Chapter 4: Chapter 4:

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Chapter 4:

Internal Loadings Developed in Structural Members

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Internal loadings at a specified point Internal loadings at a specified point

• The internal load at a specified point in a member The internal load at a specified point in a member can be determined by using the method of

can be determined by using the method of sections

sections

• This consists of: This consists of:

• N, normal force N, normal force

• V, shear force V, shear force

• M, bending moment M, bending moment

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Internal loadings at a specified point Internal loadings at a specified point

• Sign convention Sign convention

• Although the choice is arbitrary, the convention has Although the choice is arbitrary, the convention has been widely accepted in structural engineering

been widely accepted in structural engineering

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Internal loadings at a specified point Internal loadings at a specified point

• Procedure for analysis Procedure for analysis

• Determine the support reactions before the member Determine the support reactions before the member is “cut”

is “cut”

• If the member is part of a pin-connected structure, If the member is part of a pin-connected structure, the pin reactions can be determine using the

the pin reactions can be determine using the methods of section

methods of section

• Keep all distributed loadings, couple moments & Keep all distributed loadings, couple moments &

forces acting on the member in their exact location

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Internal loadings at a specified point Internal loadings at a specified point

• Pass an imaginary section through the member, Pass an imaginary section through the member, perpendicular to its axis at the point where the perpendicular to its axis at the point where the

internal loading is to be determined internal loading is to be determined

• Then draw a free-body diagram of the segment that Then draw a free-body diagram of the segment that has the least no. of loads on it

has the least no. of loads on it

• Indicate the unknown resultants N, V & M acting in Indicate the unknown resultants N, V & M acting in their positive directions

their positive directions

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Internal loadings at a specified point Internal loadings at a specified point

• Moments should be summed at the section about Moments should be summed at the section about axes that pass through the centroid of the

axes that pass through the centroid of the

member’s x-sectional area in order to eliminate N &

V, thereby solving M V, thereby solving M

• If the solution of the equilibrium eqn yields a If the solution of the equilibrium eqn yields a

quantity having a –ve magnitude, then the assumed quantity having a –ve magnitude, then the assumed

directional sense of the quantity is opposite to that directional sense of the quantity is opposite to that

shown on the free-body diagram

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Determine the internal shear & moment acting in the cantilever Determine the internal shear & moment acting in the cantilever beam at sections passing through C & D.

beam at sections passing through C & D.

Example 4.1

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If we consider free-body diagrams of segments to the right of the sections, the support reactions at A do not have to be calculated.

Solution Solution

kNm M M M

kN V

V

F S

c c

c

c c

y

50 5 ( 2 ) 5 ( 3 ) 20 0 )

1 (

5 0 in the anti - clockwise as positive, moments

With

15 0

5 5

5 0 : CB egment





       









 







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Solution Solution

kNm M M M

kN V

V

F S

D D

D

D D

y

50 5 ( 2 ) 5 ( 3 ) 20 0 )

1 (

5 0 in the anti - clockwise as positive, moments

With

20 0

5 5

0 : DB egment





       









 







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Shear & Moment Functions

• Design of beam requires detailed knowledge of the Design of beam requires detailed knowledge of the variations of V & M

variations of V & M

• Internal N is generally not considered as: Internal N is generally not considered as:

• The loads applied to a beam act perpendicular to The loads applied to a beam act perpendicular to the beam’s axis

the beam’s axis

• For design purpose, a beam’s resistance to shear & For design purpose, a beam’s resistance to shear &

bending is more important than its ability to resist bending is more important than its ability to resist

normal force normal force

• An exception is when it is subjected to compressive An exception is when it is subjected to compressive axial force where buckling may occur

axial force where buckling may occur

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Shear & Moment Functions

• In general, the internal shear & moment functions In general, the internal shear & moment functions will be discontinuous or their slope will

will be discontinuous or their slope will discontinuous at points where:

discontinuous at points where:

• The type or magnitude of the distributed load The type or magnitude of the distributed load changes

changes

• Concentrated forces or couple moments are applied Concentrated forces or couple moments are applied

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Shear & Moment Functions

• Procedure for Analysis Procedure for Analysis

• Determine the support reactions on the beam Determine the support reactions on the beam

• Resolve all the external forces into components Resolve all the external forces into components acting perpendicular & parallel to beam’s axis acting perpendicular & parallel to beam’s axis

• Specify separate coordinates x and associated Specify separate coordinates x and associated origins, extending into:

origins, extending into:

• Regions of the beam between concentrated forces Regions of the beam between concentrated forces and/or couple moments

and/or couple moments

• Discontinuity of distributed loading Discontinuity of distributed loading

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Shear & Moment Functions

• Procedure for Analysis Procedure for Analysis

• Section the beam perpendicular to its axis at each Section the beam perpendicular to its axis at each distance x

distance x

• From the free-body diagram of one of the From the free-body diagram of one of the segments, determine the unknowns V & M segments, determine the unknowns V & M

• On the free-body diagram, V & M should be shown On the free-body diagram, V & M should be shown acting in their +ve directions

acting in their +ve directions

• V is obtained from V is obtained from  F

y

 0

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Shear & Moment Functions

• Procedure for Analysis Procedure for Analysis

• The results can be checked by noting that: The results can be checked by noting that:

dx w dV dx V dM



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Determine the shear & moment in the beam as a function of x.

Example 4.4

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Support reactions:

For the purpose of computing the support reactions, the distributed load is replaced by its resultant force of 30k.

However, this resultant is not the actual load on the beam

Solution

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Shear & moment functions

A free-body diagram of the beam segment of length x is shown.

Note that the intensity of the triangular load at the section is found by proportion.

With the load intensity known, the resultant of the distributed load is found in the usual manner.

Solution

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Solution Solution

results the

of check a

as

serves 3

that 10 Note

556 .

0 135

810 3 0 3

10 2

135 1 810

: ve as

moments clockwise

- anti With

667 .

1 135

3 2

0;

3 0 10 2

135 1

; 0

x w x -

V and dV/d dM/dx

x x

M

x M x x

x M

x V

F

s

y

x x V









 

 



 



 



 



 









 

 





 





 





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Shear & Moment Diagrams for a Beam

• If the variations of V & M are plotted, the graphs If the variations of V & M are plotted, the graphs are termed the shear diagram and moment

are termed the shear diagram and moment diagram

diagram

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Shear & Moment Diagrams for a Beam

• Applying the eqn of equilibrium, we have: Applying the eqn of equilibrium, we have:

)

2

( ) (

0 )

( ) (

) (

: ve as

moments clockwise

- anti With

)

( ) ( ) 0

(

; 0

x x

w x

V M

M M

x x

x w M

x V

M

x x

w

V w x x V V V

F

o

y









         









      

 





 

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Shear & Moment Diagrams for a Beam

• Dividing by Dividing by   x & taking the limit as x & taking the limit as   x x   , the , the previous eqns become:

previous eqns become:

• Integrating from one point to another between Integrating from one point to another between concentrated forces or couples in which case

concentrated forces or couples in which case

dx V x dM

dx w

dV  ( ) , 

(x)dx

, w(x)dx

V ^  ^ 

 M V

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Shear & Moment Diagrams for a Beam

• In order to account for these 2 cases, consider the In order to account for these 2 cases, consider the free-body diagrams of differential elements of the free-body diagrams of differential elements of the beam beam

• It is seen that force equilibrium requires the It is seen that force equilibrium requires the change in shear to be

change in shear to be

• Moment equilibrium requires the change in Moment equilibrium requires the change in moment to be:

moment to be:

 F



V '

M M 



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Shear & Moment Diagrams for a Beam

 F



V '

M M 



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Draw the shear & moment diagrams for the beam.

Example 4.7

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At end points:

x = 0m, V = 30kN x = 9m, V = -60kN

The load w is –ve & linearly increasing, dV/dx=w

The point of zero shear can be found by using method of sections from a beam segment of length x,

Solution Solution

x V

1 0    



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From the shear diagram, for 0<x<5.20m, the value of shear is +ve but  so dM/dx=V

At x = 5.20m, dM/dx=0

Likewise for 5.20m<x<9m, the shear & so the slope of the moment diagram are –ve 

Max M is at x = 5.20m since dM/dx =V=0

Solution

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We have,

Solution Solution

kNm M

M M

_s

104 3 0 20 . 20 5 . 9 5

20 . 20 5 2

) 1 20 . 5 ( 30

0 ise moment as - ve : Anticlockw



 



 



 



 



 



 



 



 

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Draw the shear and moment diagrams for each of the beams.

Example 4.8

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The support reactions are calculated & shown on a free-body diagram of each beam.

From the sign convention, V at the ends of each beam is plotted first.

Since dV/dx = w, the slope of the shear between these 2 ends can be determined.

Solution

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From the sign convention, M at the ends of each beam is plotted first.

Since dM/dx = V, the trend of the slope of the moment diagram between these 2 ends can be determined.