Advanced Methods in Mechanics of Materials

Homework 2 Revision
Solve problems 4.5-11 – 4.5-16 from the supplied pdf file
(Gere_Problems1):
1000
2

Introduction
› When a beam with a straight longitudinal axis is loaded by lateral forces, the axis is deformed into a curve, called the deflection curve of the beam.
› In this chapter, we will determine the equation of the deflection curve and also find deflections at specific points along the axis of the beam.

Purpose
› The calculation of deflections is an important part of structural analysis and design. For example, finding deflections is an essential ingredient in the analysis of statically indeterminate structures .
› Deflections are also important in dynamic analyses , as when investigating the vibrations of aircraft or the response of buildings to earthquakes .
› Deflections are sometimes calculated in order to verify they are within construction tolerable limits .

Differential Equations Of The
Deflection Curve
Consider a cantilever beam under the action of one concentrated load .
The axis of the beam deforms into a curve , as shown in Fig. 9-1b.
The reference axes are shown. The z axis is directed outward from the figure (toward the viewer).
The deflection the y v is the displacement in direction of any point on the axis of the beam; deflections are positive when upward .

Differential Equations Of The
Deflection Curve
The deflection deflection curve is shown in Fig. 9-2a.
Point m the origin.
v at any point m
1 is located at distance
1 on the x from
A second point distance deflection at this second point is where dv x+dx m2, is located at from the origin. The v+dv , is the increment in deflection as we move along the curve from m1 to m2.

Differential Equations Of The
Deflection Curve
When the beam is bent, there is not only deflection but also rotation.
The angle of rotation 𝜃 of the axis of the beam is the angle between the as shown.
x axis and the tangent to the deflection curve ,
The angle of rotation at point is 𝜃 + 𝑑𝜃 , where 𝑑𝜃 is the increase in angle as we move from point m 1 to point m 2.
m 2

Differential Equations Of The
Deflection Curve
It follows that if we construct lines normal to the tangents, the angle between these normals is d
θ
.
The point of intersection of these normals is the center of curvature 𝑂´ and the distance from 𝑂´ to the curve is the radius of curvature
ρ
.

Differential Equations Of The
Deflection Curve
We see that in which d
θ is in radians and ds is the distance along the deflection curve between points m1 and m2 .
Therefore, the curvature
κ is given by the equation

Differential Equations Of The
Deflection Curve
The sign convention for curvature is pictured in Fig.
9-3.
Note that curvature is positive when the angle of rotation increases as we move along the beam in the positive x direction.

Differential Equations Of The
Deflection Curve
The slope of the deflection curve is the first derivative dv/dx of the expression for the deflection v .
Since dv and dx are infinitesimally small, the slope dv / dx is equal to the tangent of the angle of rotation 𝜃 .

Differential Equations Of The
Deflection Curve
In a similar manner, we also obtain the following relationships:
Note that when the x and axes have the directions y shown in Fig. 9-2a, the slope dv / dx is positive when the tangent to the curve slopes upward to the right.

Beams with Small Angles of Rotation
The deflection curves of most beams and columns have very small angles of rotation , very small deflections, and very small curvatures .
If the angle of rotation
θ is a very small quantity, we see that the distance ds along the deflection curve is practically the same as the increment dx along the x axis.

Beams with Small Angles of Rotation
This same conclusion can be obtained directly from Eq. (9-
4a).
If the angle
θ is small, and the curvature becomes

Beams with Small Angles of Rotation
Since tan 𝜃 = 𝜃 when 𝜃 is small , we can make the following approximation to
Eq. (9-3a):
Thus, if the rotations of a beam are small , we can assume that the angle of rotation
θ and the slope dv/dx are equal .

Beams with Small Angles of Rotation
Taking the derivative of 𝜃 with respect to we get: x in Eq. (9-7),
Combining this equation with
Eq. (9-6), we obtain a relation between the curvature of a beam and its deflection :

Beams with Small Angles of Rotation
If the material of a beam is linearly elastic and follows Hooke’s law , the curvature is in which M is the bending moment and EI is the flexural rigidity of the beam.
Combining Eq. (9-9) with Eq. (9-
10) yields

Differential Equation of the Deflection Curve
This equation can be integrated in each particular case to find the deflection v, provided the bending moment M and flexural rigidity EI are known as functions of x.
the sign conventions to be used with the preceding equations are repeated here

Nonprismatic Beams
In the case of a nonprismatic beam, the flexural rigidity EI is variable, and therefore we write Eq. (9-11) in the form
Additional equations can be obtained from the relations between bending moment M, shear force V, and intensity q of distributed load.

Nonprismatic Beams
Differentiating both sides of Eq. (9-13a) and using Eqs. (9-12a) and (9-12b), we obtain

Prismatic Beams
In the case of a prismatic beam (constant EI), the differential equations become
We can express the differential equations for a prismatic beam in the following forms:

Deflections by Integration Of the Bending-moment Equation
The first equation we will use is the bending-moment equation [Eq. (9-14a)].
Since this equation is of second order, two integrations are required. The first integration produces the slope, 𝑣´ = 𝑑𝑣/𝑑𝑥 and the second produces the deflection v .

Deflections by Integration Of the Bending-moment Equation
The procedure for solving the differential equations is as follows:
› For each region of the beam, we substitute the expression for M into the differential equation and integrate to obtain the slope 𝑣´ . Each such integration produces one constant of integration.
› Next, we integrate each slope equation to obtain the corresponding deflection v . Again, each integration produces a new constant.
› Thus, there are two constants of integration for each region of the beam. These constants are evaluated from known conditions pertaining to the slopes and deflections.

Conditions
The conditions fall into three categories:
Boundary conditions pertain to the deflections and slopes at the supports of a beam.
For example, at a simple support
(either a pin or a roller) the deflection is zero (Fig. 9-5), and at a fixed support both the deflection and the slope are zero (Fig. 9-6).

Conditions
Continuity conditions occur at points where the regions of integration meet; point C in the beam of Fig. 9-7.
The deflection curve is physically continuous at point the deflection at C
C , and therefore for the left-hand part must be equal to the deflection at
C for the right-hand part.
Similarly, the slopes found for each part of the beam must be equal at C .
Each of these continuity conditions supplies an equation for evaluating the constants of integration.

Conditions
Symmetry conditions may also be available.
For instance, if a simple beam supports a uniform load throughout its length, we know in advance that the slope of the deflection curve at the midpoint must be zero.
Any symmetry conditions provide additional equations, but they are not independent of the other equations.

Mechanics of Materials, Eighth Edition, SI
Example 9.1
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Mechanics of Materials, Eighth Edition, SI
Example 9.1 continued
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Example 9.1 continued
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Example 9.1 continued
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Example 9.1 continued
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Example 9.1 continued
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Example 9.1 continued
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Example 9.1 continued
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Example 9.1 continued
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Example 9.1 continued
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Mechanics of Materials, Eighth Edition, SI
Class exercise
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Example 9.1 continued
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Mechanics of Materials, Eighth Edition, SI
Example 9.2
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Example 9.2 continued
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Example 9.2 continued
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Example 9.2 continued
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Example 9.2 continued
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Example 9.2 continued
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Example 9.2 continued
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Example 9.2 continued
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Mechanics of Materials, Eighth Edition, SI
Example 9.3
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Mechanics of Materials, Eighth Edition, SI
Example 9.3 continued
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Mechanics of Materials, Eighth Edition, SI
Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Example 9.3 continued
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Singularity Functions

Example 3-2 (
)

Example 3-2 (
)

Homework 2
Solve problems 9.2-1, 9.2-4, 9.3-2, 9.3-4, 9.3-5, 9.3-6, 9.3-7 from the supplied pdf file (Gere_Problems2):
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