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Double Integration Method | Beam Deflections
Double Integration Method | Beam Deflections
The double integration method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to get
The double integration method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to get
the equation of the elastic curve.
the equation of the elastic curve.
In calculus, the
In calculus, the
radius of c
radius of curvature of a curve y = f(x) is given by
urvature of a curve y = f(x) is given by
In the derivation of
In the derivation of
flexure formula
flexure formula
, the radius of curvature of a beam is given as
, the radius of curvature of a beam is given as
Def
Def
lection of beams is so small, such th
lection of beams is so small, such that the slope of the elastic curve dy/dx is very small, and squaring this expression the value
at the slope of the elastic curve dy/dx is very small, and squaring this expression the value
bec
becomes practically negligible, hence
omes practically negligible, hence
Thus, EI / M = 1 / y''
Thus, EI / M = 1 / y''
If EI is constant, the equation may be written as:
If EI is constant, the equation may be written as:
ρ ρ = = [[ 11 ++ (( �� � � / / �� � � ) ) 2 2 ] ] 33 // 2 2 | | � � � � / / � � | | 2 2 � � 2 2 ρ ρ = = EE I I M M ρ ρ == = = 1 1 � � / / � � � � 2 2 � � 2 2 1 1 � � ′′′ ′ == = = M M � � ′′′ ′ M M EE I I 1 1 EE I I EE I I = = M M � � ′′′ ′6
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Solution to Problem 605 | Double Integration Method
Solution to Problem 605 | Double Integration Method
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where x and y are the coordinates shown in the figure of the
where x and y are the coordinates shown in the figure of the
elastic curve of the beam under load
elastic curve of the beam under load
, y is the deflection of the beam at any
, y is the deflection of the beam at any
distance x. E is the modulus of elasticity of the beam, I represent the moment of inertia about the neutral axis, and M represents the
distance x. E is the modulus of elasticity of the beam, I represent the moment of inertia about the neutral axis, and M represents the
bending moment at
bending moment at
a distance
a distance
x from
x from
the end
the end
of the
of the
beam. The
beam. The
product EI
product EI
is called
is called
the
the
flexural rigidity
flexural rigidity of the beam.
of the beam.
The first integration y' yields the slope of the elastic curve and the second integration y gives the deflection of the beam at any distance x.
The first integration y' yields the slope of the elastic curve and the second integration y gives the deflection of the beam at any distance x.
The resulting solution must contain two constants of integration since EI y" = M is of second order. These two constants must be
The resulting solution must contain two constants of integration since EI y" = M is of second order. These two constants must be
evaluated from known conditions concerning the slope deflection at certain points of the beam. For instance, in the case of a simply
evaluated from known conditions concerning the slope deflection at certain points of the beam. For instance, in the case of a simply
supported beam with rigid supports, at x = 0 and x = L, the deflection y = 0, and in locating the point of maximum deflection, we simply
supported beam with rigid supports, at x = 0 and x = L, the deflection y = 0, and in locating the point of maximum deflection, we simply
set the slope of the elastic curve y' to zero.
set the slope of the elastic curve y' to zero.
Solution to Problem 605 | Double Integration Method
Solution to Problem 605 | Double Integration Method
Solution to Problem 606 | Double Integration Method
Solution to Problem 606 | Double Integration Method
Solution to Problem 607 | Double Integration Method
Solution to Problem 607 | Double Integration Method
Solution to Problem 608 | Double Integration Method
Solution to Problem 608 | Double Integration Method
Solution to Problem 609 | Double Integration Method
Solution to Problem 609 | Double Integration Method
Solution to Problem 610 | Double Integration Method
Solution to Problem 610 | Double Integration Method
Solution to Problem 611 | Double Integration Method
Solution to Problem 611 | Double Integration Method
Solution to Problem 612 | Double Integration Method
Solution to Problem 612 | Double Integration Method
Solution to Problem 613 | Double Integration Method
Solution to Problem 613 | Double Integration Method
Solution to Problem 614 | Double Integration Method
Solution to Problem 614 | Double Integration Method
Solution to Problem 615 | Double Integration Method
Solution to Problem 615 | Double Integration Method
Solution to Problem 616 | Double Integration Method
Solution to Problem 616 | Double Integration Method
Solution to Problem 617 | Double Integration Method
Solution to Problem 617 | Double Integration Method
Solution to Problem 618 | Double Integration Method
Solution to Problem 618 | Double Integration Method
Solution to Problem 619 | Double Integration Method
Solution to Problem 619 | Double Integration Method
Solution to Problem 620 | Double Integration Method
Solution to Problem 620 | Double Integration Method
Solution to Problem 621 | Double Integration Method
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Solution to Problem 605
Solution to Problem 605
|
|
Double Integration Method
Double Integration Method
Problem 605
Problem 605
Determine the maximum deflection δ in a simply supported beam of length L carrying a concentrated load P at midspan.
Determine the maximum deflection δ in a simply supported beam of length L carrying a concentrated load P at midspan.
Solution 605
Solution 605
Click here to show or hide the solution
Click here to show or hide the solution
At x = 0, y = 0, therefore, C
At x = 0, y = 0, therefore, C
2
2
=
=
0
0
At x = L, y = 0
At x = L, y = 0
Thus,
Thus,
EE � � = = �� � � − − � � ⟨ ⟨ � � − − � � ⟩ ⟩ � � ′′′ ′ 1 1 2 2 1 1 2 2 EE � � � � = = � � − − � � ⟨ ⟨ � � − − � � + + ′ ′ 1 1 4 4 � � 2 2 1 1 2 2 1 1 2 2 ⟩ ⟩ 2 2 C C 1 1 EE �� � � = = � � − − � � ⟨ ⟨ � � − − � � + + � � + + 1 1 11 2 2 � � 3 3 1 1 6 6 1 1 2 2 ⟩ ⟩ 3 3 C C 1 1 C C 2 2 00 = = � � − − � � ⟨ ⟨ � � − − � � + + � � 1 1 11 2 2 � � 3 3 1 1 6 6 1 1 2 2 ⟩ ⟩ 3 3 C C 1 1 00 = = � � − − � � + + � � 1 1 11 2 2 � � 33 1 1 44 8 8 � � 3 3 C C 1 1 == − − � � C C 1 1 1 1 11 6 6 � � 2 2 EE �� � � = = � � − − � � ⟨ ⟨ � � − − � � − − �� � � 1 1 11 2 2 � � 3 3 1 1 6 6 1 1 2 2 ⟩ ⟩ 3 3 1 1 11 6 6 � � 2 26
6//1
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beam deflection
elastic curve
elastic curve
maximum deflection
maximum deflection
midspan deflection
midspan deflection
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Solution to Problem 606 | Double Integration Method
Solution to Problem 606 | Double Integration Method
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Comments
Comments
Submitted by Noel Gatbonton on October 16, 2016 - 7:55pm
Submitted by Noel Gatbonton on October 16, 2016 - 7:55pm
Sa part po na
Sa part po na
x=0, y=0 therefore C
x=0, y=0 therefore C
2
2
=0
=0
Kapag ni substitute po yung 0 sa x
Kapag ni substitute po yung 0 sa x
hindi naman po mag eequal yung C
hindi naman po mag eequal yung C
2
2
sa 0 po?
sa 0 po?
Submitted by Alexander on April 2, 2017 - 8:42pm
Submitted by Alexander on April 2, 2017 - 8:42pm
Read your text book bro. Ang tanong mo ay isa sa mga basic ng
Read your text book bro. Ang tanong mo ay isa sa mga basic ng
double integration method
double integration method
.
.
Maximum deflection will occur at x = ½ L (midspan)
Maximum deflection will occur at x = ½ L (midspan)
The negative sign indicates that the deflection is below the undeformed neutral axis.
The negative sign indicates that the deflection is below the undeformed neutral axis.
Therefore,
Therefore,
answer
answer
EE � � = = � � ( ( � � − − � � ( ( � � − − � � − − � � ( ( � � ) ) � � �� �� � � 1 1 11 2 2 1 1 2 2 ) ) 3 3 1 1 6 6 1 1 2 2 1 1 2 2 ) ) 3 3 1 1 11 6 6 � � 2 2 1 1 2 2 EE � � � � = = � � −− 00 − − � � �� �� � � 1 1 99 6 6 � � 3 3 1 1 33 2 2 � � 3 3 == − − � � �� �� � � �� � � 3 3 44 8 8 EE � � = = δ δ �� �� � � �� � � 3 3 44 8 8 EE � �ht
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Solution to Problem 606 | Double Integration Method
Solution to Problem 606 | Double Integration Method
Problem 606
Problem 606
Determine the maximum deflection δ in a simply supported beam of length L carrying a uniformly distributed load of intensity w
Determine the maximum deflection δ in a simply supported beam of length L carrying a uniformly distributed load of intensity w
o
o
applied over its entire length.
applied over its entire length.
Solution 606
Solution 606
Click here to show or hide the solution
Click here to show or hide the solution
From the figure below
From the figure below
At x = 0, y = 0, therefore C
At x = 0, y = 0, therefore C
2
2
=
=
0
0
At x = L, y = 0
At x = L, y = 0
EE � � � � = = �� � � − − � � ( ( � � ) ) ′′′ ′ 1 1 2 2 � � � � � � � � 1 1 2 2 EE � � = = �� � � − − � � ′′′ ′ 1 1 2 2 � � � � 1 1 2 2 � � � � � � 2 2 EE � � = = � � −− + + � � ′ ′ 1 1 4 4 � � � � � � 2 2 1 1 6 6 � � � � � � 3 3 C C 1 1 EE �� � � = = � � −− + + � � + + 1 1 11 2 2 � � � � � � 33 1 1 22 4 4 � � � � � � 4 4 C C 1 1 C C 2 2 00 == −− ++ � � 1 1 11 2 2 � � � � � � 4 4 1 1 22 4 4 � � � � � � 4 4 C C 1 1 == − − C C 1 1 1 1 22 4 4 � � � � � � 3 36
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elastic curve
elastic curve
maximum deflection
maximum deflection
midspan deflection
midspan deflection
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Solution to Problem 607 | Double Integration Method
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Therefore,
Therefore,
Maximum deflection will occur at x = ½ L (midspan)
Maximum deflection will occur at x = ½ L (midspan)
answer
answer
Taking W = w
Taking W = w
o
o
L:
L:
answer
answer
EE �� � � = = � � −− − − � � 1 1 11 2 2 � � � � � � 3 3 1 1 22 4 4 � � � � � � 4 4 1 1 22 4 4 � � � � � � 3 3 EE � � � � = = � � ( ( � � −− ( ( � � −− ( ( � � ) ) �� �� � � 1 1 11 2 2 � � � � 1 1 2 2 ) ) 3 3 1 1 22 4 4 � � � � 1 1 2 2 ) ) 4 4 1 1 22 4 4 � � � � � � 3 3 1 1 2 2 EE � � == −− − − � � �� �� � � 1 1 99 6 6 � � � � � � 4 4 1 1 33 88 4 4 � � � � � � 4 4 1 1 44 8 8 � � � � � � 4 4 EE � � � � == − − �� �� � � 5 5 33 88 4 4 � � � � � � 4 4 = = δ δ �� �� � � 5 5 � � � � � � 4 4 33 88 4 4 EE � � = = δ δ �� �� � � 55 ( ( � � � � )) (( ) ) � � � � 3 3 33 88 4 4 EE � � = = δ δ �� �� � � 5 5 �� � � 3 3 33 88 4 4 EE � �ht
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Solution to Problem 607
Solution to Problem 607
|
|
Double Integration Method
Double Integration Method
Problem 607
Problem 607
Determine the maximum value of EIy for the cantilever beam loaded as shown in Fig. P-607. Take the origin at the wall.
Determine the maximum value of EIy for the cantilever beam loaded as shown in Fig. P-607. Take the origin at the wall.
Solution 607
Solution 607
Click her
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e to show or hide the solution
e to show or hide the solution
At x = 0, y' = 0, therefore C
At x = 0, y' = 0, therefore C
1
1
=
=
0
0
At x = 0, y = 0, therefore C
At x = 0, y = 0, therefore C
2
2
=
=
0
0
Therefore,
Therefore,
�� � � == − − �� � � + + �� � � − − � � ⟨ ⟨ � � − − � � ⟩ ⟩ � � ′′′ ′ �� � � � � == − − �� �� � � + + � � − − � � ⟨ ⟨ � � − − � � + + ′ ′ 1 1 2 2 � � 2 2 1 1 2 2 ⟩ ⟩ 2 2 � � 1 1 �� �� � � == − − �� � � + + � � − − � � ⟨ ⟨ � � − − � � + + � � + + 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 1 1 6 6 ⟩ ⟩ 3 3 � � 1 1 � � 2 2 �� �� � � == − − �� � � + + � � − − � � ⟨ ⟨ � � − − � � 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 1 1 6 6 ⟩ ⟩ 3 36
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Solution to Problem 608 | Double Integration Method
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The maximum value of EI y is at x = L (free end)
The maximum value of EI y is at x = L (free end)
answer
answer
�� � � � � == − − �� � � + + � � − − � � ( ( � � − − � � �� �� � � 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 1 1 6 6 ) ) 3 3 �� � � == − − �� � � + + � � − − � � (( −− 3 3 � � ++ 3 3 � � −− ) ) � � �� �� � � 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 1 1 6 6 � � 3 3 � � 2 2 � � 2 2 � � 3 3 �� � � == − − �� � � + + � � − − � � + + �� � � − − �� � � + + � � � � �� �� � � 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 1 1 6 6 � � 3 3 1 1 2 2 � � 2 2 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 �� � � == − − �� � � + + � � � � �� �� � � 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 �� � � == − − �� � � + + � � � � �� �� � � 1 1 2 2 � � 2 2 1 1 6 6 � � 3 3 �� � � == − − � � (( 3 3 � � − − � � ) ) � � �� �� � � 1 1 6 6 � � 2 2ht
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Solution to Problem 608
Solution to Problem 608
|
|
Double Integration Method
Double Integration Method
Problem 608
Problem 608
Find the equation of the elastic curve for the cantilever beam shown in
Find the equation of the elastic curve for the cantilever beam shown in
Fig. P-608
Fig. P-608
; it carries a load that varies from zero at the wall to w
; it carries a load that varies from zero at the wall to w
o
o
at the free end. Take the origin at the wall.
at the free end. Take the origin at the wall.
Solu
Solution
tion
608
608
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solution
solution
By ratio and proportion
By ratio and proportion
� � = = � � 1 1 2 2 � � � � � � = = � � ( ( � � ) ) 1 1 2 2 � � � � 2 2 3 3 � � = = 1 1 3 3 � � � � � � 2 2 = = � � � � � � � � � � � � = = � � � � � � � � � � = = �� � � 1 1 2 2 � � = = � � � � � � � � 1 1 2 2 � � � � � � � � = = � � � � 2 2 � � � � 2 2 �� � � == − − � � + + �� � � − − � � ( ( � � ) ) � � ′′′ ′ 1 1 3 3 �� � � == −− + + �� � � − − � � �� � � � � ′′′ ′ 1 1 3 3 � � � � � � 2 2 1 1 2 2 � � � � 1 1 3 3 � � � � 2 2 � � � � 2 2 2 26
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‹ Solution to Problem 607 | Double Integration
‹ Solution to Problem 607 | Double Integration
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Solution to Problem 609 | Double Integration Method
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At x = 0, y' = 0, therefore C
At x = 0, y' = 0, therefore C
1
1
=
=
0
0
At x = 0, y = 0, therefore C
At x = 0, y = 0, therefore C
2
2
=
=
0
0
Therefore, the equation of the elastic curve is
Therefore, the equation of the elastic curve is
answer
answer
�� � � 2 2 � � �� � � � � == −− + + � � − − ′′′ ′ � � � � � � 2 2 3 3 � � � � � � 2 2 � � � � 6 6 � � � � 3 3 �� � � � � == − − � � ++ −− + + ′ ′ � � � � � � 2 2 3 3 � � � � � � 4 4 � � 2 2 � � � � 22 4 4 � � � � 4 4 � � 1 1 �� �� � � == −− ++ −− ++ � � + + � � � � � � 2 2 6 6 � � 2 2 � � � � � � 11 2 2 � � 3 3 � � � � 11 22 0 0 � � � � 5 5 � � 1 1 � � 2 2 �� �� � � == −− ++ −− � � � � � � 2 2 6 6 � � 2 2 � � � � � � 11 2 2 � � 3 3 � � � � 11 22 0 0 � � � � 5 5ht
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Solution to Problem 609 | Double Integration Method
Solution to Problem 609 | Double Integration Method
Problem 609
Problem 609
As shown in
As shown in
Fig. P-609
Fig. P-609
, a simply supported beam carries two symmetrically placed concentrated loads. Compute the maximum
, a simply supported beam carries two symmetrically placed concentrated loads. Compute the maximum
deflection δ.
deflection δ.
So
Solution 60
lution 609
9
C
C
lick here to show or hide the solutio
lick here to show or hide the solutio
n
n
By
By
symmetry
symmetry
== = = � � � � 1 1 � � 2 2 EE � � � � = = �� � � − − � � ⟨ ⟨ � � − − � � ⟩ ⟩ − − � � ⟨ ⟨ � � − − � � + + � � ⟩ ⟩ ′′′ ′ EE � � � � = = � � − − � � ⟨ ⟨ � � − − � � − − � � ⟨ ⟨ � � − − � � + + � � + + ′ ′ 1 1 2 2 � � 2 2 1 1 2 2 ⟩ ⟩ 2 2 1 1 2 2 ⟩ ⟩ 2 2 C C 1 1 EE �� � � = = � � − − � � ⟨ ⟨ � � − − � � − − � � ⟨ ⟨ � � − − � � + + � � + + � � + + 1 1 6 6 � � 33 1 1 6 6 ⟩ ⟩ 33 1 1 6 6 ⟩ ⟩ 3 3 C C 1 1 C C 2 26
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At x = 0, y = 0, therefore C
At x = 0, y = 0, therefore C
2
2
=
=
0
0
At x = L, y = 0
At x = L, y = 0
Therefore,
Therefore,
Maximum deflection will occur at x = ½ L (midspan)
Maximum deflection will occur at x = ½ L (midspan)
answer
answer
00 = = � � − − � � ⟨ ⟨ � � − − � � + + � � 1 1 6 6 � � 3 3 1 1 6 6 ⟩ ⟩ 3 3 C C 1 1 00 = = � � � � − − � � (( −− 3 3 � � ++ 3 3 � � −− )) − − � � ++ 6 6 � � 3 3 � � 3 3 � � 2 2 � � 2 2 � � 3 3 � � 3 3 C C 1 1 00 = = � � − − � � ++ 3 3 �� � � −− 3 3 �� � � + + � � − − � � ++ 6 6 � � � � 3 3 � � 3 3 � � 2 2 � � 2 2 � � 3 3 � � 3 3 C C 1 1 00 == 3 3 �� � � � � −− 3 3 �� � � ++ 6 6 � � 2 2 � � 2 2 C C 1 1 00 == 3 3 �� �� � � ( ( � � − − � � )) ++ 6 6 C C � � 1 1 == − − �� � � ( ( � � − − � � ) ) C C 1 1 1 1 2 2 EE �� � � = = � � − − � � ⟨ ⟨ � � − − � � − − � � ⟨ ⟨ � � − − � � + + � � − − �� � � ( ( � � − − � � ) ) � � 1 1 6 6 � � 3 3 1 1 6 6 ⟩ ⟩ 3 3 1 1 6 6 ⟩ ⟩ 3 3 1 1 2 2 EE � � = = � � ( ( � � − − � � ( ( � � − − � � − − �� � � ( ( � � − − � � )) ( ( � � ) ) � � �� �� � � 1 1 6 6 1 1 2 2 ) ) 33 1 1 6 6 1 1 2 2 ) ) 33 1 1 2 2 1 1 2 2 EE � � � � = = � � − − � � [[ ( ( � � −− 2 2 � � )) − − �� � � + + �� � � �� �� � � 1 1 44 8 8 � � 3 3 1 1 6 6 1 1 2 2 ] ] 3 3 1 1 4 4 � � 2 2 1 1 4 4 � � 2 2 EE � � � � = = � � − − � � [[ −− 33 (( 22 � � )) ++ 3 3 � � (( 2 2 � � −− (( 2 2 � � ]] − − �� � � + + �� � � �� �� � � 1 1 44 8 8 � � 3 3 1 1 44 8 8 � � 3 3 � � 2 2 ) ) 2 2 ) ) 3 3 1 1 4 4 � � 2 2 1 1 4 4 � � 2 2 EE � � � � = = � � − − � � + + �� � � − − �� � � + + � � − − �� � � + + �� � � �� �� � � 1 1 44 8 8 � � 3 3 1 1 44 8 8 � � 3 3 1 1 8 8 � � 2 2 1 1 4 4 � � 2 2 1 1 6 6 � � 3 3 1 1 4 4 � � 2 2 1 1 4 4 � � 2 2 EE � � � � == − − �� � � + + � � �� �� � � 1 1 8 8 � � 2 2 1 1 6 6 � � 3 3 EE � � � � == − − �� � � (( 33 −− 44 )) �� �� � � 1 1 22 4 4 � � 2 2 � � 2 2 == −− (( 33 −− 44 )) � � �� �� � � �� � � 22 4 4 EE � � � � 2 2 � � 2 2 == (( 33 −− 44 )) δ δ �� �� � � �� � � 22 4 4 EE � � � � 2 2 � � 2 2ht
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Solution to Problem 610
Solution to Problem 610
|
|
Double Integration Method
Double Integration Method
Problem 610
Problem 610
The simply supported beam shown in
The simply supported beam shown in
Fig. P-610
Fig. P-610
carries a uniform load of intensity w
carries a uniform load of intensity w
o
o
symmetrically distributed over part of its length.
symmetrically distributed over part of its length.
Determine the maximum deflection δ and check your result by letting a = 0 and comparing with the answer to
Determine the maximum deflection δ and check your result by letting a = 0 and comparing with the answer to
Problem 606
Problem 606
.
.
So
Solution
lution
610
610
C
C
lick here to show or hide the solutio
lick here to show or hide the solutio
n
n
By
By
symmetry
symmetry
At x = 0, y = 0, therefore C
At x = 0, y = 0, therefore C
2
2
=
=
0
0
== = = � � � � 1 1 � � 2 2 � � � � EE � � � � = = �� � � − − ⟨ ⟨ � � − − � � ′′′ ′ � � � � 1 1 2 2 � � � � ⟩ ⟩ 2 2 EE � � � � = = � � − − ⟨ ⟨ � � − − � � + + ′ ′ 1 1 2 2 � � � � � � 2 2 1 1 6 6 � � � � ⟩ ⟩ 3 3 C C 1 1 EE �� � � = = � � − − ⟨ ⟨ � � − − � � + + � � + + 1 1 6 6 � � � � � � 33 1 1 22 4 4 � � � � ⟩ ⟩ 4 4 C C 1 1 C C 2 26
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Uniformly
Uniformly
Distributed
Distributed
Load
Load
maximum
maximum
deflection
deflection
midspan
midspan
deflection
deflection
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At x = a + b, y' = 0
At x = a + b, y' = 0
Therefore,
Therefore,
Maximum deflection will occur at x = a + b (midspan)
Maximum deflection will occur at x = a + b (midspan)
Therefore,
Therefore,
answer
answer
Checking:
Checking:
When a = 0, 2b = L, thus b = ½ L
When a = 0, 2b = L, thus b = ½ L
(
(okay!
okay!)
)
00 = = � � ( ( � � + + � � −− + + 1 1 2 2 � � � � ) ) 2 2 1 1 6 6 � � � � � � 3 3 C C 1 1 == − − � � ( ( � � + + � � C C 1 1 1 1 6 6 � � � � � � 3 3 1 1 2 2 � � � � ) ) 2 2 EE �� � � = = � � − − ⟨ ⟨ � � − − � � + + � � − − � � ( ( � � + + �� � � 1 1 6 6 � � � � � � 3 3 1 1 22 4 4 � � � � ⟩ ⟩ 4 4 1 1 6 6 � � � � � � 3 3 1 1 2 2 � � � � ) ) 2 2 EE � � = = � � ( ( � � + + � � −− ++ ( ( � � + + � � )) − − � � ( ( � � + + � � � � �� �� � � 1 1 6 6 � � � � ) ) 3 3 1 1 22 4 4 � � � � � � 4 4 1 1 6 6 � � � � � � 3 3 1 1 2 2 � � � � ) ) 3 3 EE � � � � == − − � � ( ( � � + + � � −− ++ ( ( � � + + � � ) ) �� �� � � 1 1 3 3 � � � � ) ) 3 3 1 1 22 4 4 � � � � � � 4 4 1 1 6 6 � � � � � � 3 3 EE � � � � == − − � � [[ 88 ( ( � � + + � � ++ −− 44 (( � � + + � � )) ] ] �� �� � � 1 1 22 4 4 � � � � ) ) 3 3 � � 3 3 � � 2 2 == [[ 88 (( � � + + � � ++ −− 44 (( � � + + � � )) ] ] δ δ �� �� � � � � � � � � 22 4 4 EE � � ) ) 3 3 � � 3 3 � � 2 2 == [[ 88 (( 00 ++ � � ++ ( ( � � −− 44 ( ( � � (( 00 + + � � )) ] ] δ δ �� �� � � ( ( � � ) ) � � � � 1 1 2 2 22 4 4 EE � � 1 1 2 2 ) ) 3 3 1 1 2 2 ) ) 3 3 1 1 2 2 ) ) 2 2 1 1 2 2 == [[ ++ −− ]] δ δ �� �� � � � � � � � � 44 8 8 EE � � � � 33 1 1 8 8 � � 33 1 1 2 2 � � 3 3 == [[ ] ] δ δ �� �� � � � � � � � � 44 8 8 EE � � 5 5 8 8 � � 3 3 = = δ δ �� �� � � 5 5 � � � � � � 4 4 33 88 4 4 EE � �ht
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Solution to Problem 611 | Double Integration Method
Solution to Problem 611 | Double Integration Method
Problem 611
Problem 611
Compute the value of EI δ at midspan for the beam loaded as shown in
Compute the value of EI δ at midspan for the beam loaded as shown in
Fig. P-611
Fig. P-611
. If E = 10 GPa, what value of I is required to limit the
. If E = 10 GPa, what value of I is required to limit the
midspan deflection to 1/360 of the span?
midspan deflection to 1/360 of the span?
So
Solution 61
lution 61
1
1
C
C
lick here to show or hide the solutio
lick here to show or hide the solutio
n
n
ΣΣ == 0 0 � � � � 2 2 44 � � == 33 00 00 (( 22 )) (( 33 )) 1 1 == 44 55 00 NN � � 1 1 ΣΣ == 0 0 � � � � 1 1 44 � � == 33 00 00 (( 22 )) (( 11 )) 2 2 == 11 55 00 NN � � 2 2 EE � � � � == 44 55 00 � � −− (( 33 00 00 )) ++ (( 33 00 00 )) ⟨ ⟨ � � −− 2 2 ′′′ ′ 1 1 2 2 � � 2 2 1 1 2 2 ⟩ ⟩ 2 2 EE � � � � == 44 55 00 � � −− 11 55 00 ++ 11 55 00 ⟨ ⟨ � � −− 2 2 ′′′ ′ � � 2 2 ⟩ ⟩ 2 2 EE � � � � == 22 22 55 −− 55 00 ++ 55 00 ⟨ ⟨ � � −− 22 + + ′ ′ � � 2 2 � � 3 3 ⟩ ⟩ 3 3 C C 1 1 EE �� � � == 77 55 � � −− 11 22 .. 55 ++ 11 22 .. 55 ⟨ ⟨ � � −− 22 + + � � + + 3 3 � � 4 4 ⟩ ⟩ 4 4 C C 1 1 C C 2 2
