7 pure bending 2015 bach khoa

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7.0 INTRODUCTION

Pure Bending: Prismatic members subjected

to equal and opposite couples acting in the same longitudinal plane

OTHER LOADING TYPES

• Eccentric Loading: Axial loading which

does not pass through section centroid produces internal forces equivalent to an axial force and a couple

• Transverse Loading: Concentrated or

distributed transverse load produces internal forces equivalent to a shear force and a couple

• Principle of Superposition: The normal

stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state

of stress

dA z

M

dA F

x z

x y

x x







0 0

• These requirements may be applied to the sums

of the components and moments of the statically indeterminate elementary internal forces

• Internal forces in any cross section are 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

BENDING DEFORMATIONS

• 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 change

• stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it

Beam with a plane of symmetry in pure bending:

• member remains symmetric

7.1 BENDNG DEFORMATIONS

STRAIN DUE TO BENDING

Consider a beam segment of length L

After deformation, the length of the neutral

surface remains L At other sections,

 

 

m x

m m

x

c y

c ρ c

y y

L

y y

L L

y L

linearly) ries

(strain va

STRESS DUE TO BENDING

• For a linearly elastic material,

linearly) varies

(stress

m

m x

x

c y

E c

y E

dA c

y dA

F

m

m x

First moment with respect to neutral

plane is zero Therefore, the neutral

surface must pass through the

section centroid

• For static equilibrium,

I My

c y S

M I

Mc

c

I dA

y c M

dA c

y y dA

y M

x

m x

m

m m

m x

2

7.2 STRESS DUE TO PURE BENDING

BEAM SECTION PROPERTIES

• The maximum normal stress due to bending,

modulus section

inertia of

moment section

I

S

M I

Mc m

h

bh c

I S

6 1 3 6 1

3 12 1

• Structural steel beams are designed to have a large section modulus

DEFORMATIONS IN A TRANSVERSE CROSS SECTION

• Deformation due to bending moment M is quantified by the curvature of the neutral surface

EI M

I

Mc Ec Ec

c

m m

7.2 STRESS DUE TO BENDING

PROPERTIES OF AMERICAN STANDARD SHAPES

EXAMPLE 7.01

A cast-iron machine part is acted upon

by a 3 kN-m couple Knowing E = 165

GPa and neglecting the effects of fillets

(đường gờ cong), determine (a) the

maximum tensile and compressive

stresses, (b) the radius of curvature

SOLUTION:

• Based on the cross section geometry, calculate the location of the section centroid and moment of inertia

A

A y

7.2 STRESS DUE TO PURE BENDING

3 2

10 114 3000

10 4 2 20

1200 30

40 2

10 90 50

1800 90

20 1

mm mm

mm Area,

A y A

A y y

4 9 - 3

2 3

12 1 2

3 12

1

2 3

12 1 2

m 10 868 mm

10 868

18 1200 40

30 12

1800 20

d A I

I x

EXAMPLE 7.01

• Apply the elastic flexural formula to find the maximum tensile and compressive stresses

4 9

4 9

mm 10

868

m 038 0 m kN 3

mm 10

868

m 022 0 m kN 3

c M I Mc

B B

A A

76

131

165

m kN 3 1

m 10 95 20

7.2 STRESS DUE TO PURE BENDING

TECHNICAL EXPRESSION

y I

x

c I

M

max max 



n compressio x

x

c I

M

max min  



 tension

tension x

x

c I

compressio

c

tension tension

c

ymax  max

7.3 COMPOSITE SECTION

BENDING MEMBERS MADE OF SEVERAL ATERIALS

• Consider a composite beam formed from

two materials with E 1 and E 2

• Normal strain varies linearly

• Elemental forces on the section are

dA y E dA

dF dA

y E dA

1 1

1

2 1

1 2

E

E n dA

n y E dA

y nE

x

n I

EXAMPLE 7.02

Bar is made from bonded pieces of

steel (E s = 29x106 psi) and brass

(E b = 15x106 psi) Determine the

maximum stress in the steel and

brass when a moment of 40 kip*in

• Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of elasticity

1 3 12

1

in 063 5

in 3 in.

25 2

933

1 psi 10 15

psi 10 29

6 6

b E

E n

• Calculate the maximum stresses

ksi 85

11 in

5.063

in 5 1 in kip 40

m b

REINFORCED CONCRETE BEAMS

• Concrete beams subjected to bending moments are reinforced by steel rods

• In the transformed section, the cross sectional area

of the steel, A s , is replaced by the equivalent area

2 2

b

x d A n

x bx

s s

s

• The normal stress in the concrete and steel

x s

x c

x

n I

7.3 COMPOSITE SECTION

REINFORCED CONCRETE BEAMS – EXAMPLE 7.03

A concrete floor slab is reinforced with

5/8-in-diameter steel rods The modulus

of elasticity is 29x106psi for steel and

3.6x106psi for concrete With an applied

bending moment of 40 kip*in for 1-ft

width of the slab, determine the maximum

stress in the concrete and steel

• Calculate the maximum stresses

in the concrete and steel

REINFORCED CONCRETE BEAMS – EXAMPLE 7.03

6 6

in 95 4 in

2 06 8

06

8 psi 10 6 3

psi 10 29

nA

E

E n

• Evaluate the geometric properties of the transformed section

 

  3  2  2 4 3

1 12 in 1 45 in 4 95 in 2 55 in 44 4 in

in 450 1 0

4 95 4 2 12

x x

• Calculate the maximum stresses

4 2

4 1

in 44.4

in 55 2 in kip 40 06 8

in 44.4

in 1.45 in

kip 40

I Mc



s



7.4 STRESS CONCENTRATIONS

STRESS CONENTRATIONS

Stress concentrations may occur:

• in the vicinity of points where the

loads are applied

I

Mc K

m 



• in the vicinity of abrupt changes

in cross section

PLASTIC DEFORMATIONS

• For any member subjected to pure bending

m x

c

y

   strain varies linearly across the section

• If the member is made of a linearly elastic material,

the neutral axis passes through the section centroid

7.5 PLASTIC ANALYSIS

PLASTIC DEFORMATIONS

• When the maximum stress is equal to the ultimate strength of the material, failure occurs and the corresponding moment M U is referred to as the

ultimate bending moment

• The modulus of rupture in bending, R B, is found from an experimentally determined value of M U

and a fictitious linear stress distribution

I

c M

R B  U

• R B may be used to determine M U of any member made of the same material and with the same cross sectional shape but different

dimensions

MEMBERS MADE OF AN ELASTOPLASTIC MATERIAL

• Rectangular beam made of an elastoplastic material

moment elastic

maximum

Y m

m Y

x

c

I M

I Mc

thickness -

half core elastic 1

2

2 3

1 2

M

• In the limit as the moment is increased further, the elastic core thickness goes to zero, corresponding to a fully plastic deformation

shape) section

cross

on only (depends

factor shape

moment plastic

2 3

Y p

M

M k

M M

R R



1

2 1



• The neutral axis cannot be assumed to pass through the section centroid

RESIDUAL STRESSES

• Plastic zones develop in a member made of an elastoplastic material if the bending moment is large enough

• Since the linear relation between normal stress and strain applies at all points during the unloading phase, it may be handled by assuming the member

7.5 PLASTIC ANALYSIS

EXAMPLE 7.04, 7.05

A member of uniform rectangular cross section is subjected to a bending moment M = 36.8 kN-m The member is made of an elastoplastic material with a yield strength of 240 MPa and a modulus

of elasticity of 200 GPa

Determine (a) the thickness of the elastic core, (b) the radius of curvature of the neutral surface After the loading has been reduced back to zero, determine (c) the distribution of residual stresses, (d) radius of curvature

EXAMPLE 7.04, 7.05

m kN 8 28

MPa 240

m 10 120

m 10 120

10 60 10

50

3 6

3 6

2 3 3

3 2 2 3

c

I

M

m m

bc

c

I



• Maximum elastic moment:

• Thickness of elastic core:

666 0 mm 60

1 m kN 28.8 m

kN 8 36

1

2

2 3

1 2

3

2

2 3

1 2

Y

Y Y

y c

y

c y c

y M

M

mm 80

2y Y 

• Radius of curvature:

3 3

3

9 6

10 2 1

m 10 40

10 2 1

Pa 10 200

Pa 10 240

Y Y

Y Y

y y E





7.5 PLASTIC ANALYSIS

EXAMPLE 7.04, 7.05

• M = 36.8 kN-m

MPa 240

mm 40

2 MPa 7

306

m 10 120

m kN 8 36

10 5 177

m 10 40

10 5 177

Pa 10 200

Pa 10 5 35

core, elastic

the of edge

x x

y E



