Beams: Pure Bending (4.1-4.5)
MAE 314 – Solid Mechanics
Yun Jing
Beams: Pure Bending
1
Beams in Pure Bending

Prismatic beams subject to equal and opposite couples acting in the
same plane are in pure bending.
Beams: Pure Bending
2
Pure vs. Non-Uniform Bending


Pure bending: Shear force (V) = 0 over the section
Non-uniform bending: V ≠ 0 over the section
Non-uniform bending
• Moment → normal stresses
• Shear force → shear stresses (Ch. 6)
Pure bending
• Moment → normal stresses
Beams: Pure Bending
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Pure Bending: Assumptions


Beam is symmetric about the x – y plane
All loads act in the x – y plane
Beams: Pure Bending
4
Pure Bending: Curvature

Sections originally perpendicular to the
axis of the member remain plane and perpendicular:
“Plane sections remain plane.”
Right angle

Sign convention


Positive bending moment:
beam bends towards +y direction
Negative bending moment:
beam bends towards -y direction
Beams: Pure Bending
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Pure Bending: Deformation







Since angles do not change
(remain plane), there is no
shear stress.
The top part of the beam contracts
in the axial direction.
The bottom part of the beam expands
in the axial direction.
There exists a line in the beam that
remains the same length called the
neutral line.
Set y = 0 at the neutral line.
ρ = radius of curvature
εx < 0 for y > 0 and εx > 0 for y < 0
Beams: Pure Bending
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Pure Bending: Axial Strain



The length of DE is LDE = ρθ
The length of JK is LJK = (ρ-y)θ
Axial strain at a distance y from
the neutral axis (εx):

LJK  LDE (   y)  
x  

L
LDE

x  


LJK
y

LDE
Maximum compressive strain occurs on the upper surface.
Maximum tensile strain occurs on the lower surface.
Beams: Pure Bending
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Pure Bending: Axial Strain

c = maximum distance between the neutral axis and the upper or
lower surface

When c is the distance to the surface in compression

When c is the distance to the surface in tension
Beams: Pure Bending
 max   c 
 max  c 
8
Pure Bending: Transverse Strain

Recall there are no transverse stresses since
the beam is free to move in the y and z directions.

However, transverse strains (in the y and z
directions) exist due to the Poisson’s ratio
of the material.
 y  x   y  y 
 z  x   z  y 

ρ' = radius of anticlastic curvature = ρ/υ
Beams: Pure Bending
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Pure Bending: Normal Stress

Let us now assume that the beam is made of a linear-elastic material.
 x  E x  

Ey

The normal stress varies linearly with the distance from the neutral
surface.
Beams: Pure Bending
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Pure Bending: Normal Stress


Recollecting that the applied loading is a pure moment, we calculate
resultant loads on the cross-section.
The resultant axial force must be equal to zero.

A
x
dA   E x dA  
A
E
ydA  0


A
  ydA  0 
A
1
ydA  0

AA
…which is the definition of the centroid, so the
neutral axis is just the centroid of the section.
Beams: Pure Bending
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Pure Bending: Normal Stress

The resultant moment about the z-axis must be equal to the applied
moment M.
  y x dA    yE x dA 
A
A
E

2
y
 dA  M
A
Definition of the second moment of inertia, I
M
EI
x  
Beams: Pure Bending

My
I
12
Example Problem
A steel bar of 0.82.5-in. rectangular cross section is subjected to two
equal and opposite couples acting in the vertical plane of symmetry of
the bar. Determine the value of the bending moment M that causes the
bar to yield. Assume m=36 ksi.
Centroids and Moments of Inertia
13
Centroids and Moments of
Inertia (A.1-A.5)
MAE 314 – Solid Mechanics
Yun Jing
Centroids and Moments of Inertia
14
Review From Last Time





Beams in pure bending (no shear forces)
Plane sections remain plane
+M, upper surface in compression
+M, lower surface in tension
Transverse strains exist due to υ
x  
y

y
y  z 

Radius of Curvature
Neutral Line
M
EI

x  
My
I
Centroids and Moments of Inertia
15
Centroids

To match with our axis convention use centroid in the y-direction.

y
A


ydA
A
Centroids for common shapes are located in the back cover of your
textbook.
If a cross section shape has an axis of symmetry, then the centroid
passes through that axis of symmetry.
Centroids and Moments of Inertia
16
Centroids for Composite Shapes




Choose a reference x-axis (normally the base of the shape).
Divide the composite shape into individual shapes.
Calculate the area for each shape.
Find the location of the centroid for each shape relative to the
reference x-axis.
y
b3
Area
A1
h1b1
A2
h2b2
A3
h3b3
0.5h1
h1+0.5h2
0.5h12b1
h3
A3
h2
A2
h1
A1
y3
h2b2(h1+0.5h2)
h1+h2+0.5h3 h3b3(h1+h2+0.5h3)
 Ai yi
y2
i
Ay

y
A
i
i
i
i
y1
i
x
y=0
Centroids and Moments of Inertia
b2
b1
17
Moments of Inertia

Moment of inertia with respect to the x-axis

Moments of inertia for common shapes are located in the back cover of
your textbook.
x
x’
x
Ix
x
Ix’
Ix
Ix
Centroids and Moments of Inertia
18
Moments of Inertia for Composite Shapes



Divide the composite shape into individual shapes.
Calculate I of each shape about its own centroid.
Calculate I of each shape about the centroid of the entire composite
shape using the parallel axis theorem: ( I )  ( I )  A d 2
x i
x' i
I x  Ad 2
I x'
A1
b1h13/12
b1h1d12
b1h1(d12+h12/12)
A2
b2h23/12
b2h2d22
b2h2(d22+h22/12)
A3
b3h33/12
b3h3d32
b3h3(d32+h32/12)
i
i
b3
h3
x’’’
C3
d3
C
h2
x
x’’
A2
d1
h1
Centroids and Moments of Inertia
x’
A1
b2
b1
19
Example Problem
Find the location of the centroid and the moment of inertia of the section
below (the moment is applied about the horizontal axis).
5 in
0.5 in
4 in
1.5 in
3 in
3 in
Centroids and Moments of Inertia
20
Stress in the Elastic Range

Revisiting the beam subject to pure bending

For positive bending moment



Maximum tensile stress = Mc2 / I
Maximum compressive stress = Mc1 / I
For negative bending moment


Maximum tensile stress = Mc1 / I
Maximum compressive stress = Mc2 / I
c1
c2


The ratio I/c1 (or I/c2) is known as S, elastic section modulus.
Larger values for S indicate a beam that is more resistant to
bending.
Centroids and Moments of Inertia
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Deformations in the Elastic Range

Which shapes are most resistant to bending?



Why?


S-beam
rectangular
Large portion of the cross section is located far from the neutral
axis, resulting in large values of S.
The deformation of a member due to bending can be calculated
by
1 M

 EI
Centroids and Moments of Inertia
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Example Problem
A beam with cross-section like the previous example is subjected to a
pure bending moment of M = -4500 in·lb. Find the maximum tensile and
compressive normal stresses.
5 in
0.5 in
4 in
1.5 in
3 in
Centroids and Moments of Inertia
3 in
23
Example Problem
Knowing that the couple shown acts in a vertical plane, determine the
stress at (a) point A, (b) point B.
Centroids and Moments of Inertia
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Example Problem
The beam shown is made of nylon with an allowable stress of 24 MPa in
tension and 30 Mpa in compression. Determine the largest couple M that
can be applied to the beam.
Centroids and Moments of Inertia
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Example Problem
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion
BC of the beam.
Centroids and Moments of Inertia
26
Bending in Composite
Materials (4.6)
MAE 314 – Solid Mechanics
Yun Jing
Centroids and Moments of Inertia
27
Normal Stress in Composite Materials



Strain is still the same as homogenous material because it is not
dependent on material properties.
The stress distribution is no longer linear.
Use different material properties in each section.
Centroids and Moments of Inertia
28
Normal Stress in Composite Materials

Top section:

Ey
dF1   1dA   1 dA

Bottom section:
dF2  
E2 y

dA  
nE1 y

dA 
E1 y

(n  dA)
E2
n
E1
Original Shape
Centroids and Moments of Inertia
New Shape
29
Bending in Composite Materials




The new “homogenous” beam produces the same response to bending
as the composite beam.
We can now calculate the centroid and moment of inertia for this
beam.
To calculate the axial stresses, use σx for material 1 and nσx for
material 2.
Curvature of the composite member is
1
M

 E1 I
Centroids and Moments of Inertia
30
Reinforced concrete beam



Beams are reinforced by steel rods
Concrete is very weak in tension, the steel rods will carry the entire
tensile load
Concrete in the beam acts effectively only in compression
Centroids and Moments of Inertia
31
Example Problem
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Determine the largest permissible
bending moment when the composite bar is bent about a horizontal
axis. Ea = 70 GPa, σall,a = 100 MPa, Eb = 105 GPa, σall,b = 160 MPa.
Centroids and Moments of Inertia
32
Example Problem
Knowing that the bending moment in the reinforced concrete beam is
+100kip.ft and that the modulus of elasticity is 3.625106 psi for the
concrete and 29 106 psi for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
Beams: Pure Bending
33
Example Problem
A flitched beam consists of two 50mm×200mm wooden beams and a 12
mm×80mm steel plate. The plate is placed centrally between the
wooden beams and recessed into each so that, when rigidly joined, the
three units form a 100mm×200mm section as shown in the figure.
Determine the bending moment the beam can withstand when the
maximum bending stress in the timber is 12MN/m2. What will then be
the maximum bending stress in the steel. (E for the timber is 10GPa,
for steel is 120GPa)
Beams: Pure Bending
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