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Bài giảng Strength of materials - Chương 7: Pure bending - Nguyễn Sỹ Lâm

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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.  
F dA 0  
x
x
M zdA 0  
y
x
M   ydA M  
z
x
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  
L
y  
  
y
    
   
(strain varies linearly)  
x
c
c
  
or ρ   
m
m
y
    
x
m
c
7.2 STRESS DUE TO PURE BENDING  
STRESS DUE TO BENDING  
• For a linearly elastic material,  
y
E  E  
x
x
m
c
y
  (stress varieslinearly)  
m
c
• For static equilibrium,  
• For static equilibrium,  
y
F 0 dA   dA  
y
x
x
m
c
M   ydA   y   
dA  
x
m
c
c
m
0    
y dA  
c
I  
2
m
m
M   
y dA   
c
First moment with respect to neutral  
plane is zero. Therefore, the neutral  
surface must pass through the  
section centroid.  
Mc M  
  
m
I
S
y
Substituting     
x
m
c
My  
   
x
I
7.2 STRESS DUE TO PURE BENDING  
BEAM SECTION PROPERTIES  
• The maximum normal stress due to bending,  
Mc M  
  
m
I
S
I section moment of inertia  
I
S   section modulus  
c
A beam section with a larger section modulus  
will have a lower maximum stress  
• Consider a rectangular beam cross section,  
3
1
bh  
I
S    
c
3
12  
1
1
bh Ah  
6
6
h 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  
1   
c  
1 Mc  
m
m
Ec Ec I  
M
EI  
• Although cross sectional planes remain planar  
when subjected to bending moments, in-plane  
deformations are nonzero,  
y  
y  
   
   
z x  
y
x
• 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.  
yA  
2
Y   
I I Ad  
x
A
• Apply the elastic flexural formula to  
find the maximum tensile and  
compressive stresses.  
Mc  
  
m
I
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.  
• 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.  
2
3
3
3
3
Area,mm y, mm  
yA, mm  
9010  
1 2090 1800  
2 4030 1200  
A 3000  
50  
20  
2410  
yA 11410  
3
yA 11410  
Y   
38 mm  
A
3000  
2
3
2
1
I I Ad   
bh Ad  
x
12  
3
2
3
2
1
1
   
9020 180012 3040 120018  
12  
12  
3
-9 4  
I 86810 mm 86810 m  
7.2 STRESS DUE TO PURE BENDING  
EXAMPLE 7.01  
• Apply the elastic flexural formula to find the  
maximum tensile and compressive stresses.  
Mc  
  
m
I
M c  
I
3 kNm0.022m  
A
 76.0 MPa  
  
A
A
9  
4
86810 mm  
3 kNm0.038m  
M c  
I
B
 131.3MPa  
   
   
B
B
9  
4
86810 mm  
• Calculate the curvature  
1 M  
EI  
1
3 kNm  
3 -1  
20.9510 m  
-9 4  
165 GPa  
86810 m  
47.7 m  
7.2 STRESS DUE TO PURE BENDING  
TECHNICAL EXPRESSION  
min  
x
z
z
max  
Mx > 0  
y
z  
z  
max  
z
z
min  
z  
z  
Mx < 0  
7.2 STRESS DUE TO PURE BENDING  
TECHNICAL EXPRESSION  
M x  
Ix  
x   
y
x
tension  
max  
tension  
max  
y
y
c  
compression  
max  
compression  
max  
c  
y
M x  
Ix  
tension  
max  
M x  
Ix  
max  
c
tension  
max  
max  
c
 tension  
For safety:  
M x  
compression  
max  
min    
c
min  
 compression  
Ix  
7.3 COMPOSITE SECTION  
BENDING MEMBERS MADE OF SEVERALATERIALS  
• Consider a composite beam formed from  
two materials with E1 and E2.  
• Normal strain varies linearly.  
y
   
x
• Piecewise linear normal stress variation.  
E y  
E y  
1
2
E    
E    
2 2 x  
1
1 x  
Neutral axis does not pass through  
section centroid of composite section.  
• Elemental forces on the section are  
E y  
E y  
1
2
dF dA    
dA dF dA    
dA  
1
1
2
2
My  
I
   
• Define a transformed section such that  
x
nE  
y
E y  
E
2
E
1
1
1
  
n  
2 x  
dF    
dA    
ndA  
n   
1
x
2
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.  
6
E
E
2910 psi  
s
n   
1.933  
6
1510 psi  
b
b 0.4in 1.9330.75in 0.4in 2.25in  
T
• Evaluate the transformed cross sectional properties  
3
3
1
1
12  
I b h   
2.25in.3in  
   
T
12  
4
5.063in  
• Calculate the maximum stresses  
Mc  
I
40 kipin1.5in  
  
11.85 ksi  
m
4
5.063in  
  
11.85 ksi  
22.9 ksi  
max  
b
s
m
b
s
max  
max  
n1.93311.85 ksi  
m
max  
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,  
x
bx  
n A  
d x 0  
s
2
2
1
2
b x n A x n A d 0  
s
s
• The normal stress in the concrete and steel  
My  
   
x
I
  
n  
s x  
c
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.  
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