2005 pearson education south asia pte ltd 1. stress 1 chapter objectives review important principles...

2005 Pearson Education South Asia Pte Ltd

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CHAPTER OBJECTIVES

• Review important principles of statics

• Use the principles to determine internal resultant loadings in a body

• Introduce concepts of normal and shear stress

• Discuss applications of analysis and design of members subjected to an axial load or direct shear


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CHAPTER OUTLINE

1. Introduction

2. Equilibrium of a deformable body

3. Stress

4. Average normal stress in an axially loaded bar

5. Average shear stress

6. Allowable stress

7. Design of simple connections


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Solid Mechanics

• A branch of mechanics

• It studies the relationship of– External loads applied to a deformable body, and– The intensity of internal forces acting within the

body

• It is used to compute deformations of a body

• Study body’s stability when external forces are applied to it

1.1 INTRODUCTION


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• Beginning of 17th century (Galileo)

• Early 18th century (Saint-Venant, Poisson, Lamé and Navier)

• In recent times, with advanced mathematical and computer techniques, more complex problems can be solved

1.1 INTRODUCTION

Historical development


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1.2 EQUILIBRIUM OF A DEFORMABLE BODY

• Surface forces– Area of contact– Concentrated force– Linear distributed force– Centroid C (or geometric

center)

• Body force (e.g., weight)

External loads


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Support reactions• for 2D problems



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– balance of forces– balance of moments

∑ F = 0

∑ MO = 0


Equations of equilibrium

Two-Dimension: Σ Fx = 0

Σ Fy = 0

Σ Mx = 0

Σ My = 0

Σ Mo = 0Three-Dimension: Σ Fx = 0

Σ Fy = 0

Σ Fz = 0

Σ Mx = 0

Σ My = 0

Σ Mz = 0

Σ Mo = 0


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• For coplanar loadings (2-D):– Normal force, N– Shear force, V– Bending moment, M


Internal resultant loadings


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• For coplanar loadings:

– Apply Fx = 0 to solve for N

– Apply Fy = 0 to solve for V

– Apply MO = 0 to solve for Mo



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1.3 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)


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Fig.(a) : A body is subjected to 4-forces and it is in equilibrium

Fig.(b) : Method of Section. Internal force distribution acting on

the exposed imaginary section.

1.3 STRESS


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1.3 STRESS

Consider ΔA in figure above

Small finite force, ΔF acts on ΔA

As ΔA → 0, ΔF → 0

But stress (ΔF / ΔA) → finite limit (∞)

Normal Stress, s : The intensity of force, or force per

unit area, acting normal to DA.

σz =limΔA →0

ΔFz

ΔA


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Normal stress• Intensity of force, or force per unit area, acting

normal to ΔA• Symbol used for normal stress, is σ (sigma)

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

• Compressive stress: normal force “pushes” or “compresses” the area element ΔA

1.3 STRESS

σz =lim

ΔA →0

ΔFz

ΔA


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1.3 STRESS

Shear stress

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

• Symbol used for normal stress is (tau)

τzx =lim

ΔA→0

ΔFx

ΔAτzy =

lim

ΔA →0

ΔFy

ΔA


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1.3 STRESS

Stress Components : sz , tzx , tzy sy , tyx , tyz

sx , txy , txz


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1.3 STRESS

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

• 1 kPa = 103 N/m2 (kilo-pascal)

• 1 MPa = 106 N/m2 (mega-pascal)

• 1 GPa = 109 N/m2 (giga-pascal)

Units (SI system)

General state of stress• Figure shows the state of stress

acting around a chosen point in a body


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1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

Examples of axially loaded bar

• Usually long and slender structural members

• Truss members, hangers, bolts, etc.

• Prismatic means all the cross sections are the same


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Assumptions

1. Uniform deformation: Bar remains straight before and after load is applied, and cross section remains flat or plane during deformation

2. In order to get uniform deformation, force P be applied along centroidal axis of cross section



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Tension


PA

σ =

Compression

PA

σ =


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EXAMPLE 1-1

Bar width = 35 mm, thickness = 10 mm

Determine maximum average normal stress in bar when subjected to loading shown.


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EXAMPLE 1-1

Internal loading

By inspection, largest loading segment is BC, where PBC = 30 kN

A B C D

PCD

PAB = 12 kN

PBC = 30 kN

PCD = 22 kN

Normal force diagram


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EXAMPLE 1-1

Average maximum normal stress

σBC =PBC

A

30(103) N

(0.035 m)(0.010 m)= = 85.7 MPa

PBC = 30 kN


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EXAMPLE 1-2

The 200 N ( 20 kg) lamp is supported by two steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take q = 60o. The diameter of each rod is given in the figure.


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EXAMPLE 1-2

60o60o

W

FACFAB

x

y

+Fx = 0, FACcos60o – FABsin60o = 0

+Fy = 0, FACsin60o + FABcos60o – W = 0

Equation of equilibrium

The 2 equations yield

FAB = 100 N

FAC = 173.2 N


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EXAMPLE 1-2

Average normal stress

AB

AAB A

F BAC

ACAC A

F

AAB = ¼ (12)2 mm2 = 113.1 mm2

AAC = ¼ (10)2 mm2 = 78.54 mm2

Substituting the appropriate values,

we havesAB = 0.884 N/mm2 = 0.884 MPa

sAC = 2.205 N/mm2 = 2.204 MPa

The greater normal stress is in rod AC.

60o60o

W

FACFAB

x

y


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1.5 AVERAGE 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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τavg = average shear stress at section, assumed to be the same at each point on the section

V = internal resultant shear force at section determined from equations of equilibrium

A = area of section

VA

τavg =

Average shear stress over each section is:


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• Case discussed above is example of simple or direct shear

• Caused by the direct action of applied load F

• Occurs in various types of simple connections, e.g., bolts, pins, welded material


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Single shear

• Steel and wood joints shown above are examples of single-shear connections, also known as lap joints.

• Since we assume members are thin, there are no moments caused by F



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Single shear

• For equilibrium, cross-sectional area of bolt and bonding surface between the two members are subjected to single shear force, V = F

• The average shear stress equation can be applied to determine average shear stress acting on colored section in (d).



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F F

Bolt

Shear area

The bolt is undergoing shear stress

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d

F

A

Vavg

d = bolt diameter


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Double shear • The joints shown below are examples of double-

shear connections, often called double lap joints.• For equilibrium, cross-sectional area of bolt and

bonding surface between two members subjected to double shear force, V = F/2

• Apply average shear stress equation to determine average shear stress acting on colored section in (d).


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The bolt is undergoing shear stress: 2

4

2

d

F

A

Vavg

d = bolt diameter

F

Bolt

F/2

F/2

Shear area


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24

2

d

F

A

Vavg

Which one stronger ??

FF

Bolt

Single shear

F

Bolt

F/2

F/2

double shear

WHY ??

Smaller shear stress !!

24

d

F

A

Vavg


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1.6 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


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1.6 ALLOWABLE STRESS

• 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. = τfail

τallow

• 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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1.7 DESIGN OF SIMPLE CONNECTIONS

• To determine area of section subjected to a normal force, use

A = P

σallow

A = V

τallow

• To determine area of section subjected to a shear force, use


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EXAMPLE 1-3

The two members pinned together at B. If the pins have an allowable shear stress of τallow = 90 MPa, and allowable tensile stress of rod CB is (σt)allow = 115 MPa

Determine to nearest mm the smallest diameter of pins A and B and the diameter of rod CB necessary to support the load.

6 kN

B

A

AB

C

2 m 1 m

5

4

36 kN

B

A

AB

C

2 m 1 m

5

4

3

6 kN

B

A

A B

C

2 m 1 m

5

4

3


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EXAMPLE 1-3

Draw free-body diagram of member AB:

Ax Bx

ByAy

Reaction forces:

∑ Fx = 0;+

+ ∑ Fy = 0;

∑ MA = 0;+ By = 4 kN

Ay = 2 kN

Ax = Bx


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EXAMPLE 1-3

C

Bx

By

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5BB Components :

Bx = B (4/5)

By = 4 kN

We get Bx = 5.32 kN

Resultant reaction force: B = 6.67 kN

= B (3/5)


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EXAMPLE 1-3

22yx AAA

Ax = 5.32 kN

Ay = 2 kNB = 6.67 kN

A

Resultant reaction force: 22yx AAA = 5.68 kN


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VA

VA

A = 5.68 kN

EXAMPLE 1-3

dA = 6.3 mm

Diameter of pin A (double shear):

AA = (dA

2/4)

AA =VA

tallow

2.84 kN

90 103 kPa = 31.56 10−6 m2 =

Choose a size larger to nearest millimeter: dA = 7 mm

Shear force at pin A :

VA = A/2 = 2.84 kN


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B = 6.67 N

VB

EXAMPLE 1-3

Diameter of pin B (single shear):

AB =VB

tallow

6.67 kN

90 103 kPa = = 74.11 10−6 m2

AB = (dB

2/4) dB = 9.7 mm

Shear force at pin B :

VB = B = 6.67 kN

Choose a size larger to nearest millimeter: dB = 10 mm

Diameter of pin B :


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EXAMPLE 1-3

dBC = 8.59 mm

B

CDiameter of rod BC:

(The analysis is based on the normal stress)

Choose a size larger to nearest millimeter: dBC = 9 mm

ABC = B

(σt)allow

6.67 kN

115 103 kPa = = 58 10−6 m2 = (dBC

2/4)

2005 pearson education south asia pte ltd 1. stress 1 chapter objectives review important principles...

Documents

stress slide

stress f

stress concept of stress

allowable stress

deformable body slide

average normal stress

average shear stress

introduction slide