Lecture Strength of Materials I: Chapter 1 - PhD. Tran Minh Tu

Lecture Strength of Materials I - Chapter 1: Introduction – Concept of stress. The following will be discussed in this chapter: Review of static, equilibrium of deformable body, concept of stress, stress under general loadings, strain, types of loading, assumptions, principle superposition.

• Ass Prof Tran Minh Tu, PhD Eng.

• Mon., Fri.: 12.15 -14.45 A.M at Room 202 H1

• Office: 1 Floor – Lab Building

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Sample reading list:

Russell C Hibbeler, Mechanics of Materials, 6/E

(required text)

Roy R Craig, Jr (1996), Mechanics and Materials

Bedford, Fowler & Liecht (2003), Statics and

Mechanics and Materials

Beer & Johnson, Mechanics and Materials

James H Gere, Mechanics and Materials, 6th

edition

Strength of Materials

This course discusses stress calculation due to loads.

Description includes stress-strain concept, tensile test, stress

and strain due to axial loading, statically indeterminate case

CHAPTER

1.1 Introduction1.2 Review of Static1.3 Equilibrium of deformable body1.4 Concept of Stress

1.5 Stress Under General Loadings1.6 Strain

1.7 Types of loading1.8 Assumptions1.9 Principle Superposition

Contents

1.1 Introduction

- Consider a diving board as an example of a deformable body

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1.1 Introduction

1.1.1 Strength of materials

• 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

• Are used to compute the deformations of a body

1.1 Introduction

Fluid Mechanics

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1.1.2 Classification of Structural element

1.1 Introduction

1.1 Introduction

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1.1 Introduction

Classification of Structural elements

Structural elements compose a structure and can be classified as bytheir forms (shapes and dimensions)

• Three dimensional body

• Plates and Shells

1.1 Introduction

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1.1 Introduction

• Rods, Bars

1.1 Introduction

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1.1 Introduction

1.1.3 External loads

• Surface forces

– Area of contact

– Concentrated force

– Linear distributed force

– Centroid C (or geometric

center)

• Body force (e.g., weight)

1.1 Introduction

1.1.4 Support reactions

• for 2D problems

1.2 Review of Static

1.2.1 Structure Free-Body Diagram

The first step towards solving an engineering problems is drawing thefree body diagram of the element/structure considered

• The structure is designed tosupport a 30 kN load

• Perform a static analysis to

• The structure consists of aboom and a rod joined by

connections) at the junctionsand supports

1.2 Review of Static

• The structure is detached from thesupports and the loads and reactionforces are indicated

• A y and C y can not be determined fromthese equations

• Conditions for static equilibrium:

1.2 Review of Static

• In addition to complete the structure,each component must satisfy theconditions for static equilibrium

• Consider a free-body diagram for theboom:

kN30

Internal forces: The forces set up within a body to balance the effect ofthe externally applied forces.

1.3 Equilibrium of deformable body

1.3.1 Method of Sections

To obtain the internal loading acting on specific region – “cut” byimaginarily section through the region where internal loadings are to bedetermined

1.3 Equilibrium of deformable body

Two parts of the body are separated => A free-body diagram of one ofthe parts is drawn

Internal loadings

The Internal force is distributed on the area of section and representeffects of the material of the top part of the body acting on the adjacentmaterial of the bottom part

The Resultant force FR and moment MR0 at any specific point O isshown in the fig c

If a member is long and slender (rod, beam), the section considered isgenerally taken perpendicular to the longitudinal axis – is called a cross

Define resultant force (FR) and moment (MRo) in 3D:

1.3 Equilibrium of deformable body

- Coplanar Loadings If the body is subjected to a coplanar system offorces (fig a), then only normal-force, shear-force, and bending-momentcomponents will exist at the section (fig b)

- If we use the x, y, z coordinate axes, as shown on the left segment

1.3.2 Internal resultant loadings

• For coplanar loadings:

– Apply ∑ Fx = 0 to solve for N

– Apply ∑ Fy = 0 to solve for Q

– Apply ∑ MO = 0 to solve for M

1.3 Equilibrium of deformable body

Procedure for Analysis

The resultant internal loadings at a point located on the section of abody can be obtained by using the method of section This requires thefollowing steps

1.3 Equilibrium of deformable body

Procedure for analysis

1 Keep all external loadings in exact locations before “sectioning”

2 Indicate the unknown resultants, N, Q, M, and T at the section,

normally at the centroid C of the sectioned area

3 Coplanar system of forces only include N, Q, and M

4 Establish x, y, z coordinate axes with origin at the centroid

3 Resultant force with negative value implies that the assumed

direction in reality is opposite to that shown on the free-bodydiagram

Example 1.1

Example 1.1

1.4 Concept of Stress

• The main objective of the study of mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load bearing structures.

• Both the analysis and design of a given structure involve the determination of

stresses and deformations. This chapter

is devoted to the concept of stress.

1.4 Concept of Stress

Stress: The distribution of the internal force over the area on which itacts is expressed as the force intensity, that is force per unit area

There are two types of stress: normal stress and shear stress

Consider the sectioned area subdivided in

to small area such as DA

Very small force DF acting on DA, will be

replaced by three components

1.4.1 Stress Analysis

1.4 Concept of Stress

The sign convention for the normal stress is as follows:

- A positive value for s indicates tensile stress, that is, the stress due

to a force DF that pulls on the area on which it acts

- A negative value for s indicates compressive stress – push on

1.4.3 Shear stress

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

1.4.2 Normal stress

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

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

1.4 Concept of Stress

Units (SI system)

• Newtons per square meter (N/m2)

1.4 Concept of Stress

1.4.4 Average shear stress:

1.4 Concept of Stress

A

F A

ave

t

Single Shear

A

F A

1.4 Concept of Stress

• Bolts, rivets, and pins createstresses on the points of

contact or bearing surfaces

of the members theyconnect

• Corresponding average

• The resultant of the forcedistribution on the surface isequal and opposite to theforce exerted on the pin

1.4.6 Bearing Stress in Connections

• The rod is in tension with an axial force of 50 kN.

• The boom is in compression with an axial force of

40 kN and average normal stress of –26.7 MPa

• The minimum area sections at the boom ends are

unstressed since the boom is in compression

MPa

167 m

10 300

10 50

m 10 300 mm

25 mm 40 mm 20

2 6

3 ,

2 6

P

A

end BC

s

• At the flattened rod ends, the smallest sectional area occurs at the pin centerline,

cross-• At the rod center, the average normal stress in

the circular cross-section (A = 314x10-6m2) is sBC

1.4 Concept of Stress

• The cross-sectional area for pins at A,

B, and C,

2 6

10 491

N 10 50

2 6

t

• The force on the pin at C is equal to the force exerted by the rod BC,

• The pin at A is in double shear with

1.4.8 Pin Shearing Stresses

1.4 Concept of Stress

• Divide the pin at B into sections to

determine the section with the largest

shear force,

(largest)

kN 25

kN 15

MPa 9

.

50 m

10 491

kN 25

2 6

• To determine the bearing stress at A in the boom

AB, we have t = 30 mm and d = 25 mm,

1.5 Stress Under General Loadings

• A member subjected to ageneral combination of loads iscut into two segments by a

plane passing through Q

A

V A

V A F

x z A

xz

x y A

xy

x A

x

D

D

 D

 D

 D

limlim

lim

0 0

0

t t

s

• The distribution of internalstress components may bedefined as,

1.5 Stress Under General Loadings

• For equilibrium, an equal and opposite internal force and stressdistribution must be exerted on the other segment of the member

• Stress components are defined for the planes

exerted on the hidden planes.

required to define the complete state of stress

• The combination of forces generated by the stresses must satisfy the conditions for equilibrium:

x

z y

x

M M

M

F F

F

yx xy

yx xy

M

t t

t t



D

 D





zy yz

zy

similarly,

• Consider the moments about the z axis:

1.5 Stress Under General Loadings

• Displacement is the relative movement of a point on a rod with

respect to a point in space

• Deformation is the relative movement of two points on a rod.

L

A

Point B displaces a distance d down

The deformation of the rod between

points A and B is d

1.6 Strain – Displacement and deformation

The strain is the geometrical expression of deformation caused by

the action of stress on a physical body The strain is calculated by firstassuming a change between two body states: the beginning state andthe final state

Basically Strain are of three types

1 Longitudinal Strain: -It is defined as the increase in length (d) perunit original length (L) when deformed by the external force

Longitudinal Strain = d / L2.Volumetric Strain:-It is defined as the change in volume (v) per unitoriginal volume (V), when deformed by external force

Volume Strain = v / V

- Normal Strain (longitudinal strain)

3 Shear Strain:-When the change takes place in the

shape of the body, the strain is called a shear strain

1.6 Strain

- Shear Strain

1.7 Types of loading

Mechanics of materials is a branch of applied mechanics that deals with

the behaviour of solid bodies subjected to various types of loading

Compression Tension (stretched) Bending Torsion (twisted) Shearing

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• Continuous

1.8 Basic Assumptions

• Homogeneous

1.8 Basic Assumptions

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• Isotropic

Isotropic Anisotropic

1.8 Basic Assumptions

1.8 Assumptions – Hooke’s Law

• "Hooke's Law" is about stretching

springs and wires.

• Hooke's Law states:- the extension is

proportional to the force

• the spring will go back to its original

length when the force is removed

• so long as we don't exceed the

elastic limit.

Hooke’s Law

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1.8 Assumptions – Hooke’s Law

1.9 Principle Superposition

Conditions must be satisfied:

1.9 Principle Superposition

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