Structural Analysis II Notes Rev1

SAS 203T

STRUCTURAL ANALYSIS II

Study Notes, 2ND Ed.

SAS 203 T STRUCTURAL ANALYSIS IIFaculty of Engineering, Tshwane University of Technology 1ST semester 2009 Lecturer: Mr S E Seanego Room 3-528, Tel. 012 3585233, [email protected]

Outcomes: 1.To understand the fundamentals of Stuctural Analysis.2.To be able to analyse and accurately compute Strucural Analysis problems correctly.3. To be able to use the computer aided design software in Structural Engineering.

Course Description:Application of combined stresses theory in the analysis of bearing pressure below footings and retaining walls. The analysis of retaining walls for overturning, and sliding. Analysis and drawing of Axial Thrust, Bending Moments and Shearforce Diagrams in Portal Frames. Analysis and drawing of Axial Thrust, Radial Shear ,Bending Moments and Shearforce Diagrams in Parabolic and Circular Arches. The analysis of load carrying capacities of struts by the theories of Euler,Rankine and Perry-Robertson. Assesment of stability and determinancy of structural models.

Tentative Schedule:Weeks Topic PAGE

1 Bases2 Retaining Walls.3 Test14 Portal Frames. 5 Parabolic Arches.6 Experiment 1 – Parabolic Arch7 Cicular Arches8 Test29 Experiment 2 - Struts10 Struts. 11 Determinancy of Structures12 Test3

Course Delivery Method: Each topic will be presented by means of a formal lecture from the courseware. It is estimated that each module will be handled in two periods indicated on your timetables. Examples will be handled during these formal lectures. The student must then individually solve the problems from the notes at home.

References: Kassamali A;Structural Analysis;Prentice Hall, Khajes Alexandra ;Structural Analysis; Prentice Hall Marshall and Nielson ;Structures; McGraw Hill,.

Evaluations: A total of three test wilbe written as shown above The final mark will be the average mark of the tests .A MINIMUM OF 40% must be obtained for your semester mark. Entrance to examinations will not be granted if your semester mark is below 40%.The final course mark is the average of your semester and mark obtained from the examination to be written at the end of the semester..

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mailto:[email protected]

1COMBINED DIRECT AND BENDING STRESS

DIRECT STRESS

When a force F is applied to the center of gravity of a cross sectional area A , the direct stress can be formulated and expressed as:

BENDING STRESSWhen a Moment M is applied to the center of gravity of a cross sectional area A , the bending stress can be formulated and expressed as:

The bending stress can also be expressed as since

If a force is applied to a cross-section at any point other than the center of gravity, the force acts with an eccentricity , ex, measured from the axes X and Y through the

2

center of gravity as shown by the figure below.It ca be seen that the eccentric load P

has an effect of inducing a direct strss equal to , as well as bending stress about

the X-X axis equal to where equal to

When the two above stresses are added the expression becomes :

bex

X

Y

F

Y

d X

Similarly if a load F is applied a distance from the Y axes shown below the

combined stress is:

NB! The section modulus is now around the Y axes.

When the Force F, IS applied eccentric about both axes as shown below: it produces

direct stress bending stress , as well as

The maximum stress at the corners can be expressed as:

3

Example 1

Y A 3m B

4m CG X 0.2 K F D C 0.15 not to scale

Calculate the ground pressure at A B C and D if;i) a Load of 2000kN is applied at the centre of gravity

kN/m2

ii) a Load of 2000kN is applied at F

kN/m2

kN/m2

4

iii) If the load is applied at K

kN/m2 ; kN/m2 ; kN/m2 ; kN/m2

iv) If a clockwise moment of 1000 kNm is applied around the Y axes

kN/m2 ; kN/m2

Example 2

5

Example 3

MAT FOUNDATIONSTRAP FOOTING

COMBINED FOOTINGCOMBINED FOOTING

STRAP BEAM

PROPERTY LINE

ISOLATED FOOTINGS

FIGURE 1. TYPES OF FOOTINGS

8

Calculate the maximum and minimum stress for the base as shown in the sketchIf a load of 2000 kN is applied at C 1.5m from A and a load of 2500 kN is applied at D3.5m from A i ) Centre of gravity of the section The section is divided onto two triangles and a rectangle as shown

m2

m from B The second moment of area around a axes through B

m4

=30.093 m4 The position of the resultant force: m from A ( 2.389 m from B )

m (to the left of the CG )

1000

2000

1000

5000

A BC D

CGxRx

1.5m

2m 1.5m

e

9

The Middle Third Rule

For zero stress to develop at the one side:

( as long as the eccentricity E falls within the middle third of the

base the stress below the base will be positive )

With eccentricity more than a negative stress develops .This is not applicable for

bases

Example 4

i ) Calculate the eccentricity measured from the centre of the round base to produce a zero stress at the edge of the base, if the load applied is 2000 kN and the diameter of the base is 2.4m

e

2d

F

2d

Z

M

A

FZ

M

A

F

10

mm

ii ) Calculate the max and min stress if the eccentricity is 60 mm

= 530.5 or 353.7 kN/m2

Consider the following base

R is known

D

11

The maximum ground pressure will occur at A and the minimum at B

For the min pressure to be zero

This means that the ground pressure below the base will always be positive providedThe resultant force R falls within the “ middle third “ of the base.

Regard the scenario where

The “ theoretical “ ground pressure will be the following:

The section presents a negative pressure ( suction/tension in joints ) which of coarse is not possible in

the case of foundations.

The ground pressure distribution is as follows and the max. pressure is corrected to be positive only as shown:

d

R

e

A B

A B C

12

The position of the Resultant must coincide with the centre of gravity if the triangle

is the average stress and

These equations are the appropriate to be used whenever

1.The trapezoidal combined footing shown below weighs and is 0.5 m thick.It carries column point loads of 200kN and 150kN at A and B respectively.

(a)Locate the centroid of the base (6)

(b) Compute the moment of inertia of the base about its centroid (10)

m

2ssR

CG

13

( c ) Plot the pressure distribution under the base (14)

2. The figure below shows the plan of a 0.8 m thick footing which carries three column loads. Each load lies 0.8m from the adjacent edges .Plot the pressure distribution underneath the base, take the density of concrete as 2 240kg/m . (total 20 marks)

14

4. For the trapezoidal combined footing shown belowtha weighs and is 0.8 m thick and loaded as shown, calculate the dimensions B1 and B2 if the bearing pressure under the base must uniformly distributed and mus not exceed 250KPa .

3. For both the circle and the rectangle shown below, calculate the minimum eccentricity (e) that will produce no tensile cracks in the joints at the extreme fibres. P IS AN APPLIED LOAD (10 MARKS)

15

6 0 0

2 5 0

B

2 0 0

6 0 0

A

D

C

y

x

1 0 0 k N

x

y

1 5 0 k N

CD

7 00

Y

XX

A

2 0

2 2 0 0 k N

2100

5 0 0

B

Y

1 5 0 0

5.A concrete column 600mm X600mm in cross section supports a 150kN load on a bracket 200mm from the face of the column and 250mmfrom the corner of the bracket, as well as an axial load of 100Kn. Ignore the mass of the column and:

a) Calculate the stresses at points A,B,C and D of the column (16)

b) Plot the stess distribution underneath the column (4)

6.The figure below shows a concrete footing. Calculate the magnitude and nature of the stresses imposed along A-B and along C-D, when a load of 2200kN is applied at 20mm from the edge as shown on the diagram. .Note that all dimensions are in mm.

16

2 RETAINING WALLS

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RANKNE`S THEORYThe assumptions made in this theory is that theRETAINED material is incompressible, homogeneous .granular and cohesion less and that it possesses internal frictional resistance to movement between the grains Thus if a quantity of such material were tipped onto a flat surface it would flow out to form a conical shape. The material within the cone supports itself due to the internal friction between the grains.The angle that the side of the cone makes with horizontal is known as the “ angle of repose ”Should the material be restrained by a vertical surface the active lateral pressure is : where

in terms of the angle of shearing resistance

and is the “weight density” of the material ( kN/m3 )Therefore at depth H the active pressure:

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=

Since varies uniformly with depth in a homogeneous material the distribution diagram for the lateral pressure is a triangle and the resultant Force resulting from the

triangular distribution of pressure 2 acting at distance above. surface of

the Soil

H

A

The moment around point A is referred to as the “ OVERTURNING MOMENT “

The overturning moment is resisted by the “ RESISTING MOMENT “ created by the weight of, the retaining wall and all the material on the base of the retaining wall.The resisting moment are generated about the toe A, of the retaining wall.

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N G L

W c 1

s u r c h a r g e

P a

P s u

h/2

h/3

P p

N G L

d 1

W c 2

d 5A

d 2

W c 3

d 3

d 4

W s

W s u

The safety factor against overturning =

SURCHARGEWhen a superimposed pressure is applied at the surface of the retained soil the activelateral pressure is increased by a proportional amount. The pressure distribution ofwhich is rectangular and the resultant force acts at half the depth.for a surcharge of kN/m2 the pressure

and the force acting from the surface

GROUND PRESSUREThe stress below the base is the combined stress of the direct stress and the bending stress caused by the eccentricity of the resultant load R on the base

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C L

A

R

E

B / 2

s e

B / 2

B

The eccentricity can be found by

The pressures at point A and B respectively are

and

since we regard I m length of the wall Z becomes

For a zero stress to develop behind the wall the eccentricity becomes

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In order to prevent negative stresses to develop below the base the resultant must fall within the middle third of the base.

Example 1 Calculate the ground pressure below the base of the retaining wall shown The soil carries a surcharge of 4 kN/m2 ( Regard 1m length of the wall only )And show a diagram of the moments on the wall, heel and toe.

; = 18 kN/m3 ; = 24kN/m3

0.3m 4kN/m2 diagram of lateral pressure xxxxxxxxxxxxxxxxxxxxxx soil surcharge

4m w2 w1 ps

Pa

0.8m A w3

0.6m

2.7m

Overturning moment:

soil: kN.m

surcharge: kN.m

Mo = 125.84 kN.m

Resisting moment:w1 1.8 x 18 x 4 = 129.60 x 1.8 233.28w2 0.3 x 4 x 24 = 28.8 x .75 21.60w3 0.8 x 2.7 x 24 = 51.84 x 1.35 69.98 S 4 x 1.8 = 7.20 x 1.80 12.96

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W = 217.44 kN MR = 337.82 kN.m

Safety factor @ overturning =

Eccentricity of the resultant force from the center of the base:

e = 1.35 - = 0.375 m

for the base Z = m3

Ground pressure

kN/m2

toe kN/m2

13.42 kN/m2

147.64 kN/m2

Passive pressure:

surface of the soil behind the wall

H soil in front

HP PP

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The soil in front of the wall does not actively exerts a pressure against the wall instead the wall is applying a pressure to the soil. The soil induces a passive pressure to resist the pressure applied by the wall with magnitude:

where

Because it is doubt full whether this pressure is sustainable. Most engineers do not take it in account when determining the moment of resistance of the wallHowever the passive pressure plays an important role in resisting sliding of the wall.

The active force pushes the wall foreword, and if this force is larger than

the esisting force sliding will occur. The resisting force comprises: a )The friction between the base and the soil, equaling where is the resultant downward force applied by the wall as well as the soil on top of the base

is the coefficient of friction between the concrete of the base and the

Soil, and

b) The passive force ; with the reciprocal of

The safety factor against sliding =

Inclined backfill: The soil surface makes a angle with the horizontal

H

The active pressure at any depth acts parallel to the surface of the fill in accordance with the following equation.

and the resultant force:

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, acting at height from the base

This force can now be compounded into vertical and horizontal componentsThe horizontal component causes overturning while the vertical component adds to the weight resistance. ( this value must if applicable be added to w1 ,w2 ,w3 and S in example 1 to form the total resisting force W, and MR )Example2

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Bending moments on retaining walls and retaining wall bases.a) The wall:

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the bending moment at the base of the wall =

where is the height of the wall above the base.

b ) The toe of the base:

soil

base

B A

d Pressure due to weight of the soil = 1 2 Pressure due to weight of the base =

4 3

is derived at by linear interpolation between and The bending moment at A is calculated as the bending moment caused at A by the load distribution 1, 2, 3, 4 on cantilever AB, fixed at AThe area 1,2,3,4 is divided onto a square with “arm” and a triangle with “arm“

c ) The heel of the base:

surcharge xxxxxxxxxxxxxxxxxxxxxxxxx surface of the soil

soil

X Y base

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X

The heel of the base is regarded as a cantilever XY fixed at XThe load on the cantilever is the weight of the soil plus the base plus the surchargeacting downwards, minus the ground pressure acting upwards.Again the area is divided onto a square with “ arm” and a triangle with “arm“ The deflected form of the retaining wall

vvvvvvvvvv

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1.The trapezoidal retaining wall with a heel shown below weighs and retains on its vertical face soil with an equivalent density of and an angle of internal friction (Ǿ) of 30 . The retained soil also carries a surcharge of .acting horizontal at the surface

(a)Check the stability of the wall against overturning (15)

(b) Check the stability of the wall against sliding (5)

( c ) Plot the bearing pressure distribution under the base (10)

The factors of safety against overturning and sliding may be taken as 2.5 and 1.5 respectively

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2. The trapezoidal retaining wall shown below weighs and retains on its vertical face soil with an equivalent density of and an angle of internal friction (Ǿ) of 30 . The imposed soil carries a surcharge of .




30

600

3600

400

1 2 0 06 0 0

4 0 0

1 2 0 0

4.The masonry retaining wall shown below supports a cohesionless soil applied to the surface of which there is a surcharge pressure of 60 . The soil has a unit weight of 19.1 and an angle of shearing resistance of 28 , the unit weiht of the masonry is 24.5 .Determine:

(a)The position of the resultant ground reaction

(b)The bearing pressure underneath the retaining wall.

( c ) The factor of safety against overturning

(d) The factor of safety against sliding

s u r c h a r g e

1.5m

5m5 m

6 m2 m

2 . 4 m 6 m

A

1 m

1 m

1 m

B

N G L

3.A concrete retaining wallretains water to a depth of 3.6m in a reservoir as shown below




The factors of safety against overturning and sliding may be taken as 2.5 and 1.5 respectively. Take the density of water as 1000 , the density of concrete as 2500

and the coefficient of friction between the wall and the bearing soil is 0.655.Bearing pressure will only be concentrated underneath the retaining wall.

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5.The concretey retaining wall shown below supports a cohesionless soil,The soil has a unit weight of 18.1 and an angle of shearing resistance of 35 , the unit weight of the concrete is 24.5 .Determine:





N G L

0 . 5 m

6m0.

5m1 m

3 . 5 m

N G L

0 . 5 m 1 . 5 m 0 . 5 m

6.The concretey retaining wall shown below supports a cohesionless soil ,The soil has a unit weight of 16.5 and an angle of shearing resistance of 38 , the unit weight of the concrete is 24.5 .Determine:





6 m

3 . 4 m

N G L

N G L

5m1m

0 . 5 m0 . 5 m

3 PORTAL FRAMES

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Three pinned Portal Frames

Example 1 . For the structure shown below , draw the axial thrust, shear force and bending moment diagrams

6 k N

3 m3 m

A

B

1 0 k N

D

5 m

E

C

Solving the reactions:

6 k N

3 m3 m

V a

H aA

B

1 0 k N

D

H e

V e

5m

E

C

It is important to note that the third pin at point C makes the arch statically determinate

The sum of the moments around C ( or any other point ) = 0

The Bending moment at point C = 0 and Moments around C of section CE kNThe sum of the horizontal forces on the frame as a whole = 0

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kNThe sum of the moments of all the forces around A

kN The sum of the vertical forces on the frame as a whole = 0 kN

1 0 k N

B

3 0 k N m 0 k N m

0 k N0 k N

0 k NO

1 0 k N

DC

BB

E

C

1 0 k N

0 k N

1 0 k N

5m

0 k N

0 k N m

A

B

0 k N

6 k N

6 k N

0 k N

B 3 0 k N m

6 k N

0 k N6 k N

0 k NB

0 k N

3 0 k N m

3 0 k N m

B D C

A

C

E

B

A X I A L T R H U S T D I A G R A M S H E A F O R C E D I A G R A M

B E N D I N G M O M E N T D I A G R A M

F B D 1

F B D 2F B D 3

F B D 4

1 0 k N

D

B

A

3 0 k N m3 0 k N m

1 0 k NE

D

E

C

6 k NA

Remember: The Bending moment at ant point on the structure is the sum of the moments of all the forces to one side of that point on the structure

Note: It is custom to draw the BM diagram on the tension side of members.

Example 2

34

Note that in this example , the positive Bending Moment is drawn on the compression side

36

Example3

37

reactions

38

1. For the structure shown below , a) Calculate the reactions (4)b) draw the combined axial thrust diagram (5)c) Draw the combined shear force diagram (5)d) Draw the combined bending moment diagram (6)

8m

5 m

4m4m

A

B

3 0 k N / m

D

C p i n

6 k NE

1 6m

5 m

3 0 m

A x B x

2 0 k N / m

2 . 5 k N / m

A

A y

D

B y

C

B

E


41

A B

C

9 m

9m

3 0 °

2 k N / m

9 m


e) .

A

2 0 k N

4 m 6 m3 0 k N

6m

CB

10m

E

C


.

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4 PARABOLIC ARCHES

1. Arch ActionAn arch is a curved beam with both ends fixed in position that is in the X and Y direction and may be fixed against rotation. This means that the supports may be pinned or fixed(built-in) but never on roller supports.

H a

V b

H b

P

s p a nV a

y

Any load placed on the arch causes both horizontal and vertical reaction at the supports. The horizontal forces exerted by the supports is known as Arching Action and greatly reduces the bending moments in the arch. The parabolic arch acts as the inverse of a flexible cable which hangs in pure tension to form a parabola when a uniform distributed load is suspended from it. The parabolic arch consequently is in pure compression when carrying a uniformly distributed load For the arch above, we have the following BM diagrams

P L / 4

H y

F r e e B M d i a g r a m f o r H t a k i n g a r c h a s s i m p l y s u p p o t e d b e a m

F r e e B M d i a g r a m f o r P t a k i n g a r c h a s s i m p l y s u p p o t e d b e a m

C o m b i n e d B M d i a g r a m

- v e

+ v e

- v e

- v e

+ v e

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2 . Terminology

2.1 Symmetrical Arch

s p a n

r i s e

c r o w n

s p r i n g l i n e

2.2 Unsymmetrical Arch

s p a n

s p r i n g l i n e

r i s e

c r o w n

r o a d w a ys p a n d r i l

i n t r a d o s

e x t r a d o s

Roadway suspended from Arch eg Footbridge on Apies River near Mayville, Akasia Golf Course ,Tollgate Bridge, Berea, Durban

44

r o a d w a y

3. Types of Arches

3.1 Fixed Ended Arch

In this case we have 4 reactions and 2 reactive moments at the supports. One can still remove three reactions and the structure will still stand. This type of structure is therefore statically indeterminate to the 3rd degree .

3.2 Two Pinned Arch

The removal of two moment reactions that were in the fixed arch, leavs us with 4 reactions at the supports. One horizotal reaction can still be removed, the arch would spread but still remain a structure This type of structure is therefore statically indeterminate to the 1st degree .

3.3 Three Pinned Arch

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The introduction of an additional pin on the arch creates an additional equation(The sum of all the moments as well as the bending moment is zero at the pin), to the three equations of statics( V = 0 ; H = 0 ; M = 0 ( of the structure as a whole ))

, we now have 4 unknowns and 4 Equations, making the arch to be statically determinate. The Storms River Bridge is an example of this type of structure

3 Geometry of the Parabolic Arch

4.1 The symmetrical Parabolic Arch

X

Y

Xlr

A B

1 Taking the origin at A ( 0,0) c =0 @ B

2

@ C

3

46

4.2 Unsymmetrical Parabolic Arch

Y

Y

X X

Y a

Y b

A

B

C

L 2 L 1

L

O R I G I N

Taking the origin at C ( 0,0) c =0

Point C is the turning point and

@ C @ A

47

………………….1

@ B

……………… …..2 Solving for a from 1

Substitute in 2

AXIAL FORCE AND RADIAL SHEAR FORCE

E

c e n t r e l i n e o f a r c h

h o r i z o n t a l

t a n g e n t t h r o u g h D

N o r n a l o r p e r p e n d i c u l a r t o X - Xc o m p o n e n t x o f H

c o m p o n e n t y o f H

D

X

A Y

D

Y

C

X

B

Regard a section of the parabola.At any point D on the curve the AXIAL THRUST acts in the direction of the tangent to the curve at that point and is the sum of the components of all horizontal components the forces ( including the reactions ) to the left ( or right ) of that point in the direction of the tangent.

The component of H in the plane Y-Y = and thee component of H in the X-X plane =

48

E

c e n t r e l i n e o f a r c h

h o r i z o n t a l

t a n g e n t t h r o u g h D

N o r n a l o r p e r p e n d i c u l a r t o X - X

D

c o m p o n e n t x o f V

V

X

A Y

c o m p o n e n t y o f V

D

Y

C

X

B

The component of V in the plane Y-Y = and the component of H in the X-X plane =

Adding both components in the direction of the tangent:

Similarly the RADIAL SHEAR force at any point on the curve is the component of all the forces to the left ( or right ) of that point perpendicular to the tangent at that point.Adding in the direction perpendicular to the tangent:

S = V cos - H sin ------------B

NB! Formulas A & B applies only for section AB with,

upwards and from left to right.

It is recommended that all inclined forces be resolved into horizontal and vertical

components to calculate and ; the sum of all the vertical and horizontal

forces to the left of point D

again H is taken as positive from left to right and V positive upwards

F = H cos + V sin ... ------------ A

49

For section CB above the formulas change to,

F = H Cos - V Sin and, S = V Cos + H Sin

To calculate the maximum and minimum bending moments, the bending moment must

be written in terms of the variable .

giving the values for maximum and minimum bending moments

eg. With udl

Example 1 A three pinned parabolic arch spans 10m between supports, and an a height of 2.5m at the crown. It carries a u.d.l. of 15kN/m (15 kN per horizontal meter ) over the left half of the span.Calculate the bending moment , axial thrust force and the radial shear force at a point D

from the left hand support

Y. 15kN/m xxxxxxxxxxxx C D

50

2.5m HA HB X

VA VB

10m

Taking the origin at A ( 0,0) …………………1

…………………..2

………………3

and

= tan = - 0.2x+1

at = 2.5 m ; tan = 0.5 and = 26.57

= 1.875 m

Reactions:

V = 0 VA + VB = 5 x 15 = 75 kN

MB = 0 10VA - = 0

VA = 56.25 kN

and VB = 75 - 56.25 = 18.75 kN

MC = 0 and HA = HB = H

51

- - H ( 2.5 ) = 0

H = 37.5 kN

Left of D 15kN/m xxxxxxxx D 37.5 kN 2.5 m 56.25 kN

H = 37.5 kN and V = 56.25 – ( 2.5 x 15 ) = 18.75 kN

apply formulas A and B

F = 37.5 cos 26 , 570 + 18.75 sin 26.570 = 41.926 kN ( compression )

S = 18.75 cos 26.570 - 37.5 sin 26.570

= 0 kN

BMD = 56.25 x 2.5 -37.5 x 1.875 - 15 x ( as before.)

= 23.4375 kN.m

a formula for the bending moment would read:

BM = 56.25 x - 37.5 - 15

= 56.25 x - 37.5 - 15

= - 3.75 + 18.75 x

52

At the point of maximum bending moment

= -7.5x + 18.75

giving, x = 2.5 m and y = 2.5 - = 1.875 m

The bending moment at this point is MD = 23.4375 kN.m

Example 2

A Three-pinned parabolic arch is loaded as shown below.

The reaction at A is elevated 30m above reaction B and the 10 kN point load at D has

inclination of 36.87 0 to the vertical

C 36.870

10kN 10m A D 40m

B 40m

120m

If we take the horizontal distance between A and C as x

=

A(-40,-10) -10 ………….1B( 80 , -40) -40 = 6400a…………..…2

Equation of arch is then With the origin C (0,0)

You can test this by rechecking the co-ordinates of A and B

53

at point D

= -10 m

Resolving the inclined force into components:

10 kN 8 kN V = 10 cos 36.870

36.870

= 8 kN

6 kN H = 10 sin 36.870

= 6 kN

Reactions:

= 500

kN

kN

Axial force and radial shear force ( just ) left of point D

54

Slope at D tan = = - 0.0125(+ 40 )

= 26.565 0

V = 2.083 kN H = 8.333 kN

F = 8.333 Cos 26.5650 - 2.083 Sin 26.5650

= 6.522 kN

S = 2.083 Cos 26.5650 + 8.333 Sin 26.5650

= 5.59 kN

Axial force and radial shear force ( just ) right of point D.

V = 2.083 – 8 = - 5.917 kN

H = 8.333 - 6 = 2.333 kN

F = 2.333 Cos 26.5650 - ( - 5.917 ) Sin 26.5650

= 4.733 kN

S = - 5.917 Cos 26.5650 + 2.333 Sin 26.5650

= - 4.249 kN

an expression for the bending moment between A and C is: = but,

and the maximum bending moment occurs at

= 0.10416x + 2.08333

x = -20 m

The bending moment at x = -20 m

= - 20.821 kN.m

55

2. A parabolic arch ABC is hinged at the supports and the apex as shown on the next page.The arch carries a uniformly distributed load of 0.7 KN/m horizontally across span AC , and two point loads as shown below:The reactions have been calculated and are show on the diagram.

(a) Derive the equation(s) for the geometry of the arch (5)

(b) Use the equation(s) derived above to calculate the heights of C,D,E and F above point A (8)

(c ) Calculate the bending moment, axial thrust ,and radial shear at a point F (7) ( total 20 marks )

A

0 . 7 k N / m

14

2 1 k N

1 7 5

1 5 0

1 2 5

B

5 0

1 0 0

F

4 2 k N

DC

E

1.A parabolic ABC arch is hinged at the supports and the apex as shown on the figure beloww.The arch carries six 20 KN point loads as wel as a udl of 10 KN/m horizontally.Find:

(a) The axial thrust and radial shear at a point 15m horizontally to the right of support A(just left of the point load) (10)

(b) The bending moment in the arch 5m horizontally to left of B, UNDER THE POINT LOAD (5)

56

3.For the parabolic arch below:a) Calculate the height of point D above the X – axes (3)b) Calculate the slope ( gradient ) at point D ( 3 ) c) Calculate the reactions (4)d) Calculate the Radial Shear force (only ), fractionally to the right of point D (5)e) Derive an expression for the Bending moment between points A and C (3)f) Calculate the value of the maximum Bending Moment between A and C (2)

.

6.6m4.4mm

xxxxxxxxxxxxxxxxxxxxxxx

60m

12m

A

B

C

D

4kN/m

20kN

450

X

Y

57

5 CIRCULAR /SEGMENTAL ARCHES

1.Geometry of the circular arch

C

h

X X

Y

YH a H b

O 1V a

A

a

( x , y )

R

O 2

b

V b

B

From O1

From O2

and

where a = half the span and h = the height of the arch

Calculation of tangent angle for segmental arches

58

C

Y

X X

R

x

Y

y

ExampleCalculate the Bending Moment, Axial Thrust and Radial Shear at point P, WHICH IS 10 m, horizontally from A.

C

A

X2 0

1 2 mY1 6 m

8m

4m

2 5 k N / m

P

Y

O

B

X

59

Example

C

A

X2 0

1 2 mY1 6 m

4m

2 5 k N / m

P

8m

H a

V a

H b

V b

Y

O

B

X

ReactionsHorizontal Equilibrium

MOMENTS ABOUT B ……………..1MOMENTS ABOUT C ………………2

Vertical Equilibrium

At point P

60

as before

Axial Thrust = 493.7 KnRadial Shear = 15.2 kN

61

B

C

D

AO

1 0 m

Y

Y

XX

5 0 k N

1..The figure below shows a segmental three-pinned circular arch, with points B and D at a third of the span. The 50kN point load is applied at 45degrees.

a) Calculate the support reactions (5)

b) Determine the tangent angle at D (5) (total 10 marks)

2. The segmental arch ABC is hinged at the supports and the apex as shown on the figure below.The arch carries a uniformly distributed load of 5 KN/m horizontally across span BC , and a point load at D as shown below:

(a) Derive the equation for the geometry of the arch (4)

(b) Use the equation(s) derived above to calculate the heights of D, B and C above point A (6)

(c)Calculate all the support reactions (8)

(d ) Calculate the bending moment, axial thrust ,and radial shear at a point D (12)

62

H a

H b

V b

R 2 5 m

8 k N / m

4 2 °

V a

A

2 3 °

C

B

4.The figure below shows a three-pinned arch consisting of two quardrants of circles and carrying a concentrated load of 40Kn. Find the maximum positive and negative bending moments0n the arch. Using a as the origin ,also find the equations for the linear arch.

4 O k N

5 m5 m

AA

5 m

5m

CC

B

6 STRUTS

3. For the three-pinned segmental arch shown below:

1. Calculate the reactions (8)

2. Calculate the axial thrust and radial shear at support A.(4)

3. Calculate the maximum bending moment in section AC (8)

63

When we speak of columns (and buckling) we are talking about members loaded in compression, often axially loaded, although columns may be loaded eccentrically. We also tend to think of columns as vertical members, however, the formulas we will utilize also apply to horizontal compression members, or to compression members in general. For instance, compression members of a truss may be considered to be columns pinned at each end point.

Columns may be divided into three general types: Short Columns, Intermediate Columns, and Long Columns. The distinction between types of columns is not well defined, but a generally accepted measure is based on the Slenderness Ratio. The Slenderness Ratio is the (effective) length of the column divided by its radius of gyration.

The radius of gyration is the distance from an axis which, if the entire cross sectional area of the object (beam) were located at that distance, it would result in the same moment of inertia that the object (beam) possesses. Or, it may be expressed as: Radius of Gyration: rxx = (Ixx/A)1/2 (radius of gyration about xx-axis)

So, Slenderness Ratio = Le / r.

Notice that we have put a subscript "e" by the length of the column. This is to indicate that, depending on how the column is supported, we do not use the actual length but an ‘effective length’. .Where L is the clear HEIGHT of a column.

The factor K is called the effective length factor. This factor depends on the end boundary of the column. Pinned-Pinned: K= 1, Fixed-Pinned: K= .7, Fixed-Fixed: K = .5, Fixed-Free: K = 2

A generally accepted relationship between the slenderness ratio and the type of Column is as follows:

64

Short Column: 0< Le / r < 60Intermediate Column: 60< Le / r <120Long Column: 120< Le / r < 300

I. Short Columns: When the slenderness ratio is less than 60, a column will not fail due to buckling, as the ratio of the column length to the effective cross sectional area is too small. Rather a short, 'thick' column, axially loaded, will fail in simple compressive failure: that is when the load/area of the column exceeds the allowable stress, P/A > s allow.

Eccentrically loaded short columns have a slightly more complicated result for compression failure, which we will look at later in this section.

We also will put off discussion of intermediate length columns until we have discussed long columns. Idealized buckling in long columns was first treated by the famous mathematician Leonard Euler.

II. Long Columns & Euler's Equation: In 1757, Leonard Euler (pronounced Oiler) developed a relationship for the critical column load which would produce buckling. A very brief derivation of Euler's equation goes as follows:

Column under pin supports

A loaded pinned-pinned column is shown in the diagram. A top section of the diagram is shown with the bending moment indicated, and in terms of the load P, and the deflection distance y,V we can write:1. M = - P y.

We also can write that for beams/columns the bending moment is proportional to the curvature of the beam, which, for small deflection can be expressed as:2. (M /EI) = (d2y /dx2)(See Strength of Material text chapters on beams and beam deformations.) Where E = Young's modulus, and I = moment of Inertial. Then substituting from EQ. 1 to EQ. 2, we obtain:

65

3. (d2y /dx2) = -(P/EI)y or (d2y /dx2) + (P/EI)y = 0

This is a second order differential equation, which has a general solution form of

4. :

We next apply boundary conditions: y = 0 at x = 0, and y = 0 at x = L. That is, the deflection of the column must be zero at each end since it is pinned at each end. Applying these conditions (putting these values into the equation) gives us the following results: For y to be zero at x =0, the value of B must be zero (since cos (0) = 1). While for y to be zero at x = L, then either A must be zero (which leaves us with no equation at all, if A and B are both zero), or

. Which results in the fact that

And we can now solve for P and find:

5. , where Pcr stands for the critical load which can be applied before buckling is initiated.

6. By replacing L with the effective length, Le, which was defined above, we can generalize the formula to:

, which then applies to Pinned-Pinned, Fixed-Pinned, Fixed-Fixed, and Fixed-Free columns.

This equation is a form of Euler's Equation. Another form may be obtained by solving

for the critical stress:

and then remembering that the Radius of Gyration: rxx = (Ixx/A)1/2 , and substituting we can obtain:

which gives the critical stress in terms of Young's Modulus of the column material and the slenderness ratio.

Let us at this time also point out that Euler's formula applies only while the material is in the elastic/proportional region. That is, the critical stress must not exceed the proportional limit stress. If we now substitute the proportional limit stress for the critical

66

stress, we can arrive at an equation for the minimum slenderness ratio such that Euler's equation will be valid.

It should also be pointed out that while these formulas give reasonable values for critical loads causing buckling, it should not be assumed the values are completely accurate. Buckling is a complicated phenomena, and the buckling in any individual column may be influenced by misalignment in loading, variations in straightness of the member, presence of initial unknown stresses in the column, and defects in the material. Several examples of buckling calculations follow.

Comments of Column Section

o Short columns have more capacity to resist buckling than long columns

o Euler’s Load is valid for long columns. For very short columns, the capacity of the column will be governed by the strength of the material

o Columns with higher flexural rigidity (EI) will be able to resist buckling better than columns having small flexural rigidity

o Columns with (yz is the plane perpendicular to the column axis), it will buckle in the direction of smaller I

o It is then more economical to design columns with the same than columns with different moment of inertia for the same cross-sectional area (This may not be true if the columns subjected to addition bending)

o To increase the moment of inertia by increasing the distribution of material to be further away from the centroid of the section ( ) will increase the capacity to resist buckling

o Hollow tubular section is generally more economical for use as a column than solid section having the same cross-sectional area

o Although theoretically the thinner the wall of the hollow tubular is, the more economical the column will be, too thin wall will lead to local buckling (Only overall buckling of column is discussed in this course)

Critical Stress

Critical stress can be obtained by dividing the critical load by the column cross-sectional area

Defining radius of gyration , we have

67

The ratio L/r is called the slenderness ratio. The above equation is also known as Euler’s stress.

68

Example2

71

Example3

73

Eccentric Loading

In practical, almost all of the columns are subjected to eccentric loadings

When we have eccentric loading, the bending moment in the column will be increased

74

It can be concluded that Perry developed his formula based on the initial curvature of the column, from

which Robertson (and Godfrey) later developed the known Perry-Robertson formula

= -

where: = Euler stress = Yield stress

Godfrey proposed the value of which was later included in the British design code BS

449

The South African Steel design code SABS 0162 uses a empirical formula developed in

Canada which is in line with present design codes in developed countries.

The compression resistance of a column is derived from:

where ; n = 1.34 and = 0.9

Safety factors

Regard the following formula

= -

x A = Pp

Pp represents the ultimate load on a column . Because it is difficult to predict the

load on a column with accuracy a load factor can be applied to Pp ( )

represents the yield stress of the material and similarly because of the difficulty

in predicting the strength of the material a material resistance factor can be applied to

77

The South African Steel design code SANS 10162 uses a material resistance factor of

0.9 for steel ( x resistance factor )

SECTION PROPERTIES

To calculate allowable stresses in columns it is essential to calculate the section

properties of the column. For a rectangular section and axes through the center of gravity b

CG d X X

X1 X1

Y

The moment of inertia ; ; the area A = bd

and the radius of gyration r = = =

where b is the width parallel to the axes and d is the depth at right angles to the axes

For round columns: I = and r = where d is the diameter of the circle.

To calculate the moment of inertia of compound columns (made up of multiple sections )

it is handy to use the “parallel axes” theorem I = I + Ay

where I is the moment of inertia around the center of gravity and I is the moment

of inertia around a axes distance y from the axes through the center of gravity.

The radius of gyration is rx =

78

I represents the moment of inertia around a axes through the center of gravity of the

compound section.

For triangles:

XAPEX IAPEX =

XCG

ICG =

XBASE

IBASE =

=

D

Example 1

a Column of height 6m is made up from two 250mm x 12mm and one

79

300mm x 10mm plates, to form a H–section, as shown in the sketch.

250

12mm

10 mm X 300mm X

12mm

Calculate : A ; Ixx ; I yy ; r xx and r yy (NB! units are important )

A = = 9 x 103 mm2

Ixx = + = 168.588 x 106 mm 4

Iyy = = 31.275 x 106 mm4

rxx = = 123.2 mm

ry = = 58.95 mm

At the top as well as the bottom the column is supported against lateral movement, in

the Y direction ( NB this prevents buckling around the X-axes )

80

X

In the X direction the column is supported against lateral movement at 2m intervals.

The column is not restrained against rotation anywhere. (pinned, not fixed )

X XCalculate :

= 48.7

= 34.4

Calculate the Euler buckling stress ; E = 200 Gpa

= = 832.3 N/mm2 (max. slenderness ratio is critical)

Calculate the maximum allowable load on the column using the Perry-Robertson

equation. Use a material resistance factor of 0.9

= = .071

= -

with the material resistance factor of 0.9 = 0.9x300 = 270 MPa

=

= 245.3 MPa

PP = 245.3 x 9 x 103 N

= 2207.8 kN

81

( This value compares with the ultimate resistance of the column. which according to

SANS 10162 has to be compared to the ultimate load, being the calculated load

multiplied by a load factor depending on the nature of the load )

For the above example:

; n = 1.34 and = 0.9

= 48.7 x 0.01233 = 0.60

= 2052 kN ( 2208 kN )

Columns are often compounded from hot rolled sections e.g. two channels compounded

to form a box profile. In cases where the effective lengths are the same around both axes

the channels can be separated by a distance sufficient to render Ixx = Iyy

Example 2

Y s YCG YCG from steel tables:

Ixx = 19.1 x 106 mm4

Iyy = 1.67 x 106 mm4

X CG X A = 3.09 x 103 mm2

ay = 22.5 mm ay a

82

Two 200x10 mm plates are welded to two PFC 200x75 parallel flange channels to form

a compound column.

Ixx =

= 82.35 x 106 mm4

Iyy = 2

For Ixx = Iyy

s = 103 09 mm

a = 2 (103.09 + 22.5 – 75 ) = 101.18 mm.

Rankine-Gordon

To predict failure in the intermediate region Rankine developed the following formula

P =

83

Where: a = (Rankine`s constant ) which is approximately for steel

P r A

= allowable load = radius of gyration = area of cross section = allowable compressive stress

.

1.. .A 5-m long pin-ended column with the cross section shown is constructed from four pieces of timber. The timbers are nailed together so that they act as a unit. Determine:

(a) the slenderness ratio (2)

(b) the Euler’s buckling load. Use E = 14 GPa for the timber. (5)

(c) The axial stress in the column when the Euler’s load is applied (3)

84

3. The H-SECTION column shown below h is to be used as a pin connected member, use the Euler equation to determine the maximum load it can withstand before it begins buckle or the steel yields at 250MPa.The sectional properties of the column are:

4.The Aluminium I-SECTION column shown below is fixed at the bottom and braced at the top by two rods as to prevent movement at the top in the X direction. If it is assumed to be fully fixed at at its base, use the Euler equation to determine the largest allowable load P that can be resisted if the factor of safety is 3 ,and aluminium yields at 215MPa. Young’s Modulus for Aluminium is 70GPaThe sectional properties of the I- SECTION are:

85

5. The I-SECTION column shown below has been strengthened by welding 300x15 plates to each flange and is to is to be used as a 9metre long pin- connected member, use the Euler Equation to calculate the safe Working load if the factor of safety is 3, and the steel yields at 250MPa.The sectional properties of the I- SECTION are shown below:

X X

Y

3 0 0 X 1 5 P L A T E S

2 0 3 X 2 0 3 X 4 6 H

Y

Designation

Mass Per

metre

Depth of Section

Width of Section

Thickness

Root Radius

Depth between

fillets

Second Moment of Area

Area of Section

Web FlangeAxisx-x

Axisy-y

h b s t R d Ix Iy A

kg/m mm mm mm mm Mm mm cm4 cm4 cm2

203x203x46 46.1 203.2 203.6 7.2 11 10.2 160.8 4568 154858.7

86

y

x

152

6 . 4

5 4 . 2

7 6y

x

Y

X

Y

X

where ; n = 1.34 and = 0.9

= -

6.A 2550mm long compound column is constructed by welding two 152x76 x17.9 tapered channels front to front, together as shown below.The material has a compressive Yield Strength of 320MPa,Young’s Modulus of 200 000Mpa and Rankine’s constant may be take as 1/7500. Calcutate:

a) The Euler load (12)

b) The Perry-Robertson load (10)

c) The Compressive resistance according to SANS10162 (8)

87

STABILITY AND DETERMINACY

88

A Reactions;

i) Roller

One reaction perpendicular to the rolling surface

ii) Hinge / pin

Two perpendicular reactions (vertical and horizontal ) iii) Fixed reaction

Three reactions, two preventing movement (displacement ) in both vertical and

horizontal directions, and one moment reaction preventing the rotation of point A

iv) Link

One independent reaction only, in the direction of the member (although there seems to

be two reactions at the hinge these reactions are geometrically dependant ) i) External stability ( reactions )

Classification of structures in terms of stability and determinacy.

For a structure to be stable all the external forces and moments acting on the structure

must be in balance or is said to be in equilibrium. That means in lay mans terms that the

A

90

forces going up must be balanced by the forces going down and the forces from left to

right must balance the forces from right to left. Similarly the clockwise moments must

balance the anti-clockwise moments.

Hence the equilibrium Equations:

: ;

a Structure with less than three reactions can not meet the above requirements and is

therefore unstable.

a Structure with three reactive forces can be determined using solely the three

equilibrium equations, and is said to be determinate. A Structure with four or more

reactions requires further equations over and above the three equilibrium equations

to solve the reactions, and is referred to as indeterminate, with the degree of

indeterminacy being the number of reactions minus three. The additional reactions are

also referred to as redundant reactions.

a Structure with reactions is indeterminate, with ( ) redundant reactions.

Condition Equations

Special conditions apply where the three equilibrium equations can be extended

to allow for further equations to solve additional reactions or unknowns.

These are referred to as condition equations.

Examples of situations that allows for additional equations:

One part of a structure connected to the rest of the structure by means of one pin only.

This allows the normal as well as BM = 0 (beams )

91

Similarly when one part of a structure is joined to the other by parallel members only.

a force component perpendicular to the members cannot be transferred.

Both scenarios implies that one portion of the structure must be self sufficient in terms of

stability thus providing an additional equation.

Geometric Instability

It is possible that three or more reactions can be arranged in such a ineffective way

that external stability cannot be assured under all loading conditions. Structures in this

category are termed geometrically unstable.

Examples of geometric instability:

i ) All reactions parallel

ii) The working line of all reactions intersects in one common point.

iii) Three hinges in a row

External classification of Structures

92

Structures with Condition Equations

93

ii) Internal stability and determinacy a) Plain frames

Trusses similar to that shown on the sketch present a two fold problem:

a) External stability and determinacy as discussed above, as well as

b) There must be sufficient members (bars) not to form a collapse mechanism

under any applied load system.

The simplest plane frame is the triangle shown in full lines in the sketch above

The first assumption with such trusses is that the joints are pinned and consequently, if the loads are applied at the joints no shear forces or bending moments will develop

leaving us with the axial force as only unknown per member.

The above triangle is stable since it will not change its shape when loads are applied to it

95

and can be extended as shown in dotted lines to form a frame or truss.

Since all member are joined by pins and loads are applied at joints only, each member

has one unknown (axial force ) and each joint supplies two equations:

structure is stable and determinate

structure is unstable

structure is stable but indeterminate

Condition equations as discussed apply.

96

b) Rigid frames

External classification: ( reactions )

The number of reactions must again be compared to three, similar to plain frames

taking account of condition equations if any.

Internal stability (rigid frames)

Frames where some or all the joints are rigid joints can transfer bending moments and

shear forces and therefore three unknowns per member, making the total number of

unknowns

To solve this we have available three equations per joint plus the number of condition

equations ; where t is the number of condition equations.

Condition equations: Two members joining at a pin gives one condition, three

members joining at one pin gives two condition equations and so on.

structure is stable and determinate

structure is unstable

98

structure is stable but indeterminate

99

-11.15

8.85 0.86

0.86

-5.71

-5.71

6.15

6.15

Shear Force Diagram, Comb: CASE1, Units: kN-mLinPro 2.7 | Enes Siljak | [email protected] | www.line.co.ba

0.00

9.23

24.74

9.23

0.000.00

61.54

61.54

0.00

Moment Diagram, Comb: CASE1, Units: kN-mLinPro 2.7 | Enes Siljak | [email protected] | www.line.co.ba

103

-2.62

-2.62

-9.18

-9.18-7.24

-7.24

2.62

2.62

Axial Force Diagram, Comb: CASE1, Units: kN-mLinPro 2.7 | Enes Siljak | [email protected] | www.line.co.ba

104

Structural Analysis II Notes Rev1

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