lesson 04, shearing force and bending moment 01

Lecturer; Dr. Dawood S. Atrushi

November 2014

Shearing Force and Bending Moment

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2

note that dM / dx = q L / 2 - q x / 2 = V

Mmax = q L2 / 8 at x = L / 2

several concentrated loads

for 0 < x < a1 V = RA M = RA x

M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)

similarly for others

M2 = Mmax because V = 0 at that point

Example 4-4

construct the shear-force and bending

-moment diagrams for the simple beam

AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

10

for a < x < a + b

V = RA - q (x - a)

M = RA x - q (x - a)2 / 2

for a + b < x < L

V = - RB M = RB (L - x)

maximum moment occurs where V = 0

i.e. x1 = a + b (b + 2c) / 2L

Mmax = q b (b + 2c) (4 a L + 2 b c + b2) / 8L2

for a = c, x1 = L / 2

Mmax = q b (2L - b) / 8

for b = L, a = c = 0 (uniform loading over the entire span)

Mmax = q L2 / 8

Example 4-5

construct the V- and M-dia for the

cantilever beam supported to P1 and P2

RB = P1 + P2 MB = P1 L + P2 b

for 0 < x < a

V = - P1 M = - P1 x

for a < x < L

V = - P1 - P2 M

= - P1 x - P2 (x - a)

Content

¢  Beams ¢  Type of beams, loadings, and

reactions ¢  Shear Forces and Bending Moments ¢  Relationships Between Loads,

Shear Forces, and Bending Moments

¢  Shear-Force and Bending-Moment Diagrams

November, 2014 Shearing Force and Bending Moment - DAT 2

Beams

Members which are subjected to loads that are transverse to the longitudinal

axis are termed as the beam.

November, 2014

The members are either subjected to the forces or the moment having their

vectors perpendicular to the axis of the bar.

Shearing Force and Bending Moment - DAT 3

Types of Beams

①  Statically Determinate beams

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316

Chapter 5 Analysis and Design of Beams for Bending

5.1 Introduction 5.2 Shear and Bending-Moment

Diagrams 5.3 Relations Among Load, Shear,

and Bending Moment 5.4 Design of Prismatic Beams for

Bending *5.5 Using Singularity Functions to

Determine Shear and Bending Moment in a Beam

*5.6 Nonprismatic Beams

5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.

Photo 5.1 Timber beams used in residential dwelling.

CB

P1

(a) Concentrated loads

w

P2

A D

(b) Distributed load

AB

C

Fig. 5.1 Transversely loaded beams.

The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is

Fig. 5.2 Common beam support configurations.

L

(a) Simply supported beam

StaticallyDeterminateBeams

StaticallyIndeterminateBeams

L2L1

(d) Continuous beam

L

(b) Overhanging beam

L

Beam fixed at one endand simply supported

at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam

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316










CB

P1


w

P2

A D


AB

C




L




L2L1

(d) Continuous beam

L


L


at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam


316










CB

P1


w

P2

A D


AB

C




L




L2L1

(d) Continuous beam

L


L


at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam



Types of Beams

②  Statically Indeterminate beams

November, 2014

316










CB

P1


w

P2

A D


AB

C




L




L2L1

(d) Continuous beam

L


L


at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam


316










CB

P1


w

P2

A D


AB

C




L




L2L1

(d) Continuous beam

L


L


at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam


316










CB

P1


w

P2

A D


AB

C




L




L2L1

(d) Continuous beam

L


L


at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam



316










CB

P1


w

P2

A D


AB

C




L




L2L1

(d) Continuous beam

L


L


at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam


Type of Loads

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316










CB

P1


w

P2

A D


AB

C




L




L2L1

(d) Continuous beam

L


L


at the other end

(e)

L

(c) Cantilever beam

L

( f ) Fixed beam


a)  Concentrated load (single force)

b)  Distributed load (measured by their intensity):

i.  Uniformly distributed load (uniform load)

ii.  Linearly varying load c)  Couple


¢ Pinned support or hinged support.

November, 2014

Type of Reactions

Solid Mechanics

force and bending moment along the length of the beam.

These diagrams are extremely useful while designing the beams for various applications.

Supports and various types of beams

(a) Roller Support – resists vertical forces only

(b) Hinge support or pin connection – resists horizontal and vertical forces

Hinge and roller supports are called as simple supports

(c) Fixed support or built-in end

Solid Mechanics









¢ Roller support

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Type of Reactions

Solid Mechanics









¢ Fixed support

November, 2014

Type of Reactions

Solid Mechanics








Solid Mechanics









1

Chapter 4 Shear Forces and Bending Moments

4.1 Introduction

Consider a beam subjected to

transverse loads as shown in figure, the

deflections occur in the plane same as

the loading plane, is called the plane of

bending. In this chapter we discuss shear

forces and bending moments in beams

related to the loads.

4.2 Types of Beams, Loads, and Reactions

Type of beams

a. simply supported beam (simple beam)

b. cantilever beam (fixed end beam)

c. beam with an overhang

Calculate the reaction forces of on of the below members

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4.1 Introduction









Type of beams




1


4.1 Introduction









Type of beams





a)

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12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0

� MA = 0 - P4 a + RB L + M1 = 0

P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC

L L 4.3 Shear Forces and Bending Moments

Consider a cantilever beam with a

concentrated load P applied at the end

A, at the cross section mn, the shear

force and bending moment are found

� Fy = 0 V = P

� M = 0 M = P x

sign conventions (deformation sign conventions)

the shear force tends to rotate the

material clockwise is defined as positive

the bending moment tends to compress

the upper part of the beam and elongate the

lower part is defined as positive

2

Type of loads

a. concentrated load (single force)

b. distributed load (measured by their intensity) :

uniformly distributed load (uniform load)

linearly varying load

c. couple

Reactions

consider the loaded beam in figure

equation of equilibrium in horizontal direction

� Fx = 0 HA - P1 cos � = 0

HA = P1 cos �

� MB = 0 - RA L + (P1 sin �) (L - a) + P2 (L - b) + q c2 / 2 = 0

(P1 sin �) (L - a) P2 (L - b) q c2 RA = CCCCCCC + CCCC + CC L L 2 L

(P1 sin �) a P2 b q c2 RB = CCCCC + CC + CC L L 2 L

for the cantilever beam

� Fx = 0 HA = 5 P3 / 13

12 P3 (q1 + q2) b � Fy = 0 RA = CC + CCCCC 13 2

3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x







2

Type of loads

a. concentrated load (single force)

b. distributed load (measured by their intensity) :

uniformly distributed load (uniform load)

linearly varying load

c. couple

Reactions

consider the loaded beam in figure

equation of equilibrium in horizontal direction

� Fx = 0 HA - P1 cos � = 0

HA = P1 cos �

� MB = 0 - RA L + (P1 sin �) (L - a) + P2 (L - b) + q c2 / 2 = 0

(P1 sin �) (L - a) P2 (L - b) q c2 RA = CCCCCCC + CCCC + CC L L 2 L

(P1 sin �) a P2 b q c2 RB = CCCCC + CC + CC L L 2 L

for the cantilever beam

� Fx = 0 HA = 5 P3 / 13

12 P3 (q1 + q2) b � Fy = 0 RA = CC + CCCCC 13 2

b)

c)


Shear Forces and Bending Moments

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

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4.5 Shear-Force and Bending-Moment Diagrams

concentrated loads

consider a simply support beam AB

with a concentrated load P

RA = P b / L RB = P a / L

for 0 < x < a

V = RA = P b / L

M = RA x = P b x / L

note that dM / dx = P b / L = V

for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L

note that dM / dx = - P a / L = V

with Mmax = P a b / L

uniform load

consider a simple beam AB with a

uniformly distributed load of constant

intensity q


¢ Consider a cantilever beam with a concentrated load P applied at the end A, at the cross section mn, the shear force and bending moment are found

November, 2014

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 ¸

¹·

¨©§ �� n� ¦ dx

dx

xdVxVdxxwxVF

y (5.1)

Dividing by dx in the limit as dxĺ0,

� � � �xwdx

xdV� (5.2)

Taking moments about the right hand edge of the element:

� � � � � � � � � �0

20 ¸

¹·

¨©§ �� ¦ dx

dx

xdMxM

dxdxxwdxxVxMM

Edge.H.R (5.3)


� � � �xVdx

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

RAY

=(1-a/L)P RBY

=Pa/L

I

I

RAY

=(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 �¸¹·

¨©§ �� ¦ xMx

L

aPM o ĺ � � x

L

aPxM ¸

¹·

¨©§ � 1 (5.5)

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

� � � �P

L

a

dx

xdMxV ¸

¹·

¨©§ � 1 (5.6)

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

++

++

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 ¸

¹·

¨©§ �� n� ¦ dx

dx

xdVxVdxxwxVF

y (5.1)


� � � �xwdx

xdV� (5.2)

Taking moments about the right hand edge of the element:

� � � � � � � � � �0

20 ¸

¹·

¨©§ �� ¦ dx

dx

xdMxM

dxdxxwdxxVxMM

Edge.H.R (5.3)


� � � �xVdx

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

RAY

=(1-a/L)P RBY

=Pa/L

I

I

RAY

=(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 �¸¹·

¨©§ �� ¦ xMx

L

aPM o ĺ � � x

L

aPxM ¸

¹·

¨©§ � 1 (5.5)

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

� � � �P

L

a

dx

xdMxV ¸

¹·

¨©§ � 1 (5.6)

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

++

++

3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x







3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x







3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x







3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x







3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x








Sign Convention

¢  The shear force tends to rotate the material clockwise is defined as positive

¢  The bending moment tends to compress the upper part of the beam and elongate the lower part is defined as positive

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� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x







3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x







3


� MA = 0 - P4 a + RB L + M1 = 0







� Fy = 0 V = P

� M = 0 M = P x








Example 1

A simple beam AB supports a force P and a couple M0, find the shear V and bending moment M at

a)  x = L/2 b)  x = (L/2)+

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Example 4-1

a simple beam AB supports a force P

and a couple M0, find the shear V and

bending moment M at

(a) at x = (L/2)_ (b) at x = (L/2)+

3P M0 P M0 RA = C - C RB = C + C 4 L 4 L

(a) at x = (L/2)_

� Fy = 0 RA - P - V = 0

V = RA - P = - P / 4 - M0 / L

� M = 0 - RA (L/2) + P (L/4) + M = 0

M = RA (L/2) + P (L/4) = P L / 8 - M0 / 2

(b) at x = (L/2)+ [similarly as (a)]

V = - P / 4 - M0 / L

M = P L / 8 + M0 / 2

Example 4-2

a cantilever beam AB subjected to a

linearly varying distributed load as shown, find

the shear force V and the bending moment M

q = q0 x / L

� Fy = 0 - V - 2 (q0 x / L) (x) = 0

V = - q0 x2 / (2 L)

Vmax = - q0 L / 2


Example 2

A cantilever beam AB subjected to a linearly varying distributed load as shown, find the shear force V and the bending moment M

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Example 4-1

a simple beam AB supports a force P

and a couple M0, find the shear V and

bending moment M at

(a) at x = (L/2)_ (b) at x = (L/2)+

3P M0 P M0 RA = C - C RB = C + C 4 L 4 L

(a) at x = (L/2)_

� Fy = 0 RA - P - V = 0

V = RA - P = - P / 4 - M0 / L

� M = 0 - RA (L/2) + P (L/4) + M = 0

M = RA (L/2) + P (L/4) = P L / 8 - M0 / 2

(b) at x = (L/2)+ [similarly as (a)]

V = - P / 4 - M0 / L

M = P L / 8 + M0 / 2

Example 4-2

a cantilever beam AB subjected to a

linearly varying distributed load as shown, find

the shear force V and the bending moment M

q = q0 x / L

� Fy = 0 - V - 2 (q0 x / L) (x) = 0

V = - q0 x2 / (2 L)

Vmax = - q0 L / 2


5

� M = 0 M + 2 (q0 x / L) (x) (x / 3) = 0

M = - q0 x3 / (6 L)

Mmax = - q0 L2 / 6

Example 4-3

an overhanging beam ABC is

supported to an uniform load of intensity

q and a concentrated load P, calculate

the shear force V and the bending

moment M at D

from equations of equilibrium, it is found

RA = 40 kN RB = 48 kN

at section D

� Fy = 0 40 - 28 - 6 x 5 - V = 0

V = - 18 kN

� M = 0

- 40 x 5 + 28 x 2 + 6 x 5 x 2.5 + M = 0

M = 69 kN-m

from the free body diagram of the right-hand part, same results can be

obtained

4.4 Relationships Between Loads, Shear Forces, and Bending Moments

consider an element of a beam of length

dx subjected to distributed loads q

equilibrium of forces in vertical direction

Example 3

An overhanging beam ABC is supported to an uniform load of intensity q and a concentrated load P, calculate the shear force V and the bending moment M at D


Relationships Between Loads, Shear Forces, and Bending Moments

a)  Consider an element of a beam of length dx subjected to distributed loads q.

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� M = 0 M + 2 (q0 x / L) (x) (x / 3) = 0

M = - q0 x3 / (6 L)

Mmax = - q0 L2 / 6

Example 4-3

an overhanging beam ABC is

supported to an uniform load of intensity

q and a concentrated load P, calculate

the shear force V and the bending

moment M at D

from equations of equilibrium, it is found

RA = 40 kN RB = 48 kN

at section D

� Fy = 0 40 - 28 - 6 x 5 - V = 0

V = - 18 kN

� M = 0

- 40 x 5 + 28 x 2 + 6 x 5 x 2.5 + M = 0

M = 69 kN-m

from the free body diagram of the right-hand part, same results can be

obtained

4.4 Relationships Between Loads, Shear Forces, and Bending Moments

consider an element of a beam of length

dx subjected to distributed loads q

equilibrium of forces in vertical direction


Equilibrium of forces in vertical direction

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� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q

integrate between two points A and B

B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)

the area of loading diagram may be positive or negative

moment equilibrium of the element

� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0

or dM / dx = V

maximum (or minimum) bending-moment occurs at dM / dx = 0,

i.e. at the point of shear force V = 0


B B Н dM = Н V dx

A A

i.e. B MB - MA = Н V dx A

= (area of the shear-force diagram between A and B)

this equation is valid even when concentrated loads act on the beam

between A and B, but it is not valid if a couple acts between A

and B

6

� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

6

� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

6

� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

6

� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

Integrate between two points A and B;

= - (area of the loading diagram between A and B) Shearing Force and Bending Moment - DAT 19

Moment equilibrium of the element

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� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

6

� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

6

� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

Maximum (or minimum) bending-moment occurs at dM/dx = 0, i.e. at the point of shear force V = 0. Shearing Force and Bending Moment - DAT 20

Integrate between two points A and B

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� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

6

� Fy = 0 V - q dx - (V + dV) = 0

or dV / dx = - q






or dM / dx = V




B B Н dM = Н V dx

A A





and B

= (area of the loading diagram between A and B)


b)  Consider an element of a beam of length dx subjected to concentrated load P.

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concentrated loads

equilibrium of force

V - P - (V + V1) = 0

or V1 = - P

i.e. an abrupt change in the shear force occurs

at any point where a concentrated load acts

equilibrium of moment

- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0

since the length dx of the element is infinitesimally small, i.e. M1

is also infinitesimally small, thus, the bending moment does not change as

we pass through the point of application of a concentrated load

loads in the form of couples

equilibrium of force V1 = 0

i.e. no change in shear force at the point of

application of a couple


- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0

the bending moment changes abruptly at a point of application of a

couple


Equilibrium of force;

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concentrated loads


V - P - (V + V1) = 0

or V1 = - P




- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0









- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0


couple

7

concentrated loads


V - P - (V + V1) = 0

or V1 = - P




- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0









- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0


couple

7

concentrated loads


V - P - (V + V1) = 0

or V1 = - P




- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0









- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0


couple

i.e. an abrupt change in the

shear force occurs at any point where a concentrated

load acts.

Since the length dx of the element is infinitesimally small, i.e. M1 is also infinitesimally small, thus, the bending moment does not change as we pass through the point of application of a concentrated load.

7

concentrated loads


V - P - (V + V1) = 0

or V1 = - P




- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0









- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0


couple

Equilibrium of moment;


c)  Consider an element of a beam of length dx subjected to concentrated loads in form of couples, M0.

November, 2014

7

concentrated loads


V - P - (V + V1) = 0

or V1 = - P




- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0









- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0


couple

Equilibrium of force; V1 = 0


¢  i.e. no change in shear force at the point of application of a couple.

November, 2014

7

concentrated loads


V - P - (V + V1) = 0

or V1 = - P




- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0









- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0


couple

7

concentrated loads


V - P - (V + V1) = 0

or V1 = - P




- M - P (dx/2) - (V + V1) dx + M + M1 = 0

or M1 = P (dx/2) + V dx + V1 dx M 0









- M + M0 - (V + V1) dx + M + M1 = 0

or M1 = - M0


couple

Equilibrium of moment;

¢  The bending moment changes abruptly at a point of application of a couple.


Shear-Force and Bending-Moment Diagrams

Concentrated loads Consider a simply support beam AB with a concentrated load P

November, 2014

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q


November, 2014

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q


8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

With Mmax = Pab/L

November, 2014

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

V - Diagram

M - Diagram


Uniform loads Consider a simply support beam AB with a uniformly distributed load of constant intensity q;

November, 2014

8


concentrated loads




for 0 < x < a

V = RA = P b / L



for a < x < L

V = RA - P = - P a / L

M = RA x - P (x - a) = P a (L - x) / L



uniform load



intensity q

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x


November, 2014

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

V - Diagram M - Diagram


Several concentrated loads Consider a simply support beam AB with a several concentrated pi;

November, 2014

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x 9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x


9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

November, 2014

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x 9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x

V - Diagram

M - Diagram


Example 4

Construct the shear force and bending moment diagrams for the simple beam AB.

November, 2014

9

RA = RB = q L / 2

V = RA - q x = q L / 2 - q x

M = RA x - q x (x/2) = q L x / 2 - q x2 / 2


Mmax = q L2 / 8 at x = L / 2



M1 = RA a1

for a1 < x < a2 V = RA - P1

M = RA x - P1 (x - a1)

M2 - M1 = (RA - P1 )(a2 - a1)



Example 4-4



AB

RA = q b (b + 2c) / 2L

RB = q b (b + 2a) / 2L

for 0 < x < a

V = RA M = RA x


Example 5

Construct the V- and M-diagram for the cantilever beam supported to P1 and P2.

November, 2014

10

for a < x < a + b

V = RA - q (x - a)

M = RA x - q (x - a)2 / 2

for a + b < x < L

V = - RB M = RB (L - x)

maximum moment occurs where V = 0

i.e. x1 = a + b (b + 2c) / 2L

Mmax = q b (b + 2c) (4 a L + 2 b c + b2) / 8L2

for a = c, x1 = L / 2

Mmax = q b (2L - b) / 8

for b = L, a = c = 0 (uniform loading over the entire span)

Mmax = q L2 / 8

Example 4-5


cantilever beam supported to P1 and P2

RB = P1 + P2 MB = P1 L + P2 b

for 0 < x < a

V = - P1 M = - P1 x

for a < x < L

V = - P1 - P2 M

= - P1 x - P2 (x - a)


Example 6

Construct the V- and M-diagram for the cantilever beam supporting to a constant uniform load of intensity q.

November, 2014

11

Example 4-6


cantilever beam supporting to a constant uniform

load of intensity q

RB = q L MB = q L2 / 2

then V = - q x M = - q x2 / 2

Vmax = - q L Mmax = - q L2 / 2

alternative method

x V - VA = V - 0 = V = -Н q dx = - q x 0 x M - MA = M - 0 = M = -Н V dx = - q x2 / 2 0

Example 4-7

an overhanging beam is subjected to a

uniform load of q = 1 kN/m on AB

and a couple M0 = 12 kN-m on midpoint

of BC, construct the V- and M-dia for

the beam

RB = 5.25 kN RC = 1.25 kN

shear force diagram

V = - q x on AB

V = constant on BC


Example 7

An overhanging beam is subjected to a uniform load of q = 1 kN/m on AB and a couple M0 = 12 kN-m on midpoint of BC. Construct the V- and M-diagram for the beam.

November, 2014

11

Example 4-6


cantilever beam supporting to a constant uniform

load of intensity q

RB = q L MB = q L2 / 2

then V = - q x M = - q x2 / 2

Vmax = - q L Mmax = - q L2 / 2

alternative method

x V - VA = V - 0 = V = -Н q dx = - q x 0 x M - MA = M - 0 = M = -Н V dx = - q x2 / 2 0

Example 4-7

an overhanging beam is subjected to a

uniform load of q = 1 kN/m on AB

and a couple M0 = 12 kN-m on midpoint

of BC, construct the V- and M-dia for

the beam

RB = 5.25 kN RC = 1.25 kN

shear force diagram

V = - q x on AB

V = constant on BC


Thank You


lesson 04, shearing force and bending moment 01

Engineering