2-8 Deformation of a Beam—Continuously Deformable Model
If you differentiate a quadratic function such as
y = ax2 + bx + c (2.8.1)
you will obtain
dy/dx = 2ax + b (2.8.2)
As shown in Fig. 2-8-1a, dy/dx represents the slope of the curve defined by the function. If you
differentiate it again, you will obtain
d2y/dx2 = 2a (2.8.3)
As shown in Fig. 2-8-1b, the radius of the curve corresponding to Eq. 2.8.3 is reduced as the
constant a increases. If a is negative, the curve is bent in the opposite direction. If the curve is
defined by trigonometric ( y sin x ) or exponential ( y e x ) functions, the term d2y/dx2 also
indicates how the curve is bent.
y
dy/dx
dy
x
0 dx
(a) Slope of a curve
y
Large a
Small a
0 Negative a x
(b) Various curves
Fig. 2-8-1 Quadratic functions Fig. 2-8-2 Cantilever beam
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The broken line in Fig. 2-8-2a shows the deflected shape of the beam axis. The deflection at a
distance x from the fixed end is denoted as v. Figure 2-8-2b shows the deflection v as a function of
x, the distance from the fixed end. The first-order differential of the deflection,
dv
(2.8.4)
dx
represents the slope of the beam axis. As we did in Section 2-6, we shall again assume that the
planes P1 and P2 in Fig. 2-8-2a remains perpendicular to the beam axis, and define the unit
curvature as
d
(2.8.5)
dx
Note that Eq. 2.8.5 is equivalent to Eq. 2.6.1 ( / a ) except that the curvature is defined in an
infinitesimal region dx rather than a finite region a. Substituting Eq. 2.8.4 into Eq. 2.8.5 leads to:
d 2v
2 (2.8.6)
dx
In other words, the second-order differential of the deflection indicates how the beam bends1 or
1
Equations 2.8.5 and 2.8.6 are approximate. The exact definition of unit curvature is
d
(2.8.7)
ds
where ds is the length of the curve shown in Fig. 2-8-3. Therefore, the curve with a constant
curvature is a circle. Using some techniques of calculus, we get:
3/ 2
d 2v dv 2
2 1 (2.8.8)
dx dx
In the case of beam deflection, however, dv/dx is very small so that squaring it makes it negligible;
and thus we can assume = d2v/dx2.
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how the slope changes over a very small distance ( d / dx ).
Recalling Eq. 2.6.4 ( M EI ), we have:
d 2v M
(2.8.9)
dx 2 EI
This result should be compared with Eq. 2.7.1, where the deformation of the beam is evaluated
within a finite region a. Equation 2.8.9 deals with the bending deformation within an infinitesimal
region dx. Solving this differential equation, we can determine the deflection of the beam. This is
equivalent to assuming an infinite number of springs representing the deformation within an
infinitesimal region dx.
Example 2-8-1: Assume that a cantilever beam is subjected to a couple M at its free end (x = l).
Determine the maximum slope and the maximum deflection of the beam.
Fig. 2-8-4 A couple at the free end
Fig. 2-8-3 Exact definition of curvature
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Solution: First, we attempt to visualize how the beam would bend. The bending moment is
distributed uniformly as shown in Fig. 2-8-5a. The section does not change. We conclude that the
curvature is uniform along the beam. Inspecting the sense of the applied moment, we conclude that
the top fiber will be in compression and that the beam will bend into a shape concave upward.
From this information, we infer that the beam will bend as sketched in broken lines and that the
maximum slope and deflection will be at the free end.
We obtain the unit-curvature distribution using the expression d2v/dx2 = M/EI (Fig. 2-8-5b).
Integrating this expression yields the slope
dv M
x C1
dx EI
where C1 is a constant of integration. Because the slope is zero at the fixed end (dv/dx = 0 at x = 0),
we have C1 = 0. The slope = dv/dx is distributed linearly along the span (Fig. 2-8-5c). Integrating
the expression for the slope and noting that the deflection is zero at the fixed end (v = 0 at x = 0),
we get the following result.
M
v x2
EI
The slope and the deflection at the free end (x = L) are
ML
(2.8.9) and
EI
ML2
v (2.8.10)
EI
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Fig. 2-8-5
Exercise: Using the last two digits of your ID ij to change the couple to C (i 2) 100
N-mm and the depth to h = (j + 5) mm, solve the example above again.
Example 2-8-2: Assume that a cantilever beam is subjected to a force F at the free end ( x = L).
Determine the slope and the deflection of the beam at the free end.
Fig. 2-8-6 A force at the free end
Solution: The bending moment varies linearly along the span as shown in Fig. 2-8-7a:
M F L x
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Use of = M/EI leads to the unit-curvature distribution shown in Fig. 2-8-7b:
d 2v F
L x
dx 2 EI
Fig. 2-8-7
Integrating this equation and noting that the slope is zero at the fixed end (dv/dx = 0 at x = 0), we
have
dv F x2
Lx (2.8.11)
dx EI 2
The slope at the free end (x = L) is given by the following equation.
FL2
(2.8.12)
2 EI
Integrating Eq. 2.8.12 and noting that the deflection is zero at the fixed end (v = 0 at x = 0), we
have
F Lx 2 x3
v (2.8.13)
EI 2 6
The deflection at the free end (x = L) is given by the following equation.
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FL3
v (2.8.14)
3EI
If you substitute the default values used in GOYA-C (F = 10 N, L = 100 mm, b = 10 mm, h = 15
mm, and E =100 N/mm2), you should get v = 11.9 mm. Check it using GOYA-C.
Exercise: Use the last two digits of your ID ij to change the force to F (i 2) N and the
depth to h = (j + 5) mm. Calculate the slope and deflection at mid-span and at the free end.
Example 2-8-3: A force F is applied at mid-span of a cantilever beam. Calculate the slope and
deflection at the free end.
Fig. 2-8-8 A load at mid-span
Solution: The bending moment varies linearly between 0 200 > 500 > 1,000 > 2,000
5,000 > 10,000 > 20,000 > 200,000
Major All Star MVP Hall of Fame
Fig. 2-8-14
Design your own beam (Part 5): We want to design a beam that can carry a nervous
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mini-elephant. The weight of the elephant is (the last two digits of your ID#/100) plus 1 kg.
Assume that the acceleration of gravity is 10 m/s2 and each leg carries the same amount of
gravitational force. The density of the beam is 5 x 10-5 kg/mm3. The beam width is 10 mm and
Young’s modulus is 100 N/mm2. He is so nervous that the free end of the beam should not deflect
more than 10 mm. What is the required beam depth, h? Check your results using GOYA-C.
x
h w
Deflection 20 20 60 mm
10 mm
Fig. 2-8-15
Hint: You may use the equation obtained in Example 2-8-5. You will have to solve a cubic
equation for the beam depth, h. Because the equation ( h3 ah2 bh c 0 ) is difficult to solve
directly, you may resort to the following scheme: input y h3 ah2 bh c into a spreadsheet
and gradually increase h until you determine a deflection not exceeding 10 mm.
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What is unit curvature?
Master, I could not understand the concept of
unit curvature.
OK. Imagine a circle which approximates the
beam axis. Unit curvature is the reciprocal of
the radius of the circle. A large curvature means a small
radius, which means that the top of the beam shortens
and the bottom lengthens considerably.
Another important point is M = EI, which indicates that the unit curvature is
proportional to the bending moment. You need more moment to bend more.
The equation, = M/EI, also has an important meaning. An oak tree is hard to bend
because it has a high value of Young’s modulus. A thick board is also hard to bend
because it has large moment of inertia.
Yes, the wood I used in Chapter 1 bent very easily. But why are you so stiff when you are
so slim?
Cough! Anyhow, unit curvature has another face: it is the second derivative of the
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deflection with respect to distance, = d2v/dx2, which leads to the most important equation.
d 2v M
dx 2 EI
Integrating the bending moment, the beam deflection can be determined by calculation.
Unit curvature represents how a beam is bent, and the beam deflects because of the unit
curvature. It sounds circular. By the way, I don’t like calculus. It’s boring. Could you tell
me how to understand the idea intuitively?
Good question. In fact, calculus is rarely required
in most structural design. Engineering intuition is
much more important than calculus. The point is that a beam
will be concave upward under positive bending moment and
vice versa. You also need to train yourself so that you can
image how the beam will deflect. Solve the problems in the
next section and look at the graphics carefully in order to
develop a good intuition.
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