RAGHU ENGINEERING COLLEGE (AUTONOMOUS ......• RAGHU ENGINEERING COLLEGE (AUTONOMOUS) DEPARTMENT OF MECHANICAL ENGINEERING Mechanics of solids UNIT –2 –Shear force and Bending

• RAGHU ENGINEERING COLLEGE

(AUTONOMOUS)

DEPARTMENT OF MECHANICAL ENGINEERING

Mechanics of solids

UNIT – 2 – Shear force and Bending moment.

Prepared by

Mr. Singuru Rajesh

Assistant Professor

1Singuru Rajesh Mechanics of Solids Mechanical Engineering REC

1

CONTENTS:

Introduction: Applied and Reactive forces

Supports

Types of supports

Beams

Types of beams

What the loads do

Types of loads

Designing beams

How we calculate the effects

Internal reactions in beams

Shear forces, bending moments sign conventions

Cantilever Beam

Simply Supporting Beams

Over Hanging Beams

Singuru Rajesh Mechanics of Solids Mechanical Engineering REC2

Syllabus

Definition of beam – Types of beams – Concept

of shear force and bending moment – S.F and

B.M diagrams for cantilever, simply supported

and overhanging beams subjected to point

loads, U.D.L, Uniformly Varying Loads and

combination of these loads – Point of contra

flexure – Relation between S.F., B.M and rate

of loading at a section of a beam.

APPLIED AND REACTIVE FORCES


Re-call of Engg Mechanics Basics:

Forces that act on a Body can be divided into two Primary types:

1. Applied and

2. Reactive

APPLIED AND REACTIVE FORCES

• In common Engineering usage, appliedforces are forces that act directly on astructure like, dead, live load etc.)

• Reactive forces are forces generated bythe action of one body on another andhence typically occur at connections orsupports.

• The existence of reactive forces followsfrom Newton’s third law, which statethat to every action , there is an equaland opposite reaction.


Re-call of Engg Mechanics Basics:

APPLIED

REACTIVE

• To bear or hold up (a load, mass, structure, part, etc.); serve asa foundation or base for any structure.

• Supports are used to connect structures to the ground or otherbodies in order to restrict (confine) their movements under theapplied loads.


SUPPORTS

• Supports are grouped into three categories, depending onthe number of reactions.

• They exert on the structures.

1) Roller support

2) Hinge support

3) Fixed support

4) Simple Support


TYPES OF SUPPORTS

ROLLER SUPPORT

• Roller supports are free to rotate and translate along thesurface upon which the roller rests.

• The surface can be horizontal, vertical, or sloped at any angle.

• The resulting reaction force is always a single force that isperpendicular to, and away from, the surface.


•Restrains the structure

from moving in one or two

perpendicular directions


ROLLER SUPPORT


HINGE SUPPORT

• A Hinge support can resist both vertical and horizontal forces butnot a moment. They will allow the structural member to rotate,but not to translate in any direction.

• Pin or hinge support is used when we need to prevent the

structure from moving or restrain its translational degrees of

freedom.

• A hinge is a type of bearing that connects two solid objects,typically allowing only a limited angle of rotation between them.Two objects connected by an ideal hinge rotate relative to eachother about a fixed axis of rotation.



Hinge Support

FIXED SUPPORT

• Fixed supports can resist vertical and horizontal forces as well asa moment. Since they restrain both rotation and translation, theyare also known as rigid supports.


SIMPLE SUPPORT

• A Simple support can be simply take the load.


Ancient

Modern

Singuru Rajesh Mechanics of Solids Mechanical Engineering REC 14

Remember Point


Remember Point

Roller Support

Roller fixed in Y direction

Hinge Support

Fixed Support


BEAM

• A beam is a structural member (horizontal) that is design tosupport the applied load (vertical). It resists the applied loadingby a combination of internal transverse shear force and bendingmoment.

• It is perhaps the most important and widely used structuralmembers and can be classified according to its supportconditions.


Structural Member

(Horizontal)

Applied Load (Vertical)

•Extremely common structural element.

•In buildings majority of loads are vertical and majority of useable

surfaces are horizontal.

1/39


BEAMS

devices for transferring

vertical loads horizontally

•Action of beams involves combination of bending and shear

2/39


are

TYPES OF BEAMS

• The following are the important types of beams:

a) Cantilever

b) Simply supported

c) Overhanging

d) Continuous beam

e) Fixed beam


Remember Point

CANTILEVER BEAM

• A beam which is fixed at one end and free at the other end Is

known as cantilever beam.


SIMPLY SUPPORTED BEAMS

• A beam supported or resting freely on the supports at its both ends.


FIXED BEAMS

• A beam whose both ends are fixed and is restrained against

rotation and vertical movement. Also known as built-in beam

or encastred beam.


OVERHANGING BEAM

• If the end portion of a beam is extended outside the supports.


CONTINUOUS BEAMS


• A beam which is provided with more than two supports.

1. Point loads, from concentrated loads or other beams

Distributed Load

10/39

Point Load

Reactions


TYPES OF LOADS ON BEAMS

2. Distributed loads, from anything continuous

TYPES OF LOADS

• Concentrated load assumed to act at a point and immediatelyintroduce an oversimplification since all practical loading systemmust be applied over a finite area.

27Singuru Rajesh Mechanics of Solids

Mechanical Engineering REC


Distributed loads

1.

2.

• The loads (& reactions) bend the beam, and try to shear through it.

11/39


WHAT THE LOADS DO????

Bending

Shear

12/39

e

Bending

ee e

C

T

Shear


WHAT THE LOADS DO????

• In architectural structures, bending moment more important

importance increases as span increases.

Short span structures with heavy loads, shear dominant

e.g. Pin connecting engine parts

beams in building

designed for bending

checked for shear

13/39


DESIGNING BEAMS

1. First, find ALL the forces (loads and reactions).

14/39


HOW WE CALCULATE THE EFFECTS

2. Make the beam into a free body (cut it out and artificially

support it).

3. Find the reactions, using the conditions of equilibrium

INTERNAL REACTIONS IN BEAMS

• The following are the internal force resultants/reactions in Beams

1. Normalforce (N),

2. Shearforce (V),

3. Bendingmoment (M),

4. Torsion (T)



• The following are the internal force resultants/reactions in Beams

1. Normalforce (N),

2. Shearforce (V),

3. Bendingmoment (M),

4. Torsion (T)



L

• At any cut in a beam, there are 3 possible internal reactions

required for equilibrium:

1. Normalforce (N),

2. Shearforce (V),

3. Bendingmoment (M).

a b


P



SHEAR FORCES, BENDING MOMENTS SIGN CONVENTIONS

left section right sectionShear forces:

Negative shear:

Bending moments:

Positive moment

Negative moment

C.W

ACW


Axis Assumed to Cut

Positive shear:Positive shear force will causethe Beam segment on which itacts to rotate clockwise

Positive bending moment willtend to bend the segment onwhich it acts in a concaveupward manner (compressionon top of section)

SHEAR FORCES, BENDING MOMENTS SIGN CONVENTIONS

Shear force

Bending moments:

Positive moment


Axis Assumed to Cut

Positive shear:

+

+

X

X

Sagging bending moment is POSITIVE

(happy)

+

Hogging bending moment is NEGATIVE

(sad)

-


Sagging and Hogging of a Ship

•Consider cantilever beam with point load on end

W

MR = -WLVertical reaction, R = -W

and Moment reaction MR = - WL

• Use the free body idea to isolate part of the beam.

• and Add in forces required for equilibrium.


CANTILEVER BEAM POINT LOAD AT END

R =- W

L

Remember Point


Finding the Reaction Forces in Beam when load is applied

𝑅𝐴 + 𝑅𝐵 = W ---- eqn.1

Finding the Moment from B end𝑅𝐴 (L)+ 𝑅𝐵 (0)= W (L/2)

𝑅𝐴 (L)+ 0= W (L/2)𝑹𝑨 = W/2

Now placing 𝑅𝐴 in 1 we get (𝑊/2) + 𝑅𝐵 = W

𝑹𝑩=W/2

LA B

W

C

L / 2𝑅𝐴 𝑅𝐵

L = 3 mA B

20 KN

C

1.5 m𝑅𝐴 𝑅𝐵

𝑅𝐴 + 𝑅𝐵 = 20 KN ---- eqn.1


𝑅𝐴 (3)+ 0= 20 (3/2)𝑹𝑨 = ((20)1.5) / 3 = 10 KN


𝑹𝑩= 10 KN


Finding the Reaction Forces in Beam when load is applied

𝑅𝐴 + 𝑅𝐵 = 100+350+150 N ---- eqn.1

Finding the Moment from B end𝑅𝐴 (6)+ 𝑅𝐵 (0)= 150 (1.5)+ 350(3)+ 100(4.5)

𝑹𝑨 = ______

Now placing 𝑅𝐴 in 1eqn we get ______+ 𝑅𝐵 = 600 N

𝑹𝑩= _____ N

L= 6 mA B

350 N

C

1.5 m𝑅𝐴 𝑅𝐵1.5 m

150 N100 N

𝑅𝐴 + 𝑅𝐵 = W x (L/2) ---- eqn.1

Finding the Moment from B end𝑅𝐴 (6)+ 𝑅𝐵 (0)= 100 (6/2)

𝑹𝑨 = ______ KN

Now placing 𝑅𝐴 in 1eqn we get ______+ 𝑅𝐵 = 300 KN

𝑹𝑩= _____ KN

100 KN per unit length

L= 6 mA

𝑅𝐴 𝑅𝐵

A B


Finding the Reaction Forces in Beam when Point load is applied.

Let Equating the forces/Load𝑅𝐴 + 𝑅𝐵 = 100+350+150 N ---- eqn.1


𝑹𝑨 = ______


𝑹𝑩= _____ N

L= 6 mA B

350 N

C

1.5 m1.5 m

150 N100 N

L= 6 mA B

350 N

C


150 N100 N

𝑭𝑩𝑫


Finding the Reaction Forces in Beam when UDL is applied

𝑅𝐴 + 𝑅𝐵 = W x (L) ---- eqn.1


𝑹𝑨 = ______ KN


𝑹𝑩= _____ KN

B

100 KN x (L)

L= 6 mA

𝑅𝐴 𝑅𝐵

A


L= 6 mA

𝑅𝐴 𝑅𝐵

A B

FBD: UDL is Converted to Point Load


LA B

W

x

ShearForceDiagram

Bending MomentDiagram

Sign Conventions:

+

Bending Moment (BM)

+

+

Base Line

WW

-W x L

Bending Moment BM = -W.x

when x = L

when x = 0

BM = -WL

BM = 0

SFD and BMD of CANTILEVER BEAM with

POINT LOAD AT END

Shear = W constant along length

0

W

W x X

x

x

For Shear Force (SF)

LoadDiagram

Singuru Rajesh Mechanics of Solids Mechanical Engineering REC

46




LA B

W

x

ShearForceDiagram


Sign Conventions:


+

For Bending Moment (BM)

+

+

Base Line

WW

-W x L

Bending Moment BM = -W.x

when x = L

when x = 0

BM = -WL

BM = 0


Shear = W constant along length

0

W

W x X

x

x

LoadDiagram


49


Finding the Reaction Forces in Beam when UDL is applied

𝑅𝐴 + 𝑅𝐵 = W x (L) ---- eqn.1


𝑹𝑨 = ______ KN


𝑹𝑩= _____ KN

B

100 KN x (L)

L= 6 mA

𝑅𝐴 𝑅𝐵

A


L= 6 mA

𝑅𝐴 𝑅𝐵

A B

FBD: UDL is Converted to Point Load


LA B

W Per Unit Length

x

ShearForceDiagram


+

Base Line

0

W x L

-

W x 𝐿2

2

Bending Moment BM = -W . 𝑋2

2

when x = L

when x = 0BM = -W x

𝐿2

2

BM = 0

SFD and BMD of CANTILEVER BEAM

WITH UDL

Shear = W x X

0

W x X

W x 𝑋2

2

x

x

A B

A B

C

A`

Sign Conventions:

+

Bending Moment (BM)

+



52



54

Difference between UDL and UVL



LA B

W x 𝑋

𝐿 x

SFD

BMD

+

Base Line

0

𝑊𝐿

2

-W x

𝐿2

6

Bending Moment BM = -𝑊𝑋3

6𝐿

when x = L

when x = 0BM = -W x

𝐿2

6

BM = 0

SFD and BMD of CANTILEVER BEAM WITH UVL

Shear = 𝑊𝑋2

2𝐿

0

𝑊𝑋2

2𝐿

𝑊𝑋3

6𝐿

x

x

A B

A B

C

A`

W

Parabolic line

Cubic line

when x = L

when x = 0

𝑊𝐿

2

0

Sign Conventions:

+

Bending Moment (BM)

+



57


Finding the Reaction Forces in Simply Supporting Beam

when load is applied 𝑅𝐴 + 𝑅𝐵 = W ---- eqn.1


𝑅𝐴 (L)+ 0= W (L/2)𝑹𝑨 = W/2


𝑹𝑩=W/2

LA B

W

C


L = 3 mA B

20 KN

C

1.5 m𝑅𝐴 𝑅𝐵

𝑅𝐴 + 𝑅𝐵 = 20 KN ---- eqn.1


𝑅𝐴 (3)+ 0= 20 (3/2)𝑹𝑨 = ((20)1.5) / 3 = 10 KN


𝑹𝑩= 10 KN


Finding the Reaction Forces in Simply Supporting Beam

when load is applied 𝑅𝐴 + 𝑅𝐵 = 100+350+150 N ---- eqn.1


𝑹𝑨 = ______


𝑹𝑩= _____ N

L= 6 mA B

350 N

C


150 N100 N

𝑅𝐴 + 𝑅𝐵 = W x (L/2) ---- eqn.1


𝑹𝑨 = ______ KN


𝑹𝑩= _____ KN


L= 6 mA

𝑅𝐴 𝑅𝐵

A B


LA B

W

SFD

BMD

+

Base Line

+𝑊𝐿

4

SIMPLY SUPPORTED BEAM POINT LOAD AT END

O

C

-

+𝑊

2

−𝑊

2

Base Line


A BC

A

B

C`

𝑅𝐴 + 𝑅𝐵 = W ---- eqn.1


𝑅𝐴 (L)+ 0= W (L/2)𝑹𝑨 = W/2


𝑹𝑩=W/2

Sign Conventions:


+


+

SIMPLY SUPPORTED BEAM POINT LOAD AT END


61

SIMPLY SUPPORTED BEAM POINT LOAD


62

SIMPLY SUPPORTED BEAM POINT TWO POINT LOADS


63


LA B

W per unit Length

SFD

BMD

+

Base Line

+𝑊 𝐿2

8

SFD and BMD for SIMPLY

SUPPORTED BEAM WITH UDL

O

C

-

+𝑊 𝐿

2

−𝑊 𝐿

2

Base Line


A BC

A

B

C`

Sign Conventions:


+


+

SIMPLY SUPPORTED BEAM with UDL


65

SIMPLY SUPPORTED BEAM with UDL


66

SIMPLY SUPPORTED BEAM with Point Load and UDL


67

SIMPLY SUPPORTED BEAM with Point Load and UDL


68


LA B

O

SFD

BMD

+

Base Line

+𝑊 𝐿2

9 √3

SFD and BMD for SIMPLY SUPPORTED BEAM WITH UVL

C

C

-

+𝑤 𝐿

6

−𝑤 𝐿

3

Base Line

x𝑅𝐴 𝑅𝐵

A BC

A

B

C`

WSign Conventions:


+


+

+

L / √3

X

X

SIMPLY SUPPORTED BEAM with UVL


70

SIMPLY SUPPORTED BEAM with UVL


71


4 mA B

2 KN per unit Length

SFD

BMD

+

Base Line

2.25 𝐾𝑁𝑚

SFD and BMD for OVER HANGING

BEAM POINT LOAD with UDL

D

-

3𝐾𝑁

5 𝐾𝑁

Base Line

2 m𝑅𝐴 𝑅𝐵

A

B

D

AB

Sign Conventions:


+


+

C

C+

+-

4KN

4 𝐾𝑁𝑚

Point of Contraflexture

C


4 mA B


SFD

BMD

+

Base Line

2.25 𝐾𝑁𝑚



D

-

3𝐾𝑁

5 𝐾𝑁

Base Line


A

B

D

AB

Sign Conventions:


+


+

C

C+

+-

4KN

4 𝐾𝑁𝑚


C




4 mA B


SFD

BMD

+

Base Line

2.25 𝐾𝑁𝑚


BEAM POINT LOAD with UDL and

Point Load

D

-

3𝐾𝑁

5 𝐾𝑁

Base Line


A

B

D

AB

Sign Conventions:


+


+

C

C+

+-

4KN

4 𝐾𝑁𝑚


C

2 KN


4 mA B


SFD

BMD

+

Base Line

2.25 𝐾𝑁𝑚


BEAM POINT LOAD with UDL and

Point Load

D

-

3𝐾𝑁

5 𝐾𝑁

Base Line


A

B

D

AB

Sign Conventions:


+


+

C

C+

+-

4KN

4 𝐾𝑁𝑚


C

2 KN


LC B

W KN

SFD

BMD

- Base Line


BEAM POINT LOAD with Point Load

A𝐾𝑁

Base Line

a𝑅𝐴 𝑅𝐵

C

BA

CB

Sign Conventions:


+


+

D

D+

-

KN

D

W KN

KNm KNm

A

a



BEAM POINT LOAD with Point Load

The rate of change of the bending moment with respect to x is equal to the shearing force, or the slope of the moment diagram at the given point is the shear at that point.

dM

dx= SF

The rate of change of the shearing force with respect to x is equal to the load or the slope of the shear diagram at a given point equals the load at that point dSF

dx=W

Relationship Between Load, Shear, and Moment



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Documents

RAGHU ENGINEERING COLLEGE (AUTONOMOUS ......• RAGHU ENGINEERING COLLEGE (AUTONOMOUS) DEPARTMENT OF MECHANICAL ENGINEERING Mechanics of solids UNIT –2 –Shear force and Bending