ME 458 Introduction to

Solid Mechanics

DEFLECTION OF BEAMS
Integration
Discontinuity Functions (Macaulay and Singularity functions)
Moment-Area

By
Prof.Dr. H. Seçil ARTEM

1
 The ELASTIC CURVE
• The deflection diagram of the longitudinal axis that passes through the
centroid of each cross-sectional area of the beam is called the “elastic
curve”. For most beams, the elastic curve can be sketched without much
difficulty. When doing so, it is necessary to know

“HOW THE SLOPE AND DISPLACEMENT IS RESTRICTED AT VARIOUS

TYPES OF SUPPORTS”
. PIN, resist a force restrict displacement
. FIXED SUPPORT, resist moment restrict rotation or
slope as well as displacement.
 • Draw the moment diagram for the beam first before
12.1 THE ELASTIC
creating CURVE
the elastic curve.

Curve changes from concave down to

concave up,
since this is a point of zero moment.
Moment-Curvature Relationship
• Consider a segment of width dx, the strain in are ds, located at a position
y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and
ds’ = (ρ-y) dθ, and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or

1 

 y

• If the material is homogeneous and behaves in a linear elastic manner,

then the Hooke’s Law applies,
ε = σ / E and the flexure formula
σ = -My/I
1 M 1 
 or 
 EI  Ey
 : radius of curvature at a specific point on elastic
curve
1/ : the curvature
 NoteTHE
12.1 thatELASTIC CURVE
• EI is the flexural rigidity and is always positive.
• Sign for  depends on the direction of the
moment.
• As shown, when M is +,  extends above the
beam. When M is –,  extends below the beam.
 SLOPE AND DISPLACEMENT BY INTEGRATION

• The elastic curve for a beam can be expressed

mathematically as v = f(x). In most calculus books it is
shown that this relationship is
d 2 A nonlinear 2nd order DE. Its
1
 dx 2 solution gives the exact shape
of the elastic curve, assuming
 3



1  d 
dx 
2 2 of course, that beam
deflections occurs only due
to bending

Slope of elastic curve determined from d/dx is very small and

therefore its square will be negligible compared with unity.
•12.2Flexural rigidity
SLOPE AND is constant along
DISPLACEMENT beam, thus
BY INTEGRATION
Boundary and Continuity Conditions:The constants for
integration are determined by evaluating the functions for shear, moment, slope or
displacement at a particular point on the beam where the value of function is known.
These values are called Boundary Conditions.
Boundary and Continuity Conditions
• If a single x coordinate cannot be used to express the eqn
for beam’s slope or elastic curve, then continuity
conditions must be used to evaluate some of the
integration constants.
 Example
Cantilevered beam shown is subjected to a vertical load P at its end.
Determine the equation of the elastic curve. EI is constant.

(1)

(2)

Equation of elastic curve:

Example
EXAMPLE 12.4
Beam is subjected to load P at its end. Determine the displacement at C.
EI is a constant.

Due to loading, two x coordinates will be

considered, 0  x1  2a and
0  x2  a,
where x2 is directed to the left from C since
internal moment is easy to formulate.

(Continuity condition)
d 21 P
for 0  x1  2a EI   x1
dx12 2
d1 P
EI   x12  C1 1
dx1 4
P 3
EI1   x1  C1x1  C2 2 
12

d 22
for 0  x2  a EI    Px2
dx2 2
d 2 P
EI   x2 2  C3 3
dx2 2
P 3
EI2   x2  C3 x2  C4 4 
6
…..The four constants of integration determined using three boundary
conditions, 1 = 12.4
EXAMPLE = 0, 1 = 0 at x1 = 2a, and 2 =0 at x2 = a.
0 at x1 (SOLN)
….Here, continuity of slope at roller requires
d1/dx1 = d2/dx2 at x1 = 2a and x2 = a.

1  0 at x1  0; 0  0  0  C2
P
1  0 at x1  2a; 0   2a 2  C12a   C2
12
P
2  0 at x2  a; 0   a3  C3a  C4
6
d12a  d a 
 2a 2  C1    a 2  C3 
P P
 2 ;
dx1 dx2 4  2 

Pa 2 7
C1  C2  0 C3  Pa2 C4   Pa 3
3 6

Displacement at point
P 7 Pa 2
Pa 3 C, is therefore Pa 3
2  
6 EI
x23 
6 EI
x2 
EI C  
EI
 DISCONTINUITY FUNCTIONS
*12.3 DISCONTINUITY FUNCTIONS
A simplified method for finding the equation of the elastic curve for a multiply
loaded beam using a single expression

1- Macaulay functions
can be used to describe distributed loadings, written generally as
n
xa { 0 for x  a
{  x  a n for x  a 12 - 11
n0
where x represents the coordinate position of a point along the beam
a is the location on the beam where a “discontinuity” occurs, or the
point where a distributed loading begins.

Integration of the Macaulay function follows the same rules as for ordinary
functions, as
n 1
xa
 xa
n
dx  C 12 - 12
n 1
13
Macaulay functions FUNCTIONS
*12.3 DISCONTINUITY

14
 2- Singularity functions
*12.3 DISCONTINUITY
are used to describe the point FUNCTIONS
location of concentrated forces or couple moments
acting on a beam.

Note that a concentrated force P can be considered as a special case of distributed

loading, where w = P/e such that its width is , →0

1
w P xa { 0 for x  a
{ P for x  a 12 - 13

2
w  M0 x  a { 0 for x  a
 { M0 for x  a 12 - 14

n 1
 xa
n
dx  x  a , n  1,2 12 - 15
15
Example
EXAMPLE
Determine the eqn.12.5
of the elastic curve for the cantilevered beam shown. EI is constant.

16
Loading functions
EXAMPLE 12.5 (SOLN)

1 0
w  52 kN x  0  258 kN  m x  0
2 0
50 kN  m x  5 m  8 kN / m x  5 m
The 12-kN load is not included, since x cannot be greater than 9 m. Because dV/dx = w(x),
then by integrating, neglect constant of integration since reactions are included in load function,
we have 0 1 1
V  52 x  0  258 x  0 8 x 0
1 1
50 x  0 8 x0
0 1 0 1
M  258 x  0  8 x  0  50 x  5  8 x  5
2 2
2 2
 2
  258  52 x  4 x 2  4 x  5  50 x  5
0
kN  m
The same result can be obtained directly from Table 12-2. 17
Slope and elastic curve
EXAMPLE
we have
12.5 (SOLN)
d 2 0 2
EI  258  52 x  4 x 2  50 x  5  4 x  5
dx 2
d 2 4 3 1 4 3
EI  258 x  26 x  x  50 x  5  x  5  C1
dx 3 3
2 26 3 1 4
EI  129 x  x  x  25 x  5
2
3 3
1 4
 x  5  C1x  C2
3
Conditions:
d/dx = 0 at x = 0 C1 = 0
 = 0 at x = 0 C2 = 0 The equation of elastic curve
1 26 1
 (129 x 2  x3  x 4
EI 3 3
2 1 4
25 x  5  x  5 ) m
3
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MOMENT AREA METHOD
Semi-graphical Method
Theorem 1
• The angle between the tangents at any two points on the
elastic curve equals the area under the M/EI diagram
between these two points.
d 2v d  dy 
EI 2  EI    M
dx dx  dx 

M 
• Since θ ≈ dv/dx, so d   dx
 EI 
• Therefore, B
M
B/ A   dx
A
EI
MOMENT AREA METHOD (cont)

Theorem 1 (cont)
• This equation forms the basis for the first moment-area theorem

B
M
B/ A   dx
A
EI
Example
Determine the slope of the beam shown at points B and C. EI is constant.

• M/EI diagram will be drawn first.

• The force P causes the beam to deflect
as shown.

B  B/ A C  C / A

• By the construction the angle between tan A and tan B

 PL  L  1  PL  L  3PL2
B  B/ A             (Ans)
 2 EI  
2 2  2 EI  
2 8 EI
1  PL  PL2
C  C / A    L   (Ans)
2  2 EI  2 EI
MOMENT AREA METHOD (cont)

Theorem 2
• The vertical deviation of the tangent at a point (A) on the elastic curve with
respect to the tangent extended from another point (B) equals the moment
of the area under the M/EI diagram between these two points (A and B).
This moment is computed about point (A) where the vertical deviation
(tA/B) is to be determined.
 MOMENT AREA METHOD (cont)
Theorem 2 (cont)
The vertical deviation of the tangent at A with respect to the tangent at B is

B
M
t A/ B  x  dx
A
EI

where x(bar) is distance measured from A to the centroid of the area under M/EI
diagram between A and B.
Procedure for analysis
M/EI Diagram
• Determine the support reactions and draw the beam’s M/EI diagram.
• If the beam is loaded with concentrated forces, the M/EI diagram will consist of a series
of straight line segments, and the areas and their moments required for the moment-
area theorems will be relatively easy to compute.
• If the loading consists of a series of distributed loads, the M/EI diagram will consist of
parabolic or perhaps higher-order curves, and we use the table on the inside front cover
to locate the area and centroid under each curve.
Elastic curve
• Draw an exaggerated view of the beam’s elastic curve.
• Recall that pts of zero slope and zero displacement always occur at a fixed support, and
zero displacement occurs at all pin and roller supports.
• If it is difficult to draw the general shape of the elastic curve, use the moment (M/EI)
diagram.
• Realize that when the beam is subjected to a +ve moment, the beam bends concave up,
whereas
-ve moment bends the beam concave down.
• An inflection pt or change in curvature occurs when the moment if the beam (or M/EI) is
zero.
• The unknown displacement and slope to be determined should be indicated on the
curve.
• Since moment-area theorems apply only between two tangents, attention should be
given as to which tangents should be constructed so that the angles or deviations
between them will lead to the solution of the problem.
• The tangents at the supports should be considered, since the beam usually has zero
displacement and/or zero slope at the supports.

24
Procedure for analysis
.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Moment-area theorems
• Apply Theorem 1 to determine the angle between any two tangents on
the elastic curve and Theorem 2 to determine the tangential deviation.
• The algebraic sign of the answer can be checked from the angle or
deviation indicated on the elastic curve.
• A positive B/A represents a counterclockwise rotation of the tangent at
B w.r.t. tangent at A, and a +ve tB/A indicates that pt B on the elastic
curve lies above the extended tangent from pt A.

25
Example
Determine the displacement of pts B and C of beam shown. EI is constant.
EXAMPLE 12.8 (SOLN)

M/EI diagram: See below.

Elastic curve:
The couple moment at C cause the beam to deflect as shown.

26
 Elastic curve:
The required displacements can be related directly to deviations between
EXAMPLE 12.8 (SOLN)
the tangents at B and A and C and A. Specifically, B is equal to deviation of
tan A from tan B,

 B  tB A C  tC A

Moment-area theorem:
Applying Theorem 2, tB/A is equal to the moment of the
shaded area under the M/EI diagram between A and B
computed about point B, since this is the point where
tangential deviation is to be determined.
 L   M 0  L  M 0 L2
 B  tB A         
 4   EI  2  8 EI
Likewise, for tC/A we must determine the moment of the
area under the entire M/EI diagram from A to C about pt
C. We have
 L   M 0   M 0 L2
C  tC A       L   
 2   EI   2 EI
27