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Bending of BeamsMECHENG242 Mechanics of Materials

Bending of BeamsNow we consider the elastic deformation of beams (bars) under bending loads.

☻

MM

Prof. AKASH MOHANTYProf. AKASH MOHANTY


Application to a Bar

Normal Force:

FnFn

Shear Force:

FtFt

Bending Moment:

MtMt

Torque or Twisting Moment:

Mn

Mn

S.B.

K.J.


Examples of Devices under Bending Loading:

Car Chassis

YachtExcavator

Atrium Structure


2.2 Stresses in Beams

2.3 Combined Bending and Axial Loading

2.0 Bending of Beams

2.4 Deflections in Beams

x

σx

Mxz Mxz

σxP

P1

P2


x

y

PA B

RByRAyMxz Mxz

Radius of Curvature, R

Deflected Shape

Consider the simply supported beam below:

Mxz MxzWhat stresses are generated within, due to bending?


Axial Stiffness

Load (W)

Extension (u)

Bending Moment

(Mxz)

Curvature (1/R)

Flexural Stiffness

PA B

RByMxzRAy

Mxz

BendingRecall: Axial Deformation

W

u


x

y

Mxz=Bending Moment

Mxz Mxz

Beam

σx (Tension)

σx (Compression)

σx=0

(i) Bending Moment, Mxz

(ii) Geometry of Cross-section

σx is NOT UNIFORM through the section depth

σx DEPENDS ON:

Axial Stress Due to Bending:

Unlike stress generated by axial loads, due to bending:


SIGN CONVENTIONS


SIGN CONVENTION ***VVIMP***


“Happy” Beam is +VE “Sad” Beam is -VE

x

yMxz=Bending Moment

+VE (POSITIVE)

Sign Conventions: Qxy=Shear Force

Mxz Mxz

Qxy Qxy

-veσx

+veσx

Be Happy No Happy


CubicParabolicLinearMoment

ParabolicLinearConstantShear

LinearConstant0Load

Common Relationships


ParabolicLinearLinearMoment

LinearConstantConstantShear

Constant00Load

Common Relationships

M


IMP

Quadratic curves of degree 3

Parabolic curves of degree 2

Inclined straight lines

Bending Moment Diagram

Parabolic curve

Inclined straight

lines

Line parallel to the

horizontal axis

Shear Force

Diagram

Type of loading


Bending Moments (Types of Loading)

Q

0

(SFD)

0

M (BMD)


;0Fy =∑;0Mz =∑

PQxy =∴

x

yExample 1: Bending Moment Diagrams P

RAy=P

A B

L

Mxz=P.L

P.L

P Qxy

Mxz

xP

MxzQxy

Mxz Mxz

Qxy Qxy

Q & M are POSITIVE ( )xLPMxz −−=∴


;PQxy =

x

yP

P

BP.L

x

( )xLPMxz −−=L

Qxy 0

Mxz 0

A Mxz

Qxy

To find σx and deflections, need to know Mxz.

Shear Force Diagram (SFD)

Bending Moment Diagram (BMD)

+veP

-ve

-P.L


x

yExample 2: Macaulay’s Notation

Qxy

Mxz

A BC

a bP

( )baaPRBy +⋅

=( )babPRAy +⋅

=x

( )babP+⋅

;0Mz =∑ xzM+

( ) ( ) ( )axPxba

PbMxz −−+

=∴

Where can only be +VE or ZERO.( )ax −

Pa

A

( )axP −+ ( ) ( ) 0xba

Pb=

+−


x

y

x

A BC

a bP

( )baPa+( )ba

Pb+

(i) When :ax ≤

(ii) When :ax >( ) ( ) ( )axPx

baPbMxz −−+

=

1

2

( ) ( ) ( )axPxba

PbMxz −−+

=0

A BC

+ve( )baPab+

Mxz

0

BMD: Eq. 1Eq. 2


A B

L


Relationship

ω−=dxdF

Fdx

dM=


;0Fy =∑;0Mz =∑

x

y

Example 3: Distributed Load

RAy=wL

A B

Lx

Qxy

Mxz

MxzQxy

Mxz=wL2

2

wL

wL2

2

Distributed Load wper unit length

wL

2wLM

2

xz +

( )xLwQxy −=⇒

wx

wx− 0Qxy =−

( )xwL− 02xwx =⎟⎠⎞

⎜⎝⎛+


;0x@ =

2wL

2wxwLxM

22

xz −−=⇒( )xLwQxy −=⇒

-ve

-wL2

2

x

Mxz

0BMD: L

2wLM

2

xz −=

;Lx@ = 0Mxz =

;2Lx@ = 8

wLM2

xz −=


Internal Reactions in Beams

• At any cut in a beam, there are 3 possible internal reactions required for equilibrium: – normal force, – shear force, – bending moment.

L

P

a b


Internal Reactions in Beams• At any cut in a beam, there are 3 possible

internal reactions required for equilibrium: – normal force, – shear force, – bending moment.

Pb/Lx

Left Side of Cut

V

M

N

Positive Directions Shown!!!


Internal Reactions in Beams• At any cut in a beam, there are 3 possible

internal reactions required for equilibrium: – normal force, – shear force, – bending moment.

Pa/LL - x

Right Side of CutVM

N

Positive Directions Shown!!!


Finding Internal Reactions• Pick left side of the cut:

– Find the sum of all the vertical forces to the left of the cut, including V. Solve for shear, V.

– Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0.

– Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M

• Pick the right side of the cut:– Same as above, except to the right of the cut.


Example: Find the internal reactions at points indicated. All axial force reactions are zero. Points

are 2-ft apart.

20 ft

P = 20 kips

12 kips8 kips12 ft

1

7

10

6

2 3 94 5 8

Point 6 is just left of P and Point 7 is just right of P.


20 ft

P = 20 kips

12 kips8 kips12 ft

1

7

10

6

2 3 94 5 8

V(kips)

M(ft-kips)

8 kips

-12 kips96

4864

4872

24

80

1632

x

x


20 ft

P = 20 kips

12 kips8 kips12 ft

V(kips)

M(ft-kips)

8 kips

-12 kips96 ft-kips

x

x

V & M Diagrams

What is the slope of this line?

a

b

c

96 ft-kips/12’ = 8 kipsWhat is the slope of this line?

-12 kips


20 ft

P = 20 kips

12 kips8 kips12 ft

V(kips)

M(ft-kips)

8 kips

-12 kips96 ft-kips

x

x

V & M Diagrams

a

b

c

What is the area of the blue rectangle?

96 ft-kipsWhat is the area of the green rectangle?

-96 ft-kips


Draw Some Conclusions

• The magnitude of the shear at a point equals the slope of the moment diagram at that point.

• The area under the shear diagram between two points equals the change in moments between those two points.

• At points where the shear is zero, the moment is a local maximum or minimum.


∫∫

=

=

=

dx)x(V)x(M

dx)x(w)x(V

functionloadthe)x(w

The Relationship Between Load, Shear and Bending Moment


Example: Draw Shear & Moment diagrams for the following beam

3 m 1 m1 m

12 kN 8 kNA C

BD

RA = 7 kN ↑ RC = 13 kN ↑


3 m 1 m1 m

12 kNA C

BD

V(KN)

M(KN-m)

7

-5

8

8 kN

7-15

8

7

-82.4 m


TYPES OF LOADS:• A load can be classified as:• (i) Concentrated: which is regarded as

acting wholly at one. Examples are loads P, P2, P3 and P4 in Figure 3.1.

• (ii) Distributed Load: A load that is spread along the axis of the beam, such as q in Figure 3.1 a. Distributed loads are measured by their intensity, which is expressed in force per unit distance e.g. kN/m.


TYPES OF LOADS CONTD.

• A uniformly distributed load, or uniform load has constant intensity, q per unit distance (Figure 3.1. a).

• A linearly varying load (Figure 3.1 b) has an intensity which changes with distance.

• (iii) Couple: This is illustrated by the couple of moment M acting on the overhanging beam in Figure 3.1 c).


SHEAR FORCES AND BENDING MOMENTS

• When a beam is loaded by forces or couples, stresses and strains are created throughout the interior of the beam.

• To determine these stresses and strains, the internal forces and internal couples that act on the cross sections of the beam must be found.


SHEAR FORCES AND BENDING MOMENTS CONTD.

• To find the internal quantities, consider a cantilever beam in Figure 3.2 .

• Cut the beam at a cross-section mn located at a distance x from the free end and isolate the left hand part of the beam as a free body (Figure 3.2 b).

• The free body is held in equilibrium by the force P and by the stresses that act over the cut cross section.


SHEAR FORCES AND BENDING MOMENTS CONTD.

• The resultant of the stresses must be such as to maintain the equilibrium of the free body.

• The resultant of the stresses acting on the cross section can be reduced to a shear force V and a bending moment M.

• The stress resultants in statically determinate beams can be calculated from equations of equilibrium.


Point of contra flexure

• it is the point at which the bending moment curve intersects with the zero line


Problem

• A horizontal beam 10m long carries a uniformly distributed load of 180N/m and in addition a concentrated load of 200N at the left end. The beam is supported at two points 7meters apart, so chosen that each support carries half the total load. Draw Shear force and bending moment diagram. Also locate the p o i n t s o f c o n t r a f l e x u r e .

Bending of BeamsMECHENG242 Mechanics of Materialsx= 1, y = 4.4 from left A


Home work


Give the values for load, distance and couple, solve it.


Summary – Is anything Necessary for RevisionGenerating Bending Moment Diagrams is a key skill you must revise. From these we will determine:

• Stress Distributions within beams,

• and the resulting Deflections

www.engineering.auckland.ac.nz/mechanical/MechEng242

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