Mechanics of structures - module3

Mechanics of structuresMechanics of structures
Stresses in beamsStresses in beams,
Inelastic bending,
Deflections
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
1

Module III
Bending stresses in beams - shear flow - shearing stress formulae for
Module III
Bending stresses in beams shear flow shearing stress formulae for
beams –
Inelastic bending of beams –Inelastic bending of beams –
Deflection of beams - direct integration method - singularity
f ti iti t h i t th dfunctions - superposition techniques - moment area method -
conjugate beam ideas –
Elementary treatment of statically indeterminate beams - fixed and
continuous beams
Dept. of CE, GCE Kannur Dr.RajeshKN
2

Important assumptions in the theory of simple bending
• Material of the beam is homogeneous and isotropic
p p y p g
• The stress is proportional to strain and stress is within elastic limit
• Modulus of elasticity is same for tension and compression• Modulus of elasticity is same for tension and compression
• Plane vertical sections remain plane after bending
• Loads are applied in the plane of bending
• Cross section is symmetrical about vertical axis
3

Theory of simple bending
r sy
p q
O
θ p q Rθ′ ′ =
( )r s R y θ′ ′ = −R ( )r s R y θ= −
p q pq′ ′ =
r’ s’
p’ q’
y
p q pq
4
p q

Strain for the fibre rs
rs r s p q r s
rs rs
′ ′ ′ ′ ′ ′− −
=
( )R R y y
R R
θ θ
θ
− −
= =
rs rs R Rθ
yσ
=
Ey
σ =
If σ is the stress in fibre rs, and
E is the Young’s modulus, strain
E R R
E is the Young s modulus, strain
O l f
. .
Ey
dA dA
R
σ =
On any cross section, normal force on an
elemental area dA,
dA
R
Moment about NA due to normal force on dA, y
NA2
. . .
Ey
dA y dA
R
σ =
Sum of all such elemental moments is the
NA
5
Sum of all such elemental moments is the
moment of resistance of the cross section.

Moment of resistance balances the bending moment at the cross section.g
2
. . .R
A A
Ey
M M dA y dA
R
σ= = =∫ ∫
2
.
A
E
M y dA
R
= ∫
2
.
A
y dA∫ is the moment of inertia of the cross section about the
neutral axis
EI
M
R
∴ = M E
I R y
σ
= = Bending equation
R I R y
6

My
σ Along a cross section, variation of bending stress is
I
σ =
For a beam with transverse downward loading, compressive force
g , g
linear.
g, p
acts above NA, tensile force acts below NA
C
T
MM
Total compressive force = Total tensile force
7

C dA
NA
y
T
0 . . 0C T dA dAσ σ+ = ⇒ + =∫ ∫
Total force on the cross section is zero
. 0dAσ =∫
1 2A A
∫ ∫
0 0
My
dA dA∫ ∫ ∫
A
∫
. 0 . 0
A A
y
dA dA
I
σ = ⇒ =∫ ∫ .
A
y dA
dA
∫
∫
Distance from a
point to
the centroidal axis
. 0
A
y dA⇒ =∫ A
dA∫ the centroidal axis
8
i.e., NA coincides with the centroidal axis

Section Modulus
( )
M
I y
σ =
M
maxZ I y=
Section modulus for various shapes of cross section
max
M
Z
σ =Section modulus
b
( )3 212bdI bd
d
( )
( )
2
max
12
2 6
bdI bd
Z
y d
= = =
d ( )4 364
2 32
d d
Z
d
π π
= =
9

B
( )
( )
3 3
3 3
1
12
2 6
BD bdI BD bd
Z
D D
− −
= = =
b
( )max 2 6y D D
D d
( ) ( )
4 4
4 4
64
2 32
D d D d
Z
D D
π
π− −
= =
2 32D D
10

2h
3
CG axis h
b
3
bh⎛ ⎞
⎜ ⎟ 2
max
36
2 24
3
bh
Z I y
h
⎜ ⎟
⎝ ⎠= = =
11

Problem 1. A beam 500mm deep of symmetrical rectangulary g
section has I = 1x108 mm4 and is simply supported over a span
of 10 m. If the beam carries a central point load of 25 kN,
calculate the bending stress at 100 mm above neutral axis andcalculate the bending stress at 100 mm above neutral axis and
the maximum bending stress on the beam.
My
σ
8 4
1 10 mmI = × 100 mmy =y
I
σ =
25 10
62 5 kN
WL
M
×
1 10 mmI = × 100 mmy
6
262.5 10 100
62 5 N/
× ×
∴62.5 kNm
4 4
M = = = 2
8
62.5 N/mm
1 10
σ∴ = =
×
max
max
My
I
σ =
max
500
250 mm
2
y = =
6
2
max 8
62.5 10 250
156.25 N/mm
1 10
σ
× ×
∴ = =
12
max 8
1 10×

Problem 2. For the above problem, calculate the UDL it mayy
carry, if the maximum bending stress is not to exceed 150
N/mm2.
I
I
M
y
σ
= max
.allow
R
I
M
y
σ
=Moment of resistance
2
150 N mmallowσ = max 250 mmy =
8 4
1 10 mmI = ×
8
6150 1 10
60 10 Nmm 60 kNm
250
RM
× ×
∴ = = × =
2
.
60 kNm
8
alloww l
= . 2
60 8
4.8 kN/m
10
alloww
×
∴ = =
13

Problem 3. Find the maximum BM that should be imposed on
hi i f il if h il i h fl ithis section of a cantilever, if the tensile stress in the top flange is
not to exceed 40 MPa. What is then the value of compressive
stress in the bottom flange?stress in the bottom flange?
200 40 260 200 40 140 120 40 20
y
× × + × × + × ×
=
200mm
200 40 200 40 120 40
y =
× + × + ×
158.5 mm=
i.e., 158.5 mmcy =
3 3
200 40 40 200
280 158.5 121.5 mmty∴ = − =
( ) ( )
( )
3 3
2 2
3
2
200 40 40 200
200 40 121.5 20 40 200 158.5 140
12 12
120 40
120 40 158.5 20
I
× ×
= + × × − + + × × − +
×
+ + × × −
14
( )120 40 158.5 20
12
+ + × ×
8 4
2.056 10 mmI = ×

.ll Iσ
2 8 4
40 N/mm 2.056 10 mm× ×
,max
.allow
t
I
M
y
σ
=Max. BM 121.5 mm
=
6
67.6 10 Nmm= × 67.6 kNm=
My
Compressive stress in the bottom flange ,maxc
c
My
I
σ =
6
67.6 10 Nmm 158.5 mm× × 2
8 4
67.6 10 Nmm 158.5 mm
2.056 10 mm
cσ =
×
2
52.19 N mm=
2
40 N mmtσ =t
40 158.5
σ
×
=
Alternatively,
121.5mm
NA
121.5
cσ =
2
52.19 N mm=
158.5mm
15
2
52.19 N mmcσ =
1

Theory of simple bending is applicable to pure bending.y g g
But since the effect of shear on bending stress is negligible, the
theory can be applied generally
The effect of shear on bending stress is not of practical
theory can be applied generally.
The effect of shear on bending stress is not of practical
importance, but shearing stresses must be considered for their
own importance.
16

Shear stresses in bending My
σc
I
σ =
( )
d
M dM y
σ
+
=
p’ q’p’ q’
c d d
I
σ
m n
y1M M dM+
yc m n
y1 y
c d
M dM+
p q
dx
p q
. .
c cy y
c
My
dA dA
I
σ =∫ ∫
( ). .
c cy y
d
M dM y
dA dA
I
σ
+
=∫ ∫
17
1 1y y
I 1 1y y
I

c cy y
dA dAσ σ−∫ ∫
( )c cy y
M dM y My
dA dA
+
∫ ∫
cy
dM
y dA= ∫
1 1
. .d c
y y
dA dAσ σ∫ ∫
( )
1 1
. .
y y
dA dA
I I
= −∫ ∫ 1
.
y
y dA
I∫
.
cy
dAσ∫
A
cy
dAσ∫1
.c
y
dAσ∫
b
d
1
.d
y
dAσ∫
. .b dxτ
NAdx NA
cy
dM 1 cy
dM d M
1
. . .
y
dM
b dx y dA
I
τ = ∫
1
1
.
y
dM
y dA
Ib dx
τ = ∫
cy
V
d M
V
d x
=
( )V A
181
.
y
V
y dA
Ib
τ = ∫
( )V Ay
Ib
τ =

Shear stress distribution Rectangular sectionShear stress distribution – Rectangular section
( )V Ay
τ =
b Ib
A
1d d
Ay b y y y
⎡ ⎤⎛ ⎞ ⎛ ⎞
= − × + −⎜ ⎟ ⎜ ⎟⎢ ⎥
d
y
y
2 2 2
Ay b y y y= − × + −⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠⎣ ⎦
1V d d⎛ ⎞ ⎛ ⎞
2
2V d⎛ ⎞
maxτ
1
2 2 2
V d d
b y y
Ib
τ ⎛ ⎞ ⎛ ⎞
= − × +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
2 4
V d
y
I
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
Hence, variation of shear stress over the cross section is parabolic
2 2
0
V d d
τ
⎛ ⎞
⎜ ⎟2 0
2 4 4
y d
I
τ =± = − =⎜ ⎟
⎝ ⎠
2 2
3V d Vd V⎛ ⎞ 3
19
( )max 0 3
3
2 4 28 12
y
V d Vd V
I bdbd
τ τ =
⎛ ⎞
= = = =⎜ ⎟
⎝ ⎠
3
2
meanτ=

Shear stress distribution I sectionShear stress distribution – I section
( )V Ay
τ =
B
Shear stress in flanges
Ib
1
2 2 2
D D
Ay B y y y
⎡ ⎤⎛ ⎞ ⎛ ⎞
= − × + −⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠⎣ ⎦
D d
b
g
2 2 2
⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠⎣ ⎦
1V D D
B y yτ
⎛ ⎞ ⎛ ⎞
= − × +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
2
2 4
V D
y
I
⎛ ⎞
= −⎜ ⎟
⎝ ⎠2 2 2
y
B
y
I
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ 2 4
y
I
⎜ ⎟
⎝ ⎠
Hence, variation of shear stress over the
flange is parabolic
2 2
2 0
2 4 4
y D
V D D
I
τ =±
⎛ ⎞
= − =⎜ ⎟
⎝ ⎠
y
2
2 4 4
y D
I
=± ⎜ ⎟
⎝ ⎠
( )
2 2
2 2V D d V
D dτ
⎛ ⎞
= − = −⎜ ⎟
20
( )/2
2 4 4 8
y d D d
I I
τ = = =⎜ ⎟
⎝ ⎠

( )V Ay
τ =Shear stress in web
Ib
( ) ( ) 1
2 4 2 2 2
D d D d d d
Ay B b y y
− + ⎛ ⎞ ⎛ ⎞
= + × − × +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
y
2 4 2 2 2
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( ) ( ) 1
2 4 2 2 2
D d D dV d d
b y y
Ib
Bτ
− +⎡ ⎤⎛ ⎞ ⎛ ⎞
= + × − × +⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠⎣ ⎦2 4 2 2 2
y y
Ib
⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠⎣ ⎦
( ) ( )2 2 2 2
4
8
V
B D d
b
b d y
I
τ ⎡ ⎤= − + −⎣ ⎦
Hence, variation of shear stress over the
web is parabolic
( ) ( ) ( )2 2 2 2 2 2V VB
⎡ ⎤
( ) ( )8 bI ⎣ ⎦
( ) ( ) ( )2 2 2 2 2 2
2
8 8
y d
V VB
B D d b d
b
d D d
I bI
τ =±
⎡ ⎤= − + − = −⎣ ⎦
( )2 2 2V
⎡ ⎤
21
( )2 2 2
max 0
8
y
V
B d
Ib
D bdτ τ =
⎡ ⎤= = − +⎣ ⎦

maxτ
Shear stress distribution
22

Shear stress distribution – Circular section
( )V Ay
Ib
τ =
R
y
R
b
dy
2 2
2b R y= −
( )2 2 2
4b R y= −
. .
y R
Ay b dy y
=
= ∫y ( )4b R y= −
2 . 8 .b db y dy= −
1
y y=
1
. .
4
y dy b db= −
When ,y y b b= =
When , 0y R b= =0
1
.
4
b
b b
Ay b b db
=
=
⎛ ⎞
∴ = −⎜ ⎟
⎝ ⎠
∫
0 3
21
.
4 12
b
b b
b
b db
=
=
= − =∫
3 2
V b Vb
τ
⎛ ⎞
= =⎜ ⎟
( ) ( )2 2 2 2
4V R y V R y× − −
= =
b b b b
23
12 12Ib I
τ = =⎜ ⎟
⎝ ⎠ 12 3I I

Hence variation of shear stress over the cross section is parabolicHence, variation of shear stress over the cross section is parabolic
( )2 2
0
V R R
τ
−
= =
m a xτ
0
3
y R
I
τ = = =
2
VR 2
4 4 4VD V V
max 0
3
y
V
I
τ τ == = 4 2
3 3 3
12
64 4
mean
V V V
AreaD D
τ
π π
= = = =
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
24

Find shear stress distribution for the sections shown below:Find shear stress distribution for the sections shown below:
25

Inelastic Bending of Beams
(Plastic Analysis)
g
ressstr
strainO
Idealised stress-strain curve of elastic-plastic materialp
AssumptionsAssumptions
• Plane sections remain plane in plastic condition
S i l i i id i l b h i i d i
• Stress-strain relation is identical both in compression and tension

• Let M at a cross-section increases gradually.
• Within elastic limit, M = σ.Z
Z i ti d l I/• Z is section modulus, I/y
• Elastic limit – yield stresses reachedElastic limit yield stresses reached
My = σy.Z
• When moment is increased, yield spreads into inner fibres.
Remaining portion still elastic
• Finally, the entire cross-section yields, at a moment of MP

yσ
σ
σ
yσ
yσ yσ yσσ
σ
yσ yσ yσσ

σy σy
σy σyσy σy y y
σy
σy

Plastic moment
• M – Moment corresponding to working load
• My – Moment at which the section yieldsMy Moment at which the section yields
• MP – Moment at which entire section is under yield stress – plastic moment
yσ
CC
T
yσ
MP

• At plastic moment the entire section is under yield stressAt plastic moment, the entire section is under yield stress
C T
A Aσ σ
=
=
A
A A⇒ = =
•NA divides cross-section into 2 equal parts
c y t yA Aσ σ=
2
c tA A⇒ = =
A
C T σ= =•NA divides cross-section into 2 equal parts
2
yC T σ= =
yσ
y
2
y
A
C = σ
yt
yc
A
T σ=
yσ
2
yT σ
31

Zσ
Similar to
•Couple due to ( )
A A
Z
⎛ ⎞
⎜ ⎟
Plastic modulus
yZσ•Couple due to ( )
2 2
y y c t y py y Zσ σ σ⎛ ⎞
= + =⎜ ⎟
⎝ ⎠
Plastic modulus
Shape factor ( )1pZ
Z
= >
b
Rectangular cross-section:
p ( )
Z b
Section modulus ( )
( )
3 212
2 6
bdI bd
Z
y d
= = =
d
( )y
( )
2
A bd d d bd
Z y y
⎛ ⎞
+ +⎜ ⎟Plastic modulus ( )
2 2 4 4 4
p c tZ y y= + = + =⎜ ⎟
⎝ ⎠
Plastic modulus
Shape factor 1.5pZ
Z
= =

Shape factor for circular section
d( )
2
p c t
A
Z y y= +
2 3
2 2
8 3 3 6
d d d d⎛ ⎞⎛ ⎞
= + =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
π
π π
1 7pZ
S∴ = =( )4 364d d
Z
π π
⎝ ⎠⎝ ⎠
1.7S
Z
∴ = =( )
2 32
Z
d
= =
Sh f t f t i ul ti
2h
Shape factor for triangular section
( )
2
p c t
A
Z y y= +
h
Equal area axis
cy
2
3
h
3
bh⎛ ⎞
⎜ ⎟
( )
2
p c t
Equal area axis
tyCG axis
2
36
2 24
3
bh
Z
h
⎜ ⎟
⎝ ⎠= =
b
S = 2.346

20mm
Shape factor for I section
10mm250mm
p
20mm
200mm
S = 1.132

Z
Load factor
P y Pcollapse load M
Load factor
working load M Z
Zσ
σ
= = =
Factor of safety and load factor
YieldLoad
Factor of Safety= =
Working Load
YieldStress
yW
W
σ
Elastic Analysis : Factor of Safety
Pl i A l i L d FYieldStress
= =
Working Stress
yσ
σ
Plastic Analysis : Load Factor

Plastic moment capacity of a rectangular cross-section:
2
P P
bd
M Zσ σ= =
2
4
p
bd
Z =Plastic modulus
Hence, plastic
t it 4
P y P yM Zσ σmoment capacity
2 2
6 1.5 6
ybd bd
M Z
σ
σ σ= = =
Elastic moment capacity
with a factor of safety of 1.5
2
2
9
2.25
4
P y P
M bd
M Z M
bd
σ= = =2
9
y
M
bd
σ∴ = Hence,
4bdbd
2 2
bd bd Elastic moment capacity
6 6
y
bd bd
M Zσ σ σ= = =
p y
with a factor of safety of 1.0
2
6 bd6M
Dept. of CE, GCE Kannur Dr.RajeshKN36
2
2
6
1.5
4
P y P
M bd
M Z M
bd
σ= = =2
6
y
M
bd
σ∴ = Hence,

Plastic hinge
• When the section is completely yielded, the section is fully plastic
• A fully plastic section behaves like a hinge – Plastic hinge
Plastic hinge is defined as an yielded zone due to bending in a
structural member, at which large rotations can occur at a section
at constant plastic moment, MP
Mechanical hinge Plastic hinge
Mechanical hinge and Plastic hinge
Mechanical hinge
Resists zero moment
Plastic hinge
Resists a constant moment MP

Mechanism of failure in a simple beamp
D t i t b &Determinate beams &
frames: Collapse after first
plastic hinge
Simple beam
MP
M
.uW l
MEquilibrium
4
u
PM =
4M
Equilibrium:
8 PM
38
4 P
u
M
W
l
∴ = 2
P
uw
l
∴ =If UDL,

fl i f bDeflection of beams
• Differential equation of the elastic curve
• Slope and deflection of beams by method of
successive integration
• Macaulay’s method
• Moment area method
• Conjugate beam method
39

Differential equation of the elastic curveDifferential equation of the elastic curve
yy
ds Rdα=
2 2
q
Rdα
ds
2 2
ds dx dy= +
tan
dy
α =
p
dy
dx
ds a
dx
α
2
2
sec
d d yα
α
α α+dα
x
2
sec
dx dx
α =
x
( )
2
2
2
1 tan
d y
d dx
dx
α α+ =
40

2 2
dy d y⎧ ⎫⎪ ⎪⎛ ⎞
2
2
d y
dx
dx
2
1
dy d y
d dx
dx dx
α
⎧ ⎫⎪ ⎪⎛ ⎞
+ =⎨ ⎬⎜ ⎟
⎝ ⎠⎪ ⎪⎩ ⎭
2
1
dxd
dy
dx
α =
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
2
2
2 2
2
d y
R dx
dxds Rd dx dy
d
α= ⇒ + =
⎛ ⎞
2
2
2
2
1
d y
R dx
dy dxdx
d
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠ ⎛ ⎞
2
1
dy
dx
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
2
1
dx dy
dx
⎜ ⎟
⎝ ⎠ ⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
2
d2
2
3
2 2
1
1
d y
dx
R
dy
=
⎡ ⎤⎛ ⎞
⎢ ⎥⎜ ⎟
2
2
1 d y
R d
=
1
dy
dx
⎛ ⎞
+⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
2
R dx
2
1M E M d 2
d2
2
1M E M d y
I R EI R dx
= ⇒ = =
2
2
d y
M EI
dx
=
I th b d i ti ff t f h d fl ti i NOT
41
•In the above derivation, effect of shear on deflection is NOT
taken into account, since it is assumed to be very small.

Method of successive integrationsMethod of successive integrations
2
d2
2
d y
M EI
dx
=
2
2
d y
EI M
dx
=
2
d y
EI M C= +∫ ∫
dy
EI M C x C= + +∫ ∫∫12
EI M C
dx
dy
EI M C
= +
= +
∫ ∫
∫
1 2
1 2
EI M C x C
dx
EIy M C x C
= + +
= + +
∫ ∫∫
∫∫
{ }
1
1
EI M C
dx
dy
M C
= +
= +
∫
∫ { }
1 2
1 2
1
y
y M C x C
EI
= + +
∫∫
∫∫{ }1M C
dx EI
= +∫ { }EI ∫∫
42

Slope at B of the deflected beam = (dy/dx) at B
( )
2
2
d y
EI M P l x= = − −l B
Deflection at B = y at B
0 0 0
dy
At x C= = ∴ =
( )2
dx
2
dy Px
l
x
10, 0 0At x C
dx
= = ∴ =
1
2
dy Px
EI Plx C
dx
= − + +
2 3
2
2 6
Plx Px
EIy C
−
= + +
2
2
dy Px
EI Plx
dx
= − +
2 62dx
2 3
Plx Px−
20, 0 0At x y C= = ∴ =
2 6
Plx Px
EIy∴ = +
43

2 2
,
2 2
dy Px dy Pl
At x l EI Plx
dx dx EI
= = − + ∴ = −
2 2x l
dx dx EI=
2 3
Plx Px−
3
,
2 6 x l
Plx Px
At x l EIy
=
= ∴ = +
3
3
Pl
y
EI
∴ = −
If when x increases, y also increases, then slope (dy/dx) is +ve.
If when x increases, y decreases, then slope (dy/dx) is -ve.
44

Slope at B of the deflected beam = (dy/dx) at B
Deflection at B = y at B
( )
22
w l xd y
EI M
−
= = −l
B
d
2
2
EI M
dx
= = −l
x
10, 0 0
dy
At x C
dx
= = ∴ =
3
2 2
1
2 3
dy w x
EI l x lx C
dx
⎛ ⎞
= − − + +⎜ ⎟
⎝ ⎠
3
2 2
2 3
dy w x
EI l x lx
dx
⎛ ⎞
∴ = − − +⎜ ⎟
⎝ ⎠
2 2 3 4
2 2 3 12
w l x lx x
EIy
⎛ ⎞
= − − +⎜ ⎟
⎝ ⎠2 3dx ⎝ ⎠
⎝ ⎠
2 2 3 4
w l x lx x⎛ ⎞
20, 0 0At x y C= = ∴ =
2 2 3 12
w l x lx x
EIy
⎛ ⎞
∴ = − − +⎜ ⎟
⎝ ⎠
45

3 3
3 3
,
2 3 6
dy w l dy wl
At x l EI l l
dx dx EI
⎛ ⎞
= = − − + ∴ = −⎜ ⎟
⎝ ⎠2 3 6dx dx EI⎝ ⎠
4 4 4 4
w l l l wl⎛ ⎞
,
2 2 3 12 8
w l l l wl
At x l EIy y
EI
⎛ ⎞
= ∴ = − − + ∴ = −⎜ ⎟
⎝ ⎠
46

Moment area methodMoment area method
Charles E. Greene
•For deflections of beams especially cantilever beams•For deflections of beams, especially cantilever beams
•Suitable when slopes and deflections at particular points are
required not the complete equation of the deflection curverequired, not the complete equation of the deflection curve
47

Moment area theoremsMoment area theorems
O
R
dαA
B
dα
ds
m
n
α
dα
x.dα y
xdx xdx
B’
48

B
O 1 M
d ds dsα = =
R
dαA
B R EI
M
ds dx d dx
EI
α≅ ⇒ =
dα
ds
m
n
EI
M
d dx
EI
α α= =∫ ∫
α
x.dα y
EI∫ ∫
M
xdx
M
xd xds
EI
M
ds dx xd xdx
α
α
=
≅ ⇒
B’
ds dx xd xdx
EI
α≅ ⇒ =
M
y xd xdx
EI
α= =∫ ∫
49

M t A Th 1Moment Area Theorem 1
Angle between tangents at A & B = (Area of BMD between A & B) /EI
Moment Area Theorem 2Moment Area Theorem 2
Deviation of B from tangent at A = (Moment of BMD between A & B,
b t B) /EIabout B) /EI
50

B
l
Slope at B of the deflected beam = Area of M/EI
( )
p /
diagram between A & B
2
l Pl Pl
= =(-)Pl
EI
2 2EI EI
= − = −
Deflection at B of the beam = Moment of M/EIDeflection at B of the beam Moment of M/EI
diagram between A & B about B 2 3
2
2 3 3
Pl l Pl
EI EI
= − = −
51

l
Slope at B of the deflected beam = Area of M/EI
l
diagram between A & B
2 3
3 2 6
l wl wl
EI EI
= − = −
2
l ( ) 3 2 6EI EI2
2
wl
EI
(-)
Deflection at B of the beam = Moment of M/EI 3 4
3l l l
/
diagram between A & B about B
3 4
3
6 4 8
wl l wl
EI EI
= − = −
52

C
C
A
B
Deflected shape
PL
4
PL
EI
M/EI diagram C
A B
g C
Slope at A of the deflected beam = Area of M/EI
di b t A & C
2
1 l Pl Pl
= =diagram between A & C
2 2 4 16EI EI
= =
Deflection at C of the beam = Moment of M/EI diagram 2 3
Deflection at C of the beam = Moment of M/EI diagram
between A & C about A
2 3
2
3 2 16 48
l Pl Pl
EI EI
= =

Deflected shape
C
B
2
8
wL
EIA B8EI
M/EI diagram C
A B
Slope at A of the deflected beam = Area of M/EI
diagram between A & C
2 3
2
3 2 8 24
l wl wl
EI EI
= =
Deflection at C of the beam = Moment of M/EI
3 4
5 5l wl wl
= =
54
/
diagram between A & C about A 8 2 24 384EI EI
= =

Conjugate beam method
Actual beam
j g
l B
Conjugate beam
(-)Pl
EI
B
A
EI
Slope at B of the deflected beam Shear force at B of the conjugateSlope at B of the deflected beam = Shear force at B of the conjugate
beam when conjugate beam is loaded with M/EI diagram 2
2 2
l Pl Pl
EI EI
= =
Deflection at B of the beam = Bending moment at B of the conjugate
beam when conjugate beam is loaded with M/EI diagram 2 3
2Pl l Pl
j g / g 2
2 3 3
Pl l Pl
EI EI
= =

Conjugate beam method - proof
2
2
d y
EI M
dx
=
3 4
In the actual beam,
3
3
, Shear force
d y dM
EI V
dx dx
= =
4
4
, Load
d y dV
EI w
dx dx
= =
load,
M
w
EI
=
0 0
Shear force,
x x
M
V wdx dx
EI
= =∫ ∫In the conjugate beam,
Bending moment,
x x x
M
M Vdx dx
EI
= =∫ ∫∫0 0 0
EI

2
i h j b i h l b
x x
M d y dy
d d∫ ∫
Hence,
2
0 0
in the conjugate beam in the actual beam
Shear force in the conjugate beam Slope in the actual b ame
M d y dy
dx dx
EI dx dx
= =
=
∫ ∫
2x x x x
M d2
2
0 0 0 0
in the conjugate beam in the actual beam
BM i th j t b D fl ti i th t l b
x x x x
M d y
dx y
EI dx
= =∫∫ ∫∫
BM in the conjugate beam Deflection in the actual e mb a=

Conjugate supportsj g pp
Actual support. Conjugate support.
58

Example 1:
l
Actual beam
B
( )
Conjugate beam
B
A
(-)Pl
EI
Slope at B of the deflected beam = Shear force at B of the conjugate
beam when conjugate beam is loaded with M/EI diagram
2
2 2
l Pl Pl
EI EI
= =
Deflection at B of the beam = Bending moment at B of the conjugate
beam when conjugate beam is loaded with M/EI diagram
2 3
2Pl l Pl
59
2 3
2
2 3 3
Pl l Pl
EI EI
= =

Example 2:
ll
( )
Conjugate beam
B
A
2
2
wl
EI
(-)
Slope at B of the deflected beam = Shear force at B of the conjugate beam
when conjugate beam is loaded with M/EI diagram 2 3
l wl wl
3 2 6EI EI
= =
Deflection at B of the beam = = Bending moment at B of the conjugate beam
when conjugate beam is loaded with M/EI diagram 3 4
3
6 4 8
wl l wl
EI EI
= =
60
6 4 8EI EI

Example 3:
4
Pl
EIA
B
C
2
1 l Pl Pl⎛ ⎞
⎜ ⎟
2
Pl
2
l
2
l
C
Conjugate beam
2 2 4 16EI EI
⎛ ⎞
=⎜ ⎟
⎝ ⎠ 16EI22
j g
Slope at A of the deflected beam = Shear force at A of the conjugate beam
when conjugate beam is loaded with M/EI diagram 2
1 l Pl Pl⎛ ⎞
⎜ ⎟
1
2 2 4 16
l Pl Pl
EI EI
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
Deflection at C of the beam = = Bending moment at C of the conjugate beam
when conjugate beam is loaded with M/EI diagram
2 2 3
1l Pl l Pl Pl⎛ ⎞
1
2 16 3 2 16 48
l Pl l Pl Pl
EI EI EI
⎛ ⎞
= − =⎜ ⎟
⎝ ⎠

Example 4:
2
8
wL
EI
A B
CC
2 3
1 2l wl wl⎛ ⎞
=⎜ ⎟
3
wl
Conjugate beam
2 3 8 24EI EI
=⎜ ⎟
⎝ ⎠ 24EI
Slope at A of the deflected beam = Shear force at A of the conjugate beam when
conjugate beam is loaded with M/EI diagram 3
lconjugate beam is loaded with M/EI diagram
fl f h b d f h b
3
24
wl
EI
=
Deflection at C of the beam = = Bending moment at C of the conjugate beam
when conjugate beam is loaded with M/EI diagram
3 2 4
3 2 5l wl l l wl wl⎧ ⎫⎛ ⎞⎛ ⎞
3 2 5
2 24 8 2 3 2 8 384
l wl l l wl wl
EI EI EI
⎧ ⎫⎛ ⎞⎛ ⎞
= − =⎨ ⎬⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎩ ⎭

Principle of SuperpositionPrinciple of Superposition
Statement: Deflection at a given point in a structure produced by
several loads acting simultaneously on the structure can beseveral loads acting simultaneously on the structure can be
found by superposing deflections at the same point produced
by loads acting individually.
Applicable when there exists a linear relationship between
external forces and corresponding structural displacements.p g p
63

SummarySummary
Bending stresses in beams - shear flow - shearing stress formulae forBending stresses in beams shear flow shearing stress formulae for
beams –
Inelastic bending of beams –Inelastic bending of beams –
Deflection of beams - direct integration method - singularity functions
iti t h i t th d j t b- superposition techniques - moment area method - conjugate beam
ideas –
l f ll d b f d dElementary treatment of statically indeterminate beams - fixed and
continuous beams
64

Mechanics of structures - module3

More Related Content

What's hot

Viewers also liked

Similar to Mechanics of structures - module3

More from SHAMJITH KM

Recently uploaded

Mechanics of structures - module3