construction n5 tutorial

1

BUILDING & STRUCTURAL CONSTRUCTION

N5

CHAPTER 1

STEEL BEAM DESIGN

2

1. CALCULATION OF REACTIONS (SUPPORTS)

1.1 Symmetrically loaded beams

In symmetrically loaded beams the loads are evenly distributed and the reactionsshare the load equally

Example 1

15 N

3m 3m

RL = 7,5 N RR = 7,5 N

In this example the reactions are calculated as follows:

RL=RR=152

=7,5kN

Example 2

30 kN 20 kN 30 kN

2 m 1 m 1 m 2 m

RL = 40 kN RR = 40 kN

In this example the reactions are calculated as follows:

RL=RR=30+20+302

=40kN

3

1.2 Unsymmetrically loaded beams

1.2.1 Reactions at the ends

Example 1

15 N

3m 2m

RL RR

Take moments about RL Take moments about RR

RR (5 )=15 (3 ) RL (5 )=15 (2 )

RR (5 )=45 RL (5 )=30

RR=9N RL=6 N

Upward forces = downward forces 9 + 6 = 15 15 = 15

Example 2

30 kN 20 kN 30 kN

2 m 1 m 1 m 3 m

RL RR

Take moments about RL Take moments about RR

RR (7 )=30 (2 )+20 (3 )+30 (4) RL (7 )=30 (3 )+20 (4 )+30(5)

RR (7 )=240 RL (7 )=320

RR=34,286 kN RL=45,714 kN

Upward forces = downward forces 34,286 + 45,714 = 30 + 20 + 30

80 = 80

4

1.2.2 Reactions not at the ends

Example 1

60 kN 40 kN 20 kN 35 kN

1 m 1 m 1 m 2m 1m

RL RR

Moments about RL

RR (5 )=60 (1 )+40 (2 )+20 (3 )+35(6)

RR (5 )=410

RR=82kN

Moments about RR

RL (5 )=20 (2 )+40 (3 )+60 ( 4 )−35(1)

RL (5 )=365

RL=73kN

Now check if sum of the upward force is equal to sum of the downward forces

82+73=60+40+20+35155=155

5

That is RL is the pivot point meaning all distance are measured from RLSince RL is at the end, ALL loads acts against RR therefore:

That is RR is the pivot point meaning all distance are measured from RLSince RR is not at the end, RR has two side and the load on the same side as RL acts against RL and those not on the same side as RL assists RL. These load are subtracted from the ones acting against RL

Example 2

100 kN 60kN 80kN 50kN

20kN/m 3kN/m

A B C D E F 2,5 m 2,6 m 1,4 m 1,5 m 2 m

RL RR

52kN 30kN100 kN 60kN 80kN 50kN

A B C D E F 2,5 m 1,3 m 1,3 m 1,4 m 1,5 m 2 m

5m 3m

RL RR

Moments about RL ↺ = ↻RR (8 )=60 (2,5 )+52 (3,8 )+30 (5 )+80 (6,5 )+50(10)

RR(8)=1517,6

RR=189,7 kN

Moments about RR↻ = ↺RL (8 )=80 (1,5 )+30 (3 )+52 (4,2 )+60 (5,5 )+100 (8 )−50(2)

RL (8 )=1458,4

RL=182,3kN

6

Before doing any calculations redraw the beam and change the distributed loads into point loads,.The 3kN/m becomes 3×10=30kN which will be at the centre of the beamThe 20kN/m becomes 20×2,6=52kN which will at the centre of 2,6 m

EXERCISE 1

Calculate the magnitude of the reaction of the following beams

1.1 15kN 8kN 12kN 25kN

1m 2m 4m 3m 2m

RL RR

1.2 40kN 10kN 50kN

20kN/m

3m 5m 3m 2m

RL RR

1.3 3kN 4kN

2kN/m

5m 2m 2m 2,4m

RL RR

1.4 12kN 20kN 35kN 25kN

3kN/m 2kN/m

2,5m 1m 2,5m 3,5m 4m 2m 1m

RL RR

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2. CALCULATING THE SHEAR FORCE

Example 160 kN 40 kN 20 kN 35 kN

A B C D E F1 m 1 m 1 m 2m 1m

RL= 73 kN RR = 82 kN

A (start point) = 0A (End point) = 0 + 73 = 73

B (start point) = 73.................( because there is NO distributed loads between A and B)B (End point) = 73 – 60 = 13

C (start point) = 13..................( because there is NO distributed loads between B and CC (End point) = 13 – 40 = – 27

D (start point) = – 27 ..............( because there is NO distributed loads between C and DD (End point) = – 27– 20 = – 47

E (start point) = – 47................ ( because there is NO distributed loads between D and EE (End point) = – 47 + 82 = 35

F (start point) = 35 .................( because there is NO distributed loads between E and FF (End point) = 35 – 35 = 0

73

35

13

A B C D E F

– 27

– 47

8

Points to remember1. Each vertical line in a shear force diagram has two point, the starting point and the end point2. Starting point (we subtract distributed loads if there is any) 3. End point (we subtract point load or add support if there is any)4. Any force pointing down is subtract 5. Any force pointing up is added

Example 2

100 kN 60kN 80kN 50kN 20kN/m

3kN/m A B C D E F

2,5 m 2,6 m 1,4 m 1,5 m 2 m

RL = 182,3 RR = 189,7

A ( start point )=0

A ( End point )=0+182,3−100=82,3

B(start point )=82,3−(3×2,5 )=74,8

B(End point )=74,8−60=14,8

C ( start point )=14,8−(20×2,6 )−(3×2,6 )=−45

C (End point)=−45 .................( There is NO point load /support there we have only one point on line C )

D(start point)=−45−(3×1,4 )=−49,2

D(End point )=−49,2−80=−129,2

E(start point )=−129,2−(3× 1,5 )=−133,7

E(End point )=−133,7+189,7=56

F (start point )=56−(3×2 )=50

F (End point)=50−50=0

A B C D E F

82,3

74,8 56

50

14,8 0

– 45 – 49,2

9

NB: put down the calculated point on each line . and then join them as show below

– 129,2 – 133,7

The complete Shear force diagram is as shown below

A B C D E F

82,3

74,8 56

50

14,8 0

– 45 – 49,2

– 129,2 – 133,7

10

3. CALCULATING THE BENDING MOMENT

Example 1

60 kN 40 kN 20 kN 35 kN

A B C D E F1 m 1 m 1 m 2m 1m

RL= 73 kN RR = 82 kN

Bending moments at A and F = 0

Bending moments at B ( that means B is the pivot point)1m

B 73kN

Therefore B M B=73×1=73kN .m

Bending moments at C ( that means C is the pivot point)

60kN1m 1m

B C 73kN

Therefore B MC=73 (2 )−60(1)=86 kN .m

Bending moments at D ( that means D is the pivot point)

35kN2m 1m

D E F 82kN

Therefore B M D=82 (2 )−35(3)=59kN .m

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1. Bending moment at both ends of the beam is zero2. Any force pointing down is negative3. Any force pointing up is positive4. Use one side of the pivot point and always try to use the side with less forces

Bending moments at E ( that means E is the pivot point)

35kN 1m

E F

Therefore B M E=−35 (1 )=−35kN .m

PLOTTING THE BENDING MOMENTS DIAGRAM

A B C D E F

86

73

56

0

– 35

12

When plotting the bending moments diagram with point loads only, the graph consists of straight lines

Example 2

100 kN 60kN 80kN 50kN

20kN/m 3kN/m

A B C D E F 2,5 m 2,6 m 1,4 m 1,5 m 2 m

RL = 182,3 kN RR = 189,7

Bending moments at A and F = 0

Bending moments at B ( Remember that there is a UDL between A & B)

100 kN 7,5 kN

1,25m 1,25m

B

182,3 kN

B M B=182,3 (2,5 )−100 (2,5 )−7,5(1,25)=196,375kN .m

Bending moments at C (It is better to use the side C to F cause it’s not complicated)

80kN 14,7 kN 50kN

C D E F 1,4m 1,5m 2m

2,45189,7kN

B MC=189,7 (2,9 )−80 (1,4 )−14,7 (2,45 )−50(4,9)=157,115kN .m

13

UDL=3× 2,5=7,5kN and it is halfway AB

UDLbetweenCF=3×(1,4+1,5+2)=14,7 kN and it is halfway CF

Bending moments at D (It is better to use the side D to F cause it’s not complicated)

10,2kN 50kN 1,75m

D E F 1,5m 2m

189,7kN

B M D=189,7 (1,5 )−10,5 (1,75 )−50 (3,5)=91,175kN .m

Bending moments at E (It is better to use the side E to F cause it’s not complicated)

6kN 50kN 1,m

E F 2m

B M E=−6 (1 )−50(2)=−106 kN .m

PLOTTING THE BENDING MOMENTS DIAGRAM

A B C D E F

196,375

157,115

91,175

14

UDLbetween DF=3×(1,5+2)=10,5 kN and it is halfway DF

UDLbetween EF=3×(2)=6kN and it is halfway EF

When plotting the bending moments diagram where there is a UDL , the graph forms a curve.

–106

EXERCISE 2

2.1 Calculate and draw the shear force and bending moment diagram

2.1.1 85kN 50kN 35kN

15kN/m 7kN/m

A B C D E F G 1m 2m 1m 1m 2m 1m

RL = 161,3kN RR = 74,7kN

2.1.2 21kN 17kN 15kN

5kN/m 7kN/m

A B C D 2,5m 4,5m 2m

RL = 55,3kN RR = 107,7kN

2.2 Calculate the reactions of the following beams, do the necessary calculations and draw the Shear force and bending moment diagram

2.2.1 50kN

15kN/m 20kN/m

A B C D 2 m 1 m 4 m

RL RR

2.2.2 10kN 16kN 9,5kN 8kN

15kN/m 3kN/m

A B C D E F 2m 3m 1m 4m 1m

15

RL RR

4. CHOOSING A SUITABLE STRUCTURAL STEEL SECTION

1. When choosing a steel section, one needs to calculate the elastic modulus (Ze) of that section.

2. This can only be done if the maximum bending stress (Pt) is known and the maximum bending moment (B Mmax) is known

3. The calculation can be done using the following formula

Ze=B Mmax

Pt

4. Use the elastic modulus to find the serial size of the beam and mass per kilogram from the hot rolled structural steel sections table BOE 8/2

16

Pascal= N

m2=N m−2∧bending moment=N .m

therefore Ze= N .mN m−2=

N .m.m2

N=m3

This means that Elastic modulus is measured in m3

Lets say maximum bending moment is 70,175 kN.m and the maximum bending stress is 165 MPa

First step is express Bending moment max in Newtons which means B Mmax=70,175×103

and the max bending stress in pascals which means Pt=165× 106

Ze=B Mmax

Pt

Ze=70,175×1 03

165×106

Ze=0,000425303m3

Ze=425,303×10−6m3

4. DETERMINING THE AVERAGE SHEAR STRESS

In order to determine the average shear stress(Pv ), the following must be known

1. The maximum shear force (Fv max)

2. The elastic modulus (Ze) which is used to find

2.1 The total height of the section under stress (h) in steel table 2.2 The Thickness of the section under stress (t 1) in the steel table

3. The area ( A) which is found from (h×t 1)

4. The formula is

Pv=Fv max

A

5. Average stress is measured in Pascal

17

EXERCISE 3

3.1 Use the calculations in EXERCISE 2.1.1 to do the following

3.1.1 Choose from the table a suitable I-section with tapered flange , if the bending stress may not exceed 165 MPa

3.1.2 Determine the average shear stress. The maximum allowable stress is 100 MPa

3.2 Use the calculations in EXERCISE 2.1.2 to do the following

3.2.1 Choose from the table a suitable I-section with parallel flange, if the bending stress may not exceed 155 MPa

3.2.2 Determine the average shear stress. The maximum allowable stress is 100 MPa

3.3 The maximum bending moment and maximum shear force of a loaded simply supported beam, is 48 KN and 12 kN respectively. Select from the BOE 8/2 table, a suitable I-section with tapered flange if the following maximum values may not be exceeded:

Bending stress = 75 MPa Shear stress = 60 MPA

3.4 The diagram below shows a simply loaded beam, with Supports RL =16,5 KN and RR = 25,5 kN

8 kN

1,2m 9 kN/m 3 kN/m

A B C 3 m 1 m

RL = 16,5 kN RR = 25,5 kN

3.4.1 Make the necessary calculations and draw to suitable scales a shear force and Bending moment diagrams

3.4.2 Select from the table the smallest suitable I section steel beam with tapered flangefor the loaded beam

investigate with regard to bending ONLY. The maximum bending stress of grade 43steel is 155 MPa

18

CHAPTER 2

CENTRIODSAND

MOMENT OF INERTIA

19

1. CENTROIDS

In geometry, the centroid, geometric centre of a plane figure or two-dimensional shape is the intersection of all straight lines that divide the shape into two parts of equal moment about the line.

1.1 Rectangle

H

Distance of centre of gravity from AB

A B

Distance of the centre of gravity of the FIGUREis y= H2

1.2 Combination of figuresFIG 1

w❑

FIG 2Distance of centre of gravity of FIG 1 from AB

H

Distance of centre of gravity of FIG 2 from AB

A B

FIG 1centre of gravity ¿ AB is y1=w2

+ H

20

http://en.wikipedia.org/wiki/Moment_(mathematics)

http://en.wikipedia.org/wiki/Straight_line

http://en.wikipedia.org/wiki/Plane_figure

http://en.wikipedia.org/wiki/Geometry

FIG 2centreof gravity is y2=H2

How to calculate the position of the neutral axis of complex figures Neutral axis is the line that cut the neutral plane of a beam where the is no stress or strainIt cut through the centroid of the profile of the beam

15 mm

60 mm

26 mmA B

65 mm

Solution 1y1=

602

+26=56 mm

y2=262

=13mm

y=A1 y1+ A2 y2

A1+ A2

y=(60×15×56 )+(26×65×13 )

(60× 15 )+(25×65 )

¿ 50400+21970900+1625

¿28,7mm

This means that the Neutral axis and position of the centroid is 28,7mm from AB

Solution 2. Table method

Area y−distance Area× y1 900 56 504002 1625 13 21970Total Σ 2525 Σ 72370

21

1

2

y= Σ AyΣ A

=723702525

=28,7mm

EXERCISE 4

4.1 55 mm

16mm

12mm 80mm

10mm A B

80mm

4.2

80

30

30 70 20

4.3 20

50

18

22

15

85

2. MOMENT OF INERTIA

The moment of inertia or second moment of area of a shape is the property whichmeasures the efficiency of that shape as regards its resistance to bending. The bigger the I value the bigger the resistance against bending

Most beams have a higher resistance about one axis than the other, and the axis withthe most resistance is the xx-axis regardless whether it is horizontally or verticallypositioned. The weaker axis is called the yy-axis regardless whether it is horizontal or vertical. Only round and square sections have axes of the same magnitude

Standard practice is that the XX –axis is always perpendicular to the longest dimensionof the section as shown below

The intersection of the two axes is the centroid of the section

Y

X X

Y

X

Y Y

23

X

2.1 Standard sections

Page in book

24

2.2 Non standard sections

A built-up beam is a beam consisting of two standard sections which are put togetherto form a section. Normally one of the axes is unknown and the position of the other one must be determined or the position of both could be unknown.

To determine the moment of inertia the Neutral axis ( xx-axis or yy-axis) must be determined.

The bending of the beam is about the main axis of the section (xx-axis or yy-axis) and the moment of inertia should be determined for the axis about which it will bend.

The formula that is used is called the parallel axis theorem

Y

Parallel axis of A1

h1

Neutral axis X X

h2

Parallel axis of A2 y1

y2

Y

I=I 1+ A1 ( h1 )2+ I 2+ A2 (h2 )2

I=¿ The moment of inertia of the built-up which can be I xx∨I yy

I 1∧I 2=¿ Is the moment of inertia of the area concerned (in this case web or flange)

A1∧A2=¿ are the areas of the web and flange

h1∧h2=¿ are the distance between the parallel axis of each area and the axis of the built-up beam

When using table method the formula to calculate I xx can be used as follows:

I xx=Σ (I NA+ A h2)

25

FlangeW

eb

Example 1

The figure below shows a built-up beam. Calculate:1.1 the moment of inertia of the profile about the neutral axis 1.2 and the profile modulus (z) about the neutral axis

10

30

15

50

Solution 1 : Table method

1.1

Area Y-distance Area × Y INA h Area× h2

1 750 7,5 5 625 14 062,5 6,5 31 687,52 300 30 9 000 22 500 16 76 800,0

Σ 1 050 Σ 14 625 Σ 36 562,5 Σ 108 487,5

calculate the moments of inertia as follows:I x− x=∑ (INA+ A h2 )¿36 562,5+108 487,5

¿145050m m4

¿0,145×1 0−6m4 ( from 145050 ÷ 100 04 )

1.2 Calculate the position of the neutral axis as it will be needed to calculate the profile modulus

y=∑ A ×Y

∑ A

y=146251050

y=13,92mm

y=14 mm

Z= Iy

26

Z=14505014

Z=10360,715mm3

Z=10,361× 10−6 m3 ( from 10360,715÷ 100 03 )

Solution 2 : Straight forward calculation

1.1 A1=50× 15=750

A2=30× 10=300

y=A1 y1+ A2 y2

A1+ A2

y=750 (7,5 )+300(30)

750+300

y=13,929mm

y=14 mm

h1=14−7,5=6,5

h2=30−14=16

I xx=[ bd3

12+ A1 (h1 )2]+[ b d3

12+ A2 (h2 )2]

¿ [ 50×153

12+750× 6,52]+[ 10×303

12+300×162]

¿ (14062,5+31687,5 )+(22500+76800 )

¿43750+99300

¿145050mm4

¿0,145×10−6 mm4

1.2

27

Z= Iy

Z=14505014

Z=10360,715mm3

Z=10,361× 10−6 m3

EXERCISE 5

Calculate the moment of inertia ( I xx ) and the profile modulus of the following FIGURES

5.1 50

15

70

10

12 A B

85

5.2

18

30

15

10

65

28

20 40 205.3 Y

40 80

20

40

Y 150

29

CHAPTER 3

FRAMED STRUCTURES

30

FRAMED STRUCTURES

Three structural members, pin- jointed at their ends to form a triangle, provides a rigid framework.This is of fundamental importance in constructional work of many types. This “triangulation” of members can be put together to build up large frameworks, consisting of a network of structural members, the most obvious example of this is a roof truss

When a structural is loaded the applied forces are transmitted through the various members of the framework so that some are subjected to tensile stress, and others to compressive stress. From the loaded framework the magnitude of the force acting in each member, whether it is tensile or compressive can be found. From the size of a cross-section each member must have to take the load concerned, can be determined. The forces can be determined by calculation or graphical methods

Frameworks are normally assumed to be pin-jointed, each joint fixed only at one point where the centre line of the members meet. The total load on the framework is discharge at a number of joints as point loads. With roof trusses the load may be discharged at the joint by means of purlins and ridge members. The whole roof truss is simply a rigid framework in equilibrium under the action of a number of external forces.

Nature of forces in reinforcing members

A reinforcing member that is in equilibrium, can either push or pull on the pin positioning the frame. The nature of the force (pull or push) in the reinforcing member is indicated with arrows near the ends of the members, as shown below

Connecting pin

Tie bar as the force pulls away from the pin

Strut as the force pushes towards the pin

If a member is subjected to a tensile force the member is called a tie bar, if the member is subjected to a compressive force the member is called a strut. The force in the member can either pull or push on the pin and cannot be subjected to a combination of the two forces. Members which are not necessary for maintaining equilibrium configuration are called redundant. These are members which are neither tie nor strut.

31

Graphical method

EXAMPLE 1The structure shown is pivoted at one end placed on at the other end . Determine the magnitude and direction of the structure if it is loaded as shown. Also determine the forces in the members and show whether the forces are under compression or tension

7,5 kN

15 kNB

A7,5 kN

G H I C F J

30 ° 60 ° 30 ° D Rollers E Hinge

3 m 3 m 3 m RR

RL

The following rules apply when solving a truss graphically

FOR THE SPACE DIAGRAM

1. Draw the truss accurately and to a suitable scale, the larger the better2. Insert all given information3. Insert Bow’s notation

FOR THE VECTOR DIAGRAM

1. Draw the vector diagram to a suitable scale2. Start by drawing the lines representing loads3. Then at a node (joint) with not more than two unknown4. Annotate each vector corresponding to that of the space diagram5. Complete one node, then insert the arrows in the space diagram, to distinguish between Tension and compression.6. Now proceed to the next node, remember not more than two unknown7. Draw up a table to tabulate your findings

32

This is how it is done

Step 1: Draw the space diagram to scale with all the information SPACE DIAGRAMScale 15 mm = 1m

7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 2: Draw the vectors representing the loads VECTOR DIAGRMScale 10 mm = 3 kN

a

b

c

33

d

Step 3: Extended the force line as shown in the space diagram

SPACE DIAGRAMScale 15 mm = 1m

7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 4: Establish point o at any distance the farther the betterJoint o to a, b, c and d as shown below

VECTOR DIAGRMScale 10 mm = 3 kN

a

b

o

c

34

d

Step 5: Transfer the line ao, bo, co and do to the space diagramStarting at the hinge side


7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL


a

b

o

c

35

d

Step 6: Draw a dotted line to joint the two point where the line meet with reaction line


7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 7: Transfer the dotted line from the space diagram to the vector diagram, also draw line ae, when the two line meet that is point e Joint point e to d


a

b e

o

36

c

d

Step 9: Starting at the pin with A,F and Eusing the information from the vector diagram insert the arrows as follows:According to the vector diagram af goes down thus AF in the space diagram pushes the pinand FE pulls the pin, as shown in the space diagram


7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 8: Now draw line af to joint line ef where the two line meet that is point fTo get direction according to the space diagram ea is a force going up therefore following the sequence in the vector diagram af goes down and fe goes to the right


a

b f e

o

37

c

d

Step 10: Pin with A,F and G According to the vector diagram FA pushes the pin and AG pushes the pin FG is a stud


7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 9 : ag and af are the same line thus g and f are on the same point Clockwise f to a goes up therefore a to g goes down f and g are on the same point and thus not under stress


a

b f g e

o

38

c

d

Step 12: Pin with A,B,I,H, and GAB is known therefore according tp the vector diagram BI pushes the pin and IH pulls the pin, HG is a and GA pushes the pin


7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 11 : Draw line ab, bi and hi where bi and hi meet that is point i For the direction ab goes down, bi goes up, ih goes down, g and h are on the same point, therfore fg is not under stress

i VECTOR DIAGRMScale 10 mm = 3 kN

a

b f g h e

o

c

39

d

Step 12: Pin with B, C, J and I BC is known, therefore according to the vector diagram CJ pushes the pin, JI pushes the pin and IB pushes the pin


7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 11 : Line bc already exists now draw cj, ji and ib already exists For the direction bc goes down, cj goes up, ji goes up, g and ib goes down point, therfore fg is not under stress


a

b j f g h e

o

c

40

d

Step 14: Pin with C, D, E and J CD is known, therefore according to the vector diagram DE pushes the pin, EJ pulls the pin and JC pushes the pin


7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Step 13 : For the direction cd goes down, de goes up, ej goes to the left and jc goes down point, therfore fg is not under stress


a

b j f g h e

o

c

41

d

NOW MEASURE THE MAGNITUDES OF THE MEMBERS FROM THE VECTOR DIAGRAM

MEMBER MAGNITUDE NATUREAF 18 kN StrutAG 18 kN StrutBI 21 kN StrutCJ 21 kN StrutEF 16 kN TieFG 0 RedundantGH 0 RedundantHI 15 kN TieIJ 15 kN StrutHE 16 kN TieJE 29 kN Tie

DE 23 kN Reaction RR at 50 ° EA 9 kN Reaction RL

42

HOW GET THE RESULTANT OF ALL THE FORCES

7,5 kN

15 kNB

A7,5 kN

G H I C F J


3 m 3 m 3 m RR

RL

Resultant of all the forces

Reaction at hinge (DE)

43

EXAMPLE 2

44

EXERCISE 6

Determine the magnitude and direction of the reaction (Where required)Also determine the magnitude and nature of each member ( Tabulate the answer)

6.1 1 kN

2kN 3 kNC D

1 kN 1 kN

B I J K EA H L

30 ° 60 ° 60 ° 30 ° F Rollers G Hinge

4 m 4 m 4 m RR

RL

6.2 12 kN 10 kN

D 8 kN 5 kN 5 kN C E

B

A J K F H L

30 ° 30 ° G

8 m

RL RR

6.3 70 kN A

4m E B F G 70 kN

45

D C 70 kN

4m 4m

6.4 120 kN 80 kN A B

F G

E C30 °

45 °D

70 kN 5,5m

6.590 kN

B C

45 kN 45 kN

G H

A F E J D

30 ° 45 °

90 kN 90 kN9,5 m

46

Analytical methodExample 2

A B

3m

E θ D θ C

40 kN 40 kN 3m 3m

Step 1: Determine the angle of inclination of all inclined members. In this case,

tanθ=33

tanθ=1

θ=tan−11

θ=45 °

Step 2: Look for a joint at which there are only two unknowns. If such joint is not available, determine the reactions at the supports, and then at the supports the unknown may reduces to only two. (with cantilever type frame its usually not needed to find reactions)

At joint C, there are only two unknown, forces in members CB and CD (see figure below)Let us call the forces FCB∧FCD .

The direction of the forces are assumed and when calculating if the answer is negative, that means the direction assumed is incorrect and should be changed.

FCB

FCD 45 ° C

40

Step 3: Resolve all force into vertical and horizontal components as shown in the figure below.

FCB FCB sin 45

47

FCB cos 45

FCD 45 ° C

40

At joint C

Σ Vc=0

40−FCB sin 45=0

−FCB sin 45=−40

FCB=40

sin 45

¿56,569kNThe answer is positive that means the direction assumed is correct.

Σ Hc=0

FCD−FCB cos 45=0

FCD=FCB cos 45

¿56,569 cos 45

¿40 kN

The answer is positive that means the direction assumed is correct.

Step 4: On the diagram of the truss, mark arrow on the member near the joint analysed to indicatethe force on the joint. At the other end mark arrow in opposite direction.

In this case, near joint C, the arrows are marked on CB and CD to indicate forces FC B∧FCD

Directions as ff joint Cound in the analysis. The opposite direction are marked in the members CB and CD near joint B and D respectively (as shown below)

A B

3m

48

Remember this: up-Forces = Down-forces

Also that: Forces to left = Forces to the right

Let up force be + , down be −¿, forces right be + and forces left −¿

E D C

40 kN 40 kN

Step 5: look for the next joint where there are only two unknown forces and analyse that joint

At joint DFDB

FDE FDC=40kN

40 kN

Σ Vc=0

FDB−40=0

FDB=40kN

Σ Hc=0

FDE−40=0

FDE=40kN

Step 6: repeat Steps 4 and 5 till forces in all members are found. In this case:

At joint B FBA B

FBE cos45 56,569 cos 45

40 kN FBE 56,569 kN

FBE sin 45 56,569 sin 45

Σ Vc=0

FBE sin 45−40−56,569 sin 45=0

FBE=40+56,569 sin 45

¿80kN

49

Σ Hc=0

FBE cos45+56,569 cos45−FBA=0

−FBA=−FBEcos 45−56,569 cos45

FBA=80 cos45+56,569 cos45

¿96,569kN

The direction of these forces are marked on the diagram. Now the analysis is complete since the forces in all the members are determined (see figure below)

A B

E D C

40 kN 40 kN

Step 7: Determine the nature of forces in each member and tabulate the resultsNote: if the arrow marks on each member are towards the joint then the member is a strut If the arrow marks are pointing away from joint the member is a tie

Member magnitude NatureAB 96,569 kN TieBC 56,569 kN TieCD 40 kN StrutDE 40 kN StrutBE 80 kN StrutBD 40 kN Tie

50

Strut = is a member under compression force

Tie = is a member under tension force

Example 2

20 kN 24 kN

B

A C 4m

RL RR

Figure above shows a loaded roof truss, with TWO forces at the apex and a pitch of 30 ° at both sidesThe roof truss is supported on a hinge at RL and rests by means of rollers at RR

1.1 Calculate the magnitude and direction of the reactions of the truss,

1.2 calculate the forces in each member and distinguish between tension and compressive forces in members

Solution 20N 20 sin30 +24kN =34 kN

20cos 30 B

Distance = 2tan 30

A C 2m 2m

RL RR

1.1 Calculate the magnitude and direction of the reactions of the truss

Moments about RL: anti-clockwise = clockwise

RR(4) = 34(2) + 20cos30(2tan30)

RR(4) = 34(2) +17,32(1,15)

RR(4) = 87,92

RR = 21,98 kN

Moments about RR: clockwise = anti-clockwise

51

RL(4) + 17,32(1,15) = 34(2)

RL(4) = 34(2) – 17,32(1,15)

RL(4) = 48,08

RL = 12,02 kN

Test : upward forces = downward forces 21,98 + 12,02 = 34 kN

RR is on rollers therefore the force act vertically upwards

But RL Is not, thus 12,02 is the vertical compenent of RL

and thus : RL=√H c2+V c2

¿√17,322+12,0 22

¿21,08kN

Direction: tanθ=12,0217,32

θ=tan−1(12,0217,32 )

θ=34,76 ° North of the horizontal

1.2 Calculating the forces in each member and their nature (strut or tie)

At joint C

Finding the magnitude of BC

∑Vc=0

21,98−BC sin 30=0

−BC sin 30=−21.98

BC=¿ 21,98sin 30

BC=43,96kN (strut )

Finding the magnitude of AC

∑ Hc=0

AC+BCcos30=0AC+43,96 cos30=0

52

Assume the direction of the members

BC BC sin30

BC cos30 AC

RR = 21,98 kN BC sin30 (Vc of BC) acting in direction opposite RRBC cos30 (HC 0f BC) acting in same directions as AC

NB: Forces going up are + and down – Force to the right + and to the left –

If the answer is negative change the direction assumed

Hc = 17,32 kN

We need 17,32 kN to the left to balance it

17,32 kN 12,02 kN θ RL

AC=−43,96 cos30AC=−38,07kN

The answer is negative, it means that the direction chosen member is incorrect. AC=38,07 kN (tie)

At joint A

Finding the magnitude of AB

∑Vc=0

ABsin 30+RL=0

12,02−AB sin 30=0

−ABsin 30=−12,02

AB=¿ 12,02sin 30

AB=24,04 kN (Strut )

Finding the magnitude of AB

∑ Hc=0

AC−17,32−ABcos 30=0

AC−17,32−24,04 cos30=0

AC−38,139=0

AC=38,139 kN

Member Magnitude NatureAB 24,04 kN Strut (Compression)BC 43,96 kN Strut (Compression)AC 38,14 kN Tie (Tension)

Complete diagram looks as follows

20N 24 kN

53

Assume the direction of the members

AB sin 30 AB

AB cos 30 AC 17,32 Kn Resultant

12,02 kN

AB sin30 is Vc of ABAB cos30 is HC of AB17,32kN is HC of Resultant RL is the Vc of Resultant

NB: Forces going up are + and down – Force to the right + and to the left –

If the answer is negative change the direction assumed

B

A C

RL RR

EXERCISE 7

Calculate the magnitude and direction of the reactions and the magnitude in each member, and also determine the nature of each member

7.125kN 30 kN

35 kN

30 ° 60 °6 m 2 m

RL RR

7.2

18 kN 22 kN

16 kN B 25 kN

C

A 30 ° DE

RL 2 m 3 m 3 m RR

54

CHAPTER 4

BOLTS AND RIVETS

55

Terms Used in Riveted or bolted Joints

Pitch

It is the distance from the centre of one rivet to the centre of the next rivet measured parallel to the seam as shown. It is usually denoted by p.(see diagram below)

Minimum pitch (SABS 0162-1984)

Minimum pitch between bolts shall not be less than 2,5 times the nominal diameter of the fastener

Maximum pitch (SABS 0162-1984)

Maximum pitch shall not exceed either 32t or 300 mm ( t = thickness of thinner plate)

Margin or marginal pitch (m)It is the distance between the centre of rivet hole to the nearest edge of the plate. It is usually denoted by m (see diagram below)

Back pitch. It is the perpendicular distance between the centre lines of the successive rows (pb)

Diagonal pitch.

It is the distance between the centers of the rivets in adjacent rows of zig-zag riveted joint (pd)

Diameter of the hole ¿ diameter of the bolt + 2mm

Edge distance of the centre of the bolt to the edge of the plate ¿ 1,75× diameter of the hole

The pitch distance of each hole drilled in a simple row of holes ¿ 2,5 × diameter of the hole

Diameter of the rivet ( given the thickness (t) of the thinner plate) D=6√t

56

One row of rivets Two rows of bolts Two rows with diagonal rivets (Chain riveting) (Zig-zag riveting)

TYPES OF RIVETED OR BOLTED JOINTS

1. LAP- JOINTIn a lap joint the bolts/rivets are in single shear and also subjected to bearing.Therefore when designing a bolted/riveted joint that is subjected to single shear, the lowestof the single shear or bearing value is used

1.1 Single bolted/riveted lap jointThis joint has one row of bolts/rivets

1.2 Double bolted/riveted lap jointThis joint has two row of bolts/rivets

2. BUTT-JOINT2.1 Single covered butt-joint

In a single covered butt-joint, the bolts/rivets are in single shear and again for the strengthof the bolt/rivet, use the least strength value between single shear and bearing

57

2.2 Double covered butt-jointIn a double covered butt-joint, the bolts/rivets are in double shear and again for the strength of the bolt/rivet, use the least strength value between single shear and bearing

FAILURES OF A RIVET OR BOLT JOINT

1. TENSILE FAILURE 1.1 Tearing of the plate at an edge.

A joint may fail due to tearing of the plate at an edge. This can be avoided by keeping the margin 1,5 times the diameter of the rivet hole

m

F d F

1.2 Tearing of the plate across a row of rivets.

Due to the tensile stresses in the main plates, the main plate or cover plates may tear off across a row of rivets as shown. The resistance offered by the plate against tearing is

known as tearing resistance or tearing strength or tearing value of the plate

F d FW

58

m=1,5d

FORMULA

area under stress→ A=(W −n .d ) t

Resistance force of the plate¿ tearing→F=F t (W −n .d ) t

F = Resistance of the tie-bar to tearingf t = Maximum tensile stress of the tie-barW = Width of the tie-bar (plate)n = number of rivets in a row (in this case there are two)d = Diameter of the holest = Thickness of the tie-bar

2. SHEAR FAILURE The plates which are connected by the rivets/bolts exert tensile stress on the rivets, and if the Rivets/bolts are unable to resist the stress, they are sheared off. The resistance offered by a rivet/bolts to be sheared off is known as shearing resistance or shearing strength or shearing value of the rive

F FW

F F

FORMULAE

2.1 to calculate resistance force of the bolt/rivet to shearing

area under shear stress → A=π D2 n4

Resistant force of the bolt ¿ shear → F=F s×π D2 n

4

F = Resistant force of the boltsF s = Maximum shear stress of the bolts

59

D = Diameter of the bolts/rivetn = Total number of bolts/rivets

2.2 If the screw threads of the bolts are outside the shearing plane

Resistant force of the bolt ¿ shear → F=F s×π D2 n

4

Where D=diameter of the hole+screw thread pitch

2.3 If the screw threads of the bolts are inside the shearing plane

F=Fs .( π ( d−0,9382 p )2

4 )nWhere d=diameter of bolt∧p=screw thread pitch

3. BEARING/ CRUSHING FAILURE Due to crushing failure, the rivet hole becomes of an oval shape and hence the joint becomes loose. The failure of rivets in such a manner is also known as bearing failure. The resistance offered by a rivet to be crushed is known as crushing/compression resistance or crushing/compression strength or bearing value of the rivet

F F

F F

Formula to calculate resistance force

F=f b . D . t . n

F = Resistant force of the bolts to crushingf b = Maximum compressive stress of the boltsD = Diameter of the holet = Thickness of the tie-barn = number of bolts/rivets in a row

60

EXAMPLE 1

10 mm 10 mm

100 KN 80 mm

∅ 16 mm in ∅ 18mm

The figure above shows a double-riveted lap joint connection with four ∅ 16 mm rivets in a tie bar.The tensile load in the tie bar is 100 kNCalculate the following

1.1 Shear stress1.2 Tensile stress1.3 Compressive stress1.4 the maximum safe stress the connection joint can withhold

SOLUTION

1.1 f s=4 F

n× π D2 or f s=

F

π D2 n4

¿ 4× 100× 103

4× π × 162

¿ 100× 103

π ×162 × 44

¿124,340 MPa

61

1.2 f t=F

(W −n .d )t

¿ 100×103

(80−2×18 )10

¿227,273 MPa

1.3 f b=F

D .t .n

¿ 100× 103

16× 10× 4

¿156,25 MPa

1.4 maximum safe stress = 124,273 MPa

EXAMPLE 2 7 mm 7 mm

80 mm

The figure above shows a single riveted lap joint connection of two 6 mm thick plates and 80 mm wide

Calculate :

2.1 The diameter of the hole to be drilled for the bolt2.2 Resistant force of the bolts to shearing.2.3 Resistant force of the tie-bar to tearing.2.4 Resistant force of the bolts to crushing.

Make use of the following allowable stresses

Tensile stress in plate = 150 MPaShearing stress in rivets = 95 MPaCompression stress in rivets = 210MPa

SOLUTION2.1 D=6√t

¿6√7

62

¿15,875mm

Say 16 mm diameter

Diameter of the hole ¿16+2=18mm

2.2 F=f s ×n×π d2

4

F=95×2×π 162

4

F=38,202 k N

2.3 F=f t (W−n .d ) tF=150× (80−18 ) ×7F=65,100kN

2.4 F=f b . D . t . n

F=210×16× 7× 2

F=52,92kN

EXERCISE 8

10 mm 10 mm

120 KN 80 mm

∅ 16 mm in ∅ 18mmQUESTION 1

The figure above shows a double-riveted lap joint connection with four ∅ 16 mm rivets in a tie barThe tensile load in the tie bar is 120 kN

Calculate the following

1.1 Shear stress1.2 Tensile stress1.3 Compressive stress1.4 the maximum safe stress the connection joint can withhold

QUESTION 2

Calculate the safe load that the connection in the figure below can withstand in shear if the ultimate shear stress is 100 MPa

M 16 mm bolt

63

M 20 mm bolt

F kN

QUESTION 3

The diameter of a bolt is 20 mm. Calculate and state:

3.1 The diameter of the hole to be drilled for the bolt3.2 The edge distance of the centre of the bolt to the edge of the plate3.3 The pitch distance of each hole which is drilled in a single row of holes

QUESTION 4

The figure above show a bolted connection between two tie-bars by means of three grade 4,6 bolts

Calculate the resistance (force) against shearing of the bolted connection if :

4.1 The screw threads of the bolts are outside the shearing plane.4.2 The screw threads of the bolts are inside the shearing plane.

Use the following specifications

Bolt = M12Maximum shear stress = 100 MPaScrew- thread pitch = 1,75 mm

QUESTION 5

80 kN 50 mm

64

The figure above shows a bolted connection with four bolts securing the tie-bar to the gusset plate.

Calculate the following:

5.1 The diameter of the bolts to withstand the load of 80 kN in the joint5.2 The edge distance of the centre of the bolt to the edge of the plate5.3 The pitch distance of each hole which is drilled in a single row of holes5.4 The resistance of the tie-bar to tearing using the calculated bolt diameter in question 5.15.5 The resistance of the bolts to crushing using the calculated bolt diameter in question 5.15.6 The maximum force the connection can withstand safely

Use the following specifications

Tie bar = 50 × 8 mm thickGusset = 10 mmMaximum shear stress of bolts = 100 MPaMaximum shear stress of tie = 155 MPaMaximum compressive stress of the bolt = 240 MPa

ISOMETRIC VIEW OF BOLT/RIVET JOINTS

65

DRAWING EXPLAINING JOINT FAILURE

66

CHAPTER 5

CONSTRUCTION

DRAWINGS67

FOUNDATIONS

68

FLOOR FINISHES

69

BRICK WALLS AND SLABS

70

BEAMS

71

PILING

72

STEEL COLUMNS

73

STAIRS

74

construction n5 tutorial

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