Double Integration Method

Problem

Determine the maximum deflection for the beam shown using the Double Integration Method.

Solution:

Step 1: Determine the change of load points and label them accordingly.

Note: In the figure, there are four change of load points: the two ends, where there are reactions;

the point 3 m from the left end, where the 60 kN concentrated load is applied and the 20 kN/m

uniformly distributed load commences, and at the point 9 m from the left end, where the 20 kN/m

uniformly distributed load terminates.

Label them as A, B, C and D, from the left.

Step 2: Determine the reactions

�M� � 0

60kN3m � 20 kNm 6m �3m � 12 6m � � R�3m � 6m � 3m � 0

R� � 75.00kN

�F� � 0

R� � R� � 60kN � 20kNm 6m � 0

R� � 75.00kN � 60kN � 20kNm 6m � 0

R� � 105.00kN

Note: This step is crucial.

Step 3: Pass a cutting plane line in between the last two change of load points.

The distance from this line to the left end of the beam is designated as x.

Step 3.a: Add an imaginary 20 kN/m uniformly distributed load from points C to D upward

and downward.

Note: If the moment equation is directly written from the given loading, the following would be the

result:

From A to B:

� � ��

From B to C:

� � �� 60� � � 3! � 20 � ! � � 3! �12 � � 3! �

From C to D:

� � �� 60� � � 3! � 20 � ! 6! �� 3! � 12 6! �

Note: The last terms of the Moment Equations from B to C and from C to D are not at all similar in

form. The objective of writing the Moment Equation is to have one expression that is applicable to

the entire span. This can be resolved by adding an imaginary 20 kN/m uniformly distributed load

from points C to D upward and downward. Note that the sum of these loads is zero; it does not add

to, subtract from, or change the structural system in anyway.

Step 4: Write the moment equation from the cutting plane line, clockwise positive.

M � R�x � 60kNx � 3m � 20 kNm x � 3m �12 x � 3m �� 20 kNm x � 3m � 6m �12 x � 3m � 6m �

M � 75.00kNx � 60kNx � 3m � 20 kNm x � 3m �12 x � 3m �� 20 kNm x � 3m � 6m �12 x � 3m � 6m �

M � 75x � 60x � 3 � 10x � 3 # � 10� � 9 #Note: The units are omitted. It is understood that all units are in kN and m.

Step 5: Change the parentheses “()” into pointed brackets “<>”

Note: When the contents of the pointed brackets is zero or negative, the value of the term within the

pointed brackets is taken to be zero.

M � 75⟨x⟩ � 60⟨x � 3⟩ � 10⟨x � 3⟩# � 10⟨� � 9⟩#

Step 6: Equate the moment equation to '()"

EIy" � MEIy" � 75⟨x⟩ � 60⟨x � 3⟩ � 10⟨x � 3⟩# � 10⟨� � 9⟩#

Step 7: Integrate once to get the Slope Equation in terms of '().

EIy. � 75 ∙ 12 ⟨x⟩# � 60 ∙ 12 ⟨x � 3⟩

# � 10 ∙ 13 ⟨x � 3⟩0 � 10 ∙ 13 ⟨x � 9⟩

0EIy. � 752 ⟨x⟩

# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩0 � 103 ⟨x � 9⟩

0 � C2Note: Do not forget that integrating adds a constant of integration.

Step 8: Integrate again to get the Deflection Equation in terms of '()

EIy � 752 ∙13 ⟨x⟩

0 � 30 ∙ 13 ⟨x � 3⟩0 � 103 ∙

14 ⟨x � 3⟩

4 � 103 ∙14 ⟨x � 9⟩

4 � C2x � C#EIy � 252 ⟨x⟩

0 � 10⟨x � 3⟩0 � 56 ⟨x � 3⟩4 � 56 ⟨x � 9⟩

4 � C2x � C#Note: Do not forget that integrating again adds another constant of integration.

Step 9: Use two boundary conditions to determine the constants of integration 56 and 57

Note: The first boundary condition is 8 � 0 when � � 0. This is because at the left support, the beam

cannot deflect. Substitute this condition to the Deflection Equation in terms of 9:8.

EI0 � 252 ⟨0⟩0 � 10⟨0 � 3⟩0 � 56 ⟨0 � 3⟩

4 � 56 ⟨0 � 9⟩4 � C20 � C#

C# � 0Note: The second boundary condition is 8 � 0 when � � ;. This is because at the right support, the

beam cannot deflect. Substitute this condition to the Deflection Equation in terms of 9:8. Substitute

the value of <# also.

EI0 � 252 ⟨12⟩0 � 10⟨12 � 3⟩0 � 56 ⟨12 � 3⟩

4 � 56 ⟨12 � 9⟩4 � C212 � 0

0 � 21600 � 7290 � 109352 � 1352 � C212 � 0C2 � �14852

Step 10: Determine where the maximum deflection is by setting ). � > and solve for ?

Note: This is from the principle of maxima-minima from Differential Calculus. The maximum

deflection is not always at the midspan, but is usually very near the midspan.

EIy. � 752 ⟨x⟩# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩

0 � 103 ⟨x � 9⟩0 � 14852

0 � 752 ⟨x⟩# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩

0 � 103 ⟨x � 9⟩0 � 14852

Note: Assume that x is less than or equal to ;/2 to eliminate the second to the last term.

0 � 752 ⟨x⟩# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩

0 � 14852 Note: UseCASIOfx-991ES’sSOLVEfunction,with6astheinitialassumption.x � 4.78Note: Therefore, the earlier assumption is correct.

Step 11: Substitute the value of ? to the Deflection Equation in terms of '() to determine the

maximum deflection.

EIy � 252 ⟨x⟩0 � 10⟨x � 3⟩0 � 56 ⟨x � 3⟩

4 � 56 ⟨x � 9⟩4 � 14852 x � 0

EIy � 252 ⟨4.78⟩0 � 10⟨4.78 � 3⟩0 � 56 ⟨4.78 � 3⟩

4 � 56 ⟨4.78 � 9⟩4 � 14852 4.78

EIy � 252 ⟨4.78⟩0 � 10⟨4.78 � 3⟩0 � 56 ⟨4.78 � 3⟩

4 � 14852 4.78 EIy � �2248.72) � �77XY. Z7'( Note: The negative sign denotes that the deflection is downwards.

Note: The solution may be checked by substituting slightly lower or higher values for x. If these

values yield lower deflections, the answer is correct.

Problem


Solution:


Note: In the figure, there are four change of load points: the two ends, where there are reactions;

the point 2 m from the left end, where the 15 kN/m uniformly varying load commences, and at the

point 7 m from the left end, where the 15 kN/m uniformly varying load terminates.

Label them as A, B, C and D, from the left.


�M� � 0

12 [15

kNm \ 5m �2m �

23 5m � � R�2m � 5m � 3m � 0

R� � 20.00kN

�F� � 0

R� � R� � 12 [15kNm \ 5m � 0

R� � 20.00kN � 12 [15kNm \ 5m � 0

R� � 17.50kN




Step 3.a: Add an imaginary uniformly varying load from points C to D upward and downward.

Note: If the moment equation is directly written from the given loading, the following would be the

result:

From A to B:

� � ��

From B to C:

� � �� 12 [15� ! \ � � 2! �

12 � � 2! �

From C to D:

� � �� 60� � � 3! � 20 � ! 6! �� 3! � 12 6! �

Note: The last terms of the Moment Equations from B to C and from C to D are not at all similar in

form. The objective of writing the Moment Equation is to have one expression that is applicable to

the entire span. This can be resolved by adding an imaginary 20 kN/m uniformly distributed load

from points C to D upward and downward. Note that the sum of these loads is zero; it does not add

to, subtract from, or change the structural system in anyway.


M � R�x � 60kNx � 3m � 20 kNm x � 3m �12 x � 3m �� 20 kNm x � 3m � 6m �12 x � 3m � 6m �

M � 75.00kNx � 60kNx � 3m � 20 kNm x � 3m �12 x � 3m �� 20 kNm x � 3m � 6m �12 x � 3m � 6m �

M � 75x � 60x � 3 � 10x � 3 # � 10� � 9 #Note: The units are omitted. It is understood that all units are in kN and m.




M � 75⟨x⟩ � 60⟨x � 3⟩ � 10⟨x � 3⟩# � 10⟨� � 9⟩#


EIy" � MEIy" � 75⟨x⟩ � 60⟨x � 3⟩ � 10⟨x � 3⟩# � 10⟨� � 9⟩#


EIy. � 75 ∙ 12 ⟨x⟩# � 60 ∙ 12 ⟨x � 3⟩

# � 10 ∙ 13 ⟨x � 3⟩0 � 10 ∙ 13 ⟨x � 9⟩

0EIy. � 752 ⟨x⟩

# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩0 � 103 ⟨x � 9⟩



EIy � 752 ∙13 ⟨x⟩

0 � 30 ∙ 13 ⟨x � 3⟩0 � 103 ∙

14 ⟨x � 3⟩

4 � 103 ∙14 ⟨x � 9⟩

4 � C2x � C#EIy � 252 ⟨x⟩

0 � 10⟨x � 3⟩0 � 56 ⟨x � 3⟩4 � 56 ⟨x � 9⟩





EI0 � 252 ⟨0⟩0 � 10⟨0 � 3⟩0 � 56 ⟨0 � 3⟩

4 � 56 ⟨0 � 9⟩4 � C20 � C#

C# � 0Note: The second boundary condition is 8 � 0 when � � ;. This is because at the right support, the



EI0 � 252 ⟨12⟩0 � 10⟨12 � 3⟩0 � 56 ⟨12 � 3⟩

4 � 56 ⟨12 � 9⟩4 � C212 � 0

0 � 21600 � 7290 � 109352 � 1352 � C212 � 0C2 � �14852

Step 10: Determine where the maximum deflection is by setting ). � > and solve for ?



EIy. � 752 ⟨x⟩# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩

0 � 103 ⟨x � 9⟩0 � 14852

0 � 752 ⟨x⟩# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩

0 � 103 ⟨x � 9⟩0 � 14852

Note: Assume that x is less than or equal to ;/2 to eliminate the second to the last term.

0 � 752 ⟨x⟩# � 30⟨x � 3⟩# � 103 ⟨x � 3⟩

0 � 14852 Note: UseCASIOfx-991ES’sSOLVEfunction,with6astheinitialassumption.x � 4.78Note: Therefore, the earlier assumption is correct.


maximum deflection.

EIy � 252 ⟨x⟩0 � 10⟨x � 3⟩0 � 56 ⟨x � 3⟩

4 � 56 ⟨x � 9⟩4 � 14852 x � 0

EIy � 252 ⟨4.78⟩0 � 10⟨4.78 � 3⟩0 � 56 ⟨4.78 � 3⟩

4 � 56 ⟨4.78 � 9⟩4 � 14852 4.78

EIy � 252 ⟨4.78⟩0 � 10⟨4.78 � 3⟩0 � 56 ⟨4.78 � 3⟩

4 � 14852 4.78 EIy � �2248.72) � �77XY. Z7'( Note: The negative sign denotes that the deflection is downwards.

Note: The solution may be checked by substituting slightly lower or higher values for x. If these

values yield lower deflections, the answer is correct.

Problem


Solution:


Note: In the figure, there are only two change of load points: the two ends, because the uniformly

distributed load is applied all throughout the beam span.

Label them as A and B from the left.


�F] � 0

R� � R^ � 12 w L R� � R^ � wL2





M � R�x � wx �12 x �M � wL2 x � w2 x

#




M � wL2 ⟨x⟩ � w2 ⟨x⟩#


EIy" � MEIy" � wL2 ⟨x⟩ � w2 ⟨x⟩

#


EIy. � wL2 ∙ 12 ⟨x⟩# �w2 ∙

13 ⟨x⟩

0 � C2EIy. � wL4 ⟨x⟩# �w6 ⟨x⟩



EIy � wL4 ∙ 13 ⟨x⟩0 �w6 ∙

14 ⟨x⟩

4 � C2x � C#EIy � wL12 ⟨x⟩

0 � w24 ⟨x⟩





EI0 � wL12 ⟨0⟩0 � w

24 ⟨0⟩4 � C20 � C#

C# � 0

Note: The second boundary condition is 8 � 0 when � � ;. This is because at the right support, the



EI0 � wL12 ⟨L⟩0 � w

24 ⟨L⟩4 � C2L � C#

0 � wL4

12 �wL424 � C2L � 0

C2 � �wL0

24 Step 10: Determine where the maximum deflection is by setting ). � > and solve for ?



EIy. � wL4 ⟨x⟩# �w6 ⟨x⟩0 �wL

0

24 0 � wL4 ⟨x⟩

# �w6 ⟨x⟩0 �wL

0

24 Note: Set the value of ; � 1 so that we can determine the value of � explicitly using the calculator.

0 � w1 4 ⟨x⟩# �w6 ⟨x⟩0 �w1

0

24 �16 x

0 � 14 x# � 1

24 � 0x � _�0.366,1.366,1/2`Note:Earlier,wesetthevalueof; � 1.Wereversetoprocessbysetting1 � ;inallroots.x � _�0.366L, 1.366L, L/2`Note: Since the equation is in cubic form, it will yield 3 roots. It does not make sense for the deflection

to be located left of A �0.366; or to the right of B 1.366; . Therefore, it can be concluded that

the only correct value for � is ;/2.


maximum deflection.

EIy � wL12 ⟨x⟩0 � w

24 ⟨x⟩4 �wL

0

24 xEIy � wL12 ⟨

L2⟩0 � w

24 ⟨L2⟩4 �wL

0

24 [L2\

EIy � wL0

96 �wL0384 �

wL048

) � � ghiXjYX'(

Note: The negative sign denotes that the deflection is downwards.

Double Integration Method

Documents

kn note

points c

distributed load

change of load points

deflection equation

kn concentrated load

terms of eiy

slope equation