Giáo trình Strength of Materials - Chapter 7: Pure bending

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 
 
 
 
MdAyM
dAzM
dAF
xz
xy
xx



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. 
7.1 BENDING DEFORMATIONS 
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, 
 
 
mx
m
m
x
c
y
c
ρ
c
yy
L
yyLL
yL













or 
linearly) ries(strain va 
7.2 STRESS DUE TO PURE BENDING 
STRESS DUE TO BENDING 
• For a linearly elastic material, 
linearly) varies(stressm
mxx
c
y
E
c
y
E




• For static equilibrium, 




dAy
c
dA
c
y
dAF
m
mxx


0
0
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
dAy
c
M
dA
c
y
ydAyM
x
mx
m
mm
mx


















 ngSubstituti
2
7.2 STRESS DUE TO PURE BENDING 
BEAM SECTION PROPERTIES 
• The maximum normal stress due to bending, 
modulussection 
inertia ofmoment section 



c
I
S
I
S
M
I
Mc
m
A beam section with a larger section modulus 
will have a lower maximum stress 
• Consider a rectangular beam cross section, 
Ahbh
h
bh
c
I
S
6
13
6
1
3
12
1
2

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 
EI
M
I
Mc
EcEcc
mm


11 

• Although cross sectional planes remain planar 
when subjected to bending moments, in-plane 
deformations are nonzero, 






yy
xzxy 
• Expansion above the neutral surface and 
contraction below it cause an in-plane curvature, 
curvature canticlasti 
1

 


7.2 STRESS DUE TO BENDING 
PROPERTIES OF AMERICAN STANDARD SHAPES 
7.2 STRESS DUE TO PURE BENDING 
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. 
  


 
2dAII
A
Ay
Y x
• Apply the elastic flexural formula to 
find the maximum tensile and 
compressive stresses. 
I
Mc
m 
• Calculate the curvature 
EI
M


1
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. 
mm 38
3000
10114 3






A
Ay
Y
 


3
3
3
32
101143000
104220120030402
109050180090201
mm ,mm ,mm Area,
AyA
Ayy
   
   
49-3
23
12
123
12
1
23
12
12
m10868 mm10868
18120040301218002090


  
I
dAbhdAIIx
7.2 STRESS DUE TO PURE BENDING 
EXAMPLE 7.01 
• Apply the elastic flexural formula to find the 
maximum tensile and compressive stresses. 
49
49
mm10868
m038.0mkN 3
mm10868
m022.0mkN 3









I
cM
I
cM
I
Mc
B
B
A
A
m



MPa 0.76A
MPa 3.131B
• Calculate the curvature 
  49- m10868GPa 165
mkN 3
1




EI
M

m 7.47
m1095.20
1 1-3

 


7.2 STRESS DUE TO PURE BENDING 
TECHNICAL EXPRESSION 
y
x z z 
Mx > 0 
z z
z z 
z z
max
min
max
min
Mx < 0 
7.2 STRESS DUE TO PURE BENDING 
TECHNICAL EXPRESSION 
y
I
M
x
x
x 
y
x
tension
x
x
c
I
M
maxmax 
ncompressio
x
x
c
I
M
maxmin 
 tension
tension
x
x
c
I
M
  maxmax
  ncompressio min
For safety: 
ncompressioncompressio cy maxmax 
tensiontension cy maxmax 
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. 


y
x 
• Piecewise linear normal stress variation. 




yE
E
yE
E xx
2
22
1
11 
Neutral axis does not pass through 
section centroid of composite section. 
• Elemental forces on the section are 
dA
yE
dAdFdA
yE
dAdF



 222
1
11 
 
 
1
211
2
E
E
ndAn
yE
dA
ynE
dF 

• Define a transformed section such that 
xx
x
n
I
My




21
7.3 COMPOSITE SECTION 
EXAMPLE 7.02 
Bar is made from bonded pieces of 
steel (Es = 29x10
6 psi) and brass 
(Eb = 15x10
6 psi). Determine the 
maximum stress in the steel and 
brass when a moment of 40 kip*in 
is applied. 
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. 
• 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 
• Evaluate the transformed cross sectional properties 
  
4
3
12
13
12
1
in 063.5
in 3in. 25.2

 hbI T
SOLUTION: 
• Transform the bar to an equivalent cross section 
made entirely of brass. 
in 25.2in 4.0in 75.0933.1in 4.0
933.1
psi1015
psi1029
6
6





T
b
s
b
E
E
n
• Calculate the maximum stresses 
  
ksi 85.11
in 5.063
in 5.1inkip 40
4



I
Mc
m
 
  ksi 85.11933.1max
max


ms
mb
n
  
  ksi 22.9
ksi 85.11
max
max


s
b


7.3 COMPOSITE SECTION 
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, As, is replaced by the equivalent area 
nAs where n = Es/Ec. 
• To determine the location of the neutral axis, 
   
0
0
2
2
2
1 

dAnxAnxb
xdAn
x
bx
ss
s
• The normal stress in the concrete and steel 
xsxc
x
n
I
My




• The steel rods carry the entire tensile load below 
the neutral surface. The upper part of the 
concrete beam carries the compressive load. 
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. 
SOLUTION: 
• Transform to a section made entirely 
of concrete. 
• Evaluate geometric properties of 
transformed section. 
• Calculate the maximum stresses 
in the concrete and steel. 
7.3 COMPOSITE SECTION 
REINFORCED CONCRETE BEAMS – EXAMPLE 7.03 
SOLUTION: 
• Transform to a section made entirely of concrete. 
  22
8
5
4
6
6
in95.4in 206.8
06.8
psi 106.3
psi 1029










s
c
s
nA
E
E
n
• Evaluate the geometric properties of the 
transformed section. 
 
      4223
3
1 in4.44in55.2in95.4in45.1in12
in450.10495.4
2
12







I
xx
x
x
• Calculate the maximum stresses. 
4
2
4
1
in44.4
in55.2inkip40
06.8
in44.4
in1.45inkip40




I
Mc
n
I
Mc
s
c

 ksi306.1c
ksi52.18s
7.4 STRESS CONCENTRATIONS 
STRESS CONENTRATIONS 
Stress concentrations may occur: 
• in the vicinity of points where the 
loads are applied 
I
Mc
Km 
• in the vicinity of abrupt changes 
in cross section 
7.5 PLASTIC ANALYSIS 
PLASTIC DEFORMATIONS 
• For any member subjected to pure bending 
mx
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 
I
My
x and 
• For a material with a nonlinear stress-strain curve, 
the neutral axis location is found by satisfying 
   dAyMdAF xxx  0
• 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 stress-
strain 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. 
I
cM
R UB 
• 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 
moment elastic maximum 

YYYm
mYx
c
I
M
I
Mc


• If the moment is increased beyond the maximum 
elastic moment, plastic zones develop around an 
elastic core. 
thickness-half core elastic 1
2
2
3
1
2
3 








 Y
Y
Y y
c
y
MM
• In the limit as the moment is increased further, the 
elastic core thickness goes to zero, corresponding to a 
fully plastic deformation. 
shape)section crosson only (dependsfactor shape 
moment plastic 
2
3


Y
p
Yp
M
M
k
MM
7.5 PLASTIC ANALYSIS 
PLASTIC DEFORMATIONS OF MEMBERS WITH A SINGLE PLANE OF 
SYMMETRY 
• Fully plastic deformation of a beam with only a 
vertical plane of symmetry. 
• Resultants R1 and R2 of the elementary 
compressive and tensile forces form a couple. 
YY AA
RR
 21
21


The neutral axis divides the section into equal 
areas. 
• The plastic moment for the member, 
 dAM Yp 2
1
• The neutral axis cannot be assumed to pass 
through the section centroid. 
7.5 PLASTIC ANALYSIS 
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 
to be fully elastic. 
• Residual stresses are obtained by applying the 
principle of superposition to combine the stresses 
due to loading with a moment M (elastoplastic 
deformation) and unloading with a moment -M 
(elastic deformation). 
• The final value of stress at a point will not, in 
general, be zero. 
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. 
7.5 PLASTIC ANALYSIS 
EXAMPLE 7.04, 7.05 
  
  
mkN 8.28
MPa240m10120
m10120
10601050
36
36
233
3
22
3
2







YY
c
I
M
mmbc
c
I

• Maximum elastic moment: 
• Thickness of elastic core: 
 
666.0
mm60
1mkN28.8mkN8.36
1
2
2
3
1
2
3
2
2
3
1
2
3



















YY
Y
Y
Y
y
c
y
c
y
c
y
MM
mm802 Yy
• Radius of curvature: 
3
3
3
9
6
102.1
m1040
102.1
Pa10200
Pa10240











Y
Y
Y
Y
Y
Y
y
y
E






m3.33
7.5 PLASTIC ANALYSIS 
EXAMPLE 7.04, 7.05 
• M = 36.8 kN-m 
MPa240
mm40
Y 


Yy
• M = -36.8 kN-m 
Y
36
2MPa7.306
m10120
mkN8.36






I
Mc
m
• M = 0 
6
3
6
9
6
105.177
m1040
105.177
Pa10200
Pa105.35
core, elastic theof edge At the










x
Y
x
x
y
E




m225