
Wb
A cl
Wab
lB
Note : Maximum shear force is maximum
support reactions in the beam
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STRENGTH OF MATERIAL
iii) Carrying many concentrated loads : < l>
W1 W2 W3
R1 ^ c ^ R2
Note : If there is vertical ordinate or suden jump ^ ^
in SFD, it indicates concentrated load at that point
whose magnitude = length of ordinate. W1 R2
^
W2
^
W3
+
iv) Carrying udl . w / unit length
Wl
RA = RB = 2 RA^< l >^RB
MC = Area under SFD between A & C wl
1 wl x l = wl2 2 ^ wl
MC = 2 2 2 8
c2
^
wl2
8
< l >
v) Simply Supported beam subjected to a couple RA^ M0
RA + RB = 0 < > ^RB
MA = 0
RB x l + M0 = 0 M0 a >< b >
RB = - M0 / l l
RA = M0 / l
Note : If there is vertical ordinate or suden jump M0a
in B.M.D., it indicates concentrated moment at
that point whose magnitude = length of ordinate +l
-M0 (l - a)
l
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VI) PURE BENDING CONDITION <l >
M <M
B
R +R =0 A
AB SFD SFD
M
RB x l + M - M = 0 M BMD
RB = 0 , RA = 0 BMD
Note :- Pure bending is not possible in realistic
beam due to self wt. of the beam
( +)
VII )
40 kN-m 20 kN-m
R +R =0 <
AB A 60 kN-m
B
MA = 0 , RB x l - 20 + 60 - 40 = 0 40 ( +)
RB = 0 , RA = 0 20 ( - ) SFD
M
Note : If the algebric sum of moments on the BMD
beam is zero & no other forces are acting,
reactions are zero & SFD is a straight line.
VIII)
A <M
M/l B
R +R =0 M/l
AB SFD
RB x l + M = 0 M
BMD
RA = M ( +)
l ( +)
RB = -M / l
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STRENGTH OF MATERIAL
Cantilever beams ^ l W
i) Carrying concentrated load at end +
A >
MA = Area under SFD between A & B <
MA = W.l. ^
SFD
W
^ (-)
BMD Wl
ii) Carrying udl < l>
w / unit length
FA =Area under loading diagram between A & B = w.l. A B
<l >
MA = Area under SFD between A & B ^ (+)
= 1 .wl x l
2 SFD wl
wl2 ^ (-)
=
wl2
2 BMD
2
iii) Subjected to moment at ends A< l >
M
SF = 0 , BM is constant & maximum SFD (-)
This is condition of “PURE BENDING” BMD <
M
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< l >
w
iv) UVL (Triangular load) B
(I) A
wl SFD wl 20 parabola
RA = 2 2
wl x l
MA = 2 3
wl2
=
6
wl2 30 parabola
BMD
6
II) < l>
A W/m
20 parabola
M = wl x 2l wl2
= 30 parabola
A2 3 3
* Ratio of maximum S.F. in case I & II wl
Wl / 2 2
=1 SFD
Wl / 2
* Ratio of maximum B.M.
Wl 2/ 6 = 1 wl2
3
Wl 2/ B 2 BMD
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STRENGTH OF MATERIAL
OVERHANGING BEAMS
Overhanging beams are used to reduce maximum B.M. value
l
<>
(I) a a = l/4 (II)
<> < > w/m
a > l/4
(-)
wa
M=0 (-) BMD
( - ) POI
(III)
a < l/4
Note : Overhanging beams are better than simply (+)
supported because maximum B.M. is distributed
at two ends ( - ) POI POI (-)
(POC)
* Out of three overhanging cases above case (POC)
III is best for having good beam.
Design Criteria
1) Maximum magnitude of hogging B.M. = maximum magnitude of sagging B.M.
or
Maximum positive B.M. = Maximum Negative B.M.
or
Maximum magnitude of B.M. should be as minimum as possible
To achieve this condition Best design :
a = 0.207 l
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MECHANICAL ENGINEERING
Beam with one overhang < l
Here a = 0.293 l >
Note : Beam with two overhangs is better
than with one overhang a
<>
+ve
-ve
POI
(POC)
Point of contraflexure (POC) & Point of inflection (POI)
Point of Inflection (POI) : The point where bending moment becomes zero (i.e. B.M. = 0 at POI)
Point of Centraflexure (POC) : Point where bending moment changes sign, B.M. becomes zero at that
point.
Note : * All point of contraflexure are point of infelction but all POI are not POC.
BEAMS WITH INTERNAL HINGE (MOMENT HINGE)
Eg. 1:- 20 KN
BC Mc = 0 ,
AD
20 x 2 = R x 3,
< 2m >< 3m >< 2m > B
RB = 13.33 KN,
RC = RB + 20
B RC = 33.33 KN,
A RB Consider AB
RB C 20 KN MA = RB x 2 = 13.33 x 2
B D = 26.66 KN - M (Sagging)
Note : * Bending moment at hinge is zero.
20 SFD
13.33
26.66 ---------- 40
BMD
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STRENGTH OF MATERIAL
Eg : 2 100 KN
BC
D
MB = 0
< 2m >< >< >< 2m > RC x 6 = 100 x 3
RC = 50 KN , RB = 50 KN
3m 3m MB = 50 x 2 = 100 KN - m (hogging)
M = 100 KN m (hogging)
R 100 KN
B C
A R E RC
B
<RB >
3m R
C
D
50
50
150
100 100
Eg : 3
100 KN 80 KN
C D RB x 2 = 80 x 4
A RB = 160 KN
MA = RB x 3 - 100 x 3
B
= 180 KN - m
< 3m >< 2m >< 4m >
RC = 80 + 160 = 240 KN
100 KN
AB C 80 KN
RB D
RB
B
80 SFD
60
160
BMD
180 320
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MECHANICAL ENGINEERING
Expression for Area & x for some figures
1. Rectangular
A = b.h.<>
G h x = b/ 2
< <x>
>
2. Triangular
b
<> >
<c
Gh <> A = 1 bh
hG 2
<> x = 1 ( b + c)
b <x> 3
< x>
A = 1/2 bh <b >
x = 1 b A = 1/2 bh
3
3. Curves
i) y = kx2
<> hG
<>
hG
< x> < 3/4 b >
<b >
x = 3/8 b < b>
A = 1/3 bh
x= 3 b
4
A= 2 bh <>
3
G
h
ii) y = kxn <> iii) sine curve
1 h G ( 2 ) b
(n + 1)
A= bh < < >
< < x
>
b
x= >
x A= 2 bh
b >
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STRENGTH OF MATERIAL
CONVERSION FROM SFD TO LOADING DIAG.
Eg. : 5.5
Note : dF = W dx 10
6 m 10 2m 3
B
A CD
dF 1.5
=W
or dx 9
i.e. slope of SFD gives intensity of loading 7 KN 0.75 KN/m
6m
W1 df 10 - 5.5 3 KN
(AC) dx = >
= = 0.75 KN/m
6
W2 9 - 1.5 = 0.75 KN/m
=
(CD) 10
<
Loading at C = 5.5 + 1.5 = 7 KN
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MECHANICAL ENGINEERING
3 COMPLEX STRESS & PRINCIPAL STRESSES
Stress in 1D system :- x
Eg : - Member of trusses, axially loaded columns etc.
x
Fig : State of stress in 1 D
y
1D stress tensor = [x ]1x1. yx
Stress in 2D system :-
Eg: - Beam, shaft, column with moment xy
Note :- In any element, Normal stress is balanced x dy
x
by force equillibrium, shear stress is balanced by xy
moment equillibrium yx dx
xy = yx y
2 D tensor = x xy
xy
y 2 x 2.
3 D System : Stresses in 3 direction
Eg. : Realistic member or system, spatial member total structure, a big vehicle.
3 D tensor = xy xz >
x >y xy
y yz >yx
yx >
zy z > >
zx yz
>
zy > x
zx xz
>
xy = yx z
zx = xz
=
yz zy
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STRENGTH OF MATERIAL
State of stress Total stress component Independent stress component
1D 1 1
2D 4 3
3D 9 6
2) Analysis of 2D stresses :-
y
xy
x -------->- R x
x
y
xy
xy
y
Stresses on inclined plane :-
+ - Cos2 + sin 2 - sin 2 - cos 2
x y x y xy x y xy
2 2 2
Resultant Stress R
Angle of R with normal stress (Obliquity) . tan =
or = tan-1 Angle of obliquity
Note : is the angle with the vertical if > or is angle with plane of maximum stress
x y
Case I : Uniaxial stress system (1D)
= = =
x y
cos2 . ( 1 + cos2).
cos2
sin2
sincos
Here applied normal stress = max, For max, = 450.
( = 450) = = ) on 450 plane.
max
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MECHANICAL ENGINEERING
Case II :- Pure shear
Eg. : shaft s.t. torsion
= 0, = xy
x y
xy = yx =
sin 2 xy
sin cos
cos 2
Here applied shear stress = max, for max = 450.
At 450 plane. = 450 =
max
Note : * If an element is subjected to pure shear, diagonals are subjected to maximum normal stress called
dignral tension & diagenal compression equal to magnitude of applied shear stress.
Note : *Amild steel specimen subjected to an axial load fails along a surface at 450 to it’s axis because M.S.
is weaker in shear than in tension & place of maximum shear is 450 to the axis.
PRINCIPAL STRESSEES
Principal stress are direct normal stresses acting on mutually prependicular planes on which shear stress
are zero
dn = 0
dv
+ - Cos2 + xy sin 2
x y x y
n2 2
dn = 0- - sin 2 + xy cos 2
dv x y
2
tan 2p = xy
-
xy
p1 -Angle of major Principal plane with vertical
p2 - p1 + 90 -Angle of minor principal plane with vertical
Putting value of pl in (1)
+ - 2 x2y
1, 2 x y x y
2 2
Major & Minor principal stresses respectively.
1,
2
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STRENGTH OF MATERIAL
Principal planes :-
* The plane on which principal stress acts
* On prinicpal plane, shear stress is 0 i.e. = 0
* In 2D stress, there will be two principal planes
* Generally the angle between prinicpal planes is right angle (900)
Maximum Shear Stress :-
d
d = 0
- . sin2 - xy cos 2 (ii)
x y
2
d - ) cos 2 + 2 sin 2
d x y xy
- 2 xy s - Angle of maximum shear plane with vertical ]
Cot 2 s = -
xy
tan (90 - 2s) = - tan 2p
p + s = 450 or s = p + 450
Putting value of s in equation (ii) =+ - 22
max x y xy
2
Note :- *Angle between principal plane & maximum shear stress plane is 450.
Maximum shear planes : -
* The plane on which is acting
* max
*
There will two max planes in 2-D system
On the plane of is not zero
max normal stress
Normal stress on plane will be
max
= x + y
2
* Maximum shear planes are also separated by 900.
* In the material, angle beetween any principal plane & plane will be 450.
* max
Resultant stress on max plane.
= + .2
R max
Obliquity on plane.
max
= tan-1
max
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MECHANICAL ENGINEERING
Representation of Tensor :-
’
x xy 1 max
= = ’
yx y 2 max
by rotating p by rotating p + 450) or s
Eg. - = 400 NPa, = -200 NPa,
x y
xy = 400 NPa,
ii) max
i) iii) Principal plane, shear plane
1 2
, 2 = 400 - 200 + 600 2
1 2 2 + 4002
= 402.62 MPa, - 202.62 MPa
= 302.65 MPa
1 , 2
max
’ = + = 100 MPa
x y
2
tan 2p = 2 x 40
600
p = 3.790 = 3.80.
Princpal planes - = 3.80 = 93.80
p1 p2 2 1
Shear plane = 48.80 = 138.80
s1 s2
40 p
40
= =
1
2
Given tensor By rotating p by rotating p + 450
400 - 200 = 402.6 - 202.6 = 100 + 100 = 200
Note : The algebric sum of normal stresses on any two mutually Prependicular planes should be same.
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STRENGTH OF MATERIAL
Graphical Method (Mohr Circle)
H
max
G J (,) OD =
x
Resulant
R OB =
xy y
y 2p OE =
2 1
2 D
xy O A B <c E OA =
2
x
2p1 xy CH = max
y
xx
xy F
2
I
y <>
1
= OC + CE
1
-
x y , BC = CD = x y
OC =
2 2
CE = CF = CD2 + DF2
or = - 2
max xy
2 + x2 y
+ - 2 x2 y
1, 2 x x y
= y+ +
2 2
= OC = +
x y
2
tan 2 p = xy = x y
- ) / 2 x - y
x y
Note :- * Any pt. on the Mohr Circle will have two coordinates Normal and shear coordinates.
* Each radius drawn to Mohr circle is a plane in the material.
x coordinate :- Normal stress
y coordinate :- Shear stress
*All the angles taken in the material will be doubled in Mohr circle.
* The ends of Mohr circle on x - axis represents Principal stresses.
CH & CI are shear stress planes.
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MECHANICAL ENGINEERING
SPECIAL CASE :-
Case I :- Uniaxial stress system (1D) D
Eg. : Elements of stress
=
x
xy
= A
y
xy = 0 o C x =
t
* In 1 D system, there will be only one Principal stress,
= OA =
x
= CD = x
2max
Note : * In 1D stress system there is only one Principal stress
* Radius of any Mohr Circle is equal to in 2D system.
max
Case II : - Pure shear stress >
Eg. : Torsion on circular shaft D
B o F >
C
E
A
Radius = OA = max =
= OC = (tension)
1
= OB = (Compression)
2
= OF, = FE,
R =
= OE, max = radius
Note :- In this case origin coincides with centre of Mohr Circle ’ = 0.
* If the centre of Mohr Circle concides with origin then resultant stress on any inclined plane = radius of
Mohr circle or in 2D system.
max
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STRENGTH OF MATERIAL
Case III :-
>
>
Bo A
= (T) = OA Centre = O
x
= -(C) = OB Radius = OA or OB
y
in 2D = +
xy = 0 max
Case IV :- Spherical stress
Mohr circle
(Points)
o >
A
0
< >
= OA =
x
y =
=0
xy
= radius of Mohr Circle = 0
max
x = OA =
= 0
Note :- In real isotropic material, Mohr circle is a point.
Case V :- Hydrostatic stress
=- Mohr circle
1
=- >
2
xyz = 0
Note : If any material subject to hydrostatic presssure then Mohr circle is a point.
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MECHANICAL ENGINEERING
Case VI :- Beams or shafts subjected to normal and shear stress
C
OA = =
x
F oD
y =0 A
= = AB = OC
xy
B
max in 2D = CD = DB
Note : In beams, the principal stresses are always with opposite nature.
Case VII :-
= = OA C
x
= = OA AD
y oE B
xy = = AB = AC
OD = = OA + AD = ’ +
1 max
’
OE = 2 = OA - AE = - max
Hook’s Law for plane stresses :-
= 1 - )
x E x y
1
E y
= - )
y y x
= + )
z E x y x x
=
xy xy
G
= E + )
x - 2 x y y
1
= E + )
y 1 - 2 y x
Note :- In plane stress system, stress in Z direction is zero but strain in Z direction 0. This is due to lateral
strain of x & y direction.
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STRENGTH OF MATERIAL
Triaxial Stresses
y
In this case
z
=
x x =
z y =
x
3D Stress System :- 2
y
* 3 Principal planes & 3 Principal stress >
= Major Principal Stress >
= Intermediate Principal Stress
= Minor Principal Stress C <> B < A
<
-
max 1 1
=
- In plane
max 2 2 max
=
-
max 3
=
3
(in 3D) is maximum of
max
, &
max 1 max 1 max 3.
or = 1 x maximum of [ - 2 - 3 -
max 2 1
Principal stresses of 3 D system
y
Principal stress are given by real roots of following cubic equation :-
yx
3 - - C = 0 xy
there A = + y+z
x yz x
y+x
B= x z + y z - xy - xz - yz
zy xz
C= z+xyxz -xyz - -
x y yz y xz z xy
z zx
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MECHANICAL ENGINEERING
Eg. 1 : - 50 KN
Calculate : , 5 KN
1 5 KN
25 KN
Sol :
50
50
>
50
= Px 100 KN <>
x A = 10 MPa < <
>
= Py
y A = 2 MPa
z= Pz
A = 40 MPa
= 5000 = 2 MPa
xy 2500
+ -
1 x y x y
, = +
x2y
= 20, 10 MPa
Third plane Z is Principal plane Because no shear force is acting on this plane.
3rd Principal stress = 40 MPa
z
= 40, 20, 10.
1 ,
Strain Energy due to Principal stress
U= 1 + 1 + 1
2 1 1 2 2 2 2 33
1 1 1
2 1 2 - 3 2 2 - 3 - 1 2 3 3 - 1- 2
U= 1 - E + 2 E E + E
E E E E E
U= 1 2 + + +3
2E 1 E 1 2
of
U= 1 2 + 1 .
2E 1
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STRENGTH OF MATERIAL
Hook’s Law for Triaxial Stresses :-
= - + )
x x E y z
E
- + )
=y E x z
yE
= - + )
z y E x y
E
OR
= E [(1 - ) + (y + )]
x (1 + ) (1 - 2) x z
= E [(1 - ) + (z + )]
y (1 + ) (1 - 2) y x
= (1 + E - 2) [(1 - ) + (z+ )]
z ) (1 z y
Strain on inclined plane
= + - cos2 + xy sin 2
x y x y 2
+
2 2
- sin 2 - xy cos 2
2 x y 2
=
2
Principal strains in 2D
= + - 2 xy 2
12 x y x y 2
+ +
2
2
Note :- Principal planes calculated by stress will be same as that calculated by strains
Principal Planes :-
tan 2p = xy
-
xy
Maximum shear strain (2D)
max -
12
2 =2
In 3D (1 3)
2
max - - -
2 = Maximum of 1 2 3 1
, 2 3,
222
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MECHANICAL ENGINEERING
Strain gages
* It is strain measuring device Strain gage can be into 3 types :-
i) Mechanical strain gages
Eg. : Dial gage, deflectometer, classified extensometer
ii) Electrical strain gage
iii) Digital strain gage
No. of strain gages required = No. of independent stress compoment
State of stress Independent stress component
1D 1
2D 3
3D 6
Strain Rosette :- Arrangement of strain gages
= cos2 + sin2 + xy sincos.
x y
1350
900 1350 450 1350
<<
450 2700 < 00
450 2700 < <
450 <4 5 0 3 50
1
00 00
Fig : - Star rosette
Fig : - Rectangular rosette. Fig. : Delta rosette
Ques. : The strains measured in a rectangle strain rosette are as shown in the figure of
E = 200 GPa & = 0.3, determing major principal stress if strength of material used is 400 MPa, check
for safety
= 400 ,0 = 650 , 0 = 520
0
Sol. : = 0 ,
0 = 400
= 450 = 45 cos2 + sin2 +xy sincos.
0x x
= 900 90 520
gy
45 650 = 400 cos2 45 + 520 sin2 45 + xy sin 5 cos
xy = (650 - 200 - 260) x 2
xy = 380 2
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STRENGTH OF MATERIAL
Principal strains
= + + - 2 xy 2
1, xy x y 2
2 2 +
2
= 659
1
y = 260
We get .
= 162.02 MPa (T)
1
2 = 100.77 MPa (T)
< strength (400 MPa) material safe.
1
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MECHANICAL ENGINEERING
4 BENDING STRESS IN BEAMS
Simple bending or Pure bending : < M
B
A B A
M
(-) +
<
SFD
BMD
Between A & B there is no shear force & B.M. is const. i.e. between A & B beam is absolutely free from
shear but is subjected to B.M.M. This condition of beam between A & B is called pure bending or
simple bending.
Note : In real beam self weight will also be acting over the beam in the form of udl pure bending is not
possible in parctice.
Assumptions made in Bending Theory :
1) Euler Bernaulli’s assumption-plane cross section remains plane after bending.As per this
assumption, there is no twisting or warping in this cross section.
max
Linear upto failure
strain dist. in cross section
The Baunaullis assumption is related with strains which indicates strain distribution in cross section
is linear upto failures
Validity :
i) Valid for all type of loads
ii) Valid for all theories (Elastic, Plastic, all failure theories)
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STRENGTH OF MATERIAL<
Not Valid :
1) i) Long (slender) columns (Because of buckling)
ii) Sinking of supports
iii) Torsion of non circular shafts (In case of circular members subjected to torsion, assumption is
valid)
iv) Deep Beam
2) Material is homogenous, isotropic & follows hooks law
3) Every layer is free to expand & contract. i.e. no shear force between layer.
4) E has same value in tension & compression
E) tension = E) compression = E (const.)
i.e. Material properties are not changnig
5) Beam is subject to pure bending i.e. it bends in an arc of a circle
6) Radius of curvature is very large i.e. deflection is very small.
Bending / flexural equation
ME
I =R= y
M = Moment of resistance or Bending moment ; I = Moment of Inertia about N.A. ; E = Young’s
Modulus ;
R = Radius of curvature
= Bending stress; y = Distance from neutral axis
Bending stress Distribution across cross section :-
MM
------------------------------
max
<<
<
Linear up to proportaional limit
<
<
<
= M . y
I
At N.A. y = 0, = 0
At y = y , =
max max
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MECHANICAL ENGINEERING
Limitations : The equation is useful for homogenous & prismatic beam. For composite beam, equation
cannot be used.
Note : At neutral axis bending stresses are zero.
Section Modulus
= M . y
I
max= M . ymax
I
I
ymax = z = section modulus
Or max= M
Z
Note : More section modulus, more will be the strngth of beam
M.I. and Section Modulus of various section
I
Z= y
max
for rectangle
I= bd3 , ymax = d b
12 2 d
bd3
I 12 = bd2
d6
yZ = =
max
2
For solid circular section
IN.A. = d4 , ymax = d
64 2
<>
d4
Z= 64 d
d
Copyright : Ascent Gate Academy 42
2
z = d3
32
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STRENGTH OF MATERIAL
Triangular section
max
<>
bh3 2 <>
h
Z= 36 I h
2 = ymax 3
h >
3 <b
Z = bh2
24
Note : - In triangular section maximum stress is at top fibress.
Square section : - Diamond Section (Square with one
diagonal horizontal and one vertical)
a4 I = a4
12
Z= 12 a
a a
a4 <>
2 a
Z = 12
a3 a 2
Z= 2
6
z = a3
6 2
>
>
BD3 bd3
( D4 - d4 )
D d Z = 12 12 < d>
b D < > Z = 64
D D
<>
< 22
<
<B > =2 = 1.414
(Strength ) square Zsq a3 / 6
(Strength ) diamond = Zdia = a3 / 6 2
or Str ) sq = 1.414 Str ) Dia
Square is 41.4% more stranger than diamond
Note : Sections having particles distributed away from N.A. has more strength
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MECHANICAL ENGINEERING
Flexural Rigidity (EI) : Here EI - Flexural Rigidity
M E +M > +Ve B.M. +M
I R
=
1 E
R = EI
1 = Curvature (K)
R
M -Ve B.M.
K=
EI ( N/mm2 or KN/mm2)
Significance :- -M > -M
K 1
EI
More E I , more rigid will be the beam and less will be deflection
E - material property
I - Cross setional property
Note : The strenght (z) is a cross sectional property where as stiffness depends on both xnal & material
property.
Beam of Uniform Strength
If Maximum stress anywhere along a beam is const. then it is called Beam of Uniform strength
M Or M = x z Should be const.
I =y
M Z.
A prismatic beam cannot be a beam of uniform strength. This is because moment is changing in a beam
but z remains const. To make beam of uniform strength.
ZM
a) Beam of uniform strength - constant depth W
= M y
I
-----
I= bxd3 y= d bx
12 2
bxd2 <x>-----
Z= 6 -----
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Wx STRENGTH OF MATERIAL ---------
= 2 xW l
b d2 l 2
2
x
< x>
6
W/2
bx = 3W x W/2
d2 x
Maximum width at centre where x = l / 2
3Wl
b.max= 2d2
b) Beam of uniform strength - const. width
= M <x > b
Z dx
W/2
M= Wx xW -----
2
x
Z = bdx2 ---------
6
Wx
= 2
bdx2
6
W/2
dx = 3Wl
2b
Note : The beam of uniform strength is not practicaly feasible as in the real beam there will be shear force
in the beam in addition to B.M.
Make beam of uniform strength for the loading shown below
(I) M = Z W/m
wl2 = bd2
8 6
3wl2 <> wl2
b = 4d2 8
3wl2
d = 4d2
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MECHANICAL ENGINEERING
(II) M
y=I
wl2 = x bd2
2 6
3wl2 wl2
b = 6d2 2
d = 3wl2
4d
Strength of Solid & loose member
Strength ) loose = Z loose <b> <> <b>
Strength ) solid Z solid nt <>
_________________
Z) loose = n bt2 _________________ t
_________________
6 _________________
_________________
Z )solid = b (nt)2
6
Str. ) loose 1
Str. ) solid = n
Note : * Solid member is n times stronger than loose member
Ratio of Curvature of loose to solid member
M = E or 1 = M =K
I R R EI
K loose = ( M / EI ) loose = I solid = b (nt)3
K solid (M / EI ) solid I loose 12
bt3
K loose
K solid nx
12
= n2
Note : * Curvature of loose member is n2 times that of solid member.
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STRENGTH OF MATERIAL
Strength of solid & loose member vertical cut.
(nt) d2 t
<>
Str ) loose Z loose 6 =1 _____________________
Str ) solid = Z solid = d _____________________
nx td2 < nt > _____________________
_____________________
6 _____________________
<>
(nt) d3
K loose I solid 12
K solid
= I loose = =1
n
td3
12
Note : In vertical cutting depth is not changing it will not affect strength
Stress distribution in solid & loose member :-
Loose
1
Solid d1 ___ . ___ . ___ . ___ . ___. ___ . ___ .< >< >< >
max
2
N.A. d2 ___ . ___ . ___ . ___ . ___. ___ . ___ .
max 3
d3 ___ . ___ . ___ . ___ . ___. ___ . ___ .
Note : * In loose beam for any part Radius of curvature is same i.e. R = R = R centre of Curvature is
1 2 3
different [fig (a)]
* In Solid beam, for different layers, centre of curvature is same but radius of curvature is
different (b).
C C1 , C2
1
R C2 R1
1 R2
R (b)
2
(a)
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MECHANICAL ENGINEERING
Strongest rectangle cut from a circular cross section : - b
Dh
For strongest dimension
Z = Z max
for Z to be maximum dz
db = 0
we get b
D
= 1 (I)
3
Also dz From (I) & (II) , we get
dh = 0
we get h 2 ( II ) b = 1
D= 3 h 2
Ratio of weights of circular square & rectangular section beams : -
a 2b
d a b3
<> 2
1
W=Al
W1 = 1.118 , W3 = 1.26
W2 W2
FLITCHED BEAMS :
Flitched beam is a composite section consisting of a wooden beam strengthened by metal plates.
The arrangement is so connected that the components act together as one beam.
In order slip may not occur, strains in steel and <>
wood at any distance y must be equal y
N ----------------------------- A
Strain = w = s
steel plate
Ew Es
& - stress at distance y from N.A. of wood & steel wooden joist
w s
Ew & Es, - Mod. of elasticity of wood & steel
= Es
s E w
w
or = m .
s w
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STRENGTH OF MATERIAL
Where m= Es = Modular ratio between steel & wood
Ew
or Estrong , m>1
m = E weak
Moment of Resistance of a Section :
Mr = Mw + Ms
or Mr = x bd2 + 2 x x td2
w 6 s 6
= m.
s w
1 ( b + 2mt) d2
Mr = 6 w
Hence, Modular Ratio of a section is same as that of wooden member of width b + m (2t) and depth d.
This rectangular section b + m (2t) units wide & d units deep is called Equivalent Wooden Section
< b + m (2t) >
t bt
<>
< >< >< >
----------------
d ----------------
< >< >< >
mt b mt
Flitched beam secn Equivalent wooden secn
Note : The moment of resistance of a flitched beam is determined by considering equivalent wooden steel
section
(1)
___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _
Timber
steel t <>
< > Strain
mb distribution
Equivalent Timber Section
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MECHANICAL ENGINEERING
b/m
< >
b
Equivalent Steel Section
Note :- While converting into equivalent areas, dont change the depth as it may change the stress values
(2)
b/m
<>t Equivalent Steel
< >
mb
Equivalent Timber
(3)
___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _
< >< > mt >
bt < >< b
Equivalent Timber
Note : - In flitched beams strain distribution is linear (bernaullis assumption is valid). The stress distribu-
tion may not be linear (as bending eqution is not valid for flitched beams).
After converting to equivalent beam with the help of modular ratio then we can use bending eqution.
Note : Of 3 cases seen above, case (1) is the best (i.e. having highest strength
( its equivalent is I - section)
(1) > (2) > (3)
“Whatever you believe with feeling becomes your reality”
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