bending moment diagram

7/21/2019 Bending Moment Diagram

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BMD show how the applied loads to a beam create

a moment variation along the length of the beam.

BENDING MOMENT

DIAGRAM



MtMt

Bending Moment

Bendi

ng Moment = moments of reactions – moments of

loads



Distributed load acts downward on beam.



Internal shear force causes a clockwise rotation of the beam

segment; and the internal moment causes compression in the top

fibers of the segment.



SIGN CONVENTIONS

• A force that tends to bend the beam

downward is said to produce a

positive bending moment. A forcethat tends to shear the left portion of

the beam upward with respect to the

right portion is said to produce apositive shearing force.

Positive Negative

Bending Bending

Positive NegativeShear Shear



PROCEDURE

1. Draw the free-body-diagram of the beam with sufficient room

under it for the shear and moment diagrams

if needed! solve for support reactions first".

#. Draw the shear diagram under the free-body-diagram.

• $he change in shear ∆% e&uals the negative area under the distributed

loading.

• 'abel all the loads on the shear diagram ( )dx xwV ∫ −=∆



PROCEDURE

(. Draw the moment diagram below the shear diagram.

• $he shear load is the slope of the moment and point moments result in

)umps in the moment diagram.

• $he area under the shear diagram e&uals the change in moment over the

segment considered up to any )umps due to point moments".

• 'abel the value of the moment at all important points on the moment

diagram.

( )dx xV M ∫ =∆



Relations etween Distri!te"

#oa"$ S%ear an" Moment

Distributed Load

V dx

dM =

" xwdx

dV −=*lope of the shear

diagram

+egative of distributed

load intensity

*lope of the

shear diagram*hear moment diagram

∫ =∆Vdx M

BC

∫ −=∆ dx xwV BC

"

,hange in shear Area under shear

diagram

,hange in momentArea under shear

diagram



Support Reactions• ind all reactive forces and couple moments acting on the beam

• esolve them into components

Shear and Moment Reactions• *pecify separate coordinates /

• *ection the beam perpendicular to its a/is

• % obtained by summing the forces perpendicular to the beam

• 0 obtained by summing moments about the sectioned end

Shear and Moment Reactions

• lot % versus /" and 0 versus /"

• ,onvenient to plot the shear and the bending moment diagrams below the 2D

of the beam

Proce"!re &or Anal'sis



S%ear ( Moment Dia)ram



Load

0 Constant Linear

Shear

Constant Linear Parabolic

Moment

Linear Parabolic Cubic

Common Relations%ips



Load

0 0 Constant

Shear

Constant Constant Linear

Moment

Linear Linear Parabolic

Common Relations%ips

M



R!les o& T%!m*Re+iew

• 0oment is dependent upon the shear diagram

the area under the shear diagram 3 change in the moment i.e. A shear

diagram 3 40"

• *traight lines on shear diagrams create sloping lines on moment diagrams

• *loping lines on shear diagrams create curves on moment diagrams

• ositive shear 3 increasing slope

• +egative shear 3 decreasing slope



P1 P2 P3

Ra Rb

w

SFD

BMD

Point of zero

Point of maximum

Point of maximum

Point of contra-flexure

BEAMS IN BENDING



Concentrate" #oa"

• ind reactions

• ,ut through beam to the left of the

load a distance / from the left

end"! 2D – 5&uilibrium yields % and 0 for

the left side of the beam

• ,ut through the beam to the right of

! 2D – 5&uilibrium yields % and 0 for

the right side of the beam.



Concentrate" #oa" Moment

Dia)ram

• $he bending moment in the leftside increases linearly from 6eroat the support to ab7'" at the

concentrated load /3a

• In the right side! the bendingmoment is again a linear functionof /! varying from ab7'" at /3a

to 6ero at the support /3'.

• $he ma/imum bending moment istherefore ab7'"! which occurs atthe concentrated load.

In this example a=b=L!

Bending Moment Diagram



x

yExample 1 :

xy

Mx!

" B#

a b

P

( )ba

aPRBy

+

⋅=( )ba

bPR"y

+

⋅=

x

( )ba

bP

+

⋅

$%M! =∑ x!M+

( ) ( ) ( )axPx

ba

PbMx! −−

+=∴

&'ere can only be ()E or *ER+,( )ax −

P

a

"

( )axP −+ ( ) ( ) %xba

Pb

=+−



x

y

x

" B#

a b

P

( )ba

Pa

+( )ba

Pb

+

i. &'en :ax ≤

ii. &'en :ax >( )

( ) ( )axPxba

PbMx! −−

+=

1

2

( )

( ) ( )axPx

ba

PbMx! −−

+=

%

" B#

(/e

( )ba

Pab

+

Mx!

%

BMD:

E0, 1 E0, 2



$%Fy =∑$%M! =

∑

Pxy =∴

x

yExample 2:

P

R"yP

" B

Mx!P,

P,

P xy

Mx!

xP

Mx!

xy

Mx! Mx!

xy xy

3 M are P+S454)E ( )xPMx! −−=∴

( )



$Pxy =

x

yP

P

B

P,

x

( )xPMx! −−=

xy %

Mx! %

"Mx!

xy

S'ear ForceDiagram SFD.

BendingMoment

Diagram BMD.

(/eP

-/e

-P,



20 ft

P = 20 ips

!2 ips" ips!2 ft

#$ips%

M

$ft&ips%

" ips

&!2 ips

'( ft&ips

)

)

V ( M Dia)rams

a

b

c

*hat is the area of theblue rectan+le,

'( ft&ips*hat is the area of

the +reen rectan+le,

&'( ft&ips



BMD &or simpl' s!pporte" eam

wit% UD#,

arabolic! ma/ moment at mid span of value "L!# !

where w is the distributed load and ' the length of the beam.



UNI-ORM #OAD

• $he beam and its loading is

symmetric! the reactions are e&ual to

w'7#

• $he slope of the shear diagram at each

point e&uals the negative distributed

load intensity at each point

( ) xw

dx

dV −=



UNI-ORM #OAD

• $herefore! the shear force and bendingmoment at a distance / from the left end are8

• $hese e&uations are valid through the lengthof the beam and can be plotted as shear and

bending moment diagrams.

$he ma/imum value at the midpointwhere

3 9.

0ma/3 w'#7:

( ) V dx

dM xw

dx

dV =−= !

V dx

dM =

Di t ib t d d



$%Fy =∑$%M! =

∑

x

y

Example : Di6tributed oad

R"y7

" B

x

xy

Mx!

Mx!

xy

Mx!

72

2

7

72

2

Di6tributed oad 7

per unit lengt'

7

2

7M

2

x! +

( )x7xy −=⇒

7x

7x− %xy =−

( )x7− %

2

x7x =

+

22



$%x8 =

2

7

2

7x7xM

22

x! −−=⇒( )x7xy −=⇒

-/e

-72

2

x

Mx!

%BMD:

2

7M

2

x! −=

$x8 = %Mx! =

$2

x8 =

9

7M

2

x! −=



Draw Some Concl!sions

• $he magnitude of the shear at a point e&uals the slope of the

moment diagram at that point.

• $he area under the shear diagram between two points e&uals the

change in moments between those two points.

• At points where the shear is 6ero! the moment is a local ma/imumor minimum-

bending moment diagram

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