Review-Chapter 9
Deflection of Beams
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Objective: Develop the governing differential
equations for the elastic curve, the basis for the
several techniques considered for determining
beam deflection.
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Equation of the Elastic Curve
• From elementary calculus the curvature of a plane curve at a point
Q (x,y) is, simplified for beam parameters,
d2y
1

=
dx 2
2 3 2
  dy 
1 +   
  dx  

d2y
dx 2
• Substituting and integrating,
Figure 9.7 Slope θ(x) of tangent to the elastic curve.
EI
1

=

= EI
d2y
2
= M ( x)
dx
dy x
EI   EI
=  M ( x ) dx + C1
dx 0
M ( x)
EI
Flexural
rigidity
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x
x
0
0
EI y =  dx  M ( x ) dx + C1x + C2
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Equation of the Elastic Curve
Constants are determined from boundary
conditions:
x
x
0
0
EI y =  dx  M ( x ) dx + C1x + C2
Three cases for statically determinant beams,
• Simply supported beam
y A = 0,
yB = 0
• Overhanging beam
y A = 0,
yB = 0
• Cantilever beam
y A = 0,  A = 0
Figure 9.8 Known boundary conditions
for statically determinate beams.
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More complicated loadings require multiple
integrals and application of requirement for
continuity of displacement and slope.
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Determination of the Elastic Curve from
the Load Distribution
• For a beam subjected to a distributed load,
d 2M
dM
= V ( x)
dx
dx 2
=
dV
= −w ( x )
dx
• Equation for beam displacement becomes,
d 2M
dx
2
= EI
d4y
dx
4
= −w ( x )
• Integrating four times yields.
EI y ( x ) = −  dx  dx  dx  w ( x ) dx
+ 1 C1x3 + 1 C2 x 2 + C3 x + C4
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Figure 9.12 Boundary conditions for (a) cantilever
beam (b) simply supported beam.
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• Constants are determined from boundary
conditions.
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Example: For the beam and loading shown, determine (a) the
equation of the elastic curve for portion AB of the beam, (b) the
slope at A, (c) the slope at B.
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Review-Chapter 7
Transformation of Stress and Strain
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Introduction
• Plane Stress - state of stress in which two
faces of the cubic element are free of stress.
For the illustrated example, the state of stress
is defined by
Figure 7.2 Non-zero stress components for state
of plane stress.
 , ,
x
y
xy
and  =  =  = 0.
z
zx
zy
• State of plane stress occurs in a thin plate
subjected to forces acting in the midplane of
the plate.
Figure 7.3 Example of plane stress: thin plate
subjected to only in-plane loads.
• State of plane stress also occurs on the free
surface of a structural element or machine
component, i.e., at any point of the surface
not subjected to an external force.
Figure 7.4 Example of plane stress: free surface
of a structural component.
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State of plane stress at Q includes 𝜎𝑥 , 𝜎𝑦 , 𝜏𝑥𝑦 (𝜎𝑧 = 𝜏𝑧𝑥 = 𝜏𝑧𝑦 = 0).
Determine stress components 𝜎𝑥′ , 𝜎𝑦′ , 𝜏𝑥′𝑦′ after the cube has been rotated through an
angle θ about z axis.
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Transformation of Plane Stress
• Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
the x, y, and x′ axes.
 Fx = 0 =  x A −  x ( A cos  ) cos  −  xy ( A cos  ) sin 
−  y ( A sin  ) sin  −  xy ( A sin  ) cos 
 Fy = 0 =  xy A +  x ( A cos  ) sin  −  xy ( A cos  ) cos 
−  y ( A sin  ) cos  +  xy ( A sin  ) sin 
• The equations may be rewritten to yield.
 x =
 y =
Figure 7.6 Stress transformation equations are
determined by considering an arbitrary
prismatic wedge element. (a) Geometry of the
element. (b) Free-body diagram.
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x + y
2
x + y
 xy = −
2
+
−
 x − y
2
 x − y
2
 x − y
2
cos 2 +  xy sin 2
cos 2 −  xy sin 2
sin 2 +  xy cos 2
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Principal Stresses
• The previous equations are combined to
yield parametric equations for a circle,
( x −  ave )2 +  x2y = R 2
where
 ave =
Figure 7.7 Circular relationship of transformed
stresses.
x + y
tan 2 p =
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• Principal stresses occur on the principal
planes of stress with zero shearing stresses.
 max,min =
Figure 7.9 Principal stresses.
2
 x − y 
2
R= 
 +  xy
2


 x + y
2
 x − y 
2
 
+  xy

 2 
2
2 xy
 x − y
Note: defines two angles separated by 90o
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Maximum Shearing Stress
Maximum shearing stress occurs for  x =  ave
 x − y 
2
 max = R = 
 +  xy
2 

 x − y
tan 2s = −
2 xy
2
Figure 7.7 Circular relationship of transformed
stresses.
Note: defines two angles separated by 90o and
offset from  p by 45o
  =  ave =
x + y
2
Figure 7.10 Maximum shearing stress.
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Mohr’s Circle for Plane Stress
• With the physical significance of Mohr’s circle
for plane stress established, it may be applied
with simple geometric considerations. Critical
values are estimated graphically or calculated.
• For a known state of plane stress σx, σy,τxy plot
the points X and Y and construct the circle
centered at C.
 ave =
x + y
2
 x − y 
2
R= 
 +  xy
2


2
• The principal stresses are obtained at A and B.
 max,min =  ave  R
tan 2 p =
Figure 7.12 (a) Plane stress element and the
orientation of principal planes. (b) Corresponding
Mohr‘s circle.
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2 xy
 x − y
The direction of rotation of Ox to Oa is the
same as CX to CA.
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Mohr’s Circle for Plane Stress
• With Mohr’s circle uniquely defined, the state
of stress at other axes orientations may be
depicted.
• For the state of stress at an angle θ with respect
to the xy axes, construct a new diameter X ′Y ′
at an angle 2θ with respect to XY.
• Normal and shear stresses are obtained from
the coordinates X ′Y ′.
Figure 7.13 (a) Stress element referenced to xy axes,
transformed to obtain components referenced to x′ y′ axes.
(b) Corresponding Mohr's circle.
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Example: For the given state of stress, determine the normal and
shearing stresses after the element shown has been rotated through
(a) 25 clockwise, (b) 10 counterclockwise.
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Example: For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b)
the maximum in-plane shearing stress, (c) the corresponding
normal stress.
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Example: For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing
stress, (b) the maximum in-plane shearing stress, (c) the
corresponding normal stress.
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Review-Chapter 6
Shearing Stress in Beams
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Shear on the Horizontal Face of a Beam
Element
• Consider prismatic beam AB.
• For equilibrium of beam element.
Figure 6.4 Transversely loaded beam
with vertical plane symmetric cross
section.
F
x
= 0 = H +  ( D −  C ) dA
A
M − MC
H = D
I
 y dA
A
• Note,
Figure 6.5 Short segment of beam with stress element
CDDC defined.
First moment with
respect to the neutral
axis of the portion w of
the cross section
Q =  y dA
A
M D − MC =
dM
x = V x
dx
• Substituting,
VQ
x
I
H VQ
q=
=
= shear flow
x
I
H =
Figure 6.6 Forces exerted on element
CCDC.
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Shearing Stresses in a Beam
Figure 6.7 Short segment of beam with smaller stress element
• The average shearing stress on the
horizontal face of the element is obtained
by dividing the shearing force ΔH on the
element by the area ΔA of the face.
CDDC defined.
 ave =
=
H q x VQ x
=
=
A
A
I t x
VQ
It
Figure 6.9 Stress element CDDC showing the
shear force on a horizontal plane.
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Shearing Stresses τxy in Common Types of
Beams
• For a narrow rectangular beam
ℎ
(𝑏 ≤ 4),
VQ 3 V  y 2 
 xy =
=
1− 2 

Ib 2 A  c 
Figure 6.13 Geometric terms
for rectangular section used to
calculate shearing stress.
Figure 6.14 Shearing stress
distribution on transverse
section of rectangular beam.
 max =
3V
2A
• For American Standard (S-beam)
and wide-flange (W-beam) beams.
VQ
It
V
=
Aweb
 ave =
Figure 6.15 Wide-flange beam. (a) Area for finding first
moment of area in flange. (b) Area for finding first moment
of area in web. (c) Shearing stress distribution.
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 max
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Review-Chapter 5
Analysis and Design of Beams for Bending
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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 bending moment
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 freebodies 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.
F
y
= 0 =  M B : RB = 46 kN RD = 14 kN
• Section the beam and apply equilibrium analyses
on resulting free-bodies.
 F = 0 −20 kN − V = 0
 M = 0 ( 20 kN )( 0 m ) + M
y
V1 = −20 kN
1
1
1
= 0 M1 = 0
 F = 0 −20 kN − V = 0
 M = 0 ( 20 kN )( 2.5 m ) + M
y
V2 = −20 kN
2
2
2
= 0 M 2 = −50 kN  m
V3 = +26 kN
M 3 = −50 kN  m
V4 = +26 kN M 4 = +28 kN  m
V5 = −14 kN
Figure 1 Six sections identified for analysis,
and the free body diagram for each section.
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M 5 = +28 kN  m
V6 = −14 kN M 6 = 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 =
1
6
( 0.080 m )( 0.250 m )
2
= 833.33 10−6 m3
50 103 N  m
m =
=
S
833.33 10−6 m3
MB
Figure 1 Shear and bending moment diagrams
result from the analysis of each section.
 m = 60.0 106 Pa
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Relations Among Load, Shear, and
Bending Moment
• Relationship between load and shear:
F
y
= 0 : V − (V + V ) − w x = 0
V = − w x
dV
= −w
dx
xD
VD − VC = −  w dx
xC
= −(area under load curve between C and D )
• Relationship between shear and bending moment:
M
C
= 0:
( M + M ) − M − V x + wx
M = V x − 12 w ( x )
x
=0
2
2
dM
=V
dx
Figure 5.9 (a) Simply supported beam subjected to a
distributed load, with a small element between C and
Cʹ, (b) Free-body diagram of the element.
M D − MC =
xD
 V dx
xC
= area under shear curve between C and D
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