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Using slope-deflection

Structural Analysis I

Department of Civil Engineering

Using slope-deflectionCourse structure

Semester 1: Engineering Structure I (10 double-lectures) to cover elastic analysis of statically indeterminate structures and stability of a column.

Using slope-deflectionAssessment

January examination: 80%

Coursework CW1 Analysis of Frame: 10%Submission of Question solutions: 10%

Using slope-deflectionLearning objectives for semester 1

At the end of the module, students should be able to demonstrate knowledge and understanding of

• the deflections of beams and frames; • the assumptions and applicability of moment

distribution; • the stiffness matrix method and the use of

relevant software; • how axial and shear stresses are distributed

throughout the cross-section of structural members;

• stress distributions and deflections produced by torsional loading.

• buckling of columns

Using slope-deflectionRecommended text books

Structures Theory and Analysis

– M S Williams & J D Todd

- Macmillan Press Ltd., 2000

Structures

- Prab Bhatt

- Pearson Education UK, 2001

Essential course text

Highly recommended course text

Using slope-deflectionWeekly programme

Revision10

Stresses in unsymmetrical sections9Shear flow/torsion in closed and open sections8Introduction to the stiffness matrix method7

Euler buckling load for a perfect pin-ended strut and struts with other end boundary conditions

6Moment distribution for frames that sway5Moment distribution and slope deflection for frames4

Moment distribution and slope deflection for continuous beams (2)3

Moment distribution and slope deflection for continuous beams (1)2Slope deflection equations1

ContentsWeek

Using slope-deflectionYear One and Two Structures

1. Analysis of statically determinate structures

- Equilibrium: ΣM = 0, ΣF = 0

Using method of joints or method of sections


2

2

)(dx

ydEIxM −=

Direct integration / McCaulay’s method to finddeflections (& slopes)

2. Basic stiffness approach

Compatibility can then be used to solve statically

indeterminate structures

Year Three Structures

Using slope-deflectionYear Three Structures

Will extend stiffness/compatibility approach forstatically indeterminate structures.

Will study more complex structures

Will study non-elastic problems

Will study introduction to computer methods

Will study complex bending problems

- biaxial bending- open sections- varying section etc

Using slope-deflectionRevision of static analysis

Idealisation of loading:


Simplified beams:Simply supported beam


Cantilever beam

Revision of static analysis


Fixed ended beam



Boundary conditions:

3. Fixed

1. Roller

2. Pinned



Tutorial 0 – boundary conditions


Point load at centre-line

Bending moment


Point load off centre-line

Bending moment


UDL on propped cantilever

Bending moment


contraflexure2-span beam


Tutorial 1b – bending moment

Tutorial 1a – deformed shape

Using slope-deflectionSection One Using Slope deflection

To analyse statically indeterminate plane-frame Structures:

1. Write down expressions for the moments at each end of each member in a frame

2. Each expression will contain a fixed-end moment and a slope-deflection moment.


3. Impose equilibrium conditions: i.e. at any joint ΣM = 0. At any pinned end, M = 0.

5. This gives a series of simultaneous equations for the rotations and deflections of the ends… which can be solved.

4. Impose joint rotation compatibility, i.e. a rigid joint which rotates must result in the same rotations for all member ends at that point


These are the moments which are developed at the ends of a loaded beam, when the ends are fully fixed - i.e. no rotations or deflections are permitted

A consistent sign convention must be adopted.

Anti-clockwise moments are positive

Anti-clockwise rotations are positive

Upward deflections and forces are positive

Fixed end moments

Using slope-deflectionFixed end momentsA fixed ended beam is statically indeterminate, but the end moments can be found using the moments-curvature relationship:

2

2)(

dxydEIxM =

Note that this equation is sometimes quoted as:

2

2

)(dx

ydEIxM −=

The minus sign is a result of the sign convention adopted


P

a b

L

MFAB MFBA

Fixed end momentsConsider a fully fixed beam width an off-centre point loadThe end fixities will cause fixed moments MFAB and MFBA

Now take a section at a point x along the length.At x, there is a moment Mx. ΣM = 0

and reactions RAB and RBA

a b

L

A B

RBARAB

x

Mx

( ) 0=−+−+ axPxRMM AFABx ( )axPxRMM ABFABx −−+−=

…thus

Vx

Using slope-deflectionbut ( )axPxRMM ABFABx −−+−= 2

2

dxydEI=

Integrating gives:

( ) AaxPxRxMdxdyEI ABFAB +

−−+−=

22

22

At x = 0, dy/dx = 0 → A = 0 (note x=0 is not valid for 3rd term)At x = L, dy/dx = 0 → RAB = Pb2/L2 + 2MFAB/L

Integrating again gives:

( ) BaxPxL

ML

PbxMEIy FABFAB +

−−⎟⎟

⎠

⎞⎜⎜⎝

⎛++−=

662

2

33

2

22

Using slope-deflection( ) BaxPx

LM

LPbxMEIy FAB

FAB +−

−⎟⎟⎠

⎞⎜⎜⎝

⎛++−=

662

2

33

2

22

At x = 0, y = 0 → B = 0

At x = L, y = 0 → 2

2

LPabM FAB +=

B

MFBARBA

Mx

L-x

Similarly, taking moments on the other side

2

2

LPbaM FBA −=

gives:

Using slope-deflectionFixed end momentsP

L

MFAB

a b

L

A B

RBARAB

MFBA

2

2

LPabM FAB += 2

2

LPbaM FBA −=

Consider point load at centre:a = b = L/2 →

8PLM FAB +=

8PLM FBA −=

Using slope-deflectionFixed end moments

MFAB

L

A B

RBARAB

MFBA

w kN/m

A similar analysis can be carried out to show thatfor a UDL on intensity w kN/m

12

2wLM FAB +=12

2wLM FBA −=

Using slope-deflectionBending Moment Diagram

“The total effect of two different inputs to a system is equal to the sum of their effects when applied separately”

PRINCIPLE OF SUPERPOSITION

This is valid for linear elastic structuresin which changes in geometry caused by external loads is small


A B

25 kNm 15 kNm

Principal of superposition

5m

RAB = 2 kN RBA = -2 kN

Bending moment diagram(drawn on the tension side)

25

15

Using slope-deflectionPrincipal of superposition

A B

28 kN

2.5m 2.5m

RAB = 14 kN RBA = 14 kN


35


A B

28 kN25 kNm 15 kNm

RAB = 16 kN RBA = 12 kN


25

15

d

e

f

df = 35ef = 15


Now let’s try another example:

A B

25 kNm 15 kNmW = 28 kN

5m

Bending moment diagram from UDL alone(drawn on the tension side)

17.5 kNm


A B


Bending moment diagramfrom applied end moments

25

1517.5

Free BMD added

5m

Using slope-deflectionBending moment diagram

A B

W = 28 kN

5mWhat is the minimum moment ? (Note: this is not the moment at the mid-span…!)

x

Consider the general case….

RAB RBA

MABMBAw kN/m

At a distance x from the left hand side:

Maximum moment in the span is at zero shear force

Position of Mmax is at

wxRVForceShear ABx −=,

wRx AB=max


Now2

2wxMxRM ABABx −−=

Hencew

RMw

RM ABAB

AB

2

22

max −−=

ABAB Mw

RM −=2

2

max

wRx AB=max

and

A BA B


5mx

RAB RBA

MAB MBAw kN/m


We can also find the points of contraflexure (zero moment)

is a quadratic equation, which can be rewritten as

Solving for x gives:

wM

wR

wRx ABABAB 22

−⎟⎠⎞

⎜⎝⎛±=

20

2wxMxR ABAB −−=

0222 =+⎟⎠⎞

⎜⎝⎛−

wMx

wRx ABAB


Usually there are two values for x, zero moment :

but there could be just one:

or none :

week1

Documents

statically indeterminate structures

bending moment diagram

bending moment

static analysis

moment distribution

tension side

point load

continuous beams