Cec 205-Theory of Structures

8/12/2019 Cec 205-Theory of Structures

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UNESCO-NIGERIA TECHNICAL &

VOCATIONAL EDUCATION

REVITALISATION PROJECT-PHASE II

YEAR 2- SEMESTER 1 THEORY/Version 1: December 2008

NATIONAL DIPLOMA IN

CIVIL ENGINEERING TECHNOLOGY

THEORY OF STRUCTURES

COURSE CODE: CEC205


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CIVIL ENGINEERING TECHNOLOGY

THEORY OF STRUCTURES CEC 205

COURSE INDEX

WEEK 1. 1.0 STATICALLY DETERMINATE FRAMES

1.I INTRODUCTION

1.2 METHOD FOR SOLVING STATICALLY DETERMINATEFRAMES

WEEK 2 2.0 TENSION COEFFICIENT METHOD

WEEK 3 3.0 DEFLECTION OF BEAMS

3.1 INTRODUCTION3.1.1Load3.1.2Span3.1.3Size and shape of Beam3.1.4Stiffness of Material

3.2 Slope and Deflection of simple Beams3.3 DOUBLE INTEGRATION METHOD

3.3.1Computation of slope and deflection of simplebeams and cantilevers by double integration

method.

WEEK 4 4.0 SLOPE AND DEFFLECTION

WEEK 5 5.0 DERIVATION OF DEFLECTION EQUITION

WEEK 6 6.0 DEFLECTION OF A CANTILEVER BEAM

WEEK 7 7.0 MOMENT AREA METHOD


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7.1.1 Usefulness of Moment Area Method

7.1.2 Moment Area Theorem

7.1.3 Computation of slope and deflection of simple beams andcantilever by moment area method

WEEK 8 8.0 EXPRESSION FOR SIMPLY SUPPORTED BEAM

WEEK 9 9.0 PRINCIPLES FOR THE STABILITY OF DAMS,RETAINING WALLS AND CHIMNEYS

9.1 INTRODUCTION

9.2 Stability of Retaining Wall

9.3 Forces on Retaining Walls

9.3.1 Wind Pressure:

WEEK 10 10.0 Liquid (water) pressure.

10.1 Soil Pressure

WEEK 11 11.0 Modes of failure of Retaining Walls, Dams and Chimneys

11.1 Sliding:11.2 Overturning11.3 Overstressing

WEEK 12 12.0 DETERMINATION OF OVER-TURNING MOMENT,

OVERSTRESSING, SLIDING FORCES AND CENTRES FOR GIVEN

DAMS, RETAINING WALLS AND CHIMNEYS

WEEK 13 13.0 CALCULATIONS ON ECCENTRICITY

WEEK 14 14.0 INDETERMINATE STRUCTURES

14.1 INTRODUCTION

14.2 Definitions of determinate, indeterminatestructures and the concept of redundancies.

14.3 Degree of Redundancy or indeterminacy


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14.4 Degree of static Indeterminacy

WEEK 15 15.0 DEGREE OF REDUNDANCY OF THE FRAME

15.1 TEST FOR GEOMETRIC INSTABILITY

:


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WEEK 1

1.0 STATICALLY DETERMINATE FRAMES

1.1 INTRODUCTION

A statically determinate frame is a frame in which the force-actions in the members canbe obtained by the application of equations of static equilibrium. For a plane frame thereare three such equations and for a space frame six.The first consideration is whether any given frame is statically determinate.

Frame WorkThis is defined as an assemblage of bars which is able to resist geometrical distortionunder any system of applied loads.

1.2 METHOD FOR SOLVING STATICALLY DETERMINATE FRAMES

The main methods used for the solution of statically determinate frames are:1. The stress diagram2. Method of sections3. The method of inspection or resolution at joints4. The method of tension coefficientsThe method of tension coefficientsTension coefficient method can be applied to both plane and space frames. The ratio ofthe member force to the length of the member is known as the tension coefficient of thatmember and it can be plus or minus (+ or -) because the force in a member is negative orpositive (i.e. compressive or tensile force). The length is always positive.

Fig.1.0


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The fundamental principles of the method in the case of plane frames are as flows.If AB (fig 1.0) is a bar of length LAB in a frame, having a tensile force in it of TAB, then

the component of this force in the X and Y directions are TAB Cos BAXand TABsin BAY.If the co-ordinates of A and B are XA, YAand XB, YBrespectively then, component of

TABin the X- Direction = TABAB

AB

LXX = tAB (XB-XA).

Where tAB =TAB/LAB and is known as the tension coefficient of the bar AB.Similarly, the component in the Y- direction == tAB (YB-YA).

If at the joint A in the frame there are a number of bar AB, AC,AN and externalLoads XA, YAacting in the X and Y directions, then since the joint is in equilibrium thesum of the components of external and internal forces must be zero. In each of thesedirections. Expressing these relationships symbolically gives the equations:

tAB(XB XA) + (Xi XA).+ tAC(XC XA) +. + tAN(XN XA) + XA= 0(1)

And in the Y direction tAB(YB YA) + (Yi YA).+ tAC(YC YA) +.+ tAN(YN YA) + YA= 0. (2)

A similar pair of equations can be formed for each joint in the frame giving in all 2 Jequations in the case of a frame having J Joints. These equations will contain thetension coefficients as unknowns and if the frame has n members then there are nunknown tension coefficients. But for a plane frame n = 2j 3, hence there are threesuper flows equations. These can be used to determine the reactions or to check thevalues of the tension coefficients obtained from the previous equations.

In the case of space frame each joint has three o-ordinates and the force have

components in three directions, X, Y and Z. Thus if there are J-joints in a space framethe considerations of the equilibrium in the three directions produces 3J equationscontaining n 3J 6 unknown tension coefficients. But n = 3j 6, hence there are sixsuper flows equation which can be used either to determine the reactions or to check thevalues of the tension coefficients.Having found the tension coefficients tABthe force in the bar is he production tAB LAB.

The procedure in using the method is as follows:1. Draw the free body diagram of the structure2. Find out whether it is statically determinate structure3. Take positive directions for X, Y and Z4. Assume that all members are in tension5. Write down equations for each joint in the frame6. Solve equations for tab, etc7. Check values for tAB8. Calculate TAB= LABtAB


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WEEK 2

Example1: Use the method of tension coefficients to determine the forces in the members

of the frame shown in the fig. below

Fig.1.1Solution


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Free Body Diagram

We know D = M+r 2j:. M = 7 r = 3 j=5:. D = 7 + 3 2 (5)

D = 10 10 = 0

The structure is statically determinate. The reactions are first obtained. The pin hashorizontal and vertical components.

Resolving horizontally gives HA= 4KNb/c = 0 => HA= 2 = 2 = 4KN

Talking moments about A gives7.5 VB= 1 x 5 + 1 x 2.5 + 2 x 2 + 2 x 1

=> VB= KN8.15.75.13 =

And V + 0 = VA+ VB= 1+1VA= 2 1.8 = 0.2KN

Note tAB= AB = Tension coefficient.

Joint Direction Equations Tension Coefficient

A X - 2.5 AD 3.5 EA 4. 0 = 0 ......... (1) AD = - 3.90Y - 1.0AD 2.5 AE + 0.2 = 0(2) EA = 1.64

Multiply equation (2) by 2.5- 2.5 AD 3.5 AE 4.0 = 0.(3)- 2.5 AD 6.25 AE + 0.5= 0 .(4)Subtract equation (4) from (3)It gives 0+2.75AE 4.5 = 0

=> AE = 64.175.2

5.4 =

Substitute AE into equation (1)- 2.5AD 3.5 (1.64) 4.0 = 0 =>AD = -3.90

X +2.5AD 2.5CD 1.0 DE +2 =0 .... (1) DE = - 1.64DY +1.0AD 1.0 DC 1.5DE 1 = 0 (2)

Substitute AD into equation (1) & (2) and itgives-2.5CD 1.0 DE 7.75 = 0 .. (3)1.0DC 1.5DE 4.90 = 0 .(4)Multiply equation (4) by 2.5

CD = - 2.44


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- 2.5 CD 1.0DE 7.75 = 0..(5)- 2.5 CD 3.75DE 12.25 = 0 (6)Subtract (6) from (5) it gives0+ 2.75DE + 4.5 = 0 => DE = - 1.64Substitute DE into equation (5)

- 2.5 CD + 1.64 7.75 = 0=> CD = - 2.44.

X + 2.5BC + 4.0 BE = 0 (1) BC = - 2.62BY + 1.0 BC + 0.5 BE + 1.8 = 0 ..(2) BE = 1.64

Multiply equation (2) by 2.5+ 2.5BC + 4.0 BE 0......(3)+ 2.5 BC + 1.25BE + 4.5 = 0..(4)Subtract (3) from (4)0 2.75 BE + 4.5 = 0 => BE = 1.64Subtract BE into equation (1)

2.5BC + 4 (1.64) = 0 => BC = -2.62

`

X + 2.5CD 2.5BC + 1.5CE + 2 = 0 (1) CD = - 2.44CY +1.0CD 1.0BC 0.5CE 1 = 0 .(2) CE = - 1.64

Subtitle BC into equation(1) from (2) andMultiply equation (2) by 2.5+2.5CD + 1.5CE + 8.55 = 0 ..(3)+2.5CD -1.25CE + 4.05 = 0 ..(4)=> CE = - 1.64Substitute into (3) and it gives+ 2.5 CD + 1.5 (-1.64) + 8.55 = 0 =>CD = - 2.44The value of CD here is used for checking

X +3.5AE + 1.0 DE 1.5CE 4.0BE = 0EY +2.AE +1.5DE + 0.5CE 0. 5BE = 0

Joint E is used as a checkb/c all the tension coefficient valuesare known . substitute3.5 x 1.64 + 1.0 (-1.64) 1.5 (-1.64) 4 (1.64) = 05.74 1.64 + 2.46 6.56 = 0 => 0 = 0For Y direction2.5 (1.64) + 1.5 (-1.64) +0.5 (-1.64) -0.5 (1.64) = 04.1 -2.46 0.82 0.82 = 0=> 4.1 4.1 = 0

0 = 0

Member Length (L) Tension Coefficient Force (KN) (t x L)AD 2.52+ 12= 2.7 - 3.90 10.50 compressionDC = 2.7 - 2.44 6.59 compressionCB = 2.7 - 2.62 7.07 compressionAE 3.52+ 2.52= 4.3 + 1.64 7.05 tensionBE 42+ 0.52= 4.03 +1.64 6.61 tension


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CE 1.52+ 0.52= 1.58 -1.64 2.59 compressionDE 1.52+ 12= 1.8 - 1.64 2.95 compression


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Use the tension coefficient methods to determine the member forces for joint A of theshear legs shown in fig 1.2 below:

Fig.1.2Solution

The three equations for joint A the shear legs are formed as follows

(1) in direction x + 2tAB 2tAD= 0(2) in direction y + 3tAB+ 3tAC+ 3tAD+ 21 = 0(3)

in direction z + 2tAB+ 4 tAC+ 2tAD= 0then from (1) tAB = tAD, and adding (1) and (3) gives tAB = - tACand

substituting both into (2)- 3 tAC+ 3 tAC 3tAC= 21tAC= 7 and tAB= - 7 = tADThereforeTAB= - 7 x [2

2+ 22+ 32] = 28 . 86KN Strut.TAC= + 7 x [4

2+ 32] = 35.00KN TieTAD= 28.86KN Strut

Exercise1. A load of 7.2KN is suspended from a soffit by two ropes PQ and QR as shown in

fig.1.3 determine the forces in the ropes

Fig. 1.3


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2. Use the method of tension coefficients to determine the forces in the members of

the frame shown in the fig 1.4

Fig.1.4


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WEEK 3

2.0 DEFLECTION OF BEAMS

2.1 INTRODUCTION

Deflection of a beam is as the deviation of the neutral surface, or elastic curve of theloaded beam from its original Position in the unloaded beam.

Examples:

Factors affecting deflectionThe factors affecting deflection of beams are:1. Load2. Span3. Size and shape of beam4. Stiffness of Material2.1.1 Load

If AB (fig.2.1) represents a beam of span L. meters supported simply at its ends andcarrying a point load of WKN at mid-span. Let us assume that the deflection due to theload is 5mm. It is obvious that, if the load is increased, the deflection will increase. It canbe proved that the deflection is directly proportional to the load.


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2.1.2 Span

In fig. 2.2 (a) and (b) the loads are equal and the weights of the beams, which areassumed to be equal cross-section, are ignored for purpose of discussion. The span ofbeam b will be greater than that of beam a. it can be demonstrated experimentally orproved by mathematics that the deflection of a beam is proportional to the cube of thespan.Therefore L3is a term in the deflection formula.

2.1.3 Size and shape of Beam

Fig. 2.3 (a) and (b) represents two beams (their weights being ignored) of equal span andloading but the moment of inertia of beam b is twice that of beam a. It can be prove thatthe deflection is inversely proportional to the moment of inertia. Moment of inertia istherefore a term in denominator of the deflection formula.

(a) Moment of inertia of beam = 1 unit(b) Moment of inertia of beam = 2 units.

(It may be noted that, since the moment of inertia of a rectangular X- section beam is

12

3bd , doubling the breadth of a rectangular beam decreases the deflection by one half,

whereas doubling the depth of a beam decreases the deflection to one eight of theprevious value).


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2.1.4 Stiffness of Material

The stiffer the material of a beam, i.e. the greater its resistance to bending, the less will bethe deflection, other conditions such as span, load, etc. remaining constant. The measure

of the stiffness of a material is its modulus of elasticity E and deflection is inverselyproportional to the value of E.

2.2 Slope and Deflection of simple Beams

A structure will carry loads provided that the load carrying capacity of the member isnot exceeded. That is, the structure must have adequate stiffness; it should not deflectfrom its original position by more than certain amount. For this reason, codes of practicespecify maximum permissible deflection for a given span of beam, as well as maximumpermissible stresses.

Consider fig. 2.4 below:

Since beams are normally horizontal, the deflection is the vertical deviation () asindicated, and the tangent to deflection curve at point C is assumed to be the slope, whichis an angle Cwith the X axis (Horizontal axis).

Many reasons exist for determining the deflection of a beam and these are:In the design of building, for structural steelwork where the usual limitation that

a beam or joist supporting a plastered ceiling must not deflect more than360

1 of its

span length if cracking of the plaster as to be avoided, for reinforced concrete, thedeflection is generally governed by the span/depth ratio. For timber beams is 0.03 of thespan when the supporting member is fully loaded.Also, in the design of machines and airplanes.


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For analyzing indeterminate beams various method methods are available for thedetermining deflections in beams. The common ones are:

Double integration methodMoment area method

Super position methodMaxells reciprocal theorem methodWilliot Mohr and Analytical method

2.3 DOUBLE INTEGRATION METHOD

From he theory of simple Bending the radius curvature )1.......(..........//1 EIMR =

where E = is the young modulus and I = the flexural stiffness of the beam .

But R/1 =( )[ ] 2

32

22

/1

/

dxdy

dxyd

+

= d2y/dx2-------------- (2)

Assuming linear small displacement theory dy/dx


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Since WL is acting at mid-span Ray= 2

wl and Rby = 2wl by inspection

Suppose x is measured from end A, while y is deflection with respect to x.The bending moment at this distance x from A is Mx= + wl (x) wx (1/2x)

Mx=22

2wxwlx

Since the curvature is negativeEI d2y/dx2 = - m=> Eid2y/dx2= - [Mx]=> Eid2y/dx2 = - wlx + wx2

On integrating

EI d2y/dx2= - wlx + wx

2

EI d2y/dx2= wlx + wx

2

EI dy/dx = CSGcwxWlx

..64

1

32

++

And further integration yields

EI dy/dx = +42wlx

+63wlx

C1

Eiy = - +12

3wlx

+24

4wlx

C1x C2----------G.d.c

The constants of integration C1and C2are evaluation from the boundary conditions for Y= 0, x = o and x = L at the end. Than C2 = 0, x C1= 2


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Therefore, from general equation for deflection

Eiy = - +12

3wlx+

24

4wlxC1 x C2

=> 0 = - 12

)( 3lWL

+24

4wlx

C1 (L)+ 0

=> C1 =24

3wlx

The general equations for slope becomes Eidy/dx = -2464

332wlwxwlx

++

And for deflection Eiy = ( )323 21(24

xxwx

+

Then maximum deflection at mid-span i.e x = L

=> Eiy = ( ) ( ) +323

21

212

242

1LLLLLw = [ ]8432245.0

33 LLLwl +

max = ymax =EI

wl

384

5 4

Assignment:Generate slope and deflection equation for a simply supported beam, carrying aconcentrated lateral load at center as shown below.Establish the maximum deflection equation at mid-span i.e. slope is zero

Fig. 2.7


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WEEK 4

Example:2

Generate general slope and deflection equations for a simply supported beam, carrying aconcentrated lateral load at any point as shown

(+MB= 0 : Ray(2) W(L-a) = 0

=> Ray =( )

2

aLW

=> (+MA= 0 : Rby (2) + W (a) = 0

=> Rby =L

wa

Now consider a section of the beam at distance x from A

If X < a, Mx=( )L

xaLW )

and if x > a

Mx=( )

( )axWL

xaLW

For x< a but Mx= ( )L

xaLW

Mdx

yEId=

2

2

( )

=

L

xaLW

dx

EId2

2


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Integrating SEGCRayxdx

EIdy.)1(

21

12

+= for x > a

( )

==>= axwx

L

aLw

dx

yEIdM

dx

EIdy2

2

2

Therefore, ( ) ESGCaxxxWRayxdx

EIdy..)2(

22

21 1

12

+=

Integrating (1) and (2) further for general deflection equation =>

Eiy =6

1 Ray x3+ C1+ C2--------- (3) for x < a.

Eiy =6

1 Ray x3+ W 1211

2

263 CxCaxx ++ ------------- (4) for x > a.


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WEEK 5

In these equations, C1 C211C and

12C are arbitrary constants.

Now for x = a, the values of y given by equations (3) and (4) are equal, and the sloesgiven by equations (1) and (2) are also equal, since there is continuity of the deflectedform of the beam through the point D. then.

- )5(266

16

1 12

11

23

321

3++

+=++ CxCax

xWRayxCxCRayx

And )6(22

12

1 11

22

12

+

+=+ Cax

xWRayxCRayx

Solving equations (5) and (6)

)5(2612

11

2

3

21 ++

+=+ CxCaxxWCxC

)6(2

11

2

1 +

+= Cax

xWC

But x = a

)5(26

12

11

23

21 ++

+=+ CaCaa

aWCaC

)6(2

11

2

1 +

+= Caa

aWC

From equation (6) )7(2

2

111 +=

WaCC

Substituting 11C in equation (5)

This => )8(6

3

212 =

WaCC

At the extreme ends of the beam y = 0, x = 0 and C 2= 0 for equation ------ (3)i.e. 0 = 0 + 0 + C2=> C2= 0 ------------------------------------------------------- (9)and when y = 0, x = L for equation (4)

0 = -6

1 Rayl3+ W 12

11

2

263 ClCall ++ ------------- (10)

=> )11(6

31

2

= WaC from equation (8)

Substitute 12C in equation (10)

( ) )12(2

166

1 23211 += waaLL

WRayLC

Also from equation 7

( )2

]2

166

1[2232

1WawaaL

LWLRC ay +=


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( ) )13(66

1[32

1 = aLLWLRC ay

After substituting the values of Ray, C1,1

2,2

1

1 , CCC , equation (4) may be written as

Eiy = - ( ) ( ) ( ) )14(6

3266

3223+++ ax

WxaalL

L

waxaL

L

w

Equation (14) may used to define the deflected form in all parts of the beam.

On putting x = a, the deflection at the loaded point D is DYD=:3

22

EI

bWa

Similarly when, W is at the centre of the beam i.e.2

La = then = y =EI

WL

48

3


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WEEK 6

Example 3:Generate slope and deflection equations for a cantilever with a uniformly distributed load

as shown in fig 2.9. Establish max. deflection equation at free end.

Since WL acts at mid-span Ray= WL by inspection Rax = 0 and MA= -2

2WL

The bending moment at a distance x from origin A is MX= - ( )2

2Xl

W

But XMdx

yEId=2

2

Integrating Eidy/dx = W(3

322 xLxxL + )+C1-------------------------------G.S.E

And integrating again

Eiy = W (123

12

1 4322 xLxxL + )+C1 X +C2-------------------------------G.D.E

For boundary conditions at built-in end, X = 0 dy/dx = 0 and y = 0Thus C1= C2= 0Therefore at the free end (B), i.e. x = L the vertical deflection is

max =

EI

WL

8

4

Example 4Generate slope and deflection equations for a cantilever carrying a concentrated load atthe free end, as shown in fig 2.9.1. Establish max. deflection and slope equations at thefree end.


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Since W is at free end

Ray= W, Rax= 0

MA= - WL

The bending moment at a distance x from the built-in end is

MX = - W(L-x).

We know XMdx

yEId=

2

2

)(2

2

xLWdx

yEId=

Integrating ...)2

1(1

2 ESGCxLxWdx

EIdy+=

And further integration gives: Eiy = W(Lx2/2 - EDGCXCx ..)6

121

3++

Boundary conditions at end x = 0, = 0, 0=xd

dy

The G.S.E gives C1 = 0 and G.D.E gives C2 = 0


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At the three free end, x = L, max =( )

( )EI

WLLL

EI

LW

33

6

) 32

=

The slope of the beam at free end is L=EI

WL

2

2


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WEEK 7

2.4 MOMENT AREA METHOD

The moment area method is one of the common methods used in computing flexuraldisplacement of beams and frames. It is often referred to as a geometric method because

the displaced shape of a structure is a direct function of the strain in the structure, Oncethey are computed or measured the displace shape is uniquely determined by summingthe effects of the strain.

2.4.1 Usefulness of Moment Area Method

The moment area is the simplest method for beams with how degree ofredundancy.

It provides rapid techniques for computing displacements of statically determinatebeams or frames.

It is also highly useful in computing displacements of statically indeterminatestructures with known bending moment diagram (B.M.D)

2.4.2 Moment Area Theorem.There are two moment area theorems both applicable to a loaded beam which isoriginally straight.

Theorem 1

The angle between the tangents to deflection curve at two points A and B is equal to thearea of the bending moment diagram between those points divided by EI.

Theorem 2

The deflection () of point B from the tangent at point A is equal to the moment of the

area ofEI

Mdiagram between the points A and B taken about B.

B =X dxEIMb

a

Note: The area must be taken positive when the bending moment is positive andnegative when the bending moment is negative. i.e. positive are means that B is


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above the tangent at A, whereas negative moment means that B is between thetangent as shown in fig 3.2 above.

2.4.3 Computation of slope and deflection of simple beams and cantileverby moment area method

Example (1)Use the momentarea methods to generate expressions for the simply supportedbeam, carrying a uniformly distributed load, as shown in fig. 3.3 below. Establishthe maximum deflection equation at the mid-span i.e. slope = 0 and slope at theleft end of the beam.

Ray=2

,0,2

WLRR

WLR byaxby ===

By inspection and MD = +8

2WL


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=> The Maximum deflection =YAD= Ax.Where A = Area of the B.M.D and x= Centroid of the quadrant of parabola

A =EI

WLL 1

823

2 2

X = ( ).285L

=> Maximum deflection max = Ax =EI

LWLL 1

28

5

823

2 2

Therefore, max=EI

WL

384

5 4

Also, by the first moment-area theorem the slope at the left end of the beam (i.e.

A) = Area under the quadrant of parabola => A= A = ( )EI

WLL

1

82/

3

2 2

A=EI

WL24

3

Example (2)Use the moment area methods to generate the expressions for the simplysupported beam, carrying a concentrated load at center, as shown in fig. 3.4below. Establish the maximum deflection equation at mid-span (i.e. slope == 0)and slop at the left end of the beam.

Ray=2

,0,2

WRR

Wbyax == by inspection

MD = +

22

LW


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MD =4

WLi.e. the maximum deflection of the beam occur at the centre D, and a

tangent to the elastic curve at D will be horizontal, that is parallel to theundeflected axis of the beam. Hence the maximum deflection max= YAD(i.e. thedisplacement of point A from the tangent at D)which in turn equals the moment

about A of the moment area between A and D, divided by EI.=> max= YAD= Ax

A =EI

IWLL

422

1

X =

23

2 L

max = AX = 1/2

23

2

42

L

EI

WLL

max=

EI

WL

48

3

Also by the first moment area theorem, the slope at the left end of the beam (i.eA) = Area under the half of triangle.

=> A= A = EI

WLL 1

42

A =EI

WL

16

2


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WEEK 8

Example (3)Use the moment area methods to generate the expressions for the simplysupported beam, carrying a concentrated lateral load at a point from support Aand B (i.e. a < b), as shown in fig. 3.5 below. Establish the deflection equation atpoint of loading (1) and slope at the left end of the beam (ADand A)

MA= 0 : W(a) Rby(L) = 0

Rby =L

Wa

Similarly MB= 0 + RayL(a) Wb = 0

Ray=L

Wb

MD=L

Wb(a)

AD= AX

=> Ax = 1/2(a) )(3

2a

LEI

wab

The deflection = AD=LEI

bWa

3

3

Also slope A= Area under triangle AD. = A

=> A= 1/2(a)LEI

Wab


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A= 1/2LEI

bWa2

Example (4)Use the moment area methods to generate expression for a cantilever with a

uniformly distributed load, as shown in fig.3.6. Establish maximum deflectequation at free end and also the slope

MA= 0 => Ma= -2

2WL

Rax= 0

Ray= WL by inspection

max= Ba= AX =

)(

4

3

2)(

3

1 2L

EI

WLL

The deflection at free end Ba=EI

WL

8

4

Slope at free end = B =EI

WL

6

3

Example (5)Use the moment area methods to generate expressions for a cantilever withcarrying a concentrated load at point D away from the supports (i.e. a > b) asshown in fig. 3.7. Establish maximum deflection equation at free end and also theslope.


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MA= 0 => Ma= - W(a)Ray=W and Rax= 0 by inspectionThe deflection Baat the free end.Ba= Ax

A = 1/2 ( )

EI

waa

X = ( )

+ ba

3

2

Ba= Ax =1/2 ( )

EI

waa ( )

+ ba

3

2

= BA= [ ]baEI

wa 326

2+

Therefore, slope at the free end B = Area underEI

mcurve

=> Bm = A = 1/2(a)

EI

Wa

=> B =

EI

Wa

2

2

Exercise

Use the moment area methods to generate expression for the cantilever shownin fig.3.8 carrying a concentrated load at point D away from the support(i.e a < b). Establish the deflection and slope equations at the free end


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WEEK 9

4.0 PRINCIPLES FOR THE STABILITY OF DAMS, RETAINING WALLSAND CHIMNEYS

4.1 INTRODUCTION

Dams are structures that are meant to store water that may be used for different purposesuch as drinking; irrigation hydro-electricity etc. reliable records have shown that the firstdam was built on the Nile River sometimes before 4000. BC. It was used to divert theNile and to provide a site for the ancient city of Memphis. The oldest dam that is still inuse is the Almanza Dam in spain, which was constructed in the sixteenth century Dammust be able to suites the following conditions:

Karu Dam in Iran1. Foundation:

The foundation made for the construction of a dam must be able to support theweight of the dam, without excessive deformation / stress. To ensure the abovecondition is met site/soil investigation is carried out in order to determine;(a) nature of the suit in the area and its geology(b) Check if faults or other cracks exists in the site(c) Determine the characteristics of the foundation materials such as

permeability, bearing capacity, shear stress.

1.

See page:Dams are constructed to trap water; therefore, permeability soil should beavoided.

Fig.4.0 Type of Dams.


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GILBOA DAM in New York is an Hoover Dam, A concrete Arch gravityDamExample of a solid gravity dam

San Luis dam California is an embankment Dam

Materials used for the construction; Gravity Dam - concrete and rubble masonry. Arch Dam concrete. Buttress Dam concrete, timber and steel. Embankment Dam earth or rock.


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4.2 Stability of Retaining WallA retaining wall is a wall built to hold back earth or other solid material . The wallas it rests on soil or on a concrete footing and acted upon by the horizontal forces

induced by the material, transmits to the soil or footing stresses which consists of:(i) Direct stress from the walls weight and(ii) Stress due to the overturning moment.

These stresses bring about a combined action of the vertical and horizontalinclined forces on the overall behaviour of the wall. As a result of that action, thewall may fail in three modes, as follows:

(a) Sliding(b) Overturning, and(c) Overstressing

4.3 Forces on Retaining WallsForces acting on retaining walls are usually horizontal forces [P], which could bedue to wind, liquid (water), soil (Granular material) depending on the retainedmaterials and the nature of there forces exert on the wall will be studied first.

4.3.1 Wind Pressure: The action of a horizontal force like the wind pressure canbe investigated on a chimney, which is simply the tallest or protruding part ofbuilding, for letting out smokes from the kitchen. Consider the fig. 4.1 below

Fig. 4.1


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WEEK 10

4.3.2 Liquid (water) pressure.Liquid (water) pressure is more pronounced in the dams. Consider fig. 4.2

below, which is a vertical surface AB,Fig. 4.2

And the face of a wall which is retaining a liquid. It can be shown that a cubicmeter of liquid, situated at depth h (m) below the surface, exert a pressure of wh(kN) outward on all its six side surfaces. W in this case Is the equivalent densityor unit weight of the liquid in (kN/m3).

Thus, the intensity of outward pressure varies directly with the depth andwill have a maximum value of WH (kN/m2) at H (m), the maximum depth asindicated in the fig4.2 (b). at the surface of the liquid (h=0), the pressure will bezero. So, as the maximum is WH (kN/m2), the average pressure between A and B

is 1/2WH (kN/m

2

).In dealing with retaining wall generally, it is convenient, as said earlier, toconsider the forces acting on 1m length of wall, and that is an area of wall H (m)high and 1m measured perpendicular to the plane of the diagrams. Fig. 4.2(a & b).thus wetted area concerned is H (m2), the total force caused by water pressureon a 1m strip of wall is ;

wetted area x average pressure = H x 1/2WH = 1/2WH2

this is total resultant force on the walls vertical surface from the liquid is (as willbe seen from fig 4.2 b). The resultant of a large number forces, which range fromzero at the top to WH at the bases. The resultant will therefore act at a point 1/3rdof H from the base, as shown in fig. 4.2c.Note that, if liquid does not reach the top of the retaining wall, as shown infig. 4.3 below:Fig.4.3


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Then the resultant force is calculated with H as the depth of the liquid and not asthe height of the retaining wall. The force is again 1/2WH2and it acts at a point

1/3H (one-third the depth of the liquid) form the walls base.In cases where the wall in contact with the water is not vertical as shown infig. 4.4 below, the wetted area will be larger than in the case of vertical back,another resultant pressure will thus be Increased to 1/2WHL (i.e. wetted area willbe L (m2) instead of H (m2) considering 1m run of the wall.

Fig.4.4

4.3.3 Soil PressureIt is obvious that pressures on wall form retaining soils or other granular materials

cannot be determined with quite the same accuracy as with water. Soils vary inweight and character, they behave quite differently under varying conditions ofmoisture, etc and in general, the resultant pressures on vertical and non-verticalsurfaces from soils which is obtained from various soil pressure theories.Numerous theories vary in the assumptions which they make and the estimatedpressures which they determine. Ranking theory is most applicable.

Rankine Theory of Soil Pressure


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It has been seen that a cubic meter of liquid at a depth h below the surfacepressure outwards horizontally by an amount wh (kN/m2) [w being the equivalentdensity of liquid]. In the case of soil weighting W (kN/m 3), the outward pressureat a depth of h (m) below the surface will be less than wh (kN/m2), since some of

the soil is self-supporting. Consider the fig. 4.5 below, for example,

Fig.4.5

Of the soil retained by the vertical face AB. In figure 4.5, if the retaining face ABwas removed, then some of the soil would probably collapse at once, and in thecourse of time, the soil would assume a line BC, as shown. The angle madebelow the horizontal and the line BC varies for different type of soil and is call theangle of repose angle of internal friction of the soil.It can be said; therefore that only part of the soil was in fact being retained by thewall and as exerting pressure on the wall. Thus, it follows that the amount ofpressure on the wall from the soil depends upon the angle of repose for the type ofsoil concerned, and Rankin theory states in general terms that the outwardpressure per square meter at a depth of h (m) due to a level of fill of soil is:

wh

+

sin1

sin1 kN/m2

As compared with (wh) kN/m2

in the case of liquids. Thus by similar reasoningas used in the case of the liquid pressure, the maximum pressure at the bottom ofthe wall is given by:

Maximum Pressure = WH

+

sin1

sin1 kN/m2


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Average Pressure = 1/2 WH

+

sin1

sin1 kN/m2

The soil acts at this average rates on an area of H (m2) of wall, so that the totalresultant force per meter sun of wall is

P = 1/2WH2

+

sin1

sin1kN

And this acts, as shown in fig.4.6 below at 1/3H above the base of the wall.Fig.4.6


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WEEK 11

4.4 Modes of failure of Retaining Walls, Dams and Chimneys

4.4.1 Sliding:

The possibility that a wall may slide along its base exists, unless the weight of thewalls sufficient to prevent such movement. The ability of the walls to resist thissliding depends upon the interaction of the weight of the wall and the frictionbetween the material of the wall and soil directly in contact with the base of thewall. Consider a arbitrary body of weight W, resting on a level surface as shown

in fig. 4.7 (a ,b & c) below.

Fig. 4.7

Apply a small force P (not enough to move the body) in fig. 4.7 (b), leads to aresultant R1for W and P on the level surface. Hence R2will be inclined as shownat angle of Increasing p gradually would lead to an increment in , until a

certain load (which depends on the nature of the two surface in contact and on theweight w), the body will move horizontally. The angle which the resultantupward thrust makes with the vertical at the stage where the block starts to slide isknown as the angle of internal friction between the two surfaces. From fig 4.7 (c)

Tan = Pm/ w = least force that will cause slidingWeight of block


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Tan = Coefficient of friction for the two materials is denoted as . (it usuallyvaries between 0.4. to 0.7 for most materially.From equation (1)

Pm= W

Note that, in the case of retaining walls, Pm, the force which would cause sliding,can be calculated as W x coefficient of friction, and the horizontal for P of theretained material should not exceed approximately half of the force Pm. In otherwords, the factor of safety F.S = Pm/p 2 for safety.

4.4.2 OverturningA retaining wall may have quite a satisfactory resistance to sliding, but thepositive action of the horizontal force may tend to overturn it about its toe asshown in Fig 4.8 below

Fig.4.8

Assuming that sliding does not occur, equilibrium will be upset as the wall justlifts off the ground at the heel B, the turning point being the toe A.

At this time, the over turning moment due to the force P is just balanced bythe restoring or balancing moment due to the weight of the wall, the wall beingbalanced on the edge A. Taking distances assuming that wall is vertical, momentabout A is

P x distance AD = W x distance AC or overturning moment = restoring (balancing) moment. Applying a factor of

safety of 2 against overturning

Factor of safety =gMomentOverturnin

Momentbalancingstoring )(Re

And therefore, Overturning moment = 1/2 x Restoring (balancing) moment


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4.4.3 Overstressing

The maximum stress resulting from the combination of direct and bendingstresses must be kept within Limits of the safe bearing capacity of the soil

supporting the wall. Consider fig. 4.9 below when the weight of a wall per meterW, and

Fig.4.9

The resultant pressure from the soil or liquid P, have been calculated, these twoforces may be compounded to a resultant as shown above. It can be shown that

the position along the base at when this resultant cuts (i.e. at S) has an importantbearing on the stability of the wall and on the pressures exerted by the wall uponthe earth beneath.

The pressure under the wall at F is equal to sum of direct stress andbending stress.=> Pressure under wall at (F) = Direct stress + bending stress.

But direct stress =BaseofArea

wallofweight=

A

Wand

Bending stress =

Z

M

where M is moment due to the eccentricity of the resultant, and Z is the sectionmodules about an axis through the centre line of the base.

Pressure under wall at (Ftoe) =Z

M

A

W+ similarly pressure under wall at


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G (Heel) =Z

M

A

W for no tension the resultant of the applied loads P and W

must cross the base within its middle third or eccentricity e should not be greaterthan 1/6 x width of base.


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WEEK 12

4.5 DETERMINATION OF OVER-TURNING MOMENT, OVERSTRESSING,

SLIDING FORCES AND CENTRES FOR GIVEN DAMS, RETAINING

WALLS AND CHIMNEYS

Example:1

A masonry dam retains water on its vertical face. The wall is as shown below,what is about 3.7m, but the water level reaches only 0.7m from the to of the wall.What is the resultant water pressure per meter run of wall?

Fig.4.9.1

Solution:

The equivalent density of water, W = 10 KN/m3

P = 1/2WH2=> p = 1/2 (10) (32) = 45KN acting at 1/3 (3) = 1m above the base.

Example (2)A soil weigh 15KN/m3 and having = 300, extents a pressure on a 4.5m highvertical force what is the resultant horizontal fore per meter run of wall?


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Solution:

Sin = sin 300= 0.5, But W = 15KN/m3, = 300, H= 4.5m

P = 1/2WH2

+

sin1

sin1kN

= 1/2 x 15 x 4.5 x 4.5 x

+

5.01

5.01

P = 50.63KN

Example (3)

The masonry dam shown in fig 4.9.2 retains water to the full depth. Thecoefficient of friction between the base of the wall and the earth underneath is 0.7.Check if the wall is safe against sliding.

Fig.4.9.2

Solution:

P = actual horizontal pressure on side of wall = 1/2 WH2= 1/2 x 10 x 42= 80KNPm= horizontal force which would just case slidingPm = w => 0.7 x w = 0.7 x 1/2 (1 + 3) 4 x 18=> 0.7(144) = 100.8KN


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The actual pressure (80KN) exceeds half the value of Pmand the factor of safety

against sliding is F.S = 26.180

8.100==

P

w< 2

Which is undesirable.

Example (4)

Along boundary wall 2.7m high and 0.4m thick is constructed of brickworkweighing 18KN/m3as shown fig.4.9.3 If the Maximum wind pressure uniformlydistributed over the whole height of the wall is 500N/m2, Calculate the factor ofsafety against overturning, neglecting any small adhesive strength between thebrickwork and its base.

Fig. 4.9.3

Solution:

Weight of 1m sum of wall = 21.7 x 1.0 x 0.4 x 18 = 19.44KN

Wind Pressure on 1m sum of wall = 2.7 x 1.0 x 0.5 = 1.35KN

And this can be taken as acting at the centre of the height of the wall for purposesof taking moments about [0] as shown 4.9.4

Fig.4.9.4


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Restoring Moment = 19.44 x 1/2(0.7) = 3.88KNm

Overturning moment = 1.35 x 1/2x 2.7 = 1.823 KNm

Therefore FS against overturning = 13.2823.188.3 = .

It will be found that a satisfactory factor of safety against overturning is achievedif the resultant of the horizontal and vertical forces crosses the base of the wallwithin its Middle third i.e. when no tensile stresses are allowed to develop inthe wall.


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WEEK 13

Example (5)

A retaining wall of weight [W=28KN/m] and resultant pressure (p=10KN) fromliquid. The base is 2.5m long by 1m breadth as shown. The weight per meter ofthe retaining wall is acting at 1.0m from the heel. (i) Find the position where theresultant force per meter (W) cuts the base from the heel of the wall. (ii) find theeccentricity [e] of the resultant force [W] per meter from the centre line of thebase (iii) Find the Pressure under wall at th heel, and the toe respectively.


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Fig.4.9.5

Solution:

Consider the wall above, W acts through the centroid of the wall section

(i) From the similar triangles. ASC and ADE

mx

yy

5.028

4.110

28

10

4.1===>=

=> Thus the resultant force cuts the base at 1.0 +0.5 = 1.5m, from thepoint G (heel) of the wall.

(ii) The eccentricity (e) of the resultant force from the centre line of the base isthus by inspection.


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e = 1.5m 1.25m=> e = 0.25m

(iii) Pressure under at Toe (F) =Z

M

A

W+

=> Pressure at Toe (F) =

ModulesSection

tsuloftyEccentricitoDuemoment

baseofArea

wallofWeight .tanRe+

2

6

1

tan

0.15.2

28

bdZ

basetheofcentrethefromscewxdis

x=

+

=

( )( )25.2161

25.0

0.15.2

28

=

+

Z

wx

x

=04.1

25.028

105.2

8.2 x

x+

Pressure at Toe (F) 11.20 + 6.73 = 17.93KN/m2

Similarly Pressure under wall at G (heel) =Z

M

A

W

Pressure at G (heel) = 11.20 6.73 = 4.47KN/m2

Example (6)A masonry wall is shown below, and weighs 20KN/m3. It retains on its verticalface water weighing 10KN/m3. The water reaches the top of the wall. Calculatethe pressures under the wall at the heel and the toe.


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Fig .4.9.6

Solution:W = (1.0 + 3.0) x 4.5 x 20 = 180KNArea of Trapezium

P = WH2= (10) (4.5)2= 101.25KN

To determine the centroid of the wall, xA = Ax, + A2x2 and+

=A

2211 xAxAX

Where X=centroid of the wall,X = 1.0 x 4.5 x 0.5) + [ (20) (4.5) (1.7)(1.0 x 4.5) + ) x 2.0 x 4.5)

X = mx 08.150.450.4

50.725.2=>

+

+

By similar triangles

0.180

25.101

4.1=

y

Y =0.180

25.101

4.1=

y


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Thus, the resultant force cuts at 1.08 + 0.84 = 1.92m from B at S or 0.42 to thecentre line of the base. Hence, the Pressure at A (Toe) is

=>Z

M

A

W+

=> Pressure at Toe (A) =0.30.30.1642.00.180

0.10.30.180

xxxx

x+

= 60.0 + 5 + 50.4

Pressure at A = 110.4KN/m2

And the pressure at B (Heel) is =Z

M

A

W

= 60.0 50.4Pressure at B = 9.6KN/m2

The purpose of calculating distance y was to find the eccentricity of w, i.e.

distance e so that M = w x e could be determine. Distance e, however, can bedetermined directly by considering the equilibrium about the centroid of the baseof all the force acting on the wall, as shown below:

Fig . 4.9.7

Let RH and RV be, respectively, the horizontal and vertical components of the

soils reaction acting at the inter section of the resultant of P and W with the baseof the wall. Then for equilibrium, RH= P and RV= W, both of which have beencalculated previously. Hence, taking moments about [0], the centre of are (orcentroid) of the base W x a PxH + Wx e = 0

=> e = Px H Wx a/we = 101.25 x 15 180.0 (1.50 1.08)/180

=> e = 0.42m as before .


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WEEK 14

5.0 INDETERMINATE STRUCTURES.

5.1 INTRODUCTION

Structural analysis is the process of determining the response of a structure tospecified loadings in order to satisfy essential requirements for function, safety, economyand sometimes aesthetics. This response is usually measured by calculating the reactionsknown as, internal forces of members, and displacements of the structures.

Structure may be classified into two general categories: statically determinate andstatically indeterminate. A structure which can be completely analysed by means ofstatic alone is called statically determinate. It then follow that a statically indeterminatestructure is one which cannot be analysed by means of static alone,

Consider a structure in space subjected to non coplanar system of forces. For the structureto be in equilibrium, the components of the resultants in the orthogonal directions mustvanish. This condition constitutes the six equations of equilibrium in space which arewritten as:

fx= 0, fy= 0 fyz= 0

Mx= 0 my= 0 mz= 0

For a structure subjected to coplanar force system, only three of the six equationsequilibrium are applicable. The three equations of equilibrium in the xy plane are:

fx= 0, fy= 0 mz= 0

Here we are to deal with statically indeterminate structures in which the structures cannotbe analysed by the equations of equilibrium alone.

Statically indeterminate structure has certain advantages over the determinate structure inthat:

It is usually more economical in materials It has a greater margin of safety in that the removal or failure of a redundant member

or support will not cause immediate collapse It has joints that are easier and more economical to form, particularly true forreinforced concrete structures

It is generally stiffer for a given weight of material than the determinate structure. It can often furnish compensation by redistribution within the structure in the case of

overloads.


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On the other hand, changes of temperature or initial lack of fit of a member can set uphigh stresses in a redundant structure particularly trusses, this is not the case with thetruss that is simply stiff particularly foundation settlements can cause considerable stressredistribution in the fixed based portal frames or continuous beam and because of this,settlement of support should be avoided at all cost, thus a poor soils liable to settlement

under load, it is safer to design the structure as determinate.Indeterminate structures introduced computational difficulty in establishing the requiredequations.

5.2 Definitions of determinate, indeterminate structuresand the concept of redundancies.

A structure is either statically determinate or indeterminate. It is statically determinate ifit is possible to calculate or determined all the unknown forces in its various membersfrom the knowledge of the dimensions and external loading of the structure and byapplying the basic conditions of static equilibrium.A structure is statically indeterminate if it is not possible to calculate or determinate allthe unknown forces, reactions and moments by the use of the conditions of staticequilibrium alone.

`A redundant frame is a frame having one member or reaction more thannecessary to produce stability and cannot be solved by the ordinary methods of static. Itis, therefore, also called a statically indeterminate or hyper static frame. When a structureis statically indeterminate, there is some freedom of choice in selecting the member orreaction to be regarded as redundant. When the reaction is taken as the redundant, thestructure is said to be externally indeterminate. On the other hand, when the memberitself is regarded as the redundant, the structure is said to internally indeterminate.

The question of identifying external or internal indeterminacy is largely of academicinterest. What is of primary importance in the analysis of indeterminate structures is toknow the degree of total indeterminacy.

5.3 Degree of Redundancy or indeterminacy:

if a structure has R redundancies, it is said to be redundant to R degree, for beams andframe, the number of redundancies may be taken as the number of restraints that wouldhave to be removed to enable the structure to be analyzed by the use of the conditions of

static equilibrium i.e. to render the structure statically determinate.


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Fig . 5.1

I.e no horizontal and vertical translation and no rotations, therefore, 3 restraints andzero degree of freedom.

Fig .5.2

At A 3 no of restraints and At B 1 no of restraint0 Degree of freedom 2 degree of freedom

Note:

Encastre beam: is a beam which both the two ends have fixed supportPropped cantilever beam: is a beam fixed at one end and simply supported at the otherend.

5.4 Degree of static Indeterminacy

The most general method for determining degree of indeterminacy in beam frames isbased on the relation D = U e.

D = Degree of IndeterminacyWhere u = No of independent Unknown member forcesuf = 3 x no. of member for plane framesut = 1 x no. of member for plane trussesei = No. of unknown joint displacement + internal releasee = Degree of freedom of support(dofs) + 3 x degree of freedom of free

joints(doffj)


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Eaxample. 1. Find the degree of indeterminacy

Fig .5.3

Solution:

D = u eNo. of members = 3Therefore, u = 3 x 3 = 9e = dofs+ doffje = 0 + 2 x 3 = 6Because there are 2 free jointsTherefore, D = 9 6 = 3 => it is redundant to 3 rddegree.


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WEEK 15

Example 2. Find the degree of indeterminacy of the frame. structure in the fig.5.4

Fig. 5.4

Solution:

No of members = 28Therefore, u = 3 x no of members

= 3 x 28 = 84

e = 0 + 3 x 16 = 48D = u e = 84 48 = 36

This implies that it is redundant to the 36 thdegree .

Note: If all the supports are fixed, the D for multi-bay, multi-storey building is given byD = 3 x no. of beamsThis implies that D = 3 x 12 = 36But if one or more of the support is not fixed the condition will not hold.

Example (3)

Find the degree of redundancy of the frame structure in the fig. 5.5


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Fig. 5.5

Solution:

u = 3 x 3 = 9e = 0 + 1 + 3 + 2 = 6D = u e = 9 -6 =3

This implies that it is redundant to the 3rddegree

Example 4

Find the degree of indeterminacy of the frame structure in the fig. 5.6

Fig. 5.6

Solution:u = 3 x 1 = 3e = 1+1+1 = 3Therefore, D = u e = 3 3 =0


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Therefore, this member is determinate.

Exercise:Find the degree of redundancy of the arch in the fig . 5.7

Fig. 5.7

Example 5:Find the degree of indeterminacy for the truss shown in the fig. 5.8

Fig. 5.8

Note: Hinged support in trusses provides zero degree of freedom and one degree offreedom in beams and frames

Solution:

U = 1 x 15 = 15e = 1 + 0 + 2 x 6 = 13Therefore:D = 15 13 = 2

This implies that it is redundant to second degree

or D = m + r -2j


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Where m = number of members, r = number reactions, j = number of jointsTheref ore D = 15 + 3 2 x 8 = 2This implies that it is redundant to second degree

5.5 TEST FOR GEOMETRIC INSTABILITY

1. If D < 0, that structure is always unstable, because it is neither determinate norindeterminate thus no further analysis is needed, since we are not interested inanalyzing a structure that cannot stand.

Fig. 5.9

U=3, e=2+2=4

But D=u-e D= 3-4 = -1 it is unstable2. Even D 0, does not guarantee stability

Fig. 5.1.1

u = 2 x 3 = 6e = 2 (3) = 6, D = 6 6 = 0

But it is not a stable structure.

In this case a careful study of the structure is required in order to detect possibleunstable or critical form, for beam and frame structures, a common senseapproach without resulting to qualitative analytical method is useful

(1) Check if the structures configuration cannot carry a specific load.(2) If instability is suspected in a certain direction or rigid of the structure apply a

load there and check if it can be equilibrated.


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Cec 205-Theory of Structures

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