Transcript Chapter 1

12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
•
Assumptions:
– beam is initially straight,
– is elastically deformed by the loads, such that
the slope and deflection of the elastic curve are
very small, and
– deformations are caused by bending.
Theorem 1
• The angle between the tangents at any two pts on
the elastic curve equals the area under the M/EI
diagram between these two pts.
B
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
B
A
M
dx
EI
12 - 19 
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12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Theorem 1
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12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Theorem 2
• The vertical deviation of the tangent at a pt (A) on
the elastic curve w.r.t. the tangent extended from
another pt (B) equals the moment of the area
under the ME/I diagram between these two pts
(A and B).
• This moment is computed about pt (A) where the
vertical deviation (tA/B) is to be determined.
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12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Theorem 2
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12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
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.
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12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Procedure for analysis
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.
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12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Procedure for analysis
Elastic curve
• 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.
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12. Deflections of Beams and Shafts
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Procedure for analysis
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.
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12. Deflections of Beams and Shafts
EXAMPLE 12.7
Determine the slope of the beam shown at pts B and
C. EI is constant.
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12. Deflections of Beams and Shafts
EXAMPLE 12.7 (SOLN)
M/EI diagram: See below.
Elastic curve:
The force P causes the beam to deflect as shown.
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12. Deflections of Beams and Shafts
EXAMPLE 12.7 (SOLN)
Elastic curve:
The tangents at B and C are indicated since we are
required to find B and C. Also, the tangent at the
support (A) is shown. This tangent has a known zero
slope. By construction, the angle between tan A and
tan B, B/A, is equivalent to B, or
B  B
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A
and
C  C
A
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12. Deflections of Beams and Shafts
EXAMPLE 12.7 (SOLN)
Moment-area theorem:
Applying Theorem 1, B/A is equal to the area under
the M/EI diagram between pts A and B, that is,
B  B
A
PL  L  1  PL  L 

 
    
 
 2 EI  2  2  2 EI  2 
3PL2

8 EI
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12. Deflections of Beams and Shafts
EXAMPLE 12.7 (SOLN)
Moment-area theorem:
The negative sign indicates that angle measured from
tangent at A to tangent at B is clockwise. This checks,
since beam slopes downward at B.
Similarly, area under the M/EI diagram between pts A
and C equals C/A. We have
1  PL 
C  C A     L
2  EI 
PL2

2 EI
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12. Deflections of Beams and Shafts
EXAMPLE 12.8
Determine the displacement of pts B and C of beam
shown. EI is constant.
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12. Deflections of Beams and Shafts
EXAMPLE 12.8 (SOLN)
M/EI diagram: See below.
Elastic curve:
The couple moment at C cause the beam to deflect as
shown.
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12. Deflections of Beams and Shafts
EXAMPLE 12.8 (SOLN)
Elastic curve:
The required displacements can be related directly to
deviations between 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
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12. Deflections of Beams and Shafts
EXAMPLE 12.8 (SOLN)
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 pt B, since this is the pt where
tangential deviation is to be determined. Hence,
2
L
M
L
M
L


 B  t B A      0     0
8EI
 4   EI  2 
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12. Deflections of Beams and Shafts
EXAMPLE 12.8 (SOLN)
Moment-area theorem:
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
2
L
M
M
L


C  tC A      0  L    0
2 EI
 2   EI  
Since both answers are –ve, they indicate that pts B
and C lie below the tangent at A. This checks with the
figure.
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12. Deflections of Beams and Shafts
12.5 METHOD OF SUPERPOSITION
•
•
•
The differential eqn EI d4/dx4 = w(x) satisfies the
two necessary requirements for applying the
principle of superposition
The load w(x) is linearly related to the deflection
(x)
The load is assumed not to change significantly
the original geometry of the beam or shaft.
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12. Deflections of Beams and Shafts
EXAMPLE 12.16
Steel bar shown is supported by two springs at its
ends A and B. Each spring has a stiffness k = 45 kN/m
and is originally unstretched. If the bar is loaded with a
force of 3 kN at pt C, determine the vertical
displacement of the force. Neglect the weight of the
bar and take Est = 200 GPa, I = 4.687510-6 m.
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12. Deflections of Beams and Shafts
EXAMPLE 12.16 (SOLN)
End reactions at A and B are
computed and shown. Each
spring deflects by an amount
2 kN
 A 1 
 0.0444 m
45 kN/m
1 kN
 B 1 
 0.0222 m
45 kN/m
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12. Deflections of Beams and Shafts
EXAMPLE 12.16 (SOLN)
If bar is considered rigid, these
displacements cause it to move
into positions shown. For this
case, the vertical displacement
at C is
2m
 A 1   B 1
C 1   B 1 
3m
2
 0.0222 m  0.0444 m  0.0282 m
3
 0.0370 m
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12. Deflections of Beams and Shafts
EXAMPLE 12.16 (SOLN)
We can find the displacement at C caused by the
deformation of the bar, by using the table in Appendix
C. We have
C 2

Pab 2

L  b2  a 2
6 EIL



3 kN 1 m 2 m  3 m 2  2 m 2  1 m 2

6200 106  kN/m 2 4.6875 10 6  m 4 3 m 
 1.422 mm
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12. Deflections of Beams and Shafts
EXAMPLE 12.16 (SOLN)
Adding the two displacement components, we get
  C  0.0370 m  0.001422 m
 0.0384 m  38.4 mm
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12. Deflections of Beams and Shafts
12.6 STATICALLY INDETERMINATE BEAMS AND SHAFTS
•
•
•
A member of any type is classified as statically
indeterminate if the no. of unknown reactions
exceeds the available no. of equilibrium eqns.
Additional support reactions on beam that are not
needed to keep it in stable equilibrium are called
redundants.
No. of these redundants is referred to as the
degree of indeterminacy.
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12. Deflections of Beams and Shafts
12.7 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
METHOD OF INTEGRATION
•
•
•
For a statically indeterminate beam, the internal
moment M can be expressed in terms of the
unknown redundants.
After integrating this eqn twice, there will be two
constants of integration and the redundants to be
found.
The unknowns can be found from the boundary
and/or continuity conditions for the problem.
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12. Deflections of Beams and Shafts
EXAMPLE 12.17
Beam is subjected to the distributed loading shown.
Determine the reactions at A. EI is a constant.
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12. Deflections of Beams and Shafts
EXAMPLE 12.17 (SOLN)
Elastic curve:
Beam deflects as shown. Only one coordinate x is
needed. For convenience, we will take it directed to
the right, since internal moment is easy to formulate.
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12. Deflections of Beams and Shafts
EXAMPLE 12.17 (SOLN)
Moment function:
Beam is indeterminate to first degree as indicated
from the free-body diagram. We can express the
internal moment M in terms of the redundant force at
A using segment shown below.
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12. Deflections of Beams and Shafts
EXAMPLE 12.17 (SOLN)
Moment function:
1 x3
M  Ay x  w0
6
L
Slope and elastic curve:
Applying Eqn 12-10,
d 2
1 x3
EI 2  Ay x  w0
6
L
dx
4
d 1
1
x
EI
 Ay x 2  w0  C1
dx 2
24
L
5
1
1
x
EI  Ay x3 
w0  C1x  C2
6
120
L
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12. Deflections of Beams and Shafts
EXAMPLE 12.17 (SOLN)
Slope and elastic curve:
The three unknowns Ay, C1 and C2 are determined
from the boundary conditions x = 0,  = 0; x = L,
d/dx = 0; and x = L,  = 0. Applying these conditions
yields
x  0,   0;
0  0  0  0  C2
d
x  L,
 0;
dx
x  L,   0;
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1
2
3
0  Ay L  w0 L  C1
2
24
1
1
3
0  Ay L 
w0 L4  C1L  C2
6
120
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12. Deflections of Beams and Shafts
EXAMPLE 12.17 (SOLN)
Slope and elastic curve:
1
Solving,
Ay  w0 L
10
1
C1  
w0 L3
120
C2  0
Using the result for Ay, the reactions at B can be
determined from the equations of equilibrium. Show
that Bx = 0. By = 2w0L/5 and MB= w0L2/15
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12. Deflections of Beams and Shafts
12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
MOMENT-AREA METHOD
• Draw the ME/I diagrams such that the redundants
are represented as unknowns.
• Apply the 2 moment-area theorems to get the
relationships between the tangents on elastic curve
to meet conditions of displacement and/or slope at
supports of beam.
• For all cases, no. of compatibility conditions is
equivalent to no. of redundants.
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12. Deflections of Beams and Shafts
12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
MOMENT-AREA METHOD
Moment diagrams constructed by method of
superposition
• Since moment-area theorems needs calculation of
both the area under the ME/I diagram and
centroidal location of this area, the method of
superposition can be used to combine separate
ME/I diagrams for each of the known loads.
• This will be relevant if the resultant moment
diagram is of a complicated shape.
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12. Deflections of Beams and Shafts
12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
MOMENT-AREA METHOD
Moment diagrams constructed by method of
superposition
• Most loadings on beams are a
combination of the four loadings
as shown.
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12. Deflections of Beams and Shafts
12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
MOMENT-AREA METHOD
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12. Deflections of Beams and Shafts
12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
MOMENT-AREA METHOD
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12. Deflections of Beams and Shafts
EXAMPLE 12.20
Beam is subjected to couple moment at its end C as
shown. Determine the reaction at B. EI is constant.
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12. Deflections of Beams and Shafts
EXAMPLE 12.20 (SOLN)
M/EI Diagram:
Free-body diagram as shown.
By inspection, beam is
indeterminate to first degree.
To get a direct solution,
choose By as the redundant.
Using superposition,
the M/EI diagrams for
By and M0, each applied
to the simply supported
beam are shown.
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12. Deflections of Beams and Shafts
EXAMPLE 12.20 (SOLN)
Elastic curve:
Elastic curve as
shown. Tangents at
A, B and C has been
established.
Since A = B = C = 0, then tangential deviations
shown must be proportional,
tB C
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 tA C
2
1
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12. Deflections of Beams and Shafts
EXAMPLE 12.20 (SOLN)
Elastic curve:
From ME/I diagram, we have
tB C
1 1  By L    2 1   M 0  

 L    L   
  L   
 L 
 3   2  2 EI    3   2  2 EI  
L    M 0  

  
 L 
 2   2 EI  
tA C
1  By L 
 2
1   M 0 


2 L    2 L    
  L  
2 L 
  2  EI 

 2  2 EI 
 3
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12. Deflections of Beams and Shafts
EXAMPLE 12.20 (SOLN)
Elastic curve:
Substituting into Eqn (1), we have
3M 0
By 
2L
Equations of equilibrium:
Reactions at A and C can now be determined from the
eqns of equilibrium. Show that Ax = 0, Cy = 5M0/4L,
and Ay = M0/4L.
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12. Deflections of Beams and Shafts
EXAMPLE 12.20 (SOLN)
Equations of equilibrium:
From figure shown, this problem can also be worked
out in terms of the tangential deviations,
1
t B A  tC A
2
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12. Deflections of Beams and Shafts
12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
METHOD OF SUPERPOSITION
• First, identify the redundant support reactions on
the beam.
• Remove these reactions from the beam to get a
primary beam that is statically determinate and
stable and subjected to external load only.
• Add to this beam with a series of similarly supported
beams, each with a separate redundant, then by
principle of superposition, the final loaded beam is
obtained.
• After computing the redundants, the other reactions
on the beam determined from the eqns of
equilibrium.
• This method of analysis is sometimes called the
force method.
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12. Deflections of Beams and Shafts
12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
METHOD OF SUPERPOSITION
Procedure for analysis
Elastic curve
• Specify unknown redundant forces or moments that
must be removed from the beam in order to make it
statically determinate and stable.
• Use principle of superposition, draw the statically
indeterminate beam and show it to be equal to a
sequence of corresponding statically determinate
beams.
• The first beam (primary) supports the same external
loads as the statically indeterminate beam, and
each of the other beams “added” to the primary
beam shows the beam loaded with a separate
single redundant force or moment.
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12. Deflections of Beams and Shafts
12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
METHOD OF SUPERPOSITION
Procedure for analysis
Elastic curve
• Sketch the deflection curve for each beam and
indicate symbolically the displacement or slope at
the pt of each redundant force or moment.
Compatibility equations
• Write a compatibility eqn for the displacement or
slope at each pt where there is a redundant force or
moment.
• Determine all the displacements or slopes using an
appropriate method explained in chapter 12.212.5.
• Substitute the results into the compatibility eqns and
solve for the unknown redundants.
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12. Deflections of Beams and Shafts
12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
METHOD OF SUPERPOSITION
Procedure for analysis
Compatibility equations
• If a numerical value for a redundant is +ve, it has
the same sense of direction as originally assumed.
• Similarly, a –ve numerical value indicates the
redundant acts opposite to its assumed sense of
direction.
Equilibrium equations
• Once the redundant forces and/or moments have
been determined, the remaining unknown reactions
can be found from the eqns of equilibrium applied to
the loadings shown on the beam’s free-body
diagram.
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12. Deflections of Beams and Shafts
EXAMPLE 12.22
Determine the reactions on the beam shown. Due to
loading and poor construction, the roller support at B
settles 12 mm.
Take E = 200 GPa and I = 80(106) mm4.
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12. Deflections of Beams and Shafts
EXAMPLE 12.22 (SOLN)
Principle of superposition
By inspection, beam is indeterminate
to the first degree. Roller support at
B is chosen as the redundant.
Principle of superposition is shown.
Here, By is assumed to act upwards
on the beam.
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12. Deflections of Beams and Shafts
EXAMPLE 12.22 (SOLN)
Compatibility equation
With reference to pt B, we require
 
0.012 m   B   'B
1
Using table in Appendix C, displacements are
5wL4 524 kN/m 8 m 4 640 kN  m3
B 


768 EI
768 EI
EI
3
PL
 'B 

48 EI
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By 8 m 3
48 EI

10.67 m3By
EI
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12. Deflections of Beams and Shafts
EXAMPLE 12.22 (SOLN)
Compatibility equation
Thus Eqn (1) becomes
0.012 EI  640  10.67 By
Expressing E and I in units of kN/m2 and m4, we have
     640  10.67 By
0.012200  10 80 10
6
6
B y  42.0 kN
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12. Deflections of Beams and Shafts
EXAMPLE 12.22 (SOLN)
Equilibrium equations:
Applying this result to the beam, we then calculate the
reactions at A and C using eqns of equilibrium.
  M A  0;
96 kN 2 m   42.0 kN 4 m   C y 8 m   0
C y  3.00 kN

 Fy  0;
Ay  96 kN  42.0 kN  3.00 kN  0
Ay  51 kN
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12. Deflections of Beams and Shafts
EXAMPLE 12.24
Determine the moment at B for beam shown. EI is
constant. Neglect the effects of axial load.
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12. Deflections of Beams and Shafts
EXAMPLE 12.24
Principle of superposition:
Since axial load if neglected, a there
is a vertical force and moment at A
and B. Since only two eqns of
equilibrium are available, problem is
indeterminate to the second degree.
Assume that By and MB are
redundant, so that by principle of
superposition, beam is represented
as a cantilever, loaded separately by
distributed load and reactions By and
MB, as shown.
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12. Deflections of Beams and Shafts
EXAMPLE 12.24
Compatibility equations:
Referring to displacement and slope at B, we require
 
1
0   B   'B  ' 'B
 
2 
0   B   'B  ' 'B
Using table in Appendix C to compute slopes and
displacements, we have
3
wL
B 

48 EI
3


9 kN/m 4 m
12
48 EI

EI
7 wL4 79 kN/m 4 m 4 42
B 


384 EI
384 EI
EI
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12. Deflections of Beams and Shafts
EXAMPLE 12.24
Compatibility equations:
B y 4 m 
8By
PL
 'B 


2 EI
2 EI
EI
2
2
B y 4 m 
21.33 B y
PL
 'B 


3EI
3EI
EI
ML M B 4 m  4 M B
 ' 'B 


EI
EI
EI
3
3
ML2 M B 4 m 2 8M B
 ' 'B 


2 EI
2 EI
EI
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12. Deflections of Beams and Shafts
EXAMPLE 12.24
Compatibility equations:
Substituting these values into Eqns (1) and (2) and
canceling out the common factor EI, we have
 
 
0  12  8B y  4M B
0  42  21.33B y  8M B
Solving simultaneously, we get
By  3.375 kN
M B  3.75 kN  m
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12. Deflections of Beams and Shafts
CHAPTER REVIEW
• The elastic curve represents the centerline
deflection of a beam or shaft.
• Its shape can be determined using the moment
diagram.
• Positive moments cause the elastic curve to
concave upwards and negative moments cause it
to concave downwards.
• The radius of curvature at any pt is determined
from 1/ = M/EI.
• Eqn of elastic curve and its slope can be obtained
by first finding the internal moment in the member
as a function of x.
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12. Deflections of Beams and Shafts
CHAPTER REVIEW
• If several loadings act on the member, then
separate moment functions must be determined
between each of the loadings.
• Integrating these functions once using
EI(d2/dx2) = M(x) gives the eqn for the slope of the
elastic curve, and integrating again gives the eqn
for the deflection.
• The constants of integration are determined from
the boundary conditions at the supports, or in
cases where several moment functions are
involved, continuity of slope and deflection at pts
where these functions join must be satisfied.
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12. Deflections of Beams and Shafts
CHAPTER REVIEW
• Discontinuity functions allow us to express the eqn
of elastic curve as a continuous function,
regardless of the no. of loadings on the member.
• This method eliminates the need to use continuity
conditions, since the two constants of integration
can be determined solely from the two boundary
conditions.
• The moment-area method if a semi-graphical
technique for finding the slope of tangents or the
vertical deviation of tangents at specific pts on the
elastic curve.
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12. Deflections of Beams and Shafts
CHAPTER REVIEW
• The moment-area method requires finding area
segments under the M/EI diagram, or the moment
of these segments about pts on the elastic curve.
• The method works well for M/EI diagrams
composed of simple shapes, such as those
produced by concentrated forces and couple
moments.
• The deflection or slope at a pt on a member
subjected to various types of loadings can be
determined by using the principle of superposition.
The table in the back of the book can be used for
this purpose.
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12. Deflections of Beams and Shafts
CHAPTER REVIEW
• Statically indeterminate beams and shafts have
more unknown support reactions than available
eqns of equilibrium.
• To solve them such problems, we first identify the
redundant reactions, and the other unknown
reactions are written in terms of these redundants.
• The method of integration or moment-area
theorems can be used to solve for the unknown
redundants.
• We can also determine the redundants by using the
method of superposition, where we consider the
continuity of displacement at the redundant.
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12. Deflections of Beams and Shafts
CHAPTER REVIEW
• The displacement due to the external loading is
determined with the redundant removed, and again
with the redundant applied and external loading
removed.
• The tables in Appendix C of this book can be used
to determine these necessary displacements.
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