Fundamentals of Structural Analysis Episode 1 Part 7 ppsx

Analyze the loaded frame shown and draw the thrust, shear and moment diagrams... Before drawing the thrust, shear, and moment diagrams, we need to find the nodal forces that are in the d

Beam and Frame Analysis: Force Method, Part I by S T Mau

115

For the thrust diagram, we designate tension force as positive and compression force as nagative For shear and moment diagrams, we use the same sign convention for both beams and columns For the vertically orientated columns, it is customary to equate the

“in-side” of a column to the “down-side” of a beam and draw the positive and nagetive shear and moment diagrams accordingly

Thrust, shear and moment diagrams of the example problem.

Example 10 Analyze the loaded frame shown and draw the thrust, shear and moment

diagrams

3 m

3 m

3 m

3 m

3 m

3 m

T

V

M

-5 kN

-3.75 kN

15 kN-m

-15 kN-m

15 kN-m -15 kN-m

Beam and Frame Analysis: Force Method, Part I by S T Mau

Statically determinate frame example problem.

Solution. The inclined member requires a special treatment in finding its shear diagram (1) Find reactions and draw FBD of the whole structure

FBD for finding reactions.

(2) Draw the thrust, shear and moment diagrams

Before drawing the thrust, shear, and moment diagrams, we need to find the nodal forces that are in the direction of the axial force and shear force This means we need to

decompose all forces not perpendicular to or parallel to the member axes to those that are The upper part of the figure below reflects that step Once the nodal forces are properly oriented, the drawing of the thrust, shear, and moment diagrams is achieved effortlessly

5 m

3 m

4 m

15 kN/m

5 m

3 m

4 m

60 kN

R=15 kN

15 kN

60 kN

c

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Thrust, shear, and moment diagrams of the example problem.

60 kN

48 kN 36 kN

15 kN

12 kN

9 kN

7.2 kN/m 9.6 kN/m

12 kN 9 kN

39 kN

48 kN

60 kN

15 kN

15 kN

60 kN

T

V

-12 kN

-48 kN

-60 kN

75 kN-m

9 kN

-39 kN

15 kN

M

-75 kN-m

75 kN-m

48 kN/5m=9.6 kN/m

36 kN/5m=7.2 kN/m

0.94 m

5m (9)/(39+9) =0.94 m

4.22 kN-m

M(x=0.94)=(9)(0.94)-(9.6)(0.94)2 /2=4.22 kN-m 0.94 m

5 m

Beam and Frame Analysis: Force Method, Part I by S T Mau

Problem 2. Analyze the beams and frames shown and draw the thrust (for frames only), shear and moment diagrams

Problem 2 Beam Problems.

3 m

5 m

10 kN

3 m

5 m

10 kN

3 m

5 m

3 kN/m

3 m

5 m

3 kN/m

3 m

5 m

10 kN-m

3 m

5 m

10 kN-m

6 m

10 kN-m

6 m

4 m

2 kN/m

5@2m=10m

10 kN

5@2m=10m

2 kN/m

4 m

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Problem 2 Frame problems.

5 m

5 m

10 kN

5 m

5 m

10 kN-m

5 m

5 m

10 kN

5 m

5 m

10 kN-m

5 m

5 m

10 kN

5 m

5 m

10 kN-m

Beam and Frame Analysis: Force Method, Part I by S T Mau

Beam and Frame Analysis: Force Method, Part II

Deflection of beams and frames is the deviation of the configuration of beams and frames from their un-displaced state to the displaced state, measured from the neutral axis of a beam or a frame member It is the cumulative effect of deformation of the infinitesimal elements of a beam or frame member As shown in the figure below, an infinitesimal

element of width dx can be subjected to all three actions, thrust, T, shear, V, and moment,

M Each of these actions has a different effect on the deformation of the element.

Effect of thrust, shear, and moment on the deformation of an element.

The effect of axial deformation on a member is axial elongation or shortening, which is calculated in the same way as a truss member’s The effect of shear deformation is the distortion of the shape of an element that results in the transverse deflections of a

member The effect of flexural deformation is the bending of the element that results in transverse deflection and axial shortening These effects are illustrated in the following figure

dx x

Beam and Frame Analysis: Force Method, Part II by S T Mau

Effects of axial, shear, and flexural deformations on a member.

While both the axial and flexural deformations result in axial elongation or shortening, the effect of flexural deformation on axial elongation is considered negligible for

practical applications Thus bending induced shortening, ∆, will not create axial tension

in the figure below, even when the axial displacement is constrained by two hinges as in the right part of the figure, because the axial shortening is too small to be of any

significance As a result, axial and transverse deflections can be considered separately and independently

Bending induced shortening is negligible.

We shall be concerned with only transverse deflection henceforth The shear

deformation effect on transverse deflection, however, is also negligible if the length-to-depth ratio of a member is greater than ten, as a rule of thumb Consequently, the only effect to be included in the analysis of beam and frame deflection is that of the flexural deformation caused by bending moments As such, there is no need to distinguish frames from beams We shall now introduce the applicable theory for the transverse deflection

of beams

Linear Flexural Beam theory—Classical Beam Theory The classical beam theory is

based on the following assumptions:

(1) Shear deformation effect is negligible

(2) Transverse deflection is small (<< depth of beam)

Consequently:

∆

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(1) The normal to a transverse section remains normal after deformation (2) The arc length of a deformed beam element is equal to the length of the

beam element before deformation

Beam element deformation and the resulting curvature of the neutral axis(n.a).

From the above figure, it is clear that the rotation of a section is equal to the rotation of the neutral axis The rate of change of angle of the neutral axis is defined as the

curvature The reciprocal of the curvature is called the radius of curvature, denoted by ρ

ds

dè

= rate of change of angle =curvature

ds

dè

=

ρ

1

For a beam made of linearly elastic materials, the curvature of an element, represented

by the curvature of its neutral axis , is proportional to the bending moment acting on the

element The proportional constant, as derived in textbooks on Strength of Materials or

Mechanics of Deformable bodies, is the product of the Young’s modulus, E, and the

moment of inertia of the cross-section, I Collectively, EI is called the sectional flexural

rigidity

M(x)= k

) (

1

x

ρ , where k = EI.

n.a

ds

dx

dθ

dθ ρ

ds

Beam and Frame Analysis: Force Method, Part II by S T Mau

Re-arrange the above equation, we obtain the following moment-curvature formula

EI

M

=

ρ

1

(7)

Eq 7 is applicable to all beams made of linearly elastic materials and is independent of

any coordinate system In order to compute any beam deflection, measured by the deflection of its neutral axis, however, we need to define a coordinate system as shown below Henceforth, it is understood that the line or curve shown for a beam represents that of the neutral axis of the beam

Deflection curve and the coordinate system.

In the above figure, u and v are the displacements of a point of the neutral axis in the x-and y-direction, respectively As explained earlier, the axial displacement, u, is

separately considered and we shall concentrate on the transverse displacement, v At a typical location, x, the arc length, ds and its relation with its x- and y-components is

depicted in the figure below

Arc length, its x- and y-component, and the angle of rotation.

The small deflection assumption of the classical beam theory allows us to write

Tan θ ~ θ =

dx

dv

where we have replaced the differential operator

dx

d

by the simpler symbol, prime ’.

A direct substitution of the above formulas into Eq 6 leads to

y, v

x, u

v(x)

x, u

dx ds

θ

Beam and Frame Analysis: Force Method, Part II by S T Mau

125

ds

dè

=

ρ

1

=

dx

dè

which in turn leads to, from Eq 7,

EI

M

=

ρ

1

This last equation, Eq 10, is the basis for the solution of the deflection curve, represented

by v(x) We can solve for v’ and v from Eq 9 and Eq 10 by direct integration.

Direct Integration If we express M as a function of x from the moment diagram, then

we can integrate Eq 10 once to obtain the rotation

θ = v’=∫ dx

EI

M

(11)

Integrate again to obtain the deflection

v = ∫∫ dxdx

EI

M

(12)

We shall now illustrate the solution process by two examples

Example 11 The beam shown has a constant EI and a length L, find the rotation and

deflection formulas

A cantilever beam loaded by a moment at the tip.

Solution. The moment diagram is easily obtained as shown below

Moment diagram.

Mo

x

Mo

M

Beam and Frame Analysis: Force Method, Part II by S T Mau

Clearly,

M(x) = Mo

Condition 1: v’ = 0 at x = L C1= − Mo L

EIv’ = Mo (x −L)

2

2

x −Lx)+C2

2

2

L

EIv = Mo (

2

2

2 L

x + −Lx)

Rotation: θ = v’ =

EI

Mo

EI

Mo L

Deflection: v =

EI

Mo

( 2

2

2 L

EI

Mo

2

2

L

The rotation and deflection at x=0 are commonly referred to as the tip rotation and the tip

deflection, respectively

This example demonstrates the lengthy process one has to go through in order to obtain a deflection solution On the other hand, we notice the process is nothing but that of a integration, similar to that we have used for the shear and moment diagram solutions Can we devise a way of drawing rotation and deflection diagrams in much the same way

as drawing shear and moment diagrams? The answer is yes and the method is called the Conjugate Beam Method

Conjugate Beam Method In drawing the shear and moment diagrams, the basic

equations we rely on are Eq 1 and Eq 2, which are reproduced below in equivalent

forms

Clearly the operations in Eq 11 and Eq 12 are parallel to those in Eq 11 and Eq 12 If

we define

Beam and Frame Analysis: Force Method, Part II by S T Mau

127

−

EI

M

as “elastic load” in parallel to q as the real load,

then the two processes of finding shear and moment diagrams and rotation and deflection diagrams are identical

Rotation and deflection diagrams: −

EI

M

We can now define a “conjugate beam,” on which an elastic load of magnitude −M/EI is applied We can draw the shear and moment diagrams of this conjugate beam and the results are actually the rotation and deflection diagrams of the original beam Before we can do that, however, we have to find out what kind of support or connection conditions

we need to specify for the conjugate beam

This can be easily achieved by following the reasoning from left to right as illustrated in

the table below, noting that M and V of the conjugate beam corresponding to deflection

and rotation of the real beam, respectively

Support and Connection Conditions of a Conjugated Beam

At a fixed end of the original beam, the rotation and deflection should be zero and the shear and moment are not At the same location of the conjugate beam, to preserve the parallel, the shear and moment should be zero But, that is the condition of a free end Thus the conjugate bean should have a free end at where the original beam has a fixed end The other conditions are derived in a similar way

Note that the support and connection conversion summarized in the above table can be summarized in the following figure, which is easy to memorize The various quantities are also attached but the important thing to remember is a fixed support turns into a free support and vise versa while an internal connection turns into an internal support and vise versa

Beam and Frame Analysis: Force Method, Part II by S T Mau

Conversion from a real beam to a conjuagte beam.

We can now summarize the process of constructing of the conjugate beam and drawing the rotation and deflection diagrams:

(1) Construct a conjugate beam of the same dimension as the original beam

(2) Replace the support sand connections in the original beam with another set of

supports and connections on the conjugate beam according to the above table i.e fixed becomes free , free becomes fixed etc

(3) Place the M/EI diagram of the original beam onto the conjugate beam as a distributed

load, turning positive moment into upward load

(4) Draw the shear diagram of the conjugate beam, positive shear indicates

counterclockwise rotation of the original beam

(5) Draw the moment diagram of the conjugate beam, positive moment indicates upward deflection

Example 12 The beam shown has a constant EI and a length L, draw the rotation and

deflection diagrams

A cantilever beam load by a moment at the tip.

Solution.

(1) Draw the moment diagram of the original beam

Moment diagram.

θ =0

v=0

V≠0

M≠0

θ ≠0

v≠0 θ =0

v=0

V≠0

M=0 V M≠0=0

V=0 M=0

V≠0

M≠0

V=0

M=0

Real

Conjugate

V≠0

M≠0

θ ≠0

v=0

V≠0

M=0 V M≠0=0

V=0 M=0

V≠0

M≠0

V=0 M=0

θ ≠0

v=0

Mo

x

Mo

M

Beam and Frame Analysis: Force Method, Part II by S T Mau

129

(2) Construct the conjugate beam and apply the elastic load

Conjagte beam and elastic load.

(3) Analyze the conjugate beam to find all reactions

Conjagte beam, elastic load and reactions.

(4) Draw the rotation diagram ( the shear digram of the conjugate beam)

Shear(Rotation) diagram indicating clockwise rotation.

(5) Draw the deflection diagram ( the moment diagram of the conjugate beam)

Moment(Deflection) diagram indicating upward deflection.

Example 13 Find the rotation and deflection at the tip of the loaded beam shown EI is

constant

Find the tip rotation and deflection.

Solution. The solution is presented in a series of diagrams below

Mo

x

Mo/EI

x

Mo/EI

MoL /EI

MoL 2 /2EI

−MoL /EI

MoL 2 /2EI

Beam and Frame Analysis: Force Method, Part II by S T Mau

Solution process to find tip rotation and deflection.

At the right end (tip of the real beam):

Shear = 5aMo/3EI θ = 5aMo/3EI

Moment = 7a2Mo/6EI v = 7a2Mo/6EI

Example 14 Draw the rotation and deflection diagrams of the loaded beam shown EI

is constant

Beam example on rotation and deflection diagrams.

Mo

Mo/2a

Mo/2a

Mo

Mo/EI

Mo/EI

2aMo/3EI

AMo/3EI

Mo/EI

5aMo/3EI

7a 2 Mo/6EI

aMo/EI

Reactions

Moment Diagram

Conjugate Beam

Reactions

P

Beam and Frame Analysis: Force Method, Part II by S T Mau

131

Solution. The solution is presented in a series of diagrams below Readers are

encouraged to verify all numerical results

Solution process for rotation and deflection diagrams.

P Pa

Reactions

Pa/2

−Pa

Moment Diagram

Conjugate Beam

Pa/2EI Pa/EI

Reactions

Pa/2EI Pa/EI

11Pa 2 /12EI 5Pa 2 /12EI

Shear(Rotation)Diagram

( Unit: Pa 2 /EI )

−1

−1/12

5/12 1/6

Moment(Deflection) Diagram

( Unit: Pa 3 /EI )

a/6

−2701/7776=−0.35

−1/3=−0.33

Beam and Frame Analysis: Force Method, Part II by S T Mau

Problem 3 Draw the rotation and deflection diagrams of the loaded beams shown EI is

constant in all cases

(1)

(2)

(3)

(4)

(5)

Problem 3.

1 kN-m

1 kN

Mo = 1 kN-m

1 kN

2Pa

Beam and Frame Analysis: Force Method, Part II by S T Mau

133

Energy Method – Unit Load Method The conjugate beam method is the preferred

method for beam deflections, but it cannot be easily generalized for rigid frame

deflections We shall now explore energy methods and introduce the unit load method for beams and frames

One of the fundamental formulas we can use is the principle of conservation of

mechanical energy, which states that in an equilibrium system, the work done by

external forces is equal to the work done by internal forces

For a beam or frame loaded by a group of concentrated forces P i , distributed forces q j, and

concentrated moments M k , where i, j, and k run from one to the total number in the

respective group, the work done by external forces is

2

1

Pi∆i +Σ ∫

2

1

(q jdx) vj + Σ

2

1

where ∆i, v j, and θk are the deflection and rotation corresponding to P i , q j ,and M k For a concentrated load, the load-displacement relationship for a linear system is shown below and the work done is represented by the shaded triangular area Similar diagrams can be

drawn for qdx and M.

Work done by a concentrated load.

For the work done by internal forces, we shall consider only internal moments, because the effect of shear and axial forces on deflection is negligible We introduce a new entity,

strain energy, U, which is defined as the work done by internal forces Then

W int = U = Σ ∫

2

1

Mdθ = Σ ∫

2

1

EI

dx

M2

(15)

Load

Displacement

∆

P