7 pure bending 2015 bach khoa

CHAPTER 7: PURE BENDING
7.0 Introduction
7.1 Bending deformation
7.2 Stress due to pure bending
7.3 Composite section
7.4 Stress concentration

7.5 Plastic analysis


7.0 INTRODUCTION

Pure Bending: Prismatic members subjected
to equal and opposite couples acting in the
same longitudinal plane


7.0 INTRODUCTION
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.


7.0 INTRODUCTION
SYMMETRIC MEMBERS IN PURE BENDING

• 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.
• These requirements may be applied to the sums
of the components and moments of the statically
indeterminate elementary internal forces.
Fx    x dA  0
M y   z x dA  0
M z    y x dA  M


7.1 BENDING DEFORMATIONS
BENDING DEFORMATIONS

Beam with a plane of symmetry in pure
bending:
• member remains symmetric
• 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


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,
L     y 

  L  L     y      y
x 
m 





L
c


y
c

y



or

 x   m


ρ

y


c

m

(strain varies linearly)

7.2 STRESS DUE TO PURE BENDING
STRESS DUE TO BENDING

• For a linearly elastic material,
y
c

 x  E x   E m
y
   m (stress varies linearly)
c

• For static equilibrium,
y
Fx  0    x dA     m dA
c


0   m  y dA
c

First moment with respect to neutral
plane is zero. Therefore, the neutral
surface must pass through the
section centroid.

• For static equilibrium,
 y

M    y x dA    y   m  dA

 c



 I
M  m  y 2 dA  m
c
c
m 

Mc M

I
S

y
Substituti ng  x    m
c

x  

My
I


7.2 STRESS DUE TO PURE BENDING
BEAM SECTION PROPERTIES
• The maximum normal stress due to bending,
Mc M


I
S
I  section moment of inertia
I
S   section modulus
c

m 

A beam section with a larger section modulus
will have a lower maximum stress
• Consider a rectangular beam cross section,
3
1
I 12 bh
S 
 16 bh3  16 Ah
c
h2

Between two beams with the same cross
sectional area, the beam with the greater depth
will be more effective in resisting bending.
• Structural steel beams are designed to have a
large section modulus.


7.2 STRESS DUE TO PURE BENDING
DEFORMATIONS IN A TRANSVERSE CROSS SECTION
• Deformation due to bending moment M is

quantified by the curvature of the neutral surface


1 Mc
 m  m 

c
Ec Ec I
M

EI
1

• Although cross sectional planes remain planar
when subjected to bending moments, in-plane
deformations are nonzero,
 y   x 

y


 z   x 

y


• Expansion above the neutral surface and
contraction below it cause an in-plane curvature,
1 
  anticlastic curvature

 


7.2 STRESS DUE TO BENDING
PROPERTIES OF AMERICAN STANDARD SHAPES


7.2 STRESS DUE TO PURE BENDING
EXAMPLE 7.01
SOLUTION:
• Based on the cross section geometry,
calculate the location of the section
centroid and moment of inertia.
Y 

 yA
A



I x   I  A d 2



• Apply the elastic flexural formula to
find the maximum tensile and
compressive stresses.
m 

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.

Mc
I

• Calculate the curvature
1





M
EI


7.2 STRESS DUE TO PURE BENDING
EXAMPLE 7.01
SOLUTION:

Based on the cross section geometry, calculate
the location of the section centroid and
moment of inertia.
Area, mm 2
1 20  90  1800
2 40  30  1200

 A  3000

y , mm
50
20

yA, mm3
90 103
24 103
3
 yA  114 10

3

 yA 114 10
Y 

 38 mm
3000
A



 

1 bh3  A d 2
I x   I  A d 2   12







1 90  203  1800 122  1 30  403  1200  182
 12
12

I  868 103 mm  868 10-9 m 4




7.2 STRESS DUE TO PURE BENDING
EXAMPLE 7.01
• Apply the elastic flexural formula to find the
maximum tensile and compressive stresses.
Mc
I
M c A 3 kN  m  0.022 m
A 

I
868 109 mm 4
M cB
3 kN  m  0.038 m
B  

I
868 109 mm 4

m 

 A  76.0 MPa

 B  131.3 MPa

• Calculate the curvature
1






M
EI
3 kN  m

165 GPa 868 10-9 m 4 

1

 20.95 103 m-1


  47.7 m


7.2 STRESS DUE TO PURE BENDING

TECHNICAL EXPRESSION
 min

x

z

z

 max

z

Mx > 0

y

z
 max
z

Mx < 0

z

z
 min

z

7.2 STRESS DUE TO PURE BENDING
TECHNICAL EXPRESSION
x 

Mx
y
Ix

x

tension
tension
ymax
 cmax

compression
compression
ymax
 cmax

y

 max 

Mx
Ix

tension
c max

 max 
For safety:

 min  

Mx
Ix

compression
c max

Mx
Ix

tension
c max
  tension

 min   compression


7.3 COMPOSITE SECTION
BENDING MEMBERS MADE OF SEVERAL ATERIALS
• Consider a composite beam formed from
two materials with E1 and E2.
• Normal strain varies linearly.
x  

y



• Piecewise linear normal stress variation.
1  E1 x  

E1 y



 2  E2 x  

E2 y



Neutral axis does not pass through
section centroid of composite section.
• Elemental forces on the section are
Ey
E y
dF1   1dA   1 dA dF2   2dA   2 dA



x  

My
I

1   x



• Define a transformed section such that
 2  n x

dF2  

nE1  y dA   E1 y n dA




E
n 2
E1


7.3 COMPOSITE SECTION
EXAMPLE 7.02
SOLUTION:
• Transform the bar to an equivalent cross
section made entirely of brass
• Evaluate the cross sectional properties of
the transformed section
• Calculate the maximum stress in the
transformed section. This is the correct
maximum stress for the brass pieces of
the bar.

Bar is made from bonded pieces of
steel (Es = 29x106 psi) and brass
(Eb = 15x106 psi). Determine the
maximum stress in the steel and
brass when a moment of 40 kip*in
is applied.

• 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.


7.3 COMPOSITE SECTION
EXAMPLE 7.02
SOLUTION:
• Transform the bar to an equivalent cross section
made entirely of brass.
Es 29 106 psi
n

 1.933
Eb 15 106 psi
bT  0.4 in  1.933  0.75 in  0.4 in  2.25 in

• Evaluate the transformed cross sectional properties
1 b h3  1 2.25 in.3 in 3
I  12
T

12

 5.063 in 4

• Calculate the maximum stresses
m 

Mc 40 kip  in 1.5 in 

 11.85 ksi
4
I
5.063 in

 b max   m
 s max  n m  1.933 11.85 ksi

 b max  11.85 ksi
 s max  22.9 ksi


7.3 COMPOSITE SECTION
REINFORCED CONCRETE BEAMS
• Concrete beams subjected to bending moments are
reinforced by steel rods.
• The steel rods carry the entire tensile load below
the neutral surface. The upper part of the
concrete beam carries the compressive load.

• In the transformed section, the cross sectional area

of the steel, As, is replaced by the equivalent area
nAs where n = Es/Ec.
• To determine the location of the neutral axis,

bx  x  n As d  x   0
2

1 b x2
2

 n As x  n As d  0

• The normal stress in the concrete and steel
x  

My
I

c   x

 s  n x


7.3 COMPOSITE SECTION
REINFORCED CONCRETE BEAMS – EXAMPLE 7.03

SOLUTION:
• Transform to a section made entirely
of concrete.
• Evaluate geometric properties of

transformed section.
• Calculate the maximum stresses
in the concrete and steel.
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.


7.3 COMPOSITE SECTION
REINFORCED CONCRETE BEAMS – EXAMPLE 7.03
SOLUTION:
• Transform to a section made entirely of concrete.
Es 29 106 psi
n

 8.06
6
Ec 3.6 10 psi

 

2
nAs  8.06  24 85 in   4.95 in 2



• Evaluate the geometric properties of the
transformed section.
 x
12 x   4.954  x   0
 2



x  1.450 in



I  13 12 in 1.45 in 3  4.95 in 2 2.55 in 2  44.4 in 4

• Calculate the maximum stresses.
Mc1 40 kip  in 1.45 in

I
44.4 in 4
Mc
40 kip  in  2.55 in
 s  n 2  8.06
I
44.4 in 4

c 

 c  1.306 ksi
 s  18.52 ksi

7.4 STRESS CONCENTRATIONS
STRESS CONENTRATIONS

Stress concentrations may occur:
• in the vicinity of points where the
loads are applied
• in the vicinity of abrupt changes
in cross section

m  K

Mc
I


7.5 PLASTIC ANALYSIS
PLASTIC DEFORMATIONS
• For any member subjected to pure bending
y
c

 x   m

strain varies linearly across the section

• If the member is made of a linearly elastic material,
the neutral axis passes through the section centroid
and

x  

My
I

• For a material with a nonlinear stress-strain curve,
the neutral axis location is found by satisfying
Fx    x dA  0

M    y x dA

• For a member with vertical and horizontal planes of
symmetry and a material with the same tensile and
compressive stress-strain relationship, the neutral
axis is located at the section centroid and the stressstrain relationship may be used to map the strain
distribution from the stress distribution.


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 MU is referred to as the
ultimate bending moment.
• The modulus of rupture in bending, RB, is found
from an experimentally determined value of MU
and a fictitious linear stress distribution.
RB 

MU c
I

• RB may be used to determine MU of any
member made of the same material and with the
same cross sectional shape but different
dimensions.


7.5 PLASTIC ANALYSIS
MEMBERS MADE OF AN ELASTOPLASTIC MATERIAL
• Rectangular beam made of an elastoplastic material
Mc
I

 x  Y

m 

 m  Y

I
M Y   Y  maximum elastic moment
c

• If the moment is increased beyond the maximum
elastic moment, plastic zones develop around an
elastic core.
M

2

3 M 1  1 yY
2 Y
3 2





c 

yY  elastic core half - thickness

• In the limit as the moment is increased further, the
elastic core thickness goes to zero, corresponding to a
fully plastic deformation.
M p  32 M Y  plastic moment
Mp
k
 shape factor (depends only on cross section shape)
MY