Chapter

p 8:
Deflections

Structural Analysis 7th Edition in SI Units

Russell C. Hibbeler

Deflection diagrams & the elastic curve

• Deflections of structures can come from loads,

temperature,
p , fabrication errors or settlement
• In designs, deflections must be limited in order to
prevent cracking of attached brittle materials
• A structure must not vibrate or deflect severely for
the comfort of occupants
• Deflections at specified points must
m st be determined
dete mined
if one is to analyze statically indeterminate
structures

1
 Deflection diagrams & the elastic curve

• In this topic, only linear elastic material response

is considered
• This means a structure subjected to load will
return to its original undeformed position after the
load is removed
• It is useful to sketch the shape of the structure
when it is loaded in order to visualize the
computed results & to partially check the results

Deflection diagrams & the elastic curve

• This deflection diagram rep the elastic

curve for the p points at the centroids
of the cross
cross--sectional areas along each
of the members
• If the elastic curve seems difficult to
establish, it is suggested that the
moment diagram be drawn first
• From there, the curve can be constructed

2
 Deflection diagrams & the elastic curve

• Due to pin-
pin-and
and--roller support, the disp at A & D
must be zero
• Within the region of –ve moment,
the elastic curve is
concave downward
• Within the region of +ve moment,
the elastic curve is concave upward
• There must be an inflection point
where the curve changes from
concave down to concave up

Example 8.1

Draw the deflected shape of each of the beams.

3
 Solution
In (a), the roller at A allows free rotation with no deflection while
the fixed wall at B prevents both rotation & deflection. The
deflected shape is shown by the bold line.

In (b), no rotation or deflection occur at A & B

In (c), the couple moment will rotate end A. This will cause
deflections at both ends of the beam since no deflection is
possible at B & C. Notice that segment CD remains un
un-deformed
deformed
since no internal load acts within.

Solution

In (d), the pin at B allows rotation, so the slope of the deflection

curve will suddenly change at this point while the beam is
constrained by its support.
support

In (e), the compound beam deflects as shown. The slope changes

abruptly on each side of B.

In ((f),
), span
p BC will deflect concave upwards
p due to load. Since
the beam is continuous, the end spans will deflect concave
downwards.

4
 Elastic Beam theory

• To derive the DE, we look at an initially straight beam

that is elastically deformed by loads applied
perpendicular to beam’
beam’s x-
x-axis & lying in x-
x-v plane of
symmetry
• Due to loading, the beam
deforms under shear & bending
• If beam L >> d, greatest
deformation will be caused by bending
• When M deforms, the angle
between the cross sections becomes d d

Elastic Beam theory

• The arc dx that rep a portion of the elastic curve

intersects the neutral axis
• The radius of curvature for this arc is defined as
the distance, , which is measured from ctr of
curvature O’
O’ to dx
• Any arc on the element other than dx is subjected
to normal strain
• The strain in arc ds located at position y from the
neutral axis is
  (ds 'ds ) / ds

5
 Elastic Beam theory

ds  dx  d and ds '  (   y )d

(   y )d  d 1 
  
d  y
• If the material is homogeneous & behaves in a
linear manner, then Hooke’
Hooke’s law applies
  /E
• The flexure formula also applies
   My / I

Elastic Beam theory

• Combining these eqns, we have:

1 M

 EI
  the radius of curvature at a specific point on the elastic curve

M  internal moment in the beam at the point where  is to be determined

E  the material' s modulus of elasticity

I  the beam' s moment of inertia computed about the neutral axis

6
 Elastic Beam theory

EI  flexural rigidity;
g y dx  ρρdθ

M
dθ  dx
EI

1 d 2 v / dx 2
v  axis as  ve , 
 [1  (dv / dx) 2 ]3 / 2

M d 2 v / dx 2
Therefore, 
EI [1  ( dv / dx) 2 ]3 / 2

Elastic Beam theory

• This eqn rep a non-

non-linear second DE
• V
V=f(x)
f(x) gives the exact shape of the elastic curve
• The slope of the elastic curve for most structures
is very small
• Using small deflection theory, we assume dv/dx ~
0
d 2v M

dx 2 EI

7
 Elastic Beam theory

• By assuming dv/ dx ~ 0  ds
dv/dx ds,, it will approximately
equal
q to dx

ds  dx 2  dv 2  1  (dv / dx) 2 dx 2  dx
• This implies that points on the elastic curve will
only be displaced vertically & not horizontally

The double integration method

• M = f(x), successive integration of eqn 8.4 will

yyield the beam’
beam’s slope
p
•   tan  = dv/dx
dv/dx =  M/EI dx
• Eqn of elastic curve
• V = f(x) =   M/EI dx
• The internal moment in regions AB, BC & CD must
b written
be itt iin terms
t off x1, x2 and
d x3

8
 The double integration method

• Once these functions are integrated & the

constants determined,, the functions will give
g the
slope & deflection for each region of the beam
• It is important to use the proper sign for M as
established by the sign convention used in
derivation
• +ve v is upward
upward, hence angle, 
hence, the +ve slope angle
will be measured counterclockwise from the x- x-axis

The double integration method

• The constants of integration are determined by

evaluatingg the functions for slope
p or displacement
p
at a particular point on the beam where the value
of the function is known
• These values are called boundary conditions
• Here x1 and x2 coordinates are valid within the
regions AB & BC

9
 The double integration method

• Once the functions for the slope & deflections are

obtained,, theyy must give
g the same values for slope
p
& deflection at point B
• This is so as for the beam to be physically
continuous

Example 8.1

The cantilevered beam is subjected to a couple moment Mo at its

end. Determine the eqn of the elastic curve. EI is constant.

10
 Solution

By inspection, the internal moment can be represented

throughout the beam using a single x coordinate. From the free-
body diagram
diagram, with M acting in +ve direction
direction, we have: M  M o

Integrating twice yields:

d 2v
EI 2  M o
dx
ddv
EI  M o x  C1
dx
M o x2
EI v   C1 x  C2
2

Solution

Using boundary conditions, dv/dx = 0 at x = 0 & v = 0 at x = 0

then C1 = C2 =0.

Substituting these values into earlier eqns, we get:

with   dv / dx
Mox M o x2
 ; v
EI 2 EI

Max slope & disp occur at A (x = L) for which

MoL M o L2
A  ; vA 
EI 2 EI

11
 Solution

The +ve result for A indicates counterclockwise rotation & the

+ve result for vA indicates that it is upwards.

In order to obtain some idea to the actual magnitude of the slope.

Consider the beam to:

Have a length of 3.6m
Support a couple moment of 20kNm
Be made of steel having Est = 200GPa

Solution

If this beam were designed w/o a FOS by assuming the allowable

normal stress = yield stress = 250kNm/2.

A W6 x 9 would be found to be adequate

20kNm(3.6m)
A   0.0529rad
[200(10 )kN / m2 ][6.8(106 )(10 12 )m 4
6

20kNm(3.6m) 2
vA   95.3mm
2[200(106 )kN / m2 ][6.8(106 )(10 12 )m 4

12
 Solution

Since 2A = 0.00280(10-6)<<1, this justifies the use of Eq. 8-4.

Since the application is for a cantilever beam, we have obtained
larger values for max  and v than would have been obtained if
the beam were supported using pins, rollers or other supports.

Moment-Area theorems

• If we draw the moment diagram for the beam &

then divide it byy the flexural rigidity,
g y EI, the “M/EI
diagram”” is the results
diagram
M 
d   dx
 EI 

13
 Moment-Area theorems

• The change dd in the slope of the tangent on

either side of the element dx = the lighter
g shade
area under the M/EI diagram
• Integrating from point A on the elastic curve to
point B, we have
BM
B/ A   dx
A EI

• This eqn forms the basis for the first moment-

moment-area
theorem

Moment-Area theorems

• Theorem 1
• Thee cchange
a ge in sslope
ope bet
between
ee aanyy 2 po
points
ts o
on tthe
e
elastic curve equals the area of the M/EI diagram
between the 2 points
• The second moment
moment--area theorem is based on the
relative deviation of tangents to the elastic curve
• Shown in Fig 8.14(c)
8 14(c) is a greatly exaggerated view
of the vertical deviation dt of the tangents on each
side of the differential element, dx

14
 Moment-Area theorems

• Since slope of elastic curve & its deflection are

assumed to be veryy small,, it is satisfactoryy to
approximate the length of each tangent line by x &
the arc ds
ds’’ by dt
• Using s = r   dt = xd xd
• Using eqn 8.2, dd = (M/EI) dx
• The vertical
e tical deviation
de iation of the tangent at A wrtt the
tangent at B can be found by integration
B M
t A/ B   x dx
A EI

Moment-Area theorems

• Centroid of an area
x  dA   xdA

M B
t A/ B  x dx
A EI

x  distance from the vertical axis through

A to the centroid of the area between A & B.

15
 Moment-Area theorems

• Theorem 2
• Thee vertical
e t ca de
deviation
at o oof tthe
e ta
tangent
ge t at a po
pointt ((A))
on the elastic curve wrt the tangent extended from
another point (B) equals the “moment
moment”” of the area
under the M/EI diagram between the 2 points (A &
B)
• This moment is computed
p about point
p A where the
deviation is to be determined

Moment-Area theorems

• Provided the moment of a +ve M/EI area from A

to B is computed,
p , it indicates that the tangent
g at
point A is above the tangent to the curve extended
from point B
• -ve areas indicate that the tangent at A is below
the tangent extended from B

16
 Moment-Area theorems

• It is important to realise that the moment

moment--area
theorems can onlyy be used to determine the
angles & deviations between 2 tangents on the
beam’’s elastic curve
beam
• In general, they do not give a direct solution for
the slope or disp. at a point

Example 8.5

Determine the slope at points B & C of the beam. Take E =

200GPa, I = 360(106)mm4

17
 Solution

It is easier to solve the problem in terms of EI & substitute the

numerical data as a last step.
Th 10kN load
The l d causes the
th beam
b to
t deflect.
d fl t
Here the tangent at A is always horizontal.
The tangents at B & C are also indicated.
By construction, the angle between tan A and tan B (B/A) is
equivalent to B.

Solution

B  B/ A; C  C / A

Applying Theorem 1,  B / A is equal to the area under the M/EI

diagram between points A & B.

 50kNm  1  100kNm 50kNm 

 B   B / A   (5m)    (5m)
 EI  2  EI EI 
375kNm 2

EI

18
 Solution

Substituting numerical data for E & I

375kNm 2
  0.00521rad
d
[200(10 6 )kN / m 2 ][360(10 6 )(10 12 )m 4 ]

The –ve sign indicates that the angle is measured clockwise from
A.
In a similar manner, the area under the M/EI diagram between
points A & C equals (C/A).

Solution

1  100kNm  500kNm 2
C  C / A   (10m)  
2 EI  EI
Substituting numerical values of EI, we have :

 500kNm 2
 0.00694rad
[200(10 6 )kN / m 2 ][360(10 6 )(10 12 )m 4 ]

19
 Conjugate--Beam method
Conjugate

• The basis for the method comes from similarity

equations
q
• To show this similarity, we can write these eqn as
shown

dV d 2M
w w
dx dx 2
d M d 2v M

dx EI

dx 2
EI

Conjugate--Beam method
Conjugate

• Or integrating,

V   wdx M  wdx dx

M   M  
   dx v    dx  dx
 EI   EI  

20
 Conjugate--Beam method
Conjugate

• Here the shear V compares with the slope θ , the

moment M compares
p with the dispp v & the
external load w compares with the M/EI diagram
• To make use of this comparison we will now
consider a beam having the same length as the
real beam but referred to as the “conjugate beam”
beam”,

Conjugate--Beam method
Conjugate

• The conjugate beam is loaded with the M/EI

diagram
g derived from the load w on the real beam
• From the above comparisons, we can state 2
theorems related to the conjugate beam
• Theorem 1
• The slope at a point in the real beam is numerically
equal to the shear at the corresponding point in the
conjugate beam

21
 Conjugate--Beam method
Conjugate

• Theorem 2
• The
ed disp.
sp oof a po
pointt in tthe
e real
ea bea
beam iss numerically
u e ca y
equal to the moment at the corresponding point in
the conjugate beam
• When drawing the conjugate beam, it is important
that the shear & moment developed at the
supports of the conjugate beam account for the
corresponding slope & disp of the real beam at its
supports

Conjugate--Beam method
Conjugate

• Consequently from Theorem 1 & 2, the conjugate

beam must be supported
pp byy a pin
p or roller since
this support has zero moment but has a shear or
end reaction
• When the real beam is fixed supported, both beam
has a free end since at this end there is zero shear
& moment

22
 Conjugate--Beam method
Conjugate

Example 8.13

Determine the max deflection of the steel beam. The reactions

have been computed. Take E = 200GPa, I = 60(106)mm4

23
 Solution

The conjugate beam loaded with the M/EI diagram is shown.

Since M/EI diagram is +ve, the distributed load acts upward.
Th external
The t l reactions
ti on the
th conjugate
j t beam
b are determined
d t i d
first and are indicated on the free-body diagram.
Max deflection of the real beam occurs at the point where the
slope of the beam is zero.
Assuming this point acts within the region 0x9m from A’ we can
isolate the section.

Solution

Note that the peak of the distributed loading was determined from
proportional triangles,

w / x  (18 / EI ) / 9
V '0

   Fy  0
45 1  2 x 
   x 0
EI 2  EI 

x  6.71m (0  x  9m) OK

24
 Solution

Using this value for x, the max deflection in the real beam
corresponds to the moment M’.
H
Hence,

With anticlockwise moments as  ve,  M  0

45  1  2(6.71)  1
(6.71)    6.71 (6.71)  M '  0
EI  2  EI  3

Solution

The –ve sign indicates the deflection is downward .

201.2kNm 3
 max  M '  
EI
 201.2kNm 3

[200(10 6 )kN / m 2 ][60(10 6 )mm 4 (1m 4 /(10 3 ) 4 mm 4 )]
 0.0168m  16.8mm

25