Mechanics of Materials

Chapter 1

Introduction & stress

Fall Semester 2019

Instructor: Twana A. Hussein

twana.ahmad@ishik.edu.iq

Civil Engineering Dept.

Tishk International University

Course objective

To show how to determine the stress, strain, and deflection

suffered by bi-dimensional (and simple tri-dimensional)

structural elements when subjected to different loads (e.g.

normal, shear, torsion, bending and combined loads).

Once the state of stresses and strains has been established for a

particular structure type, the student will be able to evaluate the

allowable loads and associated allowable stresses before

mechanical failure

Learning Outcomes

Upon successful completion of this course students should be able to:

Understand the fundamental concepts of stress and strain and the relationship between both through the strain-stress equations in order to solve problems for simple tri-dimensional elastic solids.

Calculate and represent the stress diagrams in bars and simple structures

Solve problems relating to pure and non-uniform bending of beams and other simple structures

Solve problems relating to torsional deformation of bars and other simple tri-dimensional structures

Understand the concept of buckling and be able to solve the problems related to isolated bars

Course outline:

Fundamentals of stress and strain

Allowable stresses, allowable loads, and safety factors

Tension, Compression and Shear

Torsion

Bending

Buckling

Deflections of beams

Introduction.

Mechanics of materials is a study of the relationship between the external

loads on a body and the intensity of the internal loads within the body.

It is not possible solely on basis of force calculation to assess the

structural integrity of a component or structure.

Strength of materials is obviously a core subject for mechanical and

mechatronics engineers, since it enables us to determine by calculation, if

the components we design will function as intended or fail.

In order to do so, we define the term stress as a measure for internal force

per area acting inside a structure.

Introduction.

Converting our internal forces to stresses by calculations

provides us with a measure that contrary to forces can be

related to characteristic material values.

We will furthermore consider the problem related to

calculation of deformations in components and structures.

On basis of calculated deformations, it will be considered

how to calculate strains as a measure for how large

deformations are relative to the dimensions of the considered

component.

Recap: the concept of static equilibrium

From statics, we know that the following three relations apply

if a plane mechanical system is in static equilibrium,

∑ 𝐹𝑥 = 0 ∑ 𝐹𝑦 = 0 ∑ 𝑀𝑧 = 0 (1-1)

Equation 1-1 basically reads, that we can sum up all forces in

the directions of a cleverly chosen coordinate system and

furthermore calculate the sum of moments around a point of

our own choice.

Equilibrium of a Deformable Body

Reactions

Surface forces developed at the supports/points of contact between

bodies.

Equilibrium of a Deformable Body

Equilibrium of a body requires a balance of forces and a

balance of moments

For a body with x, y, z coordinate system with origin O,

∑ 𝐹𝑥 = 0 ∑ 𝐹𝑦 = 0 ∑ 𝐹𝑧 = 0

∑ 𝑀x = 0 ∑ 𝑀y = 0 ∑ 𝑀𝑧 = 0

Best way to account for these forces is to draw the body’s

free-body diagram (FBD).

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:

a) Normal force, N

b) Shear force, V

c) Torsional moment or torque, T

d) Bending moment, M

Example 1.1

Determine the resultant internal loadings acting on the cross

section at C of the beam.

Solution:

Free body Diagram

Distributed loading at C is found by proportion,

Magnitude of the resultant of the distributed load,

,

Solution:

Applying the equations of

equilibrium we have

Example 1.2

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

of the pipe. The pipe has a mass of 2 kg/m and is subjected to both a vertical

force of 50 N and a couple moment of 70 N·m at its end A. It is fixed to the

wall at C.

Example 1.2

Free body Diagram

Solution:

Calculating the weight of each segment of pipe,

• Applying the six scalar equations of equilibrium,

Solution:

Stress

Normal Stress- Axial Loading

• The resultant of the internal forces for

an axially loaded member is normal to a

section cut perpendicular to the member

axis.

• The force intensity on that section is

defined as the normal stress.

Both the analysis and design of a given

structure involve the determination of

stresses and deformations.

Normal Stress

The normal stress at a particular point maynot be equal to the average stress but theresultant of the stress distribution must satisfy

The detailed distribution of stress is staticallyindeterminate, i.e., can not be found from staticsalone.

A uniform distribution of stress in a sectionimplies that the line of action of the resultant ofthe internal forces passes through the centroidof the section.

Normal Stress

A uniform distribution of stress is only

possible if the concentrated loads on the end

sections of two-force members are applied at

the section centroids. This is referred to as

centric loading.

If a two-force member is eccentrically loaded,

then the resultant of the stress distribution in a

section must yield an axial force and a

moment.

The stress distributions in eccentrically

loaded members cannot be uniform or

symmetric.

Example 1.3

Two solid circular rods are welded to a plate at B to form a

single rod, as shown in the figure. Consider the 30kN force at

B to be uniformly distributed around the circumference of the

collar at B and the 10 kN load at C to be applied at the

centroid of the end cross section. Determine the axial stress in

each portion of the rod.

Example 1.4

The bar has a constant width of 35 mm and a thickness of 10 mm.

Determine the maximum average normal stress in the bar when it is

subjected to the loading shown.

Solution:

By inspection, different sections have different internal forces.

Solution:

Graphically, the normal force diagram is as shown.

By inspection, the largest loading is in region BC,

Since the cross-sectional area of the bar is

constant,

the largest average normal stress is

Example 1.5

Solution:By drawing a free-body diagram of the top segment,

the internal axial force P at the section is

Shearing Stress

Forces P and P’ are applied transversely to the member

AB.

Corresponding internal forces act in the plane of section

C and are called shearing forces.

The resultant of the internal shear force distribution is

defined as the shear of the section and is equal to the

load P.

The corresponding average shear stress is,

Shear stress distribution varies from zero at the member

surfaces to maximum values that may be much larger

than the average value.

The shear stress distribution cannot be assumed to be

uniform.

Shearing Stress

The average shear stress distributed over each sectioned area that develops a shear force.

τ = average shear stress

P = internal resultant shear force

A = area at that section

Shearing Stress Examples

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 is

equal and opposite to the force

exerted on the pin.

Corresponding average force

intensity is called the bearing

stress,

Allowable Stress

Many unknown factors that influence the actual stress in a

member.

A factor of safety is needed to obtained allowable load.

The factor of safety (F.S.) is a ratio of the failure load divided

by the allowable load

Example 1.6

Solution:

For equilibrium we have

Solution:The pin at C resists the resultant force at C. Therefore,

The pin is subjected to double shear, a shear force of

15.205 kN acts over its cross sectional area between the

arm and each supporting leaf for the pin.The required area is

Example 1.7

Solution

Example 1.8

The structure is designed to support a 30 kN load. The structure consists of a

boom and rod joined by pins (zero moment connections) at the junctions and

supports Perform a static analysis to determine the internal force in each

structural member and the reaction forces at the supports

Structure Free-Body Diagram Structure is detached from supports and the

loads and reaction forces are indicated

Conditions for static equilibrium:

Ay and Cy can not be determined from these equations

Solution

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

Consider a free-body diagram for the boom:

Substitute into the structure equilibrium equation

Results:

Reaction forces are directed along boom and rod

Method of Joints The boom and rod are 2-force members,

i.e., the members are subjected to only two forces which are applied at member ends

For equilibrium, the forces must be parallel to an axis between the force application points, equal in magnitude, and in opposite directions

Joints must satisfy the conditions for static equilibrium which may be expressed in the form of a force triangle:

Stress Analysis

Can the structure safely support the 30 kN load?

From a statics analysis

At any section through member BC, the internal force is 50 kN with a force intensity or stress of

From the material properties for steel, the allowable stress is

Conclusion: the strength of member BC is adequate

Design Design of new structures requires selection of

appropriate materials and component

dimensions to meet performance requirements

For reasons based on cost, weight, availability,

etc., the choice is made to construct the rod

from aluminum. What is an appropriate choice

for the rod diameter?

• An aluminum rod 26 mm or more in diameter is

adequate

Stress Analysis & Design Example

Would like to determine the

stresses in the members and

connections of the structure

shown.

From a statics analysis:

FAB = 40 kN (compression)

FBC = 50 kN (tension)

Must consider maximum

normal stresses in AB and BC,

and the shearing stress and

bearing stress at each pinned

connection

Rod & Boom Normal Stresses

Pin Shearing Stresses

Pin Shearing Stresses