Fourth Edition
CHAPTER
9
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Deflection of Beams
Lecture Notes:
J. Walt Oler
Texas Tech University
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Deflection of Beams
Deformation of a Beam Under Transverse
Loading
Equation of the Elastic Curve
Direct Determination of the Elastic Curve
From the Load Di...
Statically Indeterminate Beams
Sample Problem 9.1
Sample Problem 9.8
Moment-Area Theorems
Application to Cantilever Beams and
Beams With Symmetric ...
Bending Moment Diagrams by Parts
Sample Problem 9.11
Sample Problem 9.3
Application of Moment-Area Theorems to
Beams With Unsymme...
Method of Superposition
Maximum Deflection
Sample Problem 9.7
Use of Moment-Area Theorems With
Statically Indeterminate...
Application of Superposition to Statically
Indeterminate ...
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9-2
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Deformation of a Beam Under Transverse Loading
• Relationship between bending moment and
curvature for pure bending remains valid for
general transverse loadings.
1
ρ
=
M ( x)
EI
• Cantilever beam subjected to concentrated
load at the free end,
1
ρ
=−
Px
EI
• Curvature varies linearly with x
1
• At the free end A, ρ = 0,
A
• At the support B,
1
ρB
ρA = ∞
≠ 0, ρ B =
EI
PL
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Fourth
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MECHANICS OF MATERIALS
9-3
Beer • Johnston • DeWolf
Deformation of a Beam Under Transverse Loading
• Overhanging beam
• Reactions at A and C
• Bending moment diagram
• Curvature is zero at points where the bending
moment is zero, i.e., at each end and at E.
1
ρ
=
M ( x)
EI
• Beam is concave upwards where the bending
moment is positive and concave downwards
where it is negative.
• Maximum curvature occurs where the moment
magnitude is a maximum.
• An equation for the beam shape or elastic curve
is required to determine maximum deflection
and slope.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Equation of the Elastic Curve
• From elementary calculus, simplified for beam
parameters,
d2y
1
ρ
=
dx 2
2 ⎤3 2
⎡ ⎛ dy ⎞
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
≈
d2y
dx 2
• Substituting and integrating,
EI
1
ρ
= EI
EI θ ≈ EI
d2y
dx 2
= M (x )
x
dy
= M ( x )dx + C1
dx ∫
0
x
x
0
0
EI y = ∫ dx ∫ M ( x ) dx + C1x + C2
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9-5
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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, θ A = 0
• More complicated loadings require multiple
integrals and application of requirement for
continuity of displacement and slope.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Direct Determination of the Elastic Curve From the
Load Distribution
• For a beam subjected to a distributed load,
dM
= V (x )
dx
d 2M
dx 2
dV
= − w( x )
dx
=
• Equation for beam displacement becomes
d 2M
d4y
=
= − w( x )
EI
dx 2
dx 4
• Integrating four times yields
EI y ( x ) = − ∫ dx ∫ dx ∫ dx ∫ w( x )dx
+ 16 C1x3 + 12 C2 x 2 + C3 x + C4
• Constants are determined from boundary
conditions.
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Statically Indeterminate Beams
• Consider beam with fixed support at A and roller
support at B.
• From free-body diagram, note that there are four
unknown reaction components.
• Conditions for static equilibrium yield
∑ Fx = 0 ∑ Fy = 0 ∑ M A = 0
The beam is statically indeterminate.
• Also have the beam deflection equation,
x
x
0
0
EI y = ∫ dx ∫ M ( x ) dx + C1x + C2
which introduces two unknowns but provides
three additional equations from the boundary
conditions:
At x = 0, θ = 0 y = 0
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At x = L, y = 0
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.1
SOLUTION:
• Develop an expression for M(x)
and derive differential equation for
elastic curve.
W 14 × 68
I = 723 in 4
P = 50 kips L = 15 ft
E = 29 × 106 psi
• Integrate differential equation twice
and apply boundary conditions to
obtain elastic curve.
a = 4 ft
For portion AB of the overhanging beam,
• Locate point of zero slope or point
(a) derive the equation for the elastic curve,
of maximum deflection.
(b) determine the maximum deflection,
• Evaluate corresponding maximum
(c) evaluate ymax.
deflection.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.1
SOLUTION:
• Develop an expression for M(x) and derive
differential equation for elastic curve.
- Reactions:
RA =
Pa
⎛ a⎞
↓ RB = P⎜1 + ⎟ ↑
L
⎝ L⎠
- From the free-body diagram for section AD,
M = −P
a
x
L
(0 < x < L )
- The differential equation for the elastic
curve,
EI
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d2y
dx
2
= −P
a
x
L
9 - 10
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.1
• Integrate differential equation twice and apply
boundary conditions to obtain elastic curve.
EI
1 a
dy
= − P x 2 + C1
2 L
dx
1 a
EI y = − P x3 + C1x + C2
6 L
EI
d2y
a
= −P x
L
dx 2
at x = 0, y = 0 : C2 = 0
1 a
1
at x = L, y = 0 : 0 = − P L3 + C1L C1 = PaL
6 L
6
Substituting,
EI
1 a
1
dy
= − P x 2 + PaL
2 L
6
dx
2
dy PaL ⎡
⎛ x⎞ ⎤
=
⎢1 − 3⎜ ⎟ ⎥
dx 6 EI ⎣⎢
⎝ L ⎠ ⎦⎥
1 a
1
EI y = − P x3 + PaLx
6 L
6
y=
3
PaL2 ⎡ x ⎛ x ⎞ ⎤
⎢ −⎜ ⎟ ⎥
6 EI ⎣⎢ L ⎝ L ⎠ ⎦⎥
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9 - 11
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.1
• Locate point of zero slope or point
of maximum deflection.
2
dy
PaL ⎡
⎛x ⎞ ⎤
=0=
⎢1 − 3⎜ m ⎟ ⎥
dx
6 EI ⎢⎣
⎝ L ⎠ ⎦⎥
y=
3
PaL2 ⎡ x ⎛ x ⎞ ⎤
⎢ −⎜ ⎟ ⎥
6 EI ⎣⎢ L ⎝ L ⎠ ⎦⎥
xm =
L
= 0.577 L
3
• Evaluate corresponding maximum
deflection.
ymax =
[
PaL2
0.577 − (0.577 )3
6 EI
]
ymax = 0.0642
ymax = 0.0642
PaL2
6 EI
(50 kips )(48 in )(180 in )2
(
)(
6 29 × 106 psi 723 in 4
)
ymax = 0.238 in
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.3
SOLUTION:
• Develop the differential equation for
the elastic curve (will be functionally
dependent on the reaction at A).
• Integrate twice and apply boundary
conditions to solve for reaction at A
and to obtain the elastic curve.
For the uniform beam, determine the
reaction at A, derive the equation for
the elastic curve, and determine the
slope at A. (Note that the beam is
statically indeterminate to the first
degree)
• Evaluate the slope at A.
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Fourth
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MECHANICS OF MATERIALS
9 - 13
Beer • Johnston • DeWolf
Sample Problem 9.3
• Consider moment acting at section D,
∑MD = 0
1 ⎛ w x2 ⎞ x
RA x − ⎜ 0 ⎟ − M = 0
2 ⎜⎝ L ⎟⎠ 3
w x3
M = RA x − 0
6L
• The differential equation for the elastic
curve,
EI
d2y
w0 x3
=
=
−
M
R
x
A
6L
dx 2
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.3
• Integrate twice
EI
1
dy
w x4
= EIθ = RA x 2 − 0 + C1
2
24 L
dx
1
w x5
EI y = RA x3 − 0 + C1x + C2
6
120 L
d2y
w x3
EI 2 = M = R A x − 0
6L
dx
• Apply boundary conditions:
at x = 0, y = 0 : C2 = 0
at x = L, θ = 0 :
1
w L3
RA L2 − 0 + C1 = 0
2
24
at x = L, y = 0 :
1
w L4
RA L3 − 0 + C1L + C2 = 0
6
120
• Solve for reaction at A
1
1
R A L3 − w0 L4 = 0
3
30
RA =
1
w0 L ↑
10
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.3
• Substitute for C1, C2, and RA in the
elastic curve equation,
1⎛ 1
w x5 ⎛ 1
⎞
⎞
w0 L3 ⎟ x
EI y = ⎜ w0 L ⎟ x3 − 0 − ⎜
6 ⎝ 10
120 L ⎝ 120
⎠
⎠
y=
(
w0
− x5 + 2 L2 x3 − L4 x
120 EIL
)
• Differentiate once to find the slope,
θ=
at x = 0,
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(
dy
w0
=
− 5 x 4 + 6 L2 x 2 − L4
dx 120 EIL
θA =
)
w0 L3
120 EI
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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Method of Superposition
Principle of Superposition:
• Deformations of beams subjected to
combinations of loadings may be
obtained as the linear combination of
the deformations from the individual
loadings
• Procedure is facilitated by tables of
solutions for common types of
loadings and supports.
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Fourth
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MECHANICS OF MATERIALS
9 - 17
Beer • Johnston • DeWolf
Sample Problem 9.7
For the beam and loading shown,
determine the slope and deflection at
point B.
SOLUTION:
Superpose the deformations due to Loading I and Loading II as shown.
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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.7
Loading I
(θ B )I
wL3
6 EI
( yB )I
=−
wL3
48 EI
( yC )II
=
=−
wL4
8 EI
Loading II
(θC )II
=
wL4
128 EI
In beam segment CB, the bending moment is
zero and the elastic curve is a straight line.
(θ B )II = (θC )II
( yB )II
=
=
wL3
48 EI
wL4
wL3 ⎛ L ⎞ 7 wL4
+
⎜ ⎟=
128 EI 48 EI ⎝ 2 ⎠ 384 EI
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 9.7
Combine the two solutions,
θ B = (θ B )I + (θ B )II = −
wL3 wL3
+
6 EI 48 EI
θB = −
7 wL3
48 EI
wL4 7 wL4
+
8 EI 384 EI
yB = −
41wL4
384 EI
y B = ( y B )I + ( yB )II = −
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Application of Superposition to Statically
Indeterminate Beams
• Determine the beam deformation
• Method of superposition may be
without the redundant support.
applied to determine the reactions at
the supports of statically indeterminate
• Treat the redundant reaction as an
beams.
unknown load which, together with
• Designate one of the reactions as
the other loads, must produce
redundant and eliminate or modify
deformations compatible with the
the support.
original supports.
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