Chapter 2

Basic Structure Concepts

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2.0 : BASIC STRUCTURE CONCEPT

2.1 Forces

2.2 Equilibrium and Reactions

2.3 Moments

2.4 Stress and Strain

2.5 Elastic and plastic range

2.6 Primary Loads & Secondary Loads

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2.1 Forces

A force is that which tends to exert motion, tension or compression on an object.

The loads acting on a structure have mass which is usually measured in kilograms(kg).

The basic unit of force is the newton (N). The force exerted on a structure by a static load is dependent upon both its mass and the force of gravity.

The force exerted by a body as a result of gravity can be described as its weight.

F =W= m x g

Therefore, the force exerted by a mass of 1 kg is:

F = m x g

= 1 x 9.81

=9.81 N @ 0.00981 kN

1 kN = 1000 N

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2.1 Forces

Scalars and Vectors

Scalars Vectors

Examples: mass, volume force, velocity

Characteristics: It has a magnitude It has a magnitude

(positive or negative) and direction

Addition rule: Simple arithmetic Parallelogram law

Special Notation: None Bold font, a line, an

arrow or a “carrot”

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2.1 Forces

Application of Vector Addition

There are four

concurrent cable

forces acting on the

bracket.

How do you

determine the

resultant force acting

on the bracket ?

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2.1 Forces

Vector Operations

Scalar Multiplication

and Division

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2.1 Forces

Vector Addition Using Either the Parallelogram or Triangle

Parallelogram Law:

Triangle method (always

‘tip to tail’):

How do you subtract a vector? How can you add

more than two concurrent vectors graphically ?

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2.1 Forces

Resolution of a vector

“Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse.

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2.1 Forces

Each component of the vector is

shown as a magnitude and a

direction.

Cartesian Vector Notation

We ‘ resolve’ vectors into components

using the x and y axes system

The directions are based on the x

and y axes. We use the “unit vectors”

i and j to designate the x and y axes.

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2.1 Forces

For example,

F = Fx i + Fy j or F' = F'x i + F'y j

The x and y axes are always perpendicular to each

other. Together, they can be directed at any

inclination.

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2.1 Forces

Addition of Several Vectors

Step 1 is to resolve each force into its

components

Step 2 is to add all the x components

together and add all the y components

together. These two totals become the

resultant vector.

Step 3 is to find the magnitude and

angle of the resultant vector.

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2.1 Forces

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2.1 Forces

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2.1 Forces

Example 1

The screw eye in Figure below is subjected to forces F1 and F2. Determine the magnitude and direction (measured from x-positive axis) of the resultant force.

F2= 150 N

F1= 100 N 10o

15o

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2.1 Forces Solution

10o

15o

15o + 90o + 10o

= 115o

R =√(1002 +1502 – (2x 100 x 150 x cos 115 )

= 212.55 N

Sin /150 = sin 115/212.55

Sin = (sin 115/212.55) x 150

= 39.76o

Direction = 15o + 39.76o

=54.76o

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2.1 Forces

Example 2 Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x-axis

Solution

a) Resolve the forces in their x-y components.

b) Add the respective components to get the resultant vector.

c) Find magnitude and angle from the resultant components.

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2.1 Forces

F1 = { 15 sin 40° i + 15 cos 40° j } kN

= { 9.642 i + 11.49 j } kN

F2 = { -(12/13)26 i + (5/13)26 j } kN

= { -24 i + 10 j } kN

F3 = { 36 cos 30° i – 36 sin 30° j } kN

= { 31.18 i – 18 j } kN

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2.1 Forces

Summing up all the i and j components respectively, we get,

FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN

= { 16.82 i + 3.49 j } kN

FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN

= tan-1(3.49/16.82) = 11.7°

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2.1 Forces

Example 3 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis

Solution

a) Resolve the forces in their x-y components.

b) Add the respective components to get the resultant vector.

c) Find magnitude and angle from the resultant components.

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2.1 Forces

F1 = { (4/5) 850 i - (3/5) 850 j } N

= { 680 i - 510 j } N

F2 = { -625 sin(30°) i - 625 cos(30°) j } N

= { -312.5 i - 541.3 j } N

F3 = { -750 sin(45°) i + 750 cos(45°) j } N

{ -530.3 i + 530.3 j } N

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2.1 Forces

Summing up all the i and j components respectively, we get,

FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N

= { - 162.8 i - 521 j } N

FR = ((162.8)2 + (521)2) ½ = 546 N

= tan–1(521/162.8) = 72.64° or

From Positive x axis

= 180 + 72.64 = 253 °

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TUTORIAL 1 (FORCES)

Question 1

Two forces are applied to an eye bolt fastened to a beam. Determine the magnitude and direction of their resultant.

6 kN

4.5 kN

25o

50o

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Question 2

Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 60 kN and Q = 100 kN. Determine the magnitude and direction of their resultant.

(Ans: R = 150 kN, 76o towards x axis positive)

P Q

15o 30o

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Question 3

The cables AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 N in AD. Determine the magnitude and direction of the resultant of the forces exerted by cables at A.

(Ans: R = 575 N, 113o towards x axis positive)

A

B C D

2 m 1.5 m

2.5 m

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Question 4

Determine the resultant and direction from x-axis positive of the five forces shown in Figure below by the graphical method.

(Ans: R = 32.5 kN, = 124o)

y

x

8 kN 9 kN

4 kN

60o

25 kN

3 kN

15o

30o

20o

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Question 5

Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force of the three acting on the ring A. Take F1 = 500 N and = 20o

(Ans: R = 1.03 kN, = 87.9o)

y

x

600 N F1

400 N

30o 3

4

5

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Question 6

Three forces are applied at the end of the boom O. Determine the magnitude and orientation of the resultant force.

(Ans: FR= 485 N, = 37.7o)

y

x

F3 = 200N F2= 250 N

F1 = 400 N

45o

3

4

5

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• Question 7

• Two cables are attached to the frame shown in Figure below. Using trigonometry, determine :

(a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical.

(b) the corresponding magnitude of R.

(Ans: 489 N, 738 N)

A

A

25o 35o

360 N

P

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2.2 Equilibrium and Reactions

Any structure subjected to loads must be provided with supports to prevent it from moving. The forces generated on the structure by these supports are called reactions. If the structure is in equilibrium (i.e. not moving) then the net forces from the loads and reactions must be zero in all directions.

Fx = 0

Fy = 0

M = 0

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Example 1

Determine the magnitudes of F1 and F2 so that particle P is in equilibrium.

(Ans: F1 = 435 N, F2 = 171 N)

P

F1 F2

400 N

3

4

5

30o

60o

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Solution

The two unknown magnitudes F1 and F2 can be obtained from the two

scalar equations of equilibrium, ∑Fx=0 and ∑Fy = 0. To apply these

equations, the x, y axes are established on the free body diagram and

forces must be resolved into its x and y components.

P

F1 F2

400 N

3

4

5

30o

60o

400 cos 30o

400 sin 30o

F2 cos 60o

F2 sin 60o

F1 (4/5)

F1 (3/5)

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Solution

∑ Fx =0 ; -400 sin 30o + F1(4/5) –F2sin 60o=0

∑ Fy =0; 400 cos 30o -F1(3/5) –F2cos 60o=0

(1)

(2)

Simplify : 0.8 F1 – 0.866 F2 = 200 (3)

-0.6 F1 – 0.5 F2 = -346.61 (4)

Solving Eqs. (3) & (4) by simultaneous equations to determine F1 & F2.

Ans : F1 = 435.08 N

F2 = 170.97 N

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Application of equilibrium concepts in truss

A truss is a structure composed of slender

members joined together at their end

points.

The members commonly used in

construction consist of wooden struts

or metal bars

The joint connections are usually formed

by bolting or welding the ends of the

members to a common plate, called a

gusset plate, as shown in Fig below.

•

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38

Timber Roof Trusses

Steel Bridge Trusses

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Example • Determine the force in each member of the Pratt bridge truss

shown. State whether each member is in tension or compression.

6 kN 6 kN 6 kN

4 m

3 m 3 m 3 m 3 m

A

B

C

D

E

F

G H

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2.3 Moments

APPLICATION

What is the net effect of the two

forces on the wheel?

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APPLICATION

What is the effect of the 30 N force on

the lug nut?

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The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

This tendency for rotation caused by force is sometimes called a torque, but most often it is called the moment of a force or simply the moment

The magnitude of a moment about a point is the value of the force multiplied by the perpendicular distance from the line of action of the force to the point

M= T = F x d

The unit is Nm or kNm

The typical sign convention for

moment is that counter-clockwise

is considered positive.

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Example 1

From Figure below, determine the moment of the force about point O. If the moment is increased to 5 kNm, what is the maximum mass can be supported by the diving board?

1.5 m

Mass = 250 kg

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Solution

F = mg

= 250 (9.81)= 2452.5 N = 2.45 kN

1.5 m

Mo = ( 2.45 x 1.5) = 3.68 kNm

If Mo = 5 kNm, then new load is

Mo = F x d

5 = F x 1.5

F = 3.33 kN = 3333.33 N

F = m x g

3333.33 = m x 9.81

m = 339.79 kg

F

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Example 2 For each case illustrated in Figure below, determine the moment of the force about point O.

5O kN

2 m

0.75 m

Solution

Mo = - (50 x 0.75) = - 37.5 kNm

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Example 3 For each case illustrated in Figure below, determine the moment of the force about point O.

6O kN

3 m

45o

Solution

1 sin 45o

Mo = 60 x (1 sin 45o) = 42.43 kNm

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Example 4 A 400 N force is applied to the frame and = 20o. Find the moment of the force at A.

Solution

400 sin 20o

400 cos 20o

MA = (400 sin 20ox 3) + (400 cos 20o x 2)

= 1162.18 Nm Jan

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Figure above shows a bridge deck, of weight 500 kN, supporting a heavy vehicle

weighing 300 kN. Find the value of the support reactions at A and B when the load

is in the position shown.

Ans: RA = 430 kN, RB = 370 kN

Example 5

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Solution

500 kN

5 m

300 kN

A B 4 m

∑ MA =0 ;

∑ Fy =0 ;

RB (10) – ( 300 x 4) – (500 x 5) = 0

RB = 370 kN

RA + RB – 300 – 500 = 0

RA = 430 kN

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Example 6

Figure shows an anchorage for the

cable of a ski- lift, The cable produce a

total pull of 220 kN at the top of the

anchorage. Determine the support

reactions.

Ans: RAH = 191 kN, RAV = -196 kN

RB = 306 kN

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Solution

A B

220 cos 30o

220 sin 30o

RB RAV

RAH

∑ MA =0 ;

RB (2.7) – ( 220 sin 30o x 1.8) – (220 cos 30o x

3.3) = 0

RB = 306 kN

∑ Fx =0 ;

-RAH + 220 cos 30o = 0

RAH = 191kN

RAV + RB – 220 sin 30o = 0

RAV = - 196 kN

∑ Fy =0 ;

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2.4 Stress and Strain

Stress Concept of stress

To obtain distribution of force acting over a sectioned area

Assumptions of material:

1. It is continuous (uniform distribution of matter)

2. It is cohesive (all portions are connected together)

Normal stress

Intensity of force, or force per unit area, acting normal to ΔA

Symbol used for normal stress, is σ (sigma)

σz = lim

ΔA →0

ΔFz

ΔA

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Tensile stress: normal force “pulls” or “stretches” the area element ΔA

Compressive stress: normal force “pushes” or “compresses” area

element ΔA

Shear stress

Intensity of force, or force per unit area, acting tangent to ΔA

Symbol used for normal stress is τ (tau)

τzx = lim

ΔA →0

ΔFx

ΔA

τzy = lim ΔA →0

ΔFy

ΔA

Units (SI system)

Newtons per square meter (N/m2) or a pascal (1 Pa = 1 N/m2)

kPa = 103 N/m2 (kilo-pascal)

MPa = 106 N/m2 (mega-pascal)

GPa = 109 N/m2 (giga-pascal)

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Shear in Nature

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Example 1

The 80 kg lamp is supported by two rods AB and BC as shown in

Figure below. If AB has a diameter of 10 mm and BC has a diameter

of 8 mm, determine the average normal stress in each rod.

A

B

C

60o

3

4

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A

B

C

60o

3

4

F = mg

= 80 x 9.81

= 784.8 N

= 0.78 kN

F1 F2

F1cos 60o

F1sin 60o

F2cos

F2sin

Solution

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Solution

∑ Fx =0 ; – F1cos 60o + F2(4/5) =0

∑ Fy =0; F1sin 60o+ F2(3/5) – 0.78 =0

(1)

(2)

Simplify : - 0.5 F1 + 0.8 F2 = 0 (3)

0.866 F1 + 0.6 F2 = 0.78 (4)

Solving Eqs. (3) & (4) by simultaneous equations to determine F1 & F2.

0.6 x Eqs (3) - 0.3 F1 + 0.48 F2 = 0

0.8 x Eqs (4) 0.69 F1 + 0.48 F2 = 0.624

- 0.99 F1 = - 0.624

F1 = 0.63 kN

Subs F1 into Eqs (3)

-

- 0.5 F1 + 0.8 F2 = 0

- 0.5 x 0.63 + 0.8 F2 = 0

F2 = 0.315/0.8 = 0.39 kN

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Solution

1 = F1 / A1

= 0.63 /( x 0.0052)

= 8021.41 kN/m2

2 = F2 / A2

= 0.39/( x 0.0042)

= 7758.8 kN/m2

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Shear Stress

Shear stress is the stress component that act in the plane of the sectioned area.

Consider a force F acting to the bar

For rigid supports, and F is large enough, bar will deform and fail along the planes identified by AB and CD

Free-body diagram indicates that shear force, V = F/2 be applied at both sections to ensure equilibrium

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Average shear stress over each section is:

P

A avg =

avg = average shear stress at section, assumed to

be same at each pt on the section

V = internal resultant shear force at section

determined from equations of equilibrium

A = area of section

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Example 1

The bar shown in Figure below has a square cross section for which

the depth and thickness are 40 mm. If an axial force of 800 N is applied

along the centroidal axis of the bar’s cross-sectional area, determine

average normal stress and average shear stress acting on the material

along section planes a-a.

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Internal loading

Based on free-body diagram, Resultant loading of axial force, P = 800 N

Average normal stress

= P/A = 800/(0.04)(0.04) = 500 kPa

Average shear stress

No shear stress exists on the section, since the shear force at the section is zero.

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Example 2

Three plates are held together by two cyclindrical rivets. If a direct pull

of 5 kN is applied between one plate and the other two, estimate the

diameter of the rivets. The shear stress in the rivets is not to exceed 40

N/mm2.

5 kN

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Solution

• 4 sliding areas (Double shear)

P

A =

40 = 5000 4(2/4)

= 6.3 mm

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Question 1

A 50 kN axial load is applied to a short wooden post which is supported

by a square concrete footing resting on distributed soil. Determine

(a) The maximum bearing stress on the concrete footing

(b) The size of the footing for which the average bearing stress on the

soil is 150 kPa.

(Ans: B = 4MPa, b= 577 mm)

Plan

125 mm

100 mm b

b

50 kN

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TUTORIAL 2 (STRESS

Question 2

The column is subjected to an axial force of 8 kN at its top. If the cross

sectional area has the dimensions shown in the figure, determine the

average normal stress at section a-a. (Ans: = 1.74 MPa)

Plan

Front Elevation

160 mm

160 mm

a a

10 mm

10mm

10 mm

8 kN

160 m

m

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Question 3

The 20 kg lamp is supported by two steel rods connected by a ring A.

Determine which rod is subjected to the greater average normal stress

and compute its value.

(Ans: = 2.33 N/mm2)

B

A

C

60o

45o

12 mm

10 mm

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Question 4

A square hole having 12 mm sides is to be punched out of a metal

plate 1.6 mm thick. The shear stress required to cause fracture is 350

N/mm2. What force must be applied to punch die? What would be the

compressive stress in the punch?

(Ans: 26.88 kN, 0.19 kN/mm2) Jan

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ALLOWABLE STRESS

When designing a structural member or mechanical element, the stress in it must be restricted to safe level

Choose an allowable load that is less than the load the member can fully support

One method used is the factor of safety (F.S.)

F.S. = Ffail

Fallow

If load applied is linearly related to stress developed within

member, then F.S. can also be expressed as:

F.S. = σfail

σallow F.S. =

fail

allow

In all the equations, F.S. is chosen to be greater than 1, to avoid

potential for failure

Specific values will depend on types of material used and its

intended purpose

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Strain

Loads cause bodies to deform, thus points in the body will undergo displacements or changes in position

Normal strain () is a measure of elongation or contraction of small line segment in the body

Normal Strain () = Change in length = L

Original length L

Has no unit

Shear strain () is a measure of the change in angle that occurs between two small line segments that are originally perpendicular to each other.

Shear Strain () = x

L

L

x

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Conventional stress-strain diagram Figure shows the characteristic stress-strain diagram for steel, a

commonly used material for structural members and mechanical elements

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Elastic behavior. A straight line Stress is proportional to strain, i.e., linearly elastic Upper stress limit, or proportional limit; σpl

If load is removed upon reaching

elastic limit, specimen will return to its

original shape

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Yielding. Material deforms permanently; yielding;

plastic deformation Yield stress, σY

Once yield point reached, specimen continues to elongate (strain) without any increase in load

Note figure not drawn to scale, otherwise induced strains is 10-40 times larger than in elastic limit

Material is referred to as being perfectly plastic

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Strain hardening.

Ultimate stress, σu

While specimen is elongating, its x-

sectional area will decrease Decrease in area is fairly uniform over entire

gauge length

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Necking.

At ultimate stress, x-sectional area

begins to decrease in a localized

region

As a result, a constriction or “neck” tends

to form in this region as specimen

elongates further

Specimen finally breaks at fracture stress, σf

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Hooke’s Law

When a material is worked within its elastic limit, the extension is

proportional to the force.

Strain Stress

Stress = Constant (E)

Strain

This constant is known as the modulus of elasticity or Young’s

Modulus

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Example 1

A tie-bar in a steel structure is of rectangular section 30 mm x 50

mm. The extension measured in a 250 mm length of the tie bar

when load is applied to the structure is 0.1 mm. Find :-

i) The tensile stress in the bar

ii) The tensile force

iii) The factor of safety used

Take E = 205 kN/mm2 and Ultimate stress = 460 N/mm2

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Solution

i) The tensile stress in the bar

E = stress/strain

Strain (ε) = ∆L/L

= 0.1/250

= 4 x 10-4

E = stress/strain

205 = / 4 x 10-4

= 205 x 4 x 10-4 = 0.082 kN/mm2 = 82 N/mm2

ii) The tensile force

= F/A

82 = F/(30 x 50)

F = 82 x (30 x 50) = 123,000 = 123 kN

iiii) The factor of safety used

F.S. = Ultimate stress/ Tensile stress

= 460 /82 = 5.6

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Example 2

The ultimate stress for a steel is 450 N/mm2. What is the maximum load which a rod 50 mm diameter can carry with a factor of safety of 5? If the rod is 1.5 m long, determine the extension under this loading.( E = 200 kN/mm2.)

Solution

The factor of safety

F.S. = Ultimate stress/ Tensile stress

5 = 450 / tensile stress

Tensile stress = 450/5 = 90 N/mm2

= F/A

90 = F/(π r2)

F = 90 x (π x 252) = 176,714.59 N = 176.71 kN

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Solution

The tensile stress in the bar

E = stress/strain

200 x 103 = 90/ ε

ε = 90/ 200 x 103

= 4.5 x 10-4

Strain (ε) = ∆L/L

4.5 x 10-4 = ∆L/1500

∆L = 0.675 mm

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Exercise 1

A flat steel tie-bar 4.5 m long, is found to be 2.4mm short. It is

sprung into place by means of drafts driven into holes in the end

of the bar. Determine:

(a) the stress in the bar

(b) the factor of safety if the material of the tie-bar has an

ultimate stress of 450 N/mm2. Take E for the material as

205 kN/mm2.

( Ans: 109.3 N/mm2, 4.117)

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TUTORIAL 3 (STRESS, STRAIN & FACTOR OF SAFETY)

Exercise 2

A metal tube of outside diameter 75 mm and length 1.65 m is to

carry a compressive load of 60 kN. If the allowable axial stress is

75 N/mm2 , calculate the inside diameter of the tube. If E of the

material is 90 kN/mm2., by how much will the tube shorten under

this load?

Ans: 67.87 mm, 1.375 mm)

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2.5 Primary Loads & Secondary Loads

DEADLOADS Dead Loads are those loads which are considered to act permanently; they are "dead," stationary, and unable to be removed. The self-weight of the structural members normally provides the largest portion of the dead load of a building. This will clearly vary with the actual materials chosen. Permanent non-structural elements such as roofing, concrete, flooring, pipes, ducts, interior partition walls, Environmental Control Systems machinery, elevator machinery and all other construction systems within a building must also be included in the calculation of the total dead load. These loads are represented by the red arrow in the illustration.

Primary Loads are divided into three broad categories according to the way in which they act upon the structure or structural element. These are DEAD LOADS, LIVE LOADS(IMPOSED LOADS) and WIND LOADS

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2.5 Primary Loads

Unit weights of various building materials (kN/m3)

Materials Unit Weight

Aluminium 24

Bricks 22*

Concrete 24

Concrete blocks (lightweight) 12*

Concrete blocks (dense) 22*

Glass fibre composite 18

Steel 70

Timber 6*

* Subject to considerable variation

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2.5 Primary Loads

Unit weights of various sheet materials (kN/m2)

The dead load of a floor or a roof is generally evaluated for one square meter of floor or roof area

Sheet materials kN/m2

Acoustic ceiling tiles 0.1

Asphalt (19 mm) 0.45

Aluminium roof sheeting 0.04

Glass (single glazing) 0.1

Plaster (per face of wall) 0.3

Plasterboard 0.15

Rafters, battens and felt 0.14

Sand/cement screed (25 mm) 0.6

Slates 0.6

Steel roof sheeting 0.15

Timber floorboards 0.15

Vinyl Tiles 0.05

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• LIVE LOADS

Live Loads are not permanent and can

change in magnitude. They include

items found within a building such as

furniture, pianos, safes, people, books,

cars, computers, machinery, or stored

materials, as well as environmental

effects such as loads due to the sun,

earth or weather.

• WIND LOADS

Wind and earthquakes loads are put

into the special category of lateral live

loads due to the severity of their action

upon a building and their potential to

cause failure.

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Secondary loads

Structures can be subjected to secondary loads from

temperature changes, shrinkage of members and settlement of

supports.

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Example 1

Figure shows a precast concrete

Beam which is 10.5 m long.

a) Calculate the weight of the

beam per unit length in kN/m

b) Calculate the total weight of

the beam

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Solution

a) Cross sectional area of the

beam

= (0.6 x 0.25) – (0.4 x 0.15)

= 0.09 m2

Unit weight of concrete = 24 kN/m3

Weight per unit length = 0.09 x 24

= 2.16 kN/m

b) Total weight of the beam

= 2.16 x 10.5 = 22.68 kN

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Example 2 The floor in a multi-storey office. Building consists of the following:

• Vinyl tiles

• 40 mm sand/cement screed

• 125 mm reinforced concrete

slab

• Acoustic tile suspended ceiling

Determine the dead load in kN/m2

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Solution

From the table given

• Vinyl tiles = 0.05

• 40 mm sand/cement screed = 0.6 x (40/25) = 0.96

• 125 mm reinforced concrete = 0.125 x 24 = 3.00

slab

• Acoustic tile suspended ceiling = 0.10

The dead load = 4.11 kN/m2

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Exercise 1

Figure shows the outer wall of a multi-storey

building which is supported on a beam at

each floor level. The wall consist of a 1.2 m

height of cavity wall supporting 1.3 m high

double glazing. The cavity wall construction

is 102.5 mm of brickwork, a 75 mm cavity

and 100 mm of plastered lightweight concrete

blockwork.

Determine the dead load on one beam

in kN/m of beam.

(Ans: 4.77 kN/m)

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TUTORIAL 4 (LOADS)