Bending

Shear and Moment Diagram, Graphical method to construct shear

and moment diagram, Bending deformation of a straight member, The flexure formula

1

2

Shear and moment diagram

3

Axial load diagram

Torque diagram

Both of these diagrams show the internal forces acting on the members. Similarly, the shear and moment diagrams show the internal shear and moment acting on the members

Type of Beams

Statically Determinate

4

Simply Supported Beam

Overhanging Beam

Cantilever Beam

Type of Beams

Statically Indeterminate

5

Continuous Beam

Propped Cantilever Beam

Fixed Beam

6

Example 1

Equilibrium equation for 0 x 3m:

A B

* internal V and M should be assumed +ve

kNVVF

Fy

90

0

)(90

0

kNmxMMVx

M

M

VF

x

Shear Diagram

Lecture 1 8

Sign convention: V= -9kN

M

VF

x

Shear Diagram

9

Sign convention: M= -9x kNmX=0: M= 0X=3: M=-27kNm

M=-9x

M = -9x kN.m

V = -9 kNF

x

10

V=9kN

M=X

At cross section A-A

X

At section A-A

Example 2

11

kN5

01050F

kN5

0)2()1(10;0

y

y

y

y

yA

A

A

C

CM

1) Find all the external forces

12

)(50

0

10

downkNVVF

F

mx

y

)(50

0

10

ccwkNmVxMVxM

M

mx

A

)(5010

0

21

upkNVVF

F

mx

y

)()510(10

0)1(10

021

ccwkNmxMVxM

MVx

M

mx

A

Force equilibrium

Force equilibrium

Moment equilibrium

Moment equilibrium

)(5)(5

10

ccwxMdownkNV

mx

)(510)(5

21

ccwxMupkNV

mx

M=5x M=10-5x

Boundary cond for V and M

Solve it

Draw the shear and moment diagrams for simply supported beam.

14

15

Distributed Load

16

For calculation purposes, distributed load can be represented as a single load acting on the center point of the distributed area.

Total force = area of distributed load (W : height and L: length)Point of action: center point of the area

Example

17

Example

18

Solve it

Draw the shear and moment diagrams the beam:

19

Solving all the external loads

kNWlF

48)6(8

Distributed load will be

Solving the FBD

012

364)3(480)3(4

0

xy

y

x

A

AkNA

kNB

FB

M

21

Boundary Condition 40 x

xVVx

FY

8120812

0

2

2

2

412

0412

04)812(04

0

xxM

xxM

xxxMxVxM

M A

Equilibrium eq

22

Boundary Condition 64 x

xVVx

FY

848083612

0

144484

0414448

04)4(36)848(

0)2/(8

0

2

2

2

xxM

xxM

xxxM

xxVxM

M A

Equilibrium eq

23

40 x

x=0 V= 12 kNx=4 V=-20 kN

xV 812

2412 xxM

x=0 M= 0 kNx=4 V=-16 kN

64 x xV 848

x=4 V= 16kNx=6 V= 0 kN

144484 2 xxM

x=4 V= -16kNx=6 V= 0 kN

Graph based on equations

24

y = c Straight horizontal line

y = mx + c y=3x + 3 y=-3x + 3

y=3x2 + 3 y=-3x2 + 3

y = ax2 + bx +c

25

Graphical method

26

xxwV

VVxxwV

Fy

0)(

:0

• Relationship between load and shear:

2

0:0

xkxwxVM

MMxkxxwMxVM o

• Relationship between shear and bending moment:

26

2727

Dividing by x and taking the limit as xà0, the above two equations become:

Regions of distributed load:

Slope of shear diagram at each point

Slope of moment diagram at each point

= distributed load intensity at each point

= shear at each point

)(xwdxdV

VdxdM

Example

28

29

30

dxxwV )( dxxVM )(

The previous equations become:

change in shear = Area under

distributed loadchange in moment = Area under shear

diagram

31

+ve area under shear diagram

32

33

Bending deformation of a straight member

34

Observation: - bottom line : longer - top line: shorter- Middle line: remain the same but rotate (neutral line)

35

Strain sss

s

'lim0

Before deformationxs

After deformation, x has a radius of curvature r, with center of curvature at point O’

r xs

Similarly r )(' ys

r

rrr

y

ys

)(lim0

Therefore

36

r c

maxMaximum strain will be

max

max

)(

)//(

rr

cy

cy

max)( cy

-ve: compressive state+ve: tension

The Flexure Formula

37

The location of neutral axis is when the resultant force of the tension and compression is equal to zero.

0FFR

NotingdAdF

A

A

A

ydAc

dAcy

dAdF

max

max)(

0

38

Since , therefore

Therefore, the neutral axis should be the centroidal axis

0max c

0A

ydA

39

ZZR MM )(

A

A

A A

dAyc

dAcyy

dAyydFM

2max

max)(

IMc

max

Maximum normal stress

Normal stress at y distance

IMy

Line NA: neutral axis Red Line: max normal stress c = 60 mmYellow Line: max compressive stress c = 60mm

IMc

max

IMc

max

Line NA: neutral axis Red Line: Compressive stress y1 = 30 mmYellow Line: Normal stress y2 = 50mm

IMy1

1

IMy2

2

Refer to Example 6.11 pp 289

I: moment of inertial of the cross sectional area

12

3bhI xx

644

44 DrI xx

Find the stresses at A and B

I: moment of inertial of the cross sectional area

Locate the centroid (coincide with neutral axis)

mm

AAAyAy

A

Ayy n

ii

n

iii

5.237)300)(50()300)(50()300)(50(325)300)(50(150

21

2211

1

1

I: moment of inertial of the cross sectional area

Profile I

4633

)10(5.11212

)300(5012

mmbhI I

A A

46

2623

)10(344.227

)5.87)(300)(50()10(5.11212

)(

mm

AdbhI AAI

I about Centroidal axis

I about Axis A-A using parallel axis theorem

Profile II

46

23

23

)10(969.117

)5.87)(50)(300(12

)50)(300(12

)(

mm

AdbhI AAII

Total I46

466

)10(313.345

)10(969.117)10(344.227

)()(

mm

mm

III AAIIAAIAA

* Example 6-12 to 6-14 (pp 290-292)

Solve it

If the moment acting on the cross section of the beam is M = 6 kNm, determine the maximum bending stress on in the beam. Sketch a three dimensional of the stress distribution acting over the cross sectionIf M = 6 kNm, determine the resultant force the bending stress produces on the top board A of the beam

46

32

3

)10(8.78612

)300(40])170)(40)(300(12

)40(300[2

mm

I I

Total Moment of Inertia

Max Bending Stress at the top and bottom

MPaII

McM top 45.1)190()10(6000 3

MPaM bottom 45.1

Bottom of the flange

MPaII

McM topf 14.1)150()10(6000 3

_

MPaM bottomf 14.1_

1.45MPa

1.14MPa

6kNm

Resultant F = volume of the trapezoid

1.45MPa1.14MPa

40 mm

300 mm

kN

NFR

54.15

15540)300)(40(2

)14.145.1(

Solve it

The shaft is supported by a smooth thrust load at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress on the shaft

Draw the shear and moment diagram

kNFkNF

F

M

A

D

D

A

33

)25.2(3)75.0(3)3(

0

External Forces

Absolute Bending Stress

Mmax = 2.25kNm

MPa

IMc

8.52

)2540(4

)40()10(225044

3

max