4-Bending of Straight Beams.pdf

CHAPTER 4

BENDING OF STRAIGHT BEAMS

FACULTY OF MECHANICAL ENGINEERING

DIVISION OF ENGINEERING MECHANICS

V{tÑàxÜ GV{tÑàxÜ G

Bending of Straight Beams

MEC411 – MECHANICS OF MATERIALS Ch 4 - 1

1. Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek “ Mechanics of Materials”

5th Edition in SI units

2. R.C.Hibbeler “ Mechanics of Materials “ Seventh Edition

Materials for this chapter are taken from :

CHAPTER 4




Internal forces in any cross section are

equivalent to a couple. The moment of the

Symmetric Member in Pure Bending

equivalent to a couple. The moment of the

couple is the section bending moment.

From statics, a couple M consists of two equal

and opposite forces.

The sum of the components of the forces in any

direction is zero.


The moment is the same about any axis

perpendicular to the plane of the couple and

zero about any axis contained in the plane.

CHAPTER 4




These requirements may be applied to the

sums of the components and moments of the

∫ =−=

∫ ==∫ ==

MdAyM

dAzM

dAF

xz

xy

xx

σ

σ

σ

0

0

statically indeterminate elementary internal

forces.


CHAPTER 4




Beam with a plane of symmetry in pure bending:

member remains symmetric

bends uniformly to form a circular arc

Bending Deformation

bends uniformly to form a circular arc

cross-sectional plane passes through arc center and

remains planar

length of top decreases and length of bottom increases

a neutral surface must exist that is parallel to the upper

and lower surfaces and for which the length does not


and lower surfaces and for which the length does not

change

stresses and strains are negative (compressive) above

the neutral plane and positive (tension) below it

CHAPTER 4




Consider a beam segment of length L.

Strain Due to Bending

After deformation, the length of the neutral

surface remains L. At other sections,

( )( )

x

cc

yy

L

yyLL

yL

ρρθθδ

ε

θρθθρδθρ

−=−==

−=−−=−′=

−=′

linearly) ries(strain va


mx

m

m

c

y

cρ

c

εε

ερε

−=

== or

CHAPTER 4




Stress Due to Bending

For a linearly elastic material,

mxx Ec

yE εεσ −==

linearly) varies(stressmc

y

c

σ−=

For static equilibrium,

∫

∫∫

−=

−===

dAyc

dAc

ydAF

m

mxx

σ

σσ

0

0

For static equilibrium,

( ) ( )

IdAyM

dAc

yydAyM

mm

mx

==

−−=−=

∫

∫∫σσ

σσ

2


c

First moment with respect to

neutral plane is zero. Therefore,

the neutral surface must pass

through the section centroid.

I

My

c

y

S

M

I

Mc

cdAy

cM

x

mx

m

−=

−=

==

== ∫

σ

σσ

σ

ngSubstituti

CHAPTER 4




Beam Section Properties

The maximum normal stress due to bending,

==S

M

I

Mcmσ

modulussection

inertia ofmoment section

==

=

c

IS

I

SI

A beam section with a larger section modulus

will have a lower maximum stress

Consider a rectangular beam cross section,


Ahbhh

bh

c

IS

613

61

3121

2====

Between two beams with the same cross sectional

area, the beam with the greater depth will be more

effective in resisting bending.

CHAPTER 4




Example 4.1

A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165

GPa and neglecting the effects of fillets, determine (a) the maximum tensile and

compressive stresses, (b) the radius of curvature.


CHAPTER 4




Based on the cross section geometry,

calculate the location of the section

centroid and moment of inertia.

Solution

∑ ×==∑

×=×

×=×

3

3

3

32

101143000

104220120030402

109050180090201

mm ,mm ,mm Area,

AyA

Ayy

mm 383000

10114 3

=×

=∑∑=

A

AyY


3000∑ A

( ) ( )( ) ( )

49-43

23

12123

121

23

1212

m10868 mm10868

18120040301218002090

×=×=

×+×+×+×=

+=+= ∑∑′

I

dAbhdAIIx

CHAPTER 4




Apply the elastic flexural formula to find the

maximum tensile and compressive stresses.

=Mc

mσ

49

49

m10868

m038.0mkN 3

m10868

m022.0mkN 3

−

−

×

×⋅−=−=

×

×⋅==

=

I

cM

I

cM

I

BB

AA

m

σ

σ

σ

MPa 0.76+=Aσ

MPa 3.131−=Bσ

Calculate the curvature


Calculate the curvature

( )( )49- m10868GPa 165

mkN 3

1

×

⋅=

=EI

M

ρ

m 7.47

m1095.201 1-3

=

×= −

ρρ

CHAPTER 4




Example 4.2

The simply supported beam has the cross-sectional area as shown.

Determine the absolute maximum bending stress in the beam and draw

the stress distribution over the cross section at this location.the stress distribution over the cross section at this location.


CHAPTER 4




The maximum internal moment in the beam is kNm 5.22=M

SolutionSolutionSolutionSolution: Example 4.2

Solution

By symmetry, the centroid C and thus the neutral axis pass through the mid-

height of the beam, and the moment of inertia is


( )( )( ) ( )( )( ) ( )( )

( ) 46

323

2

m 103.301

3.002.012

116.0002.025.002.025.0

12

12

−=

+

+=

+=∑ AdII

CHAPTER 4




Applying the flexure formula where c = 170 mm,

( )( ) (Ans) MPa 7.12103.301

17.05.22 ;

6maxmax === −σσI

Mc

Example 4.2

( ) (Ans) MPa 7.12103.301

;6maxmax === −σσ

I


CHAPTER 4




Transverse Shear

Beams generally support both shear and

moment loading. Shear, v is the result of a

transverse shear-stress distribution that acts

over the beam’s cross section.

Due to complementary property of shear,

associated longitudinal shear stresses also act


along longitudinal planes of beam.

CHAPTER 4




If a typical element removed from the interior

point on the cross section it will be subjected to

both transverse and longitudinal shear stress.

Transverse Shear

both transverse and longitudinal shear stress.

For transverse shear, shear-strain distribution

throughout the depth of a beam cannot be easily

expressed mathematically. Thus, we need to

develop the formula for shear stress indirectly

using the flexure formula and relationship


using the flexure formula and relationship

between moment and shear ;

V = dM/dx

CHAPTER 4




The Shear Formula

The final result is therefore

VQ

Itτ =

ItHere

τ = shear stress in member at the point located a distance y’ from the

neutral axis. Assumed to be constant and therefore averaged across

the width t of member.

V = internal resultant shear force, determined from method of sections

and equations of equilibrium

I = moment of inertia of entire cross sectional area computed about the


neutral axis

t = width of the member’s cross sectional area, measured at the point

where τ is to be determined

Q = ∫A’ y dA’ = y’A’, where A’ is the top (or bottom) portion of member’s

cross sectional area, defined from section where t is measured, and y’

is distance of centroid of A’, measured from neutral axis

CHAPTER 4




Example 4.3

The beam shown in Fig. below is

made from two boards. Determine the

maximum shear stress in the glue

Internal shear: Support reactions and

shear diagram for beam are shown

below. Maximum shear in the beam is

necessary to hold the boards together

along the seams where they are

joined. Supports at B and C exert only

vertical reactions on the beam.

below. Maximum shear in the beam is

19.5 kN.

Free Body Diagram


Shear Diagram

CHAPTER 4




Section properties: The centroid andtherefore the neutral axis will bedetermined from the reference axisplaced at bottom of the cross sectional

The top board (flange) is beingheld onto the bottom board (web)by the glue, which is applied overthe thickness t = 0.03m.

Example 4.3

placed at bottom of the cross sectionalarea. Working in units of meters, wehave,

[ ]( )( ) [ ]( )( )( )( ) ( )( )

0.075 0.150 0.03 0.165 0.03 0.15

0.15 0.03 0.03 0.150

0.120

yAy

A

m m m m m m

m m m m

m

=

+=

+

=

∑∑

the thickness t = 0.03m.Consequently A’ is defined as thearea of the top board, we have

( )( )( )3 3

' '

0.18 0.015 0.12 0.03 0.15

0.2025 10

Q y A

m−

=

= − −

= ×


0.120m=

( )( ) ( )( )( )

−+= 23075.0120.0030.0150.0150.003.0

12

1mmmmmmI

( )( ) ( )( )( )

( )

3 2

6 4

10.150 0.030 0.030 0.150 0.165 0.120

12

27.0 10

m m m m m m

m−

+ + −

=

CHAPTER 4




Shear stress: Using above data, and

applying shear formula yields,

VQτ =

( )( )( ) ( )

max

3 3

6 4

19.5 0.2025 10

27.0 10 0.030

4.88 .

It

k" m

m m

MPa Ans

τ

−

−

=

=

=

Shear stress acting at top of bottom

board is shown in the next figure. It is


board is shown in the next figure. It is

the glue’s resistance to this lateral or

horizontal shear stress that is necessary

to hold the boards from slipping at

support C.

CHAPTER 4




Example 4.4

The beam is made of wood and is subjected to a resultant

internal vertical shear force of V = 3 kN. (a) Determine the

shear stress in the beam at point P, and (b) compute theshear stress in the beam at point P, and (b) compute the

maximum shear stress in the beam.

Solution

( )( ) 4633 mm 1028.1612510012

1

12

1×=== bhI

The moment of inertia of the cross sectional

area computed about the neutral axis is


( ) ( )( ) 34 mm 1075.1810050502

1125' ×=

+== AyQ

1212

Applying the shear formula, we have

( )( )( )( )

(Ans) MPa 346.01001028.16

1075.1836

4

=×

×==

It

VQpτ

CHAPTER 4




Maximum shear stress occurs at the neutral axis, since t is constant throughout

the cross section,

Example 4.4

( )( )( )( )

(Ans) MPa 360.01001028.16

1053.1936

4

max =×

×==

It

VQτ

( )( ) 34 mm 1053.195.621002

2.65'' ×=

== AyQ

Applying the shear formula yields


( )( )1001028.16 ×It

CHAPTER 4




Deflection of Beam

When a beam with a straight longitudinal

axis is loaded by lateral forces, the axis is

deformed into a curve, called the deflection

curve of the beam. Deflection is thecurve of the beam. Deflection is the

displacement in the y-direction of any point

on the axis of the beam.

The calculation of deflections is an

important part of structural analysis and

design.

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CHAPTER 4




Deflections are essential for example in the

analysis of statically indeterminate structures

and in dynamic analysis, as when investigating

Deflection of Beam

and in dynamic analysis, as when investigating

the vibration of aircraft or response of buildings

to earthquakes.

Deflections are sometimes calculated in order to verify that they are within

tolerable limits.

Use various methods to determine the Use the various methods to solve


Use various methods to determine the

deflection and slope at specific pts on

beams and shafts:

� Integration method

� Method of superposition

� Moment-area method

Use the various methods to solve

for the support reactions on a

beam or shaft that is statically

indeterminate

CHAPTER 4




For small curvatures:

1

ds dx Rdi

di

= =

Basic Differential Equation for Deflection

1

dydx

di

dx R

But i

=

=

Therefore ( )2

2

1................... 1

d ydx R

=

Now, for simple bending theory:

1 1M M


1 1M M

I R R EI= → =

Therefore subs. Eqn. (1) :

( )2

2...................... 2

d yM EI

dx

=

CHAPTER 4




Macaulay’s Method

The Macaulay’s method involves the general method of obtaining slopes and

deflections (i.e. integrating the equation for M) will still apply provided that the

term, W (x – a) is integrated with respect to (x – a) and not x.term, W (x – a) is integrated with respect to (x – a) and not x.

Such functions can be used to describe

distributed loadings, written generally as:

{( ){0 for

for

0

n

n

x a x a

x a x a

n

< − > = <

− ≥

≥


Consider a simple beam problem as

shown and try to develop the expression

for maximum deflection??..

CHAPTER 4




Bending Moment:

w l= − < − >

General equation becomes:

3 2wx w wL x

Example of Application

Macaulay’s Method

2

2

2 2

3 3

2 2

2 2

4 4 2

12 6 2

x

w lM x w x

d y w lEI x w x

dx

dy w w lEI x x A

dx

w w lEIy x x Ax B

= − < − >

= − < − >

= − < − > +

= − < − > + +

3 23

212 6 16

Lwx w wL x

EIy x= − − −

Maximum deflection occurs at x=L/2

( ) ( )3 232 2

2 2

3

12 6 16

.

L L

L Lw wLw

EIy

wLy Ans

= − − −

= −


12 6 2

Boundary conditions:

2

0, 0 0

, 016

at x y B

wLat x L y A

= = → =

= = → = −

max .48

wLy Ans

EI= −

CHAPTER 4




Example 4.5

Determine the deflection of the beam at x = 3 (middle of beam) if E = 210 kN/mm2.

The cross-section is given below.


CHAPTER 4




Bending Moment: 25

15 36 1x

M x x= − + − −


at 1, 0;x y= =

Example 4.5

Solution

2 2

2

22 3

33 4

515 36 1

2

515 36 1

2

15 36 51

2 2 6

15 36 51

6 6 24

x

xM x x

d y xEI x x

dx

dy xEI x x A

dx

xEIy x x Ax B

= − + − −

= − + − −

= − + − − +

= − + − − + +

( )

( )

at 1, 0;

0 2.5 0.21

2.71 1

at 6, 0;

0 540 750 272.16 6

6 60 2

solve :

11.45

x y

A B

A B

x y

A B

A B

A

= =

→ = − − + +

→ + =

= =

→ = − + − + +

→ + =

=

⋯⋯

⋯⋯


11.45

8.75

A

B

=

= −


33 42.5 6 1 0.2083 11.46 8.75EIy x x x x=− + − − + −

CHAPTER 4




Deflection occurs at x = 3

10.74EIy = −

Example 4.5

Moment of inertia about the Neutral Axis

( )6 42.46 10 m calculate!!I −= ×

Deflection at x = 3

0.02079m .y = = 20 79mm


CHAPTER 4




Example 4.6 Uniform load not starting at the beginning


Determine the deflection of the beam at x = 3 (middle of beam) if E = 210

kN/mm2.

CHAPTER 4




Bending Moment Equation

2515 30.5 1 1

2xM x x x= − + − − −


at 1, 0;

0 2.5

x y

A B

= =

→ = − + +

Example 4.6

22

2

2 32

3 43

2

515 30.5 1 1

2

15 30.5 51 1

2 2 6

15 30.5 51 1

6 6 24

d yEI x x x

dx

dyEI x x x A

dx

EIy x x x Ax B

= − + − − −

= − + − − − +

= − + − − − + +

( )

( )

0 2.5

2.5 1

at 6, 0;

0 540 635.42 130.21 6

6 34.79 2

solve :

6.46

A B

A B

x y

A B

A B

A

→ = − + +

→ + =

= =

→ = − + − + +

→ + =

=

⋯⋯

⋯⋯


6.46

3.96

A

B

=

= −


( ) ( )3 432.5 5.083 1 0.2083 1 6.46 3.96EIy x x x x= − + − − − + −

CHAPTER 4




Deflection occurs at x = 3 (midspan)

314.75EIy k"m= −

Example 4.6

( )6 6

14.75

14.75

210 10 246 10

.

EIy k"m

y−

= −

−=

× ×

= 28 55mm


CHAPTER 4




Example 4.7 Uniform load not reaching end of beam.




CHAPTER 4




� Since the 5 kN/m load did not reach theend, (x-1) does not represent the actualloading. The equivalent loading is:

Bending Moment Equation:

( ) ( ) ( )

( ) ( )

115 30 1 5 1

2

55 5

x

xM x x x

xx

−= − + − − −

−+ −( ) ( )

2

2

2 32

5 52

115 30 1 5 1

2

55 5

2

15 30 51 1

2 2 6

5

x

xd yEI x x x

dx

xx

dyEI x x x

dx

+ −

−= − + − + −

−− −

= − + − − −


3

3 43

4

55

6

15 30 51 1

6 6 24

55

24

x A

EIy x x x

x Ax B

+ − +

= − + − − −

+ − + +Continue as usual to

obtain y at 3m

CHAPTER 4




Supplementary Problems 4

1. The wide-flange beam shown below is made of

high-strength, low-alloy steel for which σy = 345

MPa and σu = 450 MPa. Using a factor of safety of CMPa and σu = 450 MPa. Using a factor of safety of

3.0, determine the largest couple that can be applied

to the beam when it is bent about the z axis. Neglect

the effect of fillets.

Mz

C

2. Knowing that for the extruded beam shown the


2. Knowing that for the extruded beam shown the

allowable stress is 120 MPa in tension and 150

MPa in compression, determine the largest

couple M that can be applied.

CHAPTER 4




3. (a) Using an allowable stress of 120 MPa,

determine the largest couple M that can be

applied to a beam of the cross section shown. (b)


applied to a beam of the cross section shown. (b)

Solve part a, assuming that the cross section of

the beam is an 80 mm square.

4. For the beam and loading shown, consider section n-n and determine (a) the largest

shearing stress in that section, (b) the shearing stress at point a.


a

CHAPTER 4




5. A simply supported beam carries a uniform

distributed loading of intensity 2.5 kN/m over


entire span of 5 m. The cross-section of the beam

is a T-section having the dimensions as shown.

Find the maximum shear stress for the section of

the beam.

6. For the beam and loading shown, determine (a)


6. For the beam and loading shown, determine (a)

the deflection at end A, (b) the deflection at point

C, (c) the slope at end D.

CHAPTER 4




7. For the beam and loading shown,

determine (a) the slope at end A, (b) the


deflection at the midpoint C. Use E = 200

GPa and I = 129 x 10-6 m4.

8. The wooden beam is subjected to the load shown. Determine the equation of elastic

curve. If E = 12 GPa, determine the deflection and the slope at end B.


CHAPTER 4





8. Determine the deflection of the beam at a distance of 9 m from point A. Take E =

200 GPa and I = 129 x 10-6 m4 .

8 m 4 m

4 m 2 m 2 m

50 kN/m

90 kN

60 kN/m

AB


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4-Bending of Straight Beams.pdf