Lecture Notes of Mechanics of Solids, Chapter 5

1

Chapter 5 Bending Moments and Shear Force

Diagrams for Beams

In addition to axially loaded bars/rods (e.g. truss) and torsional shafts, the structural members may

experience some loads perpendicular to the axis of the beam and will case only shear and bending

in the beam. The current chapter together with Chapters 6 to 8 will focus on such an issue.

5.0 SHEAR FORCE AND BENDING MOMENTS DIAGRAMS FOR BEAMS

A Shear Force Diagram (SFD)indicates how a force applied perpendicular to the axis (i.e.

parallel to cross section) of a beam is transmitted along the length of that beam. A Bending

Moment Diagram (BMD) will show how the applied loads to a beam create a moment

variation along the length of the beam. These diagrams are used to determine the normal and

shear stresses as well as deflection and slopes in the following chapters.

5.1 BEAM SIGN CONVENTION

(SI&4

th

:256-257; 5

th

:256-257)

At any point along its length, a beam can transmit a bending moment M(x) and a shear force

V(x). If a loaded beam is cut, the definitions of a positive distributed load, shear force and

positive bending moment are as Fig. 5.1 below:

Positive internal

bending moment

Positive internal shear force

Positive distributed load

Fig. 5.1 Beam shear force and bending moment sign convention

Where distributed load acts downward on the beam; internal shear force causes a clockwise

rotation of the beam segment on which it acts; and the internal moment causes compression

in the top fibers of the segment, or to bend the segment so that it holds water.

5.2 RELATIONSHIP BETWEEN BEAM LOADINGS

(SI&4

th

:264-268; 5

th

:264-

268)

A beam (Fig. 5.2) is loaded with vertical forces F

i

, bending moments M

i

and distributed loads

w(x).

F

1

F

2

M

1

M

2

x

dx

w(x)

dx

M(x)

V(x)

M(x)+

dM(x)

dx

dx

M(x)+

dM(x)

dx

dx

w(x)

F.B.D. of element dx

V(x)+

dV(x)

dx

dx

V(x)+

dV(x)

dx

dx

Fig. 5.2 Transversely loaded beam and free body diagram of element dx

Lecture Notes of Mechanics of Solids, Chapter 5

2

Look at the FBD of an elemental length dx of the above loaded beam (Fig. 5.2). As it has an

infinitesimal length, the distributed load can be considered as a Uniformly Distributed Load

(UDL) with constant magnitude w(x) over the differential length dx.

It is now necessary to equate the equilibrium of the element. Starting with vertical equilibrium

( ) ( ) ( )

(

)

00 =

+−−==↑+

∑

dx

dx

xdV

xVdxxwxVF

y

(5.1)

Dividing by dx in the limit as dx→0,

( )

( )

xw

dx

xdV

−=

(5.2)

Taking moments about the right hand edge of the element:

( ) ( ) ( )

( )

(

)

0

2

0 =

+++−−==

∑

dx

dx

xdM

xM

dx

dxxwdxxVxMM

Edge.H.R

(5.3)

Dividing by dx in the limit as dx→0,

( )

( )

xV

dx

xdM

=

(5.4)

Eqs. (5.2) & (5.3) are important when we have found one and want to determine the others.

5.3 BENDING MOMENT AND SHEAR FORCE EQUATIONS

Introductionary Example - Simply Supported Beam

By using the

free body diagram technique

, the bending moment and shear force distributions

can be calculated along the length of the beam. Let’s take a

simply supported beam

, Fig. (5.3),

as an example to shown the solutions:

P

L

a

R

AY

=(1-a/L)P

R

BY

=Pa/L

I

I

R

AY

=(1-a/L)P

x

F.B.D. (method of section I-I)

o

F.B.D. (global equilibrium)

II

II M(x)

V(x)

A B

A

Fig. 5.3

FBD of beam cut before force

P

Step A:

Cut beam just before the force

P

(i.e. Section I-I)

, and draw a free body diagram

including the unknown shear force and bending moment as in Fig. 5.3.

Take moments about the right hand end (O):

( )

010 =+

−−==

∑

xMx

L

a

PM

o

→

( )

x

L

a

PxM

−= 1

(5.5)

To determine the shear force, use Eq. (5.4), giving that:

( )

(

)

P

L

a

dx

xdM

xV

−== 1

(5.6)

To verify Eq. (5.6), equate vertical equilibrium:

+

+

+

+

Lecture Notes of Mechanics of Solids, Chapter 5

3

( )

01 =−

−=↑+

∑

xV

L

a

PF

y

→

( )

−=

L

a

PxV 1

which is the same equation as (5.6). These then, are the equations for the bending moment and

shear force variation in the range of

ax

≤

≤

0

.

To find out the rest of the bending moment and shear force distributions, it is necessary to

now carry out a similar analysis, but cutting the beam just before the end (Section II-II).

Step B: Cut beam just before the right hand end (RHE)

P

x

a

R

AY

=( 1-a/L)P

II

II

F.B.D. (Section II-II)

M(x)

o

V(x)

A

Fig. 5.4

FBD of beam cut before the right hand end

Equate moments about the right side:

( ) ( )

010

0

=+−+

−−==

∑

xMaxPx

L

a

PM

giving:

( )

( )

PaPx

L

a

axPx

L

a

PxM +−=−−

−= 1

(5.7)

and using Eq. (5.4), the shear force equation is :

( )

(

)

P

L

a

dx

xdM

xV −==

(5.8)

These expressions for the bending moment and shear force can now be plotted against x to

produce the

Shear Force and Bending Moment Diagrams

as Fig. 5.5:

+

+

Lecture Notes of Mechanics of Solids, Chapter 5

4

P

L

a

(1-a/L)P

Pa/L

Loading Diagram

V(x)

+ve

-ve

x

M(x)

+ve

x

(1-a/L)P

-a/LP

(1-a/L)Pa

Shear Force Diagram

Bending Moment Diagram

Fig. 5.5

Shear Force and Bending Moment Diagrams for simply supported beam

Macauley's Notation

(4

th

:590-599; 5

th

:590-599)

The two sets of equations for V(x) and M(x), Eqs. (5.5), (5.6), (5.7) and (5.8), can be condensed

to just one set of equations if we use a special type of notation called Macauley's Notation.

The above equations would look like this (to be derived in Example 5.0)

( )

00

1 axPx

L

a

PxV −−

−=

(5.9)

( )

11

1 axPx

L

a

PxM −−

−=

(5.10)

Where the notation has the following meaning:

( )

( )

0

0

≥

≥

<

−

=− n

axfor

axfor

ax

ax

n

n

(5.11)

when differentiating:

=

=

≥

−

−

=

∂

−∂

−

0

1

1

0

0

1

nfor

nfor

nfor

ax

axn

x

ax

n

n

(5.12)

Remarks

To derive the bending moment equation by using Macauley's notation, you may need to do

the following:

1)

Determine the ground reactions from global equilibrium;

2)

Cut the beam

just before

the right hand end;

Lecture Notes of Mechanics of Solids, Chapter 5

5

3)

Equate the cut FBD to equilibrium about the right hand end;

4)

All length terms in the bending moment/shear force equations

MUST

be written using

Macauley's notation;

5)

Always indicate the powers, even if they are 0 or 1.

Example 5.0:

As in the introductory example, determine the shear force and bending moment

equations and plot them for a simply-supported beam as in the introductory example.

Step 1: Determine the ground reactions;

P

L

a

R

AY

=(1-a/L)P

R

BY

=Pa/L

F.B.D. (global equilibrium)

I

I

A B

We have

R

AY

= (1-a/L)P and R

BY

= a/LP

Step 2: Draw FBD of beam cut just before the RHS (Section I-I).

P

x

a

R

AY

=( 1-a/L)P

I

I

F.B.D. (Section I-I)

M(x)

o

V(x)

A

Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I).

Take moments about RHS:

( )

010

11

O

=+−+

−−==

∑

xMaxPx

L

a

PM

( )

11

1 axPx

L

a

PxM −−

−=

and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as:

( )

( )

00

1 axPx

L

a

P

dx

xdM

xV −−

−==

Step 4: Plotting the Shear Force and Bending Moment Diagrams

+

+

Lecture Notes of Mechanics of Solids, Chapter 5

6

According to M(x) and V(x) to depict the diagrams, Look at the equations segment by segment

When

ax ≤≤0

( )

( )

x

L

a

PPx

L

a

PxM

−=×−

−= 101

1

and

( )

( )

−=×−

−=

L

a

PPx

L

a

PxV 101

0

To plot this segment in the diagram, firstly look at the boundary points as

0)(0

=

= xM,x

and

(

)

L/aPaxM,ax −== 1)(

. Draw two points and then connect them because the equation

gives a line. Likewise, one can plot Shear Force Diagram in this region.

When

Lxa ≤≤

( )

( ) ( )

x

L

a

PPaPaPxx

L

a

PaxPx

L

a

PxM −=+−

−=−×−

−= 11

11

and

( )

( ) ( )

L

a

PP

L

a

PaxPx

L

a

PxV −=−

−=−×−

−= 11

00

Remarks

: Please draw global FBD of the beam firstly and follow by its Sear Force and Bending

Moment Diagrams. The reason for doing this is that when you get sufficient experience, you may

be able to directly plot the Shear Force Diagram by observing the external forces as well as plot

Bending Moment Diagrams by observing the Shear Force Diagram. Nevertheless you MUST still

work out and indicate the locations and values (including +ve or –ve) at all turning points in the

diagrams in detail.

It is also interesting to note that concentrated forces (e.g. reaction forces and external forces)

correspond to inclined line in BMD and horizontal line in SFD.

P

L

a

(1-a/L)P

Pa/L

Loading Diagram

V(x)

+ve

-ve

x

M(x)

+ve

x

(1-a/L)P

-a/LP

(1-a/L)Pa

Shear Force Diagram

Bending Moment Diagram

Example 5.1:

Determine the shear force and bending moment equations and plot them for a

simply-supported beam loaded with a UDL.

Lecture Notes of Mechanics of Solids, Chapter 5

7

Step 1: Determine the ground reactions;

From global equilibrium the ground reaction forces can be found to be both equal to wL/2 as,

L

wL/2

F.B.D. (global equilibrium)

I

I

wL/2

B

A

w

( )

0

2

0 =

+−==

∑

L

wLLRM

AYB

→

2/wLR

AY

=

∴

(+ upwards)

00 =−+==↑+

∑

wLRRF

BYAYy

→

2/wLR

BY

=

∴

(+ upwards)

Step 2: Draw FBD of beam cut just before the RHS (Section I-I).

x

R

AY

=wL/2

F.B.D. (Section I-I)

A

o

M(x)

V(x)

w

R

w

x/2

Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I).

As far as V(x) and M(x) are concerned the UDL can be temporarily replaced by its resultant R

w

(=wx) applied at the centroid of the UDL distribution in the moment equilibrium equation. So if

we take moments about the RHS of the beam we get:

( ) ( )

( )

( )

0

2

2

2

0

1

11

1

1

O

=++−=++−==

∑

xM

x

xwx/wLxM

x

RxRM

wAY

( )

21

22

x

w

x

wL

xM

−=∴

and differentiating

w.r.t.

'

x

', as Eq. (5.4), gives :

( )

( )

10

2

xwx

wL

dx

xdM

xV

−==

Step 4: Plotting the Shear Force and Bending Moment Diagrams

According to

M

(

x

) and

V

(

x

) to depict the diagrams

+

+

+

+

Lecture Notes of Mechanics of Solids, Chapter 5

8

L

wL/2

wL/2

B

A

w

V(x)

+ve

-ve

x

M(x)

+ve

x

wL/2

Shear Force Diagram

Bending Moment Diagram

-wL/2

Loading Diagram

wL

2

/8

Parabola

It is worth pointing out that one should not completely replace such a UDL by its

corresponding resultant concentrated force

R

w

(=

wx

) in the beginning of the solution. There is

a significant difference of the Shear Force and Bending Moment Diagrams between a

concentrated force (Example 5.0) and a UDL (Example 5.1). It is also interesting to note that

the

UDL corresponds to an inclined line in the Shear Force Diagram and a quadratic curve

(

parabola

)

in the Bending Moment Diagrams.

Lecture Notes of Mechanics of Solids, Chapter 5

9

Example 5.2:

Determine the shear force and bending moment equations and plot them for a

beam loaded with a UDL between A and B and two concentrated forces at C and E.

UDL=w=1kN/m

10kN

A

B

C

E

D

20kN

R

AY

R

DY

10m

5m

5m

10m

F.B.D. (global equilibrium)

I

I

Step 1: Determine the ground reactions;

0301020152051010 =×−×+×−××−==

∑

DYA

RM

→

N532 k.R

DY

=

∴

010201010 =−+−×−==↑+

∑

DYAYy

RRF

→

N57 k.R

AY

=

∴

Step 2: Draw FBD of beam cut just before the RHS (Section I-I).

Note:

The only problem with Macauley's Notation is that it does not work when a UDL stops.

It however does work for a UDL which starts anywhere along a beam and continues to the

end. The problem can be corrected by applying a UDL of equal magnitude but opposite sense

where the first UDL ends.

w=1kN/m

A

20kN

7.5kN

32.5kN

F.B.D. (Section I-I) and application of equivalent UDL

o

M(x)

V(x)

x

Step 3: Equilibrium for FBD of beam cut just before the RHS (

Section I-I

).

Take moments about RHS:

( )

xMx.x

xx

x.M

+−−−+

−

−+−==

∑

11

22

1

O

205321520

2

10

1

2

1570

( )

11221

20532152010

2

1

2

1

57

−+−−−+−=

x.xxxx.xM

and differentiating

w.r.t.

'

x

', as Eq. (5.4), gives the shear force equation as:

( )

( )

00110

2053215201057

−+−−−+−==

x.xxxx.

dx

xdM

xV

+

+

+

+

Lecture Notes of Mechanics of Solids, Chapter 5

10

Step 4: Plotting the Shear Force and Bending Moment Diagrams

UDL=w=1kN/m

10kN

A

B

C

E

D

20kN

7.5kN

32.5kN

10m

5m

5m

10m

V(x) kN

+ve

-ve

x

M(x) kNm

+ve

x

Shear

Force

Diagram

Bending

Moment

Diagram

-ve

+ve

28.125

25

12.5

-100

-22.5

10

-2.5

7.5

Loading

Diagram

quadratic

Again, the UDL segment corresponds to an inclined line in SFD and a quadratic curve in

BMD.

Example 5.3:

Determine the shear force and bending moment equations and plot them for a

cantilever beam loaded with a moment

M

B

= 40

k

N

m

and a force F= 10

k

N.

10kN

R

AY

A

B

C

I

I

4m

Global F.B.D.

M

A

1.5m

+

M

B

=40kNm

Step 1: Determine the ground reactions;

The cantilever beam is fully clamped in the left hand end A as shown. The ground reaction for

this point should have reaction force R

AY

and reaction moment M

A.

. So the global equilibrium

is given as

0100 =+==↑+

∑

AYy

RF

→

N10kR

AY

−

=

∴

(- downwards)

05510400 =×++==

∑

.MM

AA

→

kNmM

A

95

−

=

(-l潣歷kse)=

+

+

Lecture Notes of Mechanics of Solids, Chapter 5

11

Step 2: Draw FBD of beam cut just before the RHS (Section I-I).

10kN

A

B

4m

F.B.D. (Section I-I)

95kNm

x

+

M

B

=40kNm

M(x)

V(x)

O

Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I).

Take moments about RHS:

(

)

xMxxxM +−+−==

∑

001

O

44095100

( )

001

4409510 −−+−= xxxxM

and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as:

( )

(

)

00

100010 xx

dx

xdM

xV −=−+−==

Step 4: Plotting the Shear Force and Bending Moment Diagrams

10kN

A

B

95kNm

+

M

B

=40kNm

10kN

C

V(x) kN

+ve

-ve

x

M(x) kNm

x

Shear

Force

Diagram

Bending

Moment

Diagram

-10

95

55

15

Loading

Diagram

It is interesting to observe that due to concentrated bending moments M

A

(reaction moment)

and M

B

(external moment), there is respectively a sudden leap and drop in the Bending

Moment Diagram. In addition, such concentrated moments do not affect the Shear Force

Diagram (Note that the drop at point A (the left end) in SFD is due to the concentrated

reaction force R

AY

. In fact, both M

A

and M

B

do not appear in shear force equation V(x) at all).

+

+

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