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Qualitative Structural Analysis

2 - Qualitative Structural Analysis This section focuses on the qualitative behaviour of plane slender frameworks. The primary aim is to facilitate understanding of the response of structures under load and to help in judging the correctness of results without being burdened with insignificant details. With powerful computer methods of structural analysis readily available to most engineers today, this ability is even more important since gross errors can be generated at the press of a button and can easily go unnoticed if the engineer cannot independently interpret the output. The majority of structures designed in practice are statically indeterminate. While the analysis of such structures is more involved than the analysis of statically determinate ones they have many practical design advantages. In addition to the requirements of equilibrium, the analysis of statically indeterminate structures cannot be completed without satisfying the conditions of compatibility. Compatibility implies knowledge of the deformation response of a structure to ensure that its various parts will fit together without unintentional breaks or overlaps and that they will stay connected to the supports.

Structure Deformation

Rigid deformation

Deformation of a body comprises a rigid component and a non-rigid component. Rigid deformation components do not cause a distortion (shape change) of the body in question. A rigid deformation can be defined by specifying the displacement and rotation of a reference point. Non-rigid components can be elastic or plastic. Plastic deformations are permanent and hence the structure does not return to its original shape after removal of the forces responsible for the deformations.

Deformed body

Undeformed body

The scope of this subject is limited to plane slender frame and truss structures undergoing small linear elastic deformations. Since rigid deformations are by definition not associated with a change of shape, they have no direct relation to strains or stresses and hence also no direct relations to internal forces and moments. The effects that rigid deformations can have on forces and moments are of second order and will not be discussed here.

Relationship between Deformation and Forces (Moments) In plane slender frames and trusses, shape distortion is primarily associated with axial deformation leading to axial strains and hence axial stresses. Stresses are conveniently integrated to form stress resultants, usually referred to as internal forces and moments, and are typically used in formulating equilibrium equations. In the example below, a beam segment is distorted by pure axial deformation, which is associated with axial strain, axial stresses and the corresponding axial force.

Axial deformation

Strain

Stress

=E Hooke’s Law (Material)

48349 STRUCTURAL ANALYSIS

F Axial Force

F=∫

dA

Stress Resultant

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

In the next example the shape of a beam segment is distorted by pure flexural (bending) deformation. The linear distribution of strains and stresses shown are consistent with the assumption that plane sections remain plane during deformation. The bending moment (M) is the resultant of axial stresses obtained by integrating the stress in a differential section area (dA) at a distance (y) from the centroid. Section Area A

y

Flexural deformation (causes Curvature)

Strain

Stress

M=∫ ydA

=E Hooke’s Law (Material)

M Bending Moment

Stress Resultant

Note that flexural deformation is associated with curvature, which requires a change in the direction of the axis of the member. Rotation of the member’s axis on its own is a rigid rotation and does not distort the shape and therefore causes no strains, no stresses and no bending moments.

Rigid rotation ( No Curvature:

= Constant = d /dx = 0 )

Axis is curved = changes direction Rotation = Not constant Curvature = d /dx 0

Shear deformations and the associated secondary shear stresses are mostly insignificant in slender structures and are therefore usually ignored. Note however that this does not imply the absence of shear stresses or shear forces since these are necessary for satisfying the equilibrium for example when bending moments change along the axis of a beam. Equilibrium and Compatibility Requirements Equilibrium of every part of the structure must be satisfied. It is therefore an important task of the qualitative analysis to investigate the free-body diagram of the entire structure and its parts.

Free-body diagram Undeformed structure and applied Loading

B

P C

Equilibrium of joint B demands equal and opposite moments at the free-body diagram of the joint

B

C

B

A

Equilibrium requires that no horizontal reaction force appears at A and that the Moment caused by the eccentricity of vertical forces turns anti- clockwise

48349 STRUCTURAL ANALYSIS

MB

MB A

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

It is equally important to study the deformation and to ensure that compatibility is satisfied between parts of the structure and at supports. Compatibility Rigid joint B keeps the relative angle (90o) between BA and BC during deformation.

C

B

P Compatibility The fixed support demands that the axis of AB cannot rotate and hence remains vertical during deformation

A Example 2-1-1 / case 2 In the example below, beam ABC is subjected to a downward load P causing the deformed shape and bending moment diagram shown. Note that AB deforms elastically while BC deforms rigidely. Rigid deformation (rotation) of BC causes no internal strains or stresses and hence induces no internal forces or moments. Although BC rotates, the rotation is constant and therefore the curvature is zero. Since bending moments are proportional to the curvature (M=EI ), the moments in BC are also zero. Load P

Rotation

Displacement

A

v

B

C

Segment BC rotates rigidely but remains straight. (No Curvature / No Brending Moment)

Deformed shape Elastic Deformation

Rigid Deformation

M=0

The deformation in AB causes tension at the bottom fibre. Hence the bending moment diagram is drawn on the tension side.

M Bending Moment Diagram Example 2-1-1 / case 3 In this example, the downward force P2 applied at C causes BC P2 to curve in the opposite sense of P1 AB and hence induces tension at the top fibre. The bending moments are therefore drawn at C B the top of BC. Note that since A the curvature in BC is reversed, there is an inflection point coinciding with zero curvature MB and zero bending moment. Note M1 also that the beam is continuous over the support at B and hence the moment at B is not zero. But there is no moment in support B itself as the support allows the beam to rotate freely similar to children’s seesaw. Inflection point (Zero Curvature)

48349 STRUCTURAL ANALYSIS

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

Example 2-1-1 / case 4 In this example, the fixed support at B prevents rotation and therefore the horizontal tangent (slope) of the undeformed beam axis must be maintained during the deformation. To maintain compatibility at the support the beam’s axis must reverse its curvature at the inflection point as shown. The beam’s free body diagram shows the support reactions and that their directions are consistent with the deformed shape. For example the clock-wise moment at support B would produce curvature in the beam that causes tension of the top fibre.

P

A

Inflection Point

P

B

A

MB M

B Example 2-1-1 / case 5-6 IP

IP MA

MB M

This beam is fixed at both ends and the deformed shape shows two inflection points. Direction of support reactions and bending moment diagram must be qualitatively consistent with the deformed shape. Observe that the fixed-end moments act in opposite directions and are consistent with tension at the top fibre. The applied load P has no horizontal component and hence in this case there are no horizontal support forces.

P P/2

P/2

M

A

P/2

MA

48349 STRUCTURAL ANALYSIS

P/2

B

MB

The inflection points can be thought of as virtual pins where both curvature and bending moment vanish. If the location of the inflection points can be determined, then the analysis can be simplified by considering a simply supported beam resting on two cantilevers as shown. For the symmetrical beam shown, having length L and a point load at mid-span, the inflection points are at quarter points (L/4) and the moments at the supports and at midspan are all equal to M=PL/8.

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

Example 2-1-1 / case 7

w

A

A beam subjected to a uniformly distributed load (UDL) and fixed at both ends behaves qualitatively in the same way as the beam with a point load at mid-span. The main differences are a) The bending moment diagram is parabolic instead of linear and b) The locations of the inflection points are closer to the supports. For the UDL case the inflection point locations are 0.211324L from the supports (approximately L/5). The corresponding moment at 2 midspan is M=wl /24 while at the supports it is twice that value 2 MA=MB=wl /12.

B IP

IP MA

MB M

w

w

A

M

w

w

MA

B

MB

w

w

Example 2-1-1 / case 8 The following example is a combination of the previous ones. Note how the geometry of the deformed shape influences the bending moment diagram which changes sign between inflection points (IP). P

A

C

B

D

IP IP

IP

MD MB

MC

M Top Fibre Tension

Top Fibre Tension Bottom Fibre Tension

48349 STRUCTURAL ANALYSIS

Bottom Fibre Tension

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

Example 2-1-2 / case 1 The next example shows a continuous beam over 2 spans. Observe how continuity over the support at B is maintained as the tangents of the deformed beam’s axis BA and BC have the o same slope during deformation. Hence the angle of 180 between BA and BC is constant before and after the loading is applied. Note also how the free body diagram shows a downward direction of the reaction at A. This direction is consistent with the deformed shape as RA must cause the curvature in AB to be reversed while bending the beam down towards support A. o

=180

P

A

B

C IP

MB

A

B

C

M P

RA RB

RC

Example 2-1-2 / case 2 The behaviour of the L-frame shown below can be compared with the previous 2-span beam by considering that segment AB is rotated about point B from a horizontal to a vertical orientation. Column AB in the L-frame replaces the effect of the support at B since the axial stiffness of beams is generally much higher than the flexural (bending) stiffness. This means that, for example in the L-frame, the deformation of point B will be much smaller than transverse deformations within spans AB and BC. Note also that due to the rigidity of joint B, the angle between BA and BC o o ( =180 in the 2-span beam and =90 in the L-frame) is maintained during deformation and that in both examples the curvature reverses sign at the inflection point to the right of B. MB

P

B

C

MB

B

o

=90

M

C IP

A

48349 STRUCTURAL ANALYSIS

IP

A

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

Example 2-1-2 / case 3 In this example, the support condition of the previous L-frame has been modified by making the support at A as a roller. This has a significant influence on the deformed shape and bending moment diagram. Span BC acts now as a simply supported beam forcing joint B to rotate clocko wise. Since joint B is rigid and the 90 angle between BA and BC must be maintained after the load is applied, column BA must also rotate in a clockwise direction. Column BA remains straight without curvature due to the lack of horizontal support force at the roller A. As a consequence no bending moment can develop along the entire segment AB.

P

B

B

C

o

=90

M

C

A

A

Example 2-1-2 / case 4 A similar result is obtained if the roller support is placed at point C instead of A. Since no horizontal forces are applied, the horizontal support force at A must vanish in order to satisfy the requirement of horizontal equilibrium ( Fx=0). The only difference is that the deformed shape is shifted horizontally to the right because the base of column AB is not allowed to move away from the support. P

B

C

B

C M

o

=90

A

48349 STRUCTURAL ANALYSIS

A

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

Example 2-1-3 / case 1 The T-frame example shown below is similar to the L-frame discussed earlier except that beam BD has been added to it. The rigid joint B forces BA, BC and BD to rotate by the same amount o and hence the 90 angles, between BA and BC and between BA and BD, remain unchanged after the load is applied. Since BA and BD resist the rotation of B, a moment MBC at the end of beam BC is created. To satisfy equilibrium, the sum of moments MBA and MBD must equal MBC. Note that an Inflection Point is present in segment BC but not in BA or BD. P

B

D

C

o

= 90

IP

MBD

B

A

MBC

MBA

MBC MBD

D

C MBA

B

M IP

A

P

C D

B

A

48349 STRUCTURAL ANALYSIS

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

Example 2-1-3 / case 2 In this example a pin has been inserted in BD to the left of point B. Note that the angle between BD and BA changes after the load is applied since the pin at B allows end BD to rotate freely relative to the connection. Because of the pin, BD can not resist the rotation of the connection and remains free of any bending moments and also free of any shear forces. The horizontal support force at D however, indicates that BD is capable of providing axial restraint. If fact, to satisfy equilibrium, the sum of horizontal support forces at C and D must equal the horizontal reaction at A.

D

Pin

IP

B

P

C o

90

o

=90

A MB

D

C

MB

B

M IP

A

P

C D

B

A

48349 STRUCTURAL ANALYSIS

Summary Notes – Section 2 / Page

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Qualitative Structural Analysis

Example 2-1-3 / case 3 A variation to the previous example is to place the pin inside the span of BD. Note that the right angles of the rigid connection at B are maintained. The bending moment diagram remains unchanged and the deformation of BD consists of two straight lines free of curvature. Since deformation is small, the change in length of BD is negligible and does not induce any significant axial force. Straight Curved

B

D

P

Pin

C o

=90

IP

A MB

D

C

MB

B

M IP

A

P

C D

B

A

48349 STRUCTURAL ANALYSIS

Summary Notes – Section 2 / Page 10

Qualitative Structural Analysis

Example 2-1-4 / case 1 The frame shown below combines features discussed earlier. Due to symmetry of loading, support conditions and distribution of section properties, the responses (e.g deformed shape & bending moment diagram) are symmetrical.

P

B

MB

C

B

IP

IP 90o

MC

C M

90o

A

D

A

D

Example 2-1-4 / case 2 In the next frame the fixed support at D forces the deformation of column DC to have a vertical tangent. As a consequence there is a reversal in curvature near the base. Note that symmetry of support conditions is not present and that the horizontal member EB restrains any sidesway.

E

P

B

MB

C IP

B

IP 90o

MC

C M

90o

IP

A

48349 STRUCTURAL ANALYSIS

D

A

D

Summary Notes – Section 2 / Page 11

Qualitative Structural Analysis

Example 2-1-4 / case 3 In the next example the moment reaction at D acts anti-clockwise causing the frame to sway to the left. Since, in comparison with the previous example, side sway of BC is allowed to take place, the frame is more flexible. Consequently the inflection point in CD is lower and the moment at D is less than in the previous example.

P

B

MC

MB

C

C

B

IP

M

IP 90o

90o

IP

D

A

A

D

Example 2-1-5 / case 1 The L-frame shown is subjected to a horizontal force pushing BC sideways, moving point B to the right, rotating therefore the axis of AB and the rigid connection at B also clockwise. Rotation of B induces the curvature and bending moments in BC as shown. Equilibrium of joint B demands equal but opposite moments at BA and BC and leads to the deformation and bending moment diagram shown.

C

B

C

B

M

P M o

=90

A

48349 STRUCTURAL ANALYSIS

A

Summary Notes – Section 2 / Page 12

Qualitative Structural Analysis

Example 2-1-5 / case 2 The portal frames shown below when subjected to a sway force have a similar behaviour to the previous L-frame. The frames are symmetrical with respect to geometry, section properties and boundary supports. The side load at B forces the axes of AB and DC as well...