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 only two change of load points: the two ends, because the uniformly

varying load is applied all throughout the beam span.

Label them as A and B from the left.

Step 2: Determine the reactions

�F� � 0

V� � 12 36

kNm � �6m�

V� � 108.00kN

�M� � 0

M� � 12 36

kNm � �6m� �13 �6m�� M� � 216.00kN ∙ m

Note: It will be shown later that it is not required that the reactions be solved for the given problem,

since all forces acting to the left of the cutting plane line is known.

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: Assign any variable to denote the maximum magnitude of the uniformly varying load

at any point x.

The maximum magnitude of the uniformly varying load is designated as z.

Note: Using ratio and proportion, determine the value of z in terms of x.

zx �

366 z � 6x

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

M � �12 �z��x� �

13 �x��

M � �12 �6x��x� �

13 �x��

M � �x�

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 � �⟨x⟩�

Step 6: Equate the moment equation to !""

EIy" � MEIy" � �⟨x⟩�

Step 7: Integrate once to get the Slope Equation in terms of !"'

EIy' � �14 ⟨x⟩

) * C,Note: Do not forget that integrating adds a constant of integration.

Step 8: Integrate again to get the Deflection Equation in terms of !"

EIy � �14 ∙

15 ⟨x⟩

. * C,x * C/EIy � � 1

20 ⟨x⟩. * C,x * 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 01 and 02

Note: The first boundary condition is 3′ � 0 when 5 � 6. This is because at the right support, the

beam cannot rotate, from the definition of the fixed support. Substitute this condition to the Slope

Equation in terms of 673′.

EI�0� � �14 ⟨6⟩

) * C,C, � 324Note: The second boundary condition is 3 � 0 when 5 � 6. This is because at the right support, the

beam cannot deflect. Substitute this condition to the Deflection Equation in terms of 673. Substitute

the value of 8/ also.

EI�0� � � 120 ⟨6⟩

. * 324�6� * C/0 � �1944

5 * 1944 * C/C/ � �7776

5

Step 10: Determine where the maximum deflection is

Note: For cantilever beams, the maximum deflection is always at the free end.

x � 0

Step 11: Substitute the value of ; to the Deflection Equation in terms of !" to determine the

maximum deflection.

EIy � � 120 ⟨x⟩

. * 324x � 77765

EIy � � 120 ⟨0⟩

. * 324�0� � 77765

EIy � �77765

" � �<<<=> !

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