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• An equation for the beam shape or elastic curve is required to determine maximum deflection.[r]
(1)MECHANICS OF MATERIALS
Ferdinand P Beer
E Russell Johnston, Jr. John T DeWolf
Lecture Notes: J Walt Oler
Texas Tech University
CHAPTER
(2)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 9.3 Method of Superposition Sample Problem 9.7
Application of Superposition to Statically Indeterminate
Sample Problem 9.8 Moment-Area Theorems
Application to Cantilever Beams and Beams With Symmetric
Bending Moment Diagrams by Parts Sample Problem 9.11
Application of Moment-Area Theorems to Beams With Unsymme
Maximum Deflection
(3)Deformation of a Beam Under Transverse Loading
• Relationship between bending moment and curvature for pure bending remains valid for general transverse loadings
EI x M( ) =
ρ
• Cantilever beam subjected to concentrated load at the free end,
EI Px
− =
ρ
1
• Curvature varies linearly with x • At the free end A, = A = ∞
A
ρ ρ 0,
(4)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
EI x M( ) =
ρ
• 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
(5)Equation of the Elastic Curve
• From elementary calculus, simplified for beam parameters, 2 2 1 dx y d dx dy dx y d ≈ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ρ
• Substituting and integrating, ( )
(6)Equation of the Elastic Curve
( ) 1 2
0
C x C dx x M dx y
EI
x x
+ +
= ∫ ∫
• Constants are determined from boundary conditions
• Three cases for statically determinant beams, – Simply supported beam
0 ,
0 =
= B
A y
y
– Overhanging beam ,
0 =
= B
A y
y
– Cantilever beam ,
0 =
= A
A
y θ
(7)Direct Determination of the Elastic Curve From the Load Distribution
• For a beam subjected to a distributed load, ( ) w( )x
dx dV dx M d x V dx
dM = = = −
2
• Equation for beam displacement becomes ( )x
w dx y d EI dx M
d = = −
4 2 ( ) ( ) 2
1C x C x C x C
dx x w dx dx dx x y EI + + + + − = ∫ ∫ ∫ ∫
(8)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
0
0 ∑ = ∑ =
=
∑Fx Fy MA
The beam is statically indeterminate
( ) 1 2
0
C x C dx x M dx y
EI
x x
+ +
= ∫ ∫
• Also have the beam deflection equation,
which introduces two unknowns but provides three additional equations from the boundary conditions:
0 ,
At
0 ,
0
(9)Sample Problem 9.1
ft ft
15 kips
50
psi 10 29 in
723 68
14
= =
=
× =
= ×
a L
P
E I
W
For portion AB of the overhanging beam, (a) derive the equation for the elastic curve, (b) determine the maximum deflection,
(c) evaluate ymax
SOLUTION:
• Develop an expression for M(x) and derive differential equation for elastic curve
• Integrate differential equation twice and apply boundary conditions to obtain elastic curve
• Locate point of zero slope or point of maximum deflection
(10)Sample Problem 9.1
SOLUTION:
• Develop an expression for M(x) and derive differential equation for elastic curve
- Reactions:
↑ ⎟ ⎠ ⎞ ⎜
⎝ ⎛ + =
↓ =
L a P
R L
Pa
RA B
- From the free-body diagram for section AD, ( x L)
x L a P
M = − 0< <
x a P y
d
EI = −
2