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1/60

Shear Force & Bending Moment

(Introduction)

Beam a slender member and support loadings are applied perpendicular to its

longitudinal axis

Beams are important structural and mechanical elements in engineering

The basic problem in the mechanic of materials is to determine the relations

between the stress and deformation caused by loads applied to any structure


2/60


(Introduction)

The study of bending loads is

complicated by the fact that the

loading effects vary from section

to section of the beam

These loading effects take the

form of a shearing force, Vand a

bending moment, M

In designing beam, it is

necessary to determine the

maximum shear force and

bending moment in the beam


3/60

Express V and M as a function of arbitrary positionxalong axis

The function can be represented by graphs called Shear Force

Diagram (SFD) and Bending Moment Diagram (BMD)

Engineers need to know the variation of shear force and bending

moment along the beam in order to know where to reinforce it


(Introduction)


4/60


(Types of beams)

Simply Supported Beam

Overhanging Beam

Cantilever Beam


5/60


(Types of loadings)

Point Load Couple

Uniformly Distributed Load Linearly Varying Distributed Load


6/60


(Sign Convention)


7/60


(Sign Convention)

+M+M

+V

+V


8/60

xwV

xwVVVFy 0:0

wdxdV

Relationship between load and

shear:

221

02:0

xwxVM

x

xwxVMMMMC

Vdx

dM

Relationship between shear

and bending moment:

Peak point of moment V=0

V= constant if w= 0


(Relation among load, shear and moment)


9/60


10/60


(Examples of questions)

Determine the value of shear force and bending moment at a cross

section 0.5 m to the right of point A.

Example 1:-


11/60




12/60




13/60



M

V

kNmM

M

M

kNV

VF

AB

xx

y

5.2

05.05

0

5

050

)5.0(


14/60



Draw the shear force and bending moment diagrams for the beam shown below

Example 2:-

kN75.5

0)5(51525.34;0

kN25.34

0)10()5.25)(5(5)5(1580;0

y

yy

y

yA

A

AF

C

CM


15/60

kN.m75.108575.580kN75.5m5

kN.m80kN75.5;0

75.580

075.5800

75.5

075.5;0

50

1

1sec

1

MVx

MVx

xM

MxM

V

VF

mx

tion

y




16/60



05.921075.15105.2

kN25.3410575.15m10

kN.m75.1085.92575.1555.2

kN25.95575.15m5

5.9275.155.2

0

2

55551575.580;0

575.15

0551575.5;0

105

2

2

2

2

2

2222sec

2

2

2

M

Vx

M

Vx

xxM

Mx

xxxM

xV

VxF

mxm

tion

y


17/60


18/60


19/60



Draw complete shear force and bending moment diagrams for the beam shown

below. Determine the maximum bending moment and its location on the beam.

Example 4:-


20/60



A cantilever beam ABCD is subjected to a point load and couple at point B, and uniform

distributed load along CD, as shown below.

a) Write the shear force and bending moment equations for sections AB, BC and CD by

taking point A as the origin.

b) Draw complete shear force and bending moment diagrams for the beam and indicate all

the important points in the diagrams.

Example 5:-

1 m 2m

B C

10 kN/m20 kN

DA

1 m

30kNm


21/60



Draw the shear force and bending moment diagrams for the beam shown below

Example 6:-


22/60

kN40.4

06.17)4(2)6)(2(2

18;0

kN6.17

0)10()210)(4(263

24)6)(2(

2

1)4(8;0

y

yy

y

yA

A

AF

C

CM




23/60

kN.m6.1744.4

kN4.4m;4

004.4

kN4.4;0

4.4

04.4;0

4.4

04.4;0

m40

1

1keratan

1

M

Vx

M

Vx

xM

MxM

V

VF

x

AB

AB

AB

ABy




24/60

3

4

6

2

4

2

2

xw

x

w

kN.m16556.3510267.610667.010056.0

kN637.9267.610333.110167.0m10

kN.m6.17556.354267.64667.04056.0

kN607.3267.64333.14167.0m4

556.35267.6667.0056.0

043

14

3

4

2

1484.4;0

267.6333.1167.0

04

3

4

2

184.4;0

m10m4

23

2

23

2

23

222

22keratan

2

2

2

22

2

M

Vx

M

Vx

xxxM

Mxxx

xxM

xxV

Vxx

F

x

BC

BC

BC

BCy




25/60


26/60

0196142814

014228m14

kN.m16196102810

kN810228m10

19628

02

10102106.17862

2

1484.4

;0

228

01026.17622

184.4

;0

m14m10

2

2

2

3

33333

keratan

3

3

3

M

Vx

M

Vx

xxM

Mx

xxxxx

M

xV

Vx

F

x

CD

CD

CD

CD

y




27/60

A B C

80 kNm

4m 4m

20 kN120 kN/m



A simply supported beam ABC is subjected to a point load at point B and varyingdistributed load along BC, as shown in figure below.a) Write the shear force and bending moment equations for sections AB and BC by

taking point A as the origin.b) Draw complete shear force and bending moment diagrams for the beam and

indicate all the important points in the diagrams.c) Determine the maximum bending moment and its location

Example 7:-


28/60

Bending Stress

(Introduction)

The relations between the bending

moments and the bending stresses

Assumptions:-

The cross section of the beam remain

plane after bending

Homogeneous material

Modulus elasticity, E for tension and

compression is identical

Beam straight with constant cross-

section


29/60

In general, RE

yor

R

Ey

(m)curvatureofradiusR

)(N/melasticityofmodulusE

(m)surfaceneutralfromdistancey

)(N/msectioncrossbeamthetonormalstressorstressbending

2

2

R

Ey

E

R

y

RRyR

R

RyR

x

xs

s

ss

ABAB

AB

'

'

Bending Stress

(Simple bending theory)


30/60

Bending Stress


)(mareaofmomentsecondI

(m)surfaceneutralfromdistancey

m)(NmomentbendingM

)(N/mstressbending

4

2

R

E

I

M

R

EI

dAyR

E

ydAR

Ey

ydAM

2

yR

E

I

M

I

My

Combine with previous derivation,

Bending stress


31/60

Bending Stress


NAc1

c2

I

McI

Mc

c

t

2max

1max

maxt

maxc

c1

c2


32/60

Bending Stress

(Second moment of area)

883

3

1

33

2

2

3

2

2

2

2

2

2

ddb

yb

dyby

dAyI

d

d

d

d

d

d

z

or

12

3bd

Iz

A B

C D

dAb

d

dy

z

+d/2

-d/2

0

y

y


33/60

Bending Stress

(Second moment of area)

4

4

4

4

2sin

2

sin

sin

sin

4

0

2

0

2

00

32

2

00

22

2

00

2

2

o

r

r

r

r

z

r

r

drrd

drdrr

drdrr

dAyI

o

o

o

o

or64

4d

Iz


34/60

Bending Stress

(Parallel Axes Theorem )

2'

22

2

2

'2'

'

AhI

dAhdAyhdAy

dAhy

dAyI

x

x

INA=Ixx+Ah2

h= the distance of the centroid of each section to

the neutral axisA = the area of each sectionIxx = the second moment of area of each section


35/60

Bending Stress

(Examples of Question)

Determine the maximum bending stress in tension and compression for thebeam loaded as shown in Figure (a) and its cross section as shown inFigure (b).

Example 1:-

Figure (a)

Figure (b)

Figure (b)


36/60

mm383000

10114 3

A

AyY

3

3

3

32

101143000104220120030402

109050180090201

mm,mm,mmArea,

AyA

Ayy

Bending Stress



37/60

Bending Stress


49-43

23

12123

121

23

1212

m10868mm10868

18120040301218002090

NA

NA

I

hAbdhAII

4-7 m108.68NAI


38/60

4-7

m108.68NAI

top

layer

For M = + 17.6 kN.m:

MPa77110868

038.0106.179

3

tension

MPa446

10868

022.0106.179

3

ncompressio

bottomlayer

For M = -16 kN.m:

MPa40610868

022.010169

3

tension

MPa70010868

038.01016 9

3

ncompressio bottomlayer

toplayer

MPa700

MPa771

max

max

comp

tension


39/60

Bending Stress


Determine the bending stress at point A, B and C on the cross section forthe beam which is subjected to a pure moment 2kNm at the end of the beamas shown in this figure.

Example 2:-

M=2kNm

2m

20mm 20mm

20mm

60mm

80mm

A

B

C


40/60

(b)

20mm

40mm

60mm

20mm

20mm

100mm

Figure 1

(a)

P

2m

Bending Stress


A cantilever beam with a cross section shown in Figure 1(b) is subjected to aforce, Pas shown in Figure 1(a). Determine:-a) Neutral axis of the cross sectionb) Second moment of area of the beamc) The maximum magnitude of P so that the maximum tension and maximum

compression in the beam do not exceed 70MPa and 50MPa respectively

Example 3:-


41/60

Bending Stress

(Composite Beams)

Composite beams are made from

different materials in order to

efficiently carry loads

Application of the flexural formularequires the material to be

homogeneous

Cross section of the beam must

be transformed into a single

material if this formula is to be

used to compute the bending

stress

Steel plates

M

Wood


42/60

Bending Stress

(Composite Beams)

Consider a composite beam to be

made of two materials 1 and 2 which

have the cross sectional areas as shown

If a bending moment is applied to this

beam, like one that is homogeneous,the total cross sectional area will

remain plane after bending, hence

normal strain will vary linearly

In order to apply the normal flexural

formula, the beam needs to be

transformed to a single material


43/60

Bending Stress

(Composite Beams)

The beam is to be transformed into aless stiff material 2

The height, h of the beam remains

the same, since the strain

distribution must be preserved

The upper portion of the beammust be widened in order to carry

loads equivalent to that carried by

the stiffer material 1

To determine the new width:-

dyndzEdydzE

dyndzEdAdF

dydzEdAdF

..

.'''

.

21

2

1

2

1

E

En Transformation factor


44/60

Bending Stress

(Composite Beams)

The normal stress distribution over thetransformed cross section will be linear

The neutral axis and the second

moment of area for the transformed

area can be determined and the flexure

formula applied in the usual manner to

determine the stress at each point on

the transformed beam

For the transformed material, the

stress found is to be multiplied by the

transformation factor, n

'

'

''

n

ndzdydzdy

dAdAdF


45/60

Bending Stress

(Composite Beams)

Examples of composite beams:-


46/60

20mm

40mm

60mm

20mm

20mm

100mm

D = 15mm



The weightless beam, ABCD, is supported by a pin support at B, and a roller bearing at C.There is a P1 kN/m UDL along the whole beam, as shown. There is an P2 kNm clockwisecouple at A. The cross section of the beam is also shown below.

a) Draw the shear force and bending moment diagrams and indicate all the important

pointsb) Calculate the maximum bending stress in tension and compression for the beam

Assignment 2:-

P2kNm

UDL P1kN/m

1.8m P3m 2.2m

A B C D

Group 1P1= 4kN/mP2= 11.2kNm

P3= 2 m

Group 2P1=10kN/mP2= 13kNm

P3= 2.5 m

Group 3P1= 5kN/mP2= 10.5kNm

P3=1 m

Group 4P1= 7.5kN/mP2= 8.5kNm

P3= 3 m

h


47/60

Shear Stress in Beams

(Introduction)

The presence of shearing stress on longitudinal planes

h i


48/60


(Introduction)

Method of analysis will be

limited to the beam with:-

Prismatic cross section

Homogeneous material

Behaves in a linear-

elastic manner

Effect of wrapping is assumed

to be small for slender beams

Sh S i B


49/60


(Shear formula)

Sh S i B


50/60


(Shear formula)

Sh St i B


51/60


(Shear formula)

0

;0

dAdAH

F

Dc

x

QI

M

ydAI

MM

dAI

yM

I

yM

dAH

cy

yy

CD

CD

CD

1

areaofmomentfirst

asknownalso1

cy

yy

CD

ydAQ

MMM

where:

:0If x

I

VQq

I

Q

dx

dM

dx

dH

I

Q

x

M

x

H

(N/m)flowshearasknowndx

dHq

(N)forceshearisdxdMV

)(msectioncrossfor the

areaofmomentsecond

)(myaboveareaofmomentfirst

4

3

1

I

Q

Shear flow

Sh St i B


52/60

xt

x

I

VQ

A

xq

A

H

It

VQShear stress


(Shear formula)

Sh St i B


53/60


(Shear formula)

Consider the beam width, b and height, h

The distribution of the shear stress throughout the cross section can be determined by

computing the shear stress at an arbitrary height, yfrom the neutral axis

Applying shear stress formula:-

Vbc

yc

bbh

ycb

V

It

VQ

yc

b

ycycbyAQ areashaded

3

22

3

22

22

4

3

12

2

22

1

''

If the cross sectional area is A = b(2c),

2

2

2

22

12

3

22

3

c

y

A

VV

cbc

yc

For layer rs, y= c:

01

2

32

2

min

c

c

A

V

For layer at neutral axis, y= 0:

A

V

cA

VNA

2

301

2

32max

Sh St i B


54/60


(Shear formula)

Rectangular cross section Circular cross section

maxmax

A

V

2

3max

A

V

3

4max

Sh St i B


55/60



The beam shown below is made of wood and is subjected to a resultantinternal vertical shear force of V= 3 kN. Determine:-a) the shear stress in the beam at point Pb) the maximum shear stress in the beam.

Example 1:-

Sh St i B


56/60



46

33

1026.1612

125.01.012

mbhI

351075.1805.02

10125.01.005.0'' myAQ

MPa346.01.01028.16

1075.181036

53

It

VQPat

351053.190625.02

11.00625.0'' myAQNA

MPa360.01.01028.16

1053.19103 6

53

maxItVQ

MPa360.0125.01.02

1033 3

maxIt

VQor



57/60



The beam shown below is made from two boards. Determine the maximumshear stress in the glue necessary to hold the boards together along theseam where they are joined. The supports at Band Cexert only verticalreactions on the beam.

Example 2:-



58/60



m12.0

15.003.00.030.15

075.015.003.00.1650.030.15y

46

23

23

1027

12.0075.015.003.012

15.003.0

12.0165.003.015.012

03.015.0

m

INA

33102025.0

015.012.018.003.015.0

''

m

yAQD



59/60



MPa88.403.01027

102025.0105.196

33

It

VQ



60/60

A timber beam 4m long is simply supported at its ends and carries a uniformlydistributed load of 8kN/m over its entire length. If the beam has the cross sectionshown below, determine:-a) The maximum horizontal shearing stress in the glued joints between the web

and flanges of the beam

b) The maximum horizontal shearing stress in the beamc) Sketch the distribution of the shearing stress along the cross section

Example 3:-



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