Deflection of beam due
to bending moment
1
Introduction
Classification of loading and
Beam Supports
2
Introduction
• Beams - structural members supporting loads at various
points along the member
• Transverse loadings of beams are
concentrated loads or distributed loads
classified
as
• Applied loads result in internal forces consisting of a shear
force (from the shear stress distribution) and a bending
couple (from the normal stress distribution)
• Normal stress is often the critical design criteria
x  
My
I
m 
Mc M

I
S
Requires determination of the location and magnitude
of largest bending moment
3
Shear and Bending Moment Diagrams
• Determination of maximum normal and
shearing stresses requires identification of
maximum internal shear force and bending
couple.
• Shear force and bending couple at a point
are determined by passing a section through
the beam and applying an equilibrium
analysis on the beam portions on either side
of the section.
F  F  M
y
x
A
 MB  0
4
Shear and Bending Moment Diagrams
• Sign conventions for shear forces V and V’ and bending couples M and M’
5
Bending Deformations
Beam with a plane of symmetry in pure bending:
• member remains symmetric
• bends uniformly to form a circular arc
• cross-sectional plane passes through arc center
and remains planar
• length of top decreases and length of bottom
increases
• a neutral surface must exist that is parallel to the
upper and lower surfaces and for which the
length does not change
• stresses and strains are negative (compressive)
above the neutral plane and positive (tension)
below it
6
Strain Due to Bending
Consider a beam segment of length L.
After deformation, the length of the neutral
surface remains L. At other sections,
L     y 
  L  L     y      y
x 
m 


L
c

y

or

ρ
y

(strain varies linearly)
c
m
y
c
 x   m
7
Stress Due to Bending
• For a linearly elastic material,
y
c
 x  E x   E m
y
   m (stress varies linearly)
c
• For static equilibrium,
• For static equilibrium,
y
Fx  0    x dA     m dA
c
 y

M    y x dA    y   m  dA
 c


0   m  y dA
c

 I
M  m  y 2 dA  m
c
c
First moment with respect to
neutral plane is zero. Therefore,
the neutral surface must pass
through the section centroid.
m 
Mc M

I
S
y
Substituting  x    m
c
x  
My
I
8
Deformations in a Transverse Cross Section
• Deformation due to bending moment M is
quantified by the curvature of the neutral surface
1 m  m
1 Mc




c
Ec Ec I

M
EI
• Although cross sectional planes remain planar
when subjected to bending moments, in-plane
deformations are nonzero,
 y   x 
y

 z   x 
y

• Expansion above the neutral surface and
contraction below it cause an in-plane curvature,
1 
  anticlastic curvature
 
9
Deformation of a Beam Under Transverse Loading
• Relationship between bending moment and
curvature for pure bending remains valid for
general transverse loadings.
1


M ( x)
EI
• Cantilever beam subjected to concentrated
load at the free end,
1


Px
EI
• Curvature varies linearly with x
1
• At the free end A, ρ  0,
A
• At the support B,
1
B
ρA  
 0,  B 
EI
PL
10
Deformation of a Beam Under Transverse Loading
• Overhanging beam
• Reactions at A and C
• Bending moment diagram
• Curvature is zero at points where the bending
moment is zero, i.e., at each end and at E.
1


M ( x)
EI
• Beam is concave upwards where the bending
moment is positive and concave downwards
where it is negative.
• Maximum curvature occurs where the moment
magnitude is a maximum.
• An equation for the beam shape or elastic curve
is required to determine maximum deflection
and slope.
11
Equation of the Elastic Curve
• From elementary calculus, simplified for beam
parameters,
d2y
1


dx 2
2 3 2
  dy 
1    
  dx  

d2y
dx 2
• Substituting and integrating,
EI
1

 EI
d2y
dx
2
 M x
x
dy
EI   EI
 M  x dx  C1
dx 
0
x
x
0
0
EI y   dx  M  x  dx  C1x  C2
12
ds  d 
1


ds

 d
d
, P - P very small  ds  dx
ds
d
d d   d 2 y M



 2 
 dx
dx
dx
dx
EI
M
d 
dx
EI
2
x2
M
d


 1
x1 EI dx
1
dx
X
13
Persamaan momen;
Dimana;
Sehingga;
Jika diintegralkan didapat;
atau;
14
Jika slope di x=0 adalah 0,
Maka;
Dan,
Sehingga;
Pada tumpuan tdk
ada defleksi;
Sehingga jika x=0 disubstitusikan;
Diperoleh harga C2
Atau;
Diintegralkan lagi;
15
Defleksi yg terjadi sepanjang L adalah;
Jika harga yg diketahui dimasukan maka;
16
Batang statis tak tentu
17
18
 M  Vdx  M  dM  0
dM  Vdx
dM
V
dx
19
Perhatikan potongan dibawah dan
jumlah gaya vertikal = 0
20
Dari persamaan didepan;
Diketahui;
Sehingga;
Turunkan terhadap x pd
kedua sisi;
Turunkan terhadap x;
Atau;
Sehingga defleksi;
21
Gambarkan bending momen diagram
Dari persamaan didepan;
Integralkan dikedua sisi;
Integralkan dikedua sisi;
22
Momen ditumpuan kanan
Sehingga;
wL2
C1 L  C2 
2
Integralkan persamaan didepan;
Slope ditumpuan kiri = 0
Didapatkan;
23
Integralkan
Defleksi ditumpuan kiri = 0
Sehingga;
Diperoleh;
Defleksi ditumpuan kanan = 0
24
C1 L3 C2 L2 wL4


6
2
24
Didepan;
Ada 2 pers. Dg 2 variable yg dicari;
Sehingga;
25
Bending momen maksimum;
26
Relations Among Load, Shear, and Bending Moment
• Relationship between load and shear:
 Fy  0 : V  V  V   w x  0
V   w x
dV
 w
dx
xD
VD  VC    w dx
xC
• Relationship between shear and bending
moment:
M
C
 0:
M  M   M  V x  wx x  0
2
2
1
M  V x  2 w x 
dM
V
dx
M D  MC 
xD
 V dx
xC
27
x
dV
  w  Vx  VA    w dx
dx
0
1

Vx  VA   w dx  RA  wx  w L  x 
2

0
x
x
dM
 V  M x  M A   V dx
dx
0
1

M x  M A   V dx  M A   w L  x  dx
2

0
0
x
x

1
1

M x  M A   w L  x  dx  M A  w Lx  x 2
2
2

0
x
28

Equation of the Elastic Curve
• Constants are determined from boundary
conditions
x
x
0
0
EI y   dx  M  x  dx  C1x  C2
• Three cases for statically determinant beams,
– Simply supported beam
y A  0,
yB  0
– Overhanging beam
y A  0,
yB  0
– Cantilever beam
y A  0,  A  0
• More complicated loadings require multiple
integrals and application of requirement for
continuity of displacement and slope.
29
Direct Determination of the Elastic Curve From the Load Distribution
• For a beam subjected to a distributed load,
d 2M
dM
 V x
dx
dV

  w x 
2
dx
dx
• Equation for beam displacement becomes
d 2M
dx
2
 EI
d4y
dx
4
  w x 
• Integrating four times yields
EI y  x     dx  dx  dx  w x dx
 16 C1x3  12 C2 x 2  C3 x  C4
• Constants are determined from boundary
conditions.
30
Statically Indeterminate Beams
• Consider beam with fixed support at A and roller
support at B.
• From free-body diagram, note that there are four
unknown reaction components.
• Conditions for static equilibrium yield
 Fx  0  Fy  0  M A  0
The beam is statically indeterminate.
• Also have the beam deflection equation,
x
x
0
0
EI y   dx  M  x  dx  C1x  C2
which introduces two unknowns but provides
three additional equations from the boundary
conditions:
At x  0,   0 y  0
At x  L, y  0
31
Sample Problem 1
Re action at the sup port ;
RA  RB  1/ 2 wL
1
1
wLx  wx 2
2
2
d2y 1
1
EI 2  wLx  wx 2
dx
2
2
dy 1
1
EI
 wLx 2  wx 3  C1
dx 4
6
1
1
EIy  wLx3  wx 4  C1 x  C2
12
24
x  0  y  0  C2  0
Mx 
1
wL3
24
1
1
1
EIy  wLx3  wx 4  wL3 x
12
24
24
w
y
 x 4  2 Lx 3  L3 x
24 EI
x  L  y  0  C1  


32
Sample Problem 2
SOLUTION:
• Develop an expression for M(x)
and derive differential equation for
elastic curve.
W 14  68
I  723 in 4
E  29  106 psi
P  50 kips
L  15 ft
a  4 ft
• Integrate differential equation twice
and apply boundary conditions to
obtain elastic curve.
For portion AB of the overhanging beam,
• Locate point of zero slope or point
(a) derive the equation for the elastic curve,
of maximum deflection.
(b) determine the maximum deflection,
• Evaluate corresponding maximum
(c) evaluate ymax.
deflection.
33
Sample Problem 2
SOLUTION:
• Develop an expression for M(x) and derive
differential equation for elastic curve.
- Reactions:
RA 
Pa
 a
 RB  P1   
L
 L
- From the free-body diagram for section AD,
M  P
a
x
L
0  x  L 
- The differential equation for the elastic
curve,
EI
d2y
a


P
x
2
L
dx
34
Sample Problem 2
• Integrate differential equation twice and apply
boundary conditions to obtain elastic curve.
EI
dy
1 a
  P x 2  C1
dx
2 L
1 a
EI y   P x3  C1x  C2
6 L
2
EI
d y
a


P
x
2
L
dx
at x  0, y  0 : C2  0
1 a
1
at x  L, y  0 : 0   P L3  C1L C1  PaL
6 L
6
Substituting,
dy
1 a
1
EI
  P x 2  PaL
dx
2 L
6
1 a
1
EI y   P x3  PaLx
6 L
6
2
dy PaL 
 x 

1  3  
dx 6 EI 
 L  
PaL2  x  x 
y
  
6 EI  L  L 
35
3


Sample Problem 2
• Locate point of zero slope or point
of maximum deflection.
2
dy
PaL 
 xm  
0
1  3  
dx
6 EI 
 L  
PaL2  x  x 
y
  
6 EI  L  L 
3


xm 
L
 0.577 L
3
• Evaluate corresponding maximum
deflection.

PaL2
ymax 
0.577  0.577 3
6 EI

PaL2
ymax  0.0642
6 EI
ymax

50 kips 48 in 180 in 2
 0.0642


6 29  106 psi 723 in 4

ymax  0.238 in
36
Sample Problem 3
SOLUTION:
• Develop the differential equation for
the elastic curve (will be functionally
dependent on the reaction at A).
For the uniform beam, determine the
reaction at A, derive the equation for
the elastic curve, and determine the
slope at A. (Note that the beam is
statically indeterminate to the first
degree)
• Integrate twice and apply boundary
conditions to solve for reaction at A
and to obtain the elastic curve.
• Evaluate the slope at A.
37
Sample Problem 3
• Consider moment acting at section D,
MD  0
1  w0 x 2  x
RA x 
M 0
2  L  3
w0 x3
M  RA x 
6L
• The differential equation for the elastic
curve,
d2y
w0 x3
EI 2  M  RA x 
6L
dx
38
Sample Problem 3
• Integrate twice
4
dy
1
2 w0 x
EI
 EI  R A x 
 C1
dx
2
24 L
5
1
3 w0 x
EI y  R A x 
 C1x  C2
6
120 L
2
EI
d y
w0 x

M

R
x

A
6L
dx 2
3
• Apply boundary conditions:
at x  0, y  0 : C2  0
3
1
2 w0 L
at x  L,   0 :
RA L 
 C1  0
2
24
4
1
3 w0 L
at x  L, y  0 :
RA L 
 C1L  C2  0
6
120
• Solve for reaction at A
1
1
RA L3  w0 L4  0
3
30
RA 
1
w0 L 
10
39
Sample Problem 3
• Substitute for C1, C2, and RA in the
elastic curve equation,
5
1 1
 3 w0 x  1

EI y   w0 L  x 

w0 L3  x
6  10
120 L  120


y

w0
 x5  2 L2 x3  L4 x
120 EIL
• Differentiate once to find the slope,


dy
w0

 5 x 4  6 L2 x 2  L4
dx 120 EIL
at x = 0,

w0 L3
A 
120 EI
40

 (CC )
M
D
0
Sample Problem 4
 RA x  M D  0
P
x
2
d 2 y1 P
dy1 P 2
P 3
EI

x

EI

x

C

EIy

x  C1 x  C2
1
1
2
dx
2
dx 4
12
 (CC )  M E  0
M 1  M D  RA x 
L
 RA x  P( x  )  M E  0
2
P
L
M 2  M E  x  P( x  )
2
2
d 2 y2
P
P
dy2
P 2 P
EI


x

L

EI


x  Lx  C3
dx 2
2
2
dx
4
2
P
P
EIy2   x 3  Lx 2  C3 x  C4
12
4
x  0  y1  0
x  L  y2  0
x  L/2
dy1 dy2

dx
dx
dan
y1  y2
41
SOLUTION:
• Taking the entire beam as a free body,
determine the reactions at A and D.
• Apply the relationship between shear and
load to develop the shear diagram.
Draw the shear and bending
moment diagrams for the beam
and loading shown.
• Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
42
SOLUTION:
• Taking the entire beam as a free body, determine the
reactions at A and D.
MA  0
0  D24 ft   20 kips6 ft   12 kips14 ft   12 kips28 ft 
D  26 kips
 Fy  0
0  Ay  20 kips  12 kips  26 kips  12 kips
Ay  18 kips
• Apply the relationship between shear and load to
develop the shear diagram.
dV
 w
dx
dV   w dx
- zero slope between concentrated loads
- linear variation over uniform load segment
43
• Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
dM
V
dx
dM  V dx
- bending moment at A and E is zero
- bending moment variation between A, B,
C and D is linear
- bending moment variation between D
and E is quadratic
- net change in bending moment is equal to
areas under shear distribution segments
- total of all bending moment changes across
the beam should be zero
44
   Fy  0
 wx  V  0
V   wx
 (CC )
M
C
0
 x
wx   M  0
2
1
M   wx 2
2
45
Re action at the sup port ;
RA  RB  1/ 2 wL
x
dV
  w  Vx  VA    w dx
dx
0
1

Vx  VA   w dx  RA  wx  w L  x 
2

0
x
x
dM
 V  M x  M A   V dx
dx
0
1

M x  M A   V dx  M A   w L  x  dx
2

0
0 
x
x

1
1

M x  M A   w L  x  dx  M A  w Lx  x 2
2
2

0 
x
46

   Fy  0
RA  VD  0
VD  RA
RA  P  VE  0
VE  RA  P
 (CC )
M
D
0
 RA x  M D  0
M D  RA x
 (CC )
M
E
0
L
 RA x  P( x  )  M E  0
2
L
M E  RA x  P( x  )
2
47
1 


M

0

w
a
a   R A ( 2a )  0
 B
o 
2 
1
RA  wo a
4
Between A - C 0  x  a
1
1
V1 ( x)  wo a M 1 ( x)  wo ax
4
4
Between C - B
 (CC )
1
wo a  wo  x  a   V2  0
4
1
1

 (CC )  M E  0   wo ax  wo  x  a   x  a   M 2  0
4
2

   Fy  0 
48
Between C - B 0  x  2a
1
wo a  wo  x  a 
4
1
1
2
M 2  wo ax  wo  x  a 
4
2
V1 ( x) and V2 ( x) can be represented ;
V2 ( x) 
1
wo a  wo x  a
4
M 1 ( x) and M 2 ( x) can be represented ;
V ( x) 
M ( x) 
1
1
wo ax  wo x  a
4
2
2
Second term should be included when x≥a and ignored when x<a
Or < > should be replaced by ( ) when x≥a and by 0 when x<a
49
V ( x) 
1
wo a  wo x  a
4
x
x
1
M ( x)  M (0)   V ( x) dx   wo a  wo x  a dx
4
0
0
x
x
1
M ( x)   wo a dx   wo x  a dx
4
0
0
M ( x) 
1
1
wo ax  wo x  a
4
2
w  wo x  a
0
2
 should be replaced by ( ) when x  a
and by 0 when x  a
x
x
V ( x)  V (0)    w( x) dx    wo x  a dx
0
0
0
1
1
V ( x)  wo a   wo x  a
4
1
1
V ( x)  wo a  wo x  a
4
50
x  a , x  a , x  a  sin gularity funtion
0
xa
n
2
x  a n when x  a

when x  a
0
51
52
53
Re action at the sup port ;
 (CC )
M
B
0
 RA (3.6m)  1.2kN 3m   1.8kN 2.4m   1.44kNm  0
RA  2.60kN
w( x)  wo x  0.6  wo x  1.8
0
0
x
x
dV
0
  w  Vx  V (0)    w dx  Vx  V (0)  P x  0.6   w dx
dx
0
0
x

Vx  RA  P x  0.6   wo x  0.6  wo x  1.8
0
0
0
 dx
0
Vx  RA  P x  0.6  wo x  0.6  wo x  1.8
0
1
0
1
1
V x  2 .6  1 .2 x  0 .6  1 .5 x  0 .6  1 .5 x  1 . 8
x
1
x
dM
 V  M x  M (0)   V dx  M x  M (0)   V dx
dx
0
0
x

M x  M ( 0 )   2 . 6  1 .2 x  0 .6  1 . 5 x  0 . 6  1 .5 x  1 .8
0
1
1
 dx  M
x  2 .6
o
0
0
1 .5
1. 5
1
2

M x   2 .6 x  1 .2 x  0 .6 
x  0 .6 
x  1 .8
2
2

1 .5
1. 5
1
2

M x   2 .6 x  1 .2 x  0 .6 
x  0 .6 
x  1 .8
2
2

2
2

  M o x  2 .6


  1.44 x  2.6
0
0
54
Plat aluminium mempunyai ketebalan b digunakan utk
mendukung beban terdistribusi merata w seperti pada
gambar disamping. Tentukan (a) bentuk plat yang paling
ekonomis dan (b) jika tegangan yang diijinkan 75 Mpa ,
tebal plat b=40 mm, L=800 mm dan w=135 kN/m tentukan
ho.
Bending momen
x
V ( x)  V (0)    w( x) dx   wx
0
x
x
wx 2
M ( x)  M (0)   V ( x) dx    wx dx  
2
0
0
M
6M
bh 2
3wx 2
3w
2
jika S 
dan  
h 

h
x  h dan x
6
S
b all b all
b all
berhubunga n linier maka bentuk yang paling ekonomis adalah segitiga.
mencari harga ho, yaitu jika x  L
ho 
3w
L
b all
3135kN / m  800mm  300mm
0.04m 72MPa 
55
Using singularity functions to determine the slope and deflection of
a beam
56
 (CC )
M
D
0
 RA x  M D  0
P
x
2
d 2 y1 P
dy1 P 2
P 3
EI

x

EI

x

C

EIy

x  C1 x  C2
1
1
2
dx
2
dx 4
12
 (CC )  M E  0
M 1  M D  RA x 
L
 RA x  P( x  )  M E  0
2
P
L
M 2  M E  x  P( x  )
2
2
d 2 y2
P
P
dy2
P 2 P
EI


x

L

EI


x  Lx  C3
dx 2
2
2
dx
4
2
P
P
EIy2   x 3  Lx 2  C3 x  C4
12
4
x  0  y1  0
x  L  y2  0
x  L/2
dy1 dy2

dx
dx
dan
y1  y2
57
58
3
3
PL2  C1 
PL2  C3
128
128
59
Re action at the sup port ;
RA  2.60kN
Vx  2.6  1.2 x  0.6  1.5 x  0.6  1.5 x  1.8
0
1
1
x
dM
 V  M x  M (0)   V dx
dx
0
x

M x   2.6  1.2 x  0.6  1.5 x  0.6  1.5 x  1.8
0
1
1
 dx  M
o
x  2.6
0
0
1.5
1 .5
1
2

M x  2.6 x  1.2 x  0.6 
x  0.6 
x  1.8
2
2

1.5
1 .5
1
2

M x  2.6 x  1.2 x  0.6 
x  0.6 
x  1.8
2
2

2
2

  M o x  2 .6


  1.44 x  2.6

60
0
0
1.5
1.5
1
2
2
0

M x  2.6 x  1.2 x  0.6 
x  0.6 
x  1.8   1.44 x  2.6
2
2


d2y
EI 2  M x
dx
d2y 
1.5
1.5
1
2
2
0
EI 2  2.6 x  1.2 x  0.6 
x  0.6 
x  1.8   1.44 x  2.6
dx
2
2


dy  2.6 2 1.2
1.5
1.5
2
3
3
1
EI

x 
x  0.6 
x  0.6 
x  1.8   1.44 x  2.6  C1
dx  2
2
6
6

1.5
1.5
1.44
3
4
4
2
 2.6 3 1.2
EIy  
x 
x  0.6 
x  0.6 
x  1.8  
x  2.6  C1 x  C2
6
24
24
2
 6

x0 y0
1.5
1.5
1.44
3
4
4
2
 2.6 3 1.2
0
0 
 0.6 
 0.6 
 1.8  
 2.6  C1 0  C2
6
24
24
2
 6

x  0  y  0,  C2  0
x  3.6  y  0
 2.6
3.63  1.2 3.6  0.6 3  1.5 3.6  0.6 4  1.5 3.6  1.8 4   1.44 3.6  2.6 2  C1 3.6
0
6
24
24
2
 6

 2.6
3.63  1.2 3 3  1.5 3 4  1.5 1.8 4   1.44 1 2  C1 3.6
0
6
24
24
2
 6

C1  2.692
1.5
1.5
1.44
3
4
4
2
 2.6 3 1.2
EIy  
x 
x  0.6 
x  0.6 
x  1.8  
x  2.6  2.692 x
6
24
24
2
 6

61
x  1.8
1.5
1.5
1.44
3
4
4
2
 2.6 3 1.2
EIy  
x 
x  0.6 
x  0.6 
x  1.8  
x  2.6  2.692 x
6
24
24
2
 6

62
63
64
72
Jika tegangan yg diijinkan 12 Mpa
tentukan h
Jika tegangan tarik dan tekan yg
diijinkan 80 Mpa dan 130 tentukan w
73
74