Ch06-Deflection of Beams - Lecture
Ch06-Deflection of Beams - Lecture
Ch06-Deflection of Beams - Lecture
Engineering Mechanics
Chapter 06: Deflection of Beams
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Content
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Ordinary Bending Theory
• Consider an ordinary (uniaxial) bending,
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Deflection curve
for normal stress
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Deflection curve
• To determine the beam deflection, we have the
following assumption.
w’
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Assumptions
• Every point of a cross section undergoes the same
deflection in the z-direction. This implies that the
height of the beam does not change due to bending:
𝑤=𝑤 𝑥
𝜕𝑤
𝜀𝑧 = =0
𝜕𝑧
• Plane cross sections of the beam remain plane during
the bending. In addition to the displacement w, a
cross section undergoes a rotation. The angle of
rotation ψ = ψ(x) is a small angle
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Assumptions
The displacement u of a point P
which is located at a distance z
from the x-axis is given by
𝑢 𝑥, 𝑧 = 𝜓 𝑥 𝑧
𝜕𝑢 𝑥, 𝑧 z
=𝜓 𝑥
𝜕𝑧
• These two assumptions are
sufficiently accurate in the case
of a slender beam with a
constant cross section or with a
slight taper.
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Deflection curve
𝜕𝑢 𝜕𝜓 𝑥 𝑧
𝜎 = 𝐸𝜀 = 𝐸 =𝐸 = 𝐸𝑧𝜓 ′
𝜕𝑥 𝜕𝑥
𝜕𝑤 𝜕𝑢
𝜏 = 𝐺( + ) = 𝐺(𝑤 ′ + 𝜓)
𝜕𝑥 𝜕𝑧
𝑁 = න 𝜎𝑑𝐴 = 𝐸𝜓 ′ න 𝑧𝑑𝐴
∴ 𝑀 = න 𝑧𝜎𝑑𝐴 = 𝐸𝜓 ′ න 𝑧 2 𝑑𝐴 = 𝐸𝐼𝜓 ′
𝐸𝐼𝑤 ′′ 𝑥 = 𝑀
𝑥
𝐸𝐼𝑤 ′ (𝑥) = න 𝑀𝑑𝑥 + 𝐶1 W’
0
𝑥 𝑥
𝐸𝐼𝑤 𝑥 = න 𝑑𝑥 න 𝑀𝑑𝑥 + 𝐶1 𝑥 + 𝐶2
0 0
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Boundary conditions
• To determine the deflection, we need to make use
of the derived equations together with boundary
conditions.
• We can distinguish two types of boundary
conditions.
1) Geometrical boundary conditions are
statements concerning the geometrical
(kinematic) quantities w or w’.
2) Statical boundary conditions are statements
referring to the stress resultants V or M.
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Boundary conditions
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Single region example #1
• Consider a cantilever beam (flexural rigidity EI)
subjected to a concentrated force F. The system is
statically determinate as the bending moment can
be calculated from the equilibrium conditions
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𝑀 = −𝐹(𝑙 − 𝑥)
𝐸𝐼𝑤 ′′ = 𝐹(𝑥 − 𝑙)
2
𝑥
𝐸𝐼𝑤 ′ = 𝐹 − 𝑙𝑥 + 𝐶1
2
𝑥 3 𝑙𝑥 2
𝐸𝐼𝑤 = 𝐹 − + 𝐶1 𝑥 + 𝐶2
6 2
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Single region example #1
• The geometrical boundary conditions
′
−𝐹𝑙2 −𝐹𝑙3
𝑤𝑚𝑎𝑥 = 𝑤𝑚𝑎𝑥 =
2𝐸𝐼 3𝐸𝐼
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Multiple region example #1
• A simply supported beam is subjected to a
concentrated force F at x = a
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Multiple region example #1
• Solution:
The shear force V has a jump (discontinuity) at x =
a and the bending moment is given by
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Multiple region example #1
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Multiple region example #1
There are two boundary conditions:
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Method of Superposition
Consider the beam below. It is subjected to a line load
q1 and a force F2. The deflection can be obtained
through a superposition method.
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Method of Superposition
𝑤 = 𝑤1 + 𝑤2
′ ′ ′
𝑤 = 𝑤1 + 𝑤2
𝑀 = 𝑀1 + 𝑀2
𝑉 = 𝑉1 + 𝑉2
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