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MECH2413 Engineering Mechanics Chapter 06: Deflection of Beams

1

Content • In this chapter, we will discuss the geometric result of a beam being stressed: Deflection.

• First of all, a summary of important equations are given for the ordinary bending theory.

• Then the deflection in different cases are discussed.

1) Single region calculation (single equation) 2) Multiple region calculation (multiple equation) 3) Superposition

2

Ordinary Bending Theory • Consider an ordinary (uniaxial) bending,

• Under equilibrium conditions, we have

=− ,

= 3

Ordinary Bending Theory • The bending moment M and the shear force V are the resultants of the normal stresses σ (acting in the x-direction) and the shear stresses t (acting in the zdirection), respectively

=

;

=

;

= 4

Ordinary Bending Theory • The strain ε and the shear strain γ describe the deformation of an arbitrary element of the beam with length dx and height dz. =

=

=

+

=

= = (

+

) 5

Ordinary Bending Theory • The neutral axis is an axis in the cross section of a beam along which there are no longitudinal stresses or strains.

• If the section is symmetric, isotropic and is not curved before a bend occurs, then the neutral axis is at the geometric centroid.

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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: = =

=0

• 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

9

Assumptions The displacement u of a point P which is located at a distance z from the x-axis is given by , = , =

z

• These two assumptions are sufficiently accurate in the case of a slender beam with a constant cross section or with a slight taper.

10

Deflection curve =

= = (

= +

= ∴

=

= )= (

+ )

= =

=

This equation is called constitutive equation for the bending moment. The quantity EI is referred to as flexural rigidity or bending stiffness.

11

Deflection curve • Provided the bending moment M and the flexural rigidity EI are known, we can obtain the slope w’(x) and the deflection w(x), also called the elastic line, by integration. Since w’= whose direction is started from the neutral axis. = ( )= =

+

W’

+

+ 12

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. 13

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

15

=− ( − )

= ( − ) = =

2 6

−

+

−

+

2

+ 16

Single region example #1 • The geometrical boundary conditions lead to the constants of integration: Hence, the slope and the deflection are obtained as

( )= ( )=

2 6

− −

2 17

Single region example #1 • The maximum slope and the maximum deflection (at x = l) are

− = 2

− = 3

18

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

Two regions are separated: the regions I (0 ≤ x ≤ a) and II (a ≤ x ≤ l).

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Multiple region example #1

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Multiple region example #1 There are two boundary conditions:

There are two matching conditions. Since there are no jumps in the deflection or the slope, these quantities are continuous at x = a:

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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 = = = =

+ + + +

24