Chapter 1 mechanics of solids (09 2018)

mechanics of material, stress ,strain, axial loading......you will know about stressstrain diagram, elongation, plastic and elastic deformation .......................................................................................................................................................

MAE005-Mechanics of Solids

Mechanics of solids is the branch of continuum mechanics

that studies the behavior of solid materials, especially their

motion and deformation under the action of forces,

temperature changes, phase changes, and other external orinternal agents

Mechanics of solids

Theory ofplasticity

 Analyze forces of solid body cause by external force

 Analyze the result of solid mechanics experiments

 Apply the concepts of stress, strain, torsion and bending anddeflection of bar and beam in engineering field

 Explain the stress, strain, torsion and bending

 Calculate and determine the stress, strain and bending of

 Apply the concepts of stress, strain, torsion and bending and

deflection of bar and beam in engineering field

 Explain the stress, strain, torsion and bending

 Calculate and determine the stress, strain and bending of

solid body that subjected to external and internal load

 Use solid mechanic apparatus and analyze the experiments

result

 Work in-group that relates the basic theory with application

of solid mechanics

Learning outcome

Text books:

1 Ferdinand Beer and E Russell Johnston, “Mechanics of

Materials”, 7th edition, Mc-Graw-Hill, New York, 2012

2 Russell C Hibbeler, “Mechanics of Materials”, 8th edition,

Contents

Chapter 1: Introduction-Concept of stress

Chapter 2: Stress and Strain

Chapter 3: Mechanical properties

Chapter 4: Axial load

Chapter 5: Torsion

Chapter 6: Bending

Chapter 7: Combined loadings

Chapter 8: Stress transformation

Chapter 9: Deflection of Beams and Shaft

Chapter 10: Bulking of columns

CHAPTER 1

Introduction – Concept of Stress

The bolts used for the connections of this steel framework are subjected to stress In this chapter we will discuss how engineers design these connections and their fasteners.

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

1.7 Bearing stress in connections

1.8 Stress in two force members

1.9 Allowable stress

1.10 Chapter review

• The main objective of the study of the mechanics of

materials is to provide the future engineer with the means

of analyzing and designing various machines and bearing structures.

load-• Both the analysis and the design of a given structure

involve the determination of stresses and deformations.

This first chapter is devoted to the concept of stress.

1.1 Introduction

• Determine the internal force and the reaction forces ?

• Can the structure safely support the 30 kN load?

1.2 Support reactions

Surface forces developed at the supports/points of contactbetween bodies

 Linear distributed force

 Body Forces

 Earth’s gravitation (weight)

 Electromagnetic field

 Without direct physical contact

A system of coplanar forces

Best way to account for these forces is to draw the body’s free-body diagram (FBD).

0 M

0

0 ,

0 ,

0

0 ,

0 ,

x

z y

x

M M

M

F F

 Internal resultant loadings

 Method of sections: An imaginary section or “cut” is made through the region where the internal loadings are to be determined.

F R : resultant force at specific point O

M RO resultant moment at specific point O

 Cross section: the section is taken perpendicular to the longitudinal axis of the member

1.3 Equilibrium of a deformable body

 Internal Resultant Loadings

Objective of FBD is to determine the resultant force and moment acting within a body.

In general, there are 4 different types of resultant loadings:

 Coplanar Loadings

0 N 0

The weight of this sign and the wind loadings acting on it will causenormal and shear forces and bending and torsional moments in the

1.3 Equilibrium of a deformable body

Example 1.1

Determine the resultant internal loadings acting on the cross section at C

of the cantilevered beam shown in figure

Solution:

Free body Diagram

1.3 Equilibrium of a deformable body

Distributed loading at C is found by proportion:

Magnitude of the resultant of the distributed load: 1 200 2.4  240N

 Concept

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)

1.4 Stress

Consider ΔA in figure below: Small finite force, ΔF acts on ΔA:

As ΔA → 0, Δ F → 0, But quotient (ΔF / ΔA) → finite limit (∞)

 Stress is the intensity of the internal force acting on a specific plane

(area) passing through a point

 Normal stress is the intensity of the force acting normal to ∆A.

 Symbol used for normal stress, is σ (sigma)

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

 Shear stress is the intensity of force, or force per unit area, acting

tangent to ΔA. Symbol used for normal stress is  (tau)

0

0

lim lim

x zx

A

y zy

A

F A F A

 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)

1.4 Stress

Stress

 General state of stress

• Figure shows the state of stress acting

around a chosen point in a body

• General state of tress in three

dimensions: 9 components

• Three components acting on each face

of the element

 Assumptions

1 Uniform deformation: Bar remains straight before and after load is

applied, and cross section remains flat or plane during deformation

centroidal axis of cross section

1.5 Average normal stress in axially loaded bar

 avg : average normal stress at any point on the cross-sectional area

N = internal resultant normal force, which acts through the centroid of

the cross-sectional area

A = cross-sectional area of the bar where  is determined

 Average normal stress distribution

Let ∆A → dA and therefore ∆N → dN, then,

0; ( ) '( ) 0 '

Consider vertical equilibrium of the element

This analysis applies to members subjected to tension or compression.

1.5 Average normal stress in axially loaded bar

 Where external force P and x-sectional A were constant along the

longitudinal axis of the bar normal stress is also constant.

avg

P A

1.5 Average normal stress in axially loaded bar

Centric loading Eccentric axial loading

Uniform distribution of Non-uniform distribution of

1.5 Average normal stress in axially loaded bar

 If the bar is subjected to several external loads along its axis,

change in x-sectional area may occur.

 Thus, it is important to find the maximum average normal stress

 To determine that, we need to find the location where ratio P/A is a

maximum

Sign convention:

◦ P is positive (+) if it causes tension in the member

◦ P is negative (−) if it causes compression

1.5 Average normal stress in axially loaded bar

Procedure for Analysis

 Internal Loading :

• Section member perpendicular to its longitudinal axis at position

where normal stress is to be determined

• Draw free-body diagram

• Use equation of force equilibrium to obtain internal axial force P

at the section

 Average Normal Stress :

• Determine member’s x-sectional area at the section

• Compute average normal stress σ = P/A

1.5 Average normal stress in axially loaded bar

• Determine the internal force in each rod ?

• Can rod BC safely support the 30 kN load?

1.5 Average normal stress in axially loaded bar

Example 1.1

Rod BC is made of a steel with a maximum

allowable stress  allow=165 MPa

• For reasons based on cost, weight, availability, etc., the choice is made

to construct the rod from aluminum (σall= 100 MPa) What is anappropriate choice for the rod diameter?

• An aluminum rod 26 mm or more in diameter is adequate

3

6 2 6

2.52 10 m 25.2 mm

BC BC all

BC all

A d

Example 1.2

Bar width = 35 mm, thickness = 10 mm

Determine max average normal stress in bar when subjected to loadingshown

1.5 Average normal stress in axially loaded bar

By inspection, largest loading area is BC, where N BC = 30 kN

Internal loading

Example 1.2

Normal force diagram

1.4 Average normal stress in axially loaded bar

σBC =

NBCA

30(103) N (0.035 m)(0.010 m)

1.6 Average shear stress

Shear stress is the stress component that act in the plane of the

sectioned area

• In figure, 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.

Average shear stress over each section is:

avg = average shear stress at section, assumed to

be same at each point on the section

V = internal resultant shear force at section

determined from equations of equilibrium

A = area of section

1.6 Average shear stress

 Single shear

connections, also known as lap joints.

by F

between the two members are subjected to single shear force, V = F

average shear stress acting on colored section in (d).

1.6 Average shear stress

 Double shear

• The joints shown below are examples of double-shear connections,

often called double lap joints

• For equilibrium, x-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 sections in (d).

1.6 Average shear stress

Single shear Double shear

1.6 Average shear stress

V

V V

The pin A used to connect the linkage of this tractor is subjected to

double shear because shearing stresses occur on the surface of the pin

1.6 Average shear stress

Procedure for analysis

Internal shear

1 Section member at the point where the τ avg is to be determined

3 Calculate the internal shear force V

Average shear stress

2 Compute average shear stress avg = V/A

1.6 Average shear stress

1.6 Average shear stress

Determine the average shear stress in the 20-mm-diameter pin at A and the 30-mm-diameter pin at B that support the beam in figure.

Example 1.3

1.7 Bearing stress in connections

• Bolts, rivets, and pins create stresses on the points of contact or

bearing surfaces of the members they connect.

• The resultant of the force distribution on the surface, P, is equal and opposite to the force exerted on the pin, F.

d t

P A

P 



b



1.8 Stress in two force members

• Transverse forces on bolts and pins

result in only shear stresses on the plane

perpendicular to bolt or pin axis

• Axial forces on a two force member result

in only normal stresses on a plane cut

perpendicular to the member axis

• Will show that either axial or transverse forces may produce bothnormal and shear stresses with respect to a plane other than one cutperpendicular to the member axis

N F=P F=P

forming an angle with the normal plane.

2

cos

cos cos

• The average normal and shear stresses on

the oblique plane are:

• Resolve F into components normal and

tangential to the oblique section,

distributed forces (stresses) on the plane must be equivalent to the force F.

1.8 Stress in two force members

• The maximum normal stress occurs whenthe reference plane is perpendicular to themember axis:

m

0

; 0

P A

1.9 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

Cross-sectional area of a tension member

Condition:

The force has a line of action that passes through the centroid ofthe cross section

1.9 Allowable stress

Cross-sectional area of a connecter subjected to shear

Assumption:

If bolt is loose or clamping force of bolt is

unknown, assume frictional force between plates

to be negligible.

1.9 Allowable stress

Required area to resist bearing

Bearing stress is normal stress produced by the compression of one

surface against another

Assumptions:

1 (σb)allow of concrete < (σb)allow of base plate

2 Bearing stress is uniformly distributed

between plate and concrete

1.9 Allowable stress

Although actual shear-stress distribution

along rod difficult to determine, we

assume it is uniform.

Thus use A = V / τallow to calculate l,

provided d and τallow is known

1.9 Allowable stress

Procedure for analysis

When using average normal stress and shear stress equations, considerfirst the section over which the critical stress is acting

Internal Loading

1 Section member through x-sectional area

3 Use equations of equilibrium to determine internal resultant

1.10 Chapter review

The internal loadings in a body consist of a normal force, shear force,bending moment, and torsional moment

1.10 Chapter review

Average normal stress

Average shear stress

Factor of safety (F.S.)

Stress analysis & design example

• Would like to determine thestresses in the members andconnections of the structureshown

• Must consider maximum

normal stresses in AB and

BC, and the shearing stress

and bearing stress at eachpinned connection

• From a statics analysis:

F AB = 40 kN (compression)

F BC = 50 kN (tension)

• 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

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

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

• At the flattened rod ends, the smallest

cross-sectional area occurs at the pin centerline:

2 2

m 10

491 2

Stress analysis & design example

• Divide the pin at B into sections to

determine the section with the largestshear force:



• Evaluate the corresponding average shearingstress: