Fifth SI Edition
CHAPTER
5
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
David F. Mazurek
Analysis and Design
of Beams for Bending
Lecture Notes:
J. Walt Oler
Texas Tech University
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•Why are Cast-Iron beams made thicker on tension face ?
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Prestress in concrete
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• Wind loading (bending and shear stress)
• Self weight (buckling)
• It is stiff and does not stoop and also does
not shatter. Its properties per unit weight are
better than most manufactured materials.
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Arrangement of cells
• Its superb properties result from the
arrangement of the cells and the
microscopic structure of the cell walls.
• Over 90 percent of the cells in wood are
long, thin tubes that are closely packed
together, pointing along the branches and
trunk. As well as helping them to transport
water to the leaves, this is ideal for
providing support because they point in the
direction in which the wood is stressed.
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• Trees mainly have to resist bending forces.
• The long, thin wood cells are well suited to
resist the forces; the cells on the concave
side resist being compressed, rather like
pillars, while those on the convex side resist
being stretched, rather like ropes.
Consequently, wood is very strong along
the grain.
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Pre-stressing of wood
• Wood has just one problem; because wood
cells are long, thin-walled tubes, they are very
prone to buckling, just like drinking straws.
This means that wood is only about half as
strong when compressed as when stretched, as
the cells tend to fail along a so-called
compression crease. If a wooden rod is bent,
the compression crease will form on the
concave side. This will subsequently weaken
the rod. Trees prevent this happening to their
trunks and branches by pre-stressing them.
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• New wood cells are laid down on the outside
of the trunk in a fully hydrated state. As they
mature their cell walls dry out and this tends to
make them shorter. However, because they are
already attached to the wood inside they
cannot shrink and will be held in tension.
Because this happens to each new layer of
cells, the outer part of the trunk is held in
tension, while the inside of the trunk is held in
compression.
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• The advantage of this is that when the tree is
bent over by the wind, the wood cells on the
surface are not actually compressed but some
of the pre-tension is released. On the other,
convex side the cells will be subjected to even
greater tensile forces, but they can cope very
easily with those. The consequence is that tree
trunks can bend a long way without breaking.
This fact was exploited for centuries by
shipwrights who made their masts, as far as
possible, from complete tree trunks.
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Contents
Introduction
Shear and Bending Moment Diagrams
Sample Problem 5.1
Sample Problem 5.2
Relations Among Load, Shear, and Bending Moment
Sample Problem 5.3
Sample Problem 5.5
Design of Prismatic Beams for Bending
Sample Problem 5.8
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Introduction
• Objective - Analysis and design of beams
• Beams - structural members supporting loads at
various points along the member
• Transverse loadings of beams are classified as
concentrated loads or distributed loads
• Applied loads result in internal forces consisting
of a shear force (from the shear stress
distribution) and a bending couple (from the
normal stress distribution)
• Normal stress is often the critical design criteria
x  
My
I
m 
Mc M

I
S
Requires determination of the location and
magnitude of largest bending moment
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Introduction
Classification of Beam Supports
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Boundary conditons
•M=0 at x =0 and x = L
•Shear force is not zero at the ends.
•M=0 at x=L
•If a point load is applied at x=L the shear
force will be equal to this.
•If a linearly distributed load (going to zero at
x=L) is applied at x = L then V = 0.
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Shear and Bending Moment Diagrams
• Determination of maximum normal and
shearing stresses requires identification of
maximum internal shear force and bending
couple.
• Shear force and bending couple at a point are
determined by passing a section through the
beam and applying an equilibrium analysis on
the beam portions on either side of the
section.
• Sign conventions for shear forces V and V’
and bending couples M and M’
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Sample Problem 5.1
SOLUTION:
• Treating the entire beam as a rigid
body, determine the reaction forces
For the timber beam and loading
shown, draw the shear and bendmoment diagrams and determine the
maximum normal stress due to
bending.
• Section the beam at points near
supports and load application points.
Apply equilibrium analyses on
resulting free-bodies to determine
internal shear forces and bending
couples
• Identify the maximum shear and
bending-moment from plots of their
distributions.
• Apply the elastic flexure formulas to
determine the corresponding
maximum normal stress.
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Sample Problem 5.1
SOLUTION:
• Treating the entire beam as a rigid body, determine
the reaction forces
from
 Fy  0   M B : RB  46 kN RD  14 kN
• Section the beam and apply equilibrium analyses
on resulting free-bodies
 Fy  0
 20 kN  V1  0
V1  20 kN
 M1  0
20 kN 0 m  M1  0
M1  0
 Fy  0
 20 kN  V2  0
V2  20 kN
 M2  0
20 kN 2.5 m  M 2  0
M 2  50 kN  m
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V3  26 kN
M 3  50 kN  m
V4  26 kN
M 4  28 kN  m
V5  14 kN
M 5  28 kN  m
V6  14 kN
M6  0
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Sample Problem 5.1
• Identify the maximum shear and bendingmoment from plots of their distributions.
Vm  26 kN M m  M B  50 kN  m
• Apply the elastic flexure formulas to
determine the corresponding
maximum normal stress.
S  16 b h 2  16 0.080 m 0.250 m 2
 833.33 10 6 m3
MB
50 103 N  m
m 

S
833.33 10 6 m3
 m  60.0 106 Pa
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Sample Problem 5.2
SOLUTION:
• Replace the 45 kN load with an
equivalent force-couple system at D.
Find the reactions at B by considering
the beam as a rigid body.
• Section the beam at points near the
support and load application points.
Apply equilibrium analyses on
The structure shown is constructed of a
resulting free-bodies to determine
W 250x167 rolled-steel beam. (a) Draw
internal shear forces and bending
the shear and bending-moment diagrams
couples.
for the beam and the given loading. (b)
determine normal stress in sections just
• Apply the elastic flexure formulas to
to the right and left of point D.
determine the maximum normal
stress to the left and right of point D.
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Sample Problem 5.2
SOLUTION:
• Replace the 45 kN load with equivalent forcecouple system at D. Find reactions at B.
• Section the beam and apply equilibrium
analyses on resulting free-bodies.
From A to C :
 F  0  45x  V  0
 M  0 45x  x   M  0
y
1
2
1
V  45 x kN
M  22.5 x 2 kNm
From C to D :
F  0
M  0
y
2
 108  V  0
108x  1.2  M  0 M  129.6  108 x  kNm
From D to B :
V  153 kN
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V  108 kN
M  305.1  153x  kNm
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Sample Problem 5.2
• Apply the elastic flexure formulas to
determine the maximum normal stress to
the left and right of point D.
From Appendix C for a W250x167 rolled
steel shape, S = 2.08x10-3 m3 about the X-X
axis.
To the left of D :
226.8 103 Nm
m 

2.08 10-3 m 3
S
M
 m  109 MPa
To the right of D :
199.8 103 Nm
m 

2.08 10-3 m 3
S
M
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 m  96 MPa
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Relations Among Load, Shear, and Bending Moment
• Relationship between load and shear:
 Fy  0 : V  V  V   w x  0
V   w x
dV
 w
dx
xD
VD  VC    w dx
xC
• Relationship between shear and bending
moment:
 M C  0 :
M  M   M  V x  wx x  0
M  V x  12 w x 
2
2
dM
V
dx
M D  MC 
xD
 V dx
xC
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Sample Problem 5.3
SOLUTION:
• Taking the entire beam as a free body,
determine the reactions at A and D.
• Apply the relationship between shear and
load to develop the shear diagram.
Draw the shear and bending
moment diagrams for the beam
and loading shown.
• Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
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Sample Problem 5.3
SOLUTION:
• Taking the entire beam as a free body, determine the
reactions at A and D.
M
A
0
0  D7.2 m   90 kN 1.8 m   54 kN 4.2 m   52.8 kN 8.4 m 
D  115.6 kN
F
y
0
0  Ay  90 kN  54 kN  115.6 kN  52.8 kN
Ay  81.2 kN
• Apply the relationship between shear and load to
develop the shear diagram.
dV
 w
dx
dV   w dx
- zero slope between concentrated loads
- linear variation over uniform load segment
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Sample Problem 5.3
• Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
dM
V
dx
dM  V dx
- bending moment at A and E is zero
- bending moment variation between A, B,
C and D is linear
- bending moment variation between D
and E is quadratic
- net change in bending moment is equal to
areas under shear distribution segments
- total of all bending moment changes across
the beam should be zero
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Sample Problem 5.5
SOLUTION:
• Taking the entire beam as a free body,
determine the reactions at C.
• Apply the relationship between shear
and load to develop the shear diagram.
Draw the shear and bending moment
diagrams for the beam and loading
shown.
• Apply the relationship between
bending moment and shear to develop
the bending moment diagram.
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Sample Problem 5.5
SOLUTION:
• Taking the entire beam as a free body,
determine the reactions at C.
 Fy  0   12 w0a  RC
a

 M C  0  12 w0 a L    M C
3

RC  12 w0 a
a

M C   12 w0 a L  
3

Results from integration of the load and shear
distributions should be equivalent.
• Apply the relationship between shear and load
to develop the shear diagram.
a
2 
 
x
 x
VB  V A    w0 1   dx    w0  x  

a
2a 
 
0 
0
a
VB   12 w0 a    area under load curve
- No change in shear between B and C.
- Compatible with free body analysis
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Sample Problem 5.5
• Apply the relationship between bending moment
and shear to develop the bending moment
diagram.
a
2 

 x 2 x3 

x


M B  M A    w0  x   dx   w0   
 2 6a 


2a  

0
 0


a
M B   13 w0 a 2
L


M B  M C    12 w0 a dx   12 w0 aL  a 
a
a w0 
a
M C   16 w0 a3L  a  
L 
2 
3
Results at C are compatible with free-body
analysis
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Design of Prismatic Beams for Bending
• The largest normal stress is found at the surface where the
maximum bending moment occurs.
M max c M max
m 

I
S
• A safe design requires that the maximum normal stress be
less than the allowable stress for the material used. This
criteria leads to the determination of the minimum
acceptable section modulus.
 m   all
S min 
M max
 all
• Among beam section choices which have an acceptable
section modulus, the one with the smallest weight per unit
length or cross sectional area will be the least expensive
and the best choice.
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Sample Problem 5.8
SOLUTION:
• Considering the entire beam as a freebody, determine the reactions at A and
D.
A simply supported steel beam is to
carry the distributed and concentrated
loads shown. Knowing that the
allowable normal stress for the grade
of steel to be used is 160 MPa, select
the wide-flange shape that should be
used.
• Develop the shear diagram for the
beam and load distribution. From the
diagram, determine the maximum
bending moment.
• Determine the minimum acceptable
beam section modulus. Choose the
best standard section which meets this
criteria.
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Sample Problem 5.8
• Considering the entire beam as a free-body,
determine the reactions at A and D.
 M A  0  D5 m   60 kN 1.5 m   50 kN 4 m 
D  58.0 kN
 Fy  0  Ay  58.0 kN  60 kN  50 kN
Ay  52.0 kN
• Develop the shear diagram and determine the
maximum bending moment.
V A  Ay  52.0 kN
VB  V A  area under load curve  60 kN
VB  8 kN
• Maximum bending moment occurs at
V = 0 or x = 2.6 m.
M max  area under shear curve, A to E 
 67.6 kN
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Sample Problem 5.8
• Determine the minimum acceptable beam
section modulus.
M max 67.6 kN  m
Smin 

160 MPa
 all
 422.5 106 m3  422.5 103 mm3
Shape
S  103 mm3
W410  38.8
637
W360  32.9
474
W310  38.7
549
W250  44.8
535
W200  46.1
448
• Choose the best standard section (i.e. lightest)
which meets this criteria.
W 360 32.9
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• Reactions at the supports =430w
• For the overhanging end M= -wx2/2 Maximum bending
moment is at x=a, i.e., M= -wa2/2.
• Between the supports the B.M = 430w(x-a) - wx2/2
• The maximum of this is found by substituting dM/dx =0
and occurs at x=430 mm,
If the support is moved to the right maximum moment
between the supports will decrease but support at the
overhang will increase.
Equate maximum bending moments:
430*430/2 -430 a – a2/2 = 0
a = 178.11 mm
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• Reactions at the supports =430w
• For the overhanging end M= -wx2/2 Maximum bending
moment is at x=a, i.e., M= -wa2/2.
• Between the supports the B.M = 430w(x-a) - wx2/2
• The maximum of this is found by substituting dM/dx =0
and occurs at x=430 mm,
If the support is moved to the right maximum moment
between the supports will decrease but support at the
overhang will increase.
Equate maximum bending moments:
430*430/2 -430 a – a2/2 = 0
a = 178.11 mm
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M =15.86 x 1000 w
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Orientation
Turbines can be categorized into two overarching classes based
on the orientation of the rotor
Vertical Axis
Horizontal Axis
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• When air flows over a blade, L(Lift) and
D(Drag) forces are generated on the blade.
These can be resolved into components
normal (N) and shear A along the axis as
shown at one point on the blade. Drag and
lift forces as well as N and A are actually
distributed over the whole length of the
blade. The normal force is distributed along
the length and can be seen as a uniformly
distributed load (UDL).
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•Figure 1 shows a VAWT. The dimensions
of the blade and its cross-section are
shown in figure 2. Determine the
a) maximum bending deformation,
b) bending moment and maximum stress
in the blade.
• The blade, made of Al has an aerofoil
cross-section shown. The moment of
inertia for the aerofoil section is Iz=2.4
x 10-7 m4 and its area A=4.36x10-3 m2,
UDL acting = 10.08 kN/m.
• The lift and drag forces vary 360
degrees of rotation of the blade. But
•
here a constant uniformly distributed
load (UDL) is assumed acting along the
length of the blade.
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NACA 0015 aerofoil cross section
with chord length 206mm (symmetric
, 00 indicates no camber)
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• The beam ABCD is shown in the horizontal configuration in the above
figure. It is fixed at B and C.
• For drawing the BMD and for finding the deflections we use results
derived earlier. The beam can be thought of as made of three beams,
cantilevers AB and CD and a fixed-fixed beam BC.
• we is a fixed-fixed beam with cantilevers at both ends ,i.e, fixed at 0.56m
from either ends.
• The solution is shown on next slide.
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Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Solution
•q = 10.08 kN/m
•We analyse the blade as a fixed fixed beam in the middle and two
•Cantilever beams at the ends which are shown in the figures
•q= 10.08 kN/m
•L=0.56 m
© 2009 The McGraw-Hill Companies, Inc. All rights reserved.
•L = 1.46 m
Fifth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Solution
•q = 10.08 kN/m
•We analyse the blade as a fixed fixed beam in the middle and two
•Cantilever beams at the ends which are shown in the figures
• E-Young’s modulus
• Izz = moment of inertia
about x axis
• L=length of the beam
•Yield strength of Aluminium = 200 MPa
•y=0.0154, Izz = 2.4*10^-7 m^4
•Maximum bending moment = 1.58 kNm
•Maximum stress = M*y/Izz=102 MPa
•Factor of safety = 200/102 = 1.96
© 2009 The McGraw-Hill Companies, Inc. All rights reserved.
•q= 10.08 kN/m
•L=0.56 m
•L = 1.46 m
Fifth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
•Bending moment diagram for the
blade
© 2009 The McGraw-Hill Companies, Inc. All rights reserved.
Fifth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Horizontal axis wind turbine (HAWT)
• The blades of this turbine are analysed as
cantilever beams.
• The fixed end at the hub has the maximum
stresses and that at the free end has
maximum deflection
© 2009 The McGraw-Hill Companies, Inc. All rights reserved.
Fifth
Edition
MECHANICS OF MATERIALS
© 2009 The McGraw-Hill Companies, Inc. All rights reserved.
Beer • Johnston • DeWolf • Mazurek
Principal area moments of inertia