What is Strength???

In Material Terms

Ability to withstand stress without failure is

STRENGTH

FACT:- The limit of the strength of a material can be

measured ONLY when it undergo FAILURE

Is it power???

If YES then

What is the power of materials???

Classification

Matter is classified as solid and fluid.

The study of solid involves study of rigid and

deformable bodies.

Study of rigid bodies include statics &

dynamics (which you have already studied).

Study of deformable bodies is basically given

the name “Strength of materials” or “Mechanics

of materials”.

What is Mechanics?

Mechanics of any substance is the complete

behavior of that substance under any

circumstances.

So when we talk Mechanics of materials or

Strength of material, we mean

A branch of mechanics that studies the

internal effects of stress and strain in a solid

body that is subjected to an external loading.

Why do we need Strength of Materials:

Course Overview

Detailed Study of the Following topics

Concept of stress Concept of strain Mechanical properties of materials

Axial load Torsion

Shear Force &

Bending

Grading Policy Student’s

Evaluation by Due

Division of

Grades

Home works Every week 10%

Laboratory works Every week 15%

Quizzes 5th Week & 13th Week 10%

Project 15th Week 5%

Midterm Exam 8th week 20%

Final Exam As scheduled by the

registrar 40%

My Goals for Course

That each one of you develop an intuition for the fundamental principles of Strength of Materials

That you leave this course saying, “This course makes sense”.

That we have an fruitful semester learning together

Review of main principles of Statics

Surface Forces. Surface forces are caused by the direct contact of one body with the surface of another. If this area is small in comparison with the total surface area of the body, then the surface force can be idealized as a single concentrated force, which is applied to a point on the body. For example, the force of the ground on the wheels of a bicycle can be considered as a concentrated force.

Review of main principles of Statics

If the surface loading is applied along a narrow strip of area, the loading can be idealized as a linear distributed load, w(s). Here the loading is measured as having an intensity of force/length along the strip and is represented graphically by a series of arrows along the line s. The resultant force of w(s) is equivalent to the area under the distributed loading curve, and this resultant acts through the centroid C or geometric center of this area.

Review of main principles of Statics

Body Forces. A body force is developed when one body exerts a force on another body without direct physical contact between the bodies. Examples include the effects caused by the earth’s gravitation or its electromagnetic field. Although body forces affect each of the particles composing the body, these forces are normally represented by a single concentrated force acting on the body. In the case of gravitation, this force is called the weight of the body and acts through the body’s center of gravity.

Review of main principles of Statics

Support Reactions. The surface forces that develop at the supports or points of contact between bodies are called reactions. As a general rule, if the support prevents translation in a given direction, then a force must be developed on the member in that direction. Likewise, if rotation is prevented, a couple moment must be exerted on the member.

Review of main principles of Statics

Review of main principles of Statics

For Coplanar Forces

Review of main principles of Statics

Internal Resultant Loadings

Review of main principles of Statics Types of Internal Resultant Loadings

Review of main principles of Statics Types of Internal Resultant Loadings

Review of main principles of Statics Coplanar Loadings

Review of main principles of Statics Example 1.1

Homework:- Try solving the problem using segment AC

Example 1.2

Example 1.2

Example 1.2

Example 1.3

Example 1.3

Finding pin reaction at A

Determining IRL at point E

Homework Problem

Objectives of the Lecture Understand the concept of Stress

Study the types of stress

Mathematical representation of stress

Study Normal Stresses in detail

Able to solve problems associated with normal stresses

Stress

It is defined as the intensity of the internal force acting on a specific plane (area) passing through a point within the material assuming

the material to be continuous, that is, to consist of a continuum or uniform distribution of matter having no voids.

Also, the material must be cohesive, meaning that all portions of it are connected together, without having breaks, cracks, or separations.

Stress

Stress is similar to pressure except for the fact that

1. Stress is something that is felt internally within the material body whereas the pressure is something which is applied externally on the material.

2. Stress is internal while pressure is external as it is clear by the statement “Please don’t put too much pressure on me, I am already under stress”

There are two kinds of stresses i.e.

The Normal Stress & The Shear Stress

Stress

Stress

Stress Type Directions

Required (n) Formula Comp.

required in Cart. Coord.

Example

Scalar 0 3𝑛 = 30 = 1 1 Mass (kg)

Vector 1 3𝑛 = 31 = 3

3 Force as

Fx, Fy & Fz

Tensor 2 3𝑛 = 32 = 9

9 Stress as

𝜎𝑥𝑥 𝜏𝑥𝑦 𝜏𝑥𝑧

𝜏𝑦𝑥 𝜎𝑦𝑦 𝜏𝑦𝑧

𝜏𝑧𝑥 𝜏𝑧𝑦 𝜎𝑧𝑧

Stress

Stress

When the load P is

applied to the bar

through the centroid of

its cross-sectional area,

then the bar will deform

uniformly throughout

the central region of its

length, provided the

material of the bar is

both homogeneous

and isotropic.

As a sign convention, P will be positive if it causes tension in the member, and negative if it causes compression.

Example

The largest loading

is in region BC

Example

Example

Solving we get,

Homework Problem

Homework Problem

Homework Problem

ANSWERS

Objectives of the Lecture Understand the concept of Shear Stress

Study the types of Shear stress

Mathematical representation of Shear stress

Develop shear stress equilibrium

Able to solve problems associated with Shear stresses

Average Shear Stress

Single Shear

Double Shear

Shear Stress Equilibrium

The section plane is subjected to shear stress 𝜏𝑧𝑦.

Applying force equilibrium.

Similarly,

Shear Stress Equilibrium

The section plane is subjected to shear stress 𝜏𝑧𝑦.

Applying moment equilibrium.

So that,

Shear Stress Equilibrium

In other words, all four shear stresses must have equal magnitude and be directed either toward or away from each other at opposite edges of the element.

This is referred to as the complementary property of shear, and under the conditions shown in Figure, the material is subjected to pure shear.

Example

Example

For Plane a-a For Plane b-b

Example

Homework Prob

Homework Prob

Homework Prob

Objectives of the Lecture Describe concept of Allowable Stress

Study the Applications of Allowable Stress

Solve Problems related to Allowable Stress

Describe the Concept of Strain

Study Different types of Strain

Able to solve problems associated with Strain

Allowable Stress

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

If the member is subjected to normal load

If the member is subjected to shear load

Applications

Applications

Applications

Example

Example

Homework Prob.

Homework Prob.

Homework Prob.

Strain Strain is a measure of deformation of a body

Deformation Whenever a force is applied to a body, it will tend to change the body’s shape and size. These changes are referred to as deformation, and they may be either highly visible or practically unnoticeable. Deformation of a body can also occur when the temperature of the body is changed.

Types of Strain 1.Direct Strain 2. Lateral Strain

Normal Strain

Shear Strain

Normal Strain

Normal Strain Sign Convention

Unit

Shear Strain

Shear Strain

Small Strain Analysis Proof

Shear Strain

Sign Convention

Unit

360 Degree = 2π Radians

180 Degree = π Radians

1 Degree = π/180 Radians

x Degree = x(π/180) Radians

Conversion

Cartesian Strain Components

Example

Example

Homework Prob.

Homework Prob.

Homework Prob.

Objectives of the Lecture Illustrate Tension & Compression Test

Study the Stress-Strain Diagram

Study Mechanical Properties of Materials

Able to solve problems related to Stress-Strain Diagram

Mechanical Properties of Materials

In order to apply an axial load with no bending of the specimen, the ends are usually seated into ball-and-socket joints. A testing machine is then used to stretch the specimen at a very slow, constant rate until it fails. The machine is designed to read the load required to maintain this uniform stretching.

The Stress-Strain Diagram

A plot in which the vertical axis is the stress and the horizontal axis is the strain, the resulting curve is called a conventional stress–strain diagram.

Homework:- Having so much difference between engineering and true stress-strain Diagrams, why do we still use conventional Diagrams???

Modulus of Elasticity

Since strain is dimensionless, E will have the same units as stress, such as psi, ksi, or pascals

The modulus of elasticity is a mechanical property that indicates the stiffness of a material.

Engineers often choose ductile materials for design because these materials are capable of absorbing shock or energy, and if they become overloaded, they will usually exhibit large deformation before failing.

Brittle Materials. Materials that exhibit little or no yielding before

failure are referred to as brittle materials. Concrete is classified as a brittle material, and it also has a low strength capacity in tension. The characteristics of its stress–strain diagram depend primarily on the mix of concrete (water, sand, gravel, and cement) and the time and temperature of curing. Its maximum compressive strength is almost 12.5 times greater than its tensile strength. For this reason, concrete is almost always reinforced with steel bars or rods whenever it is designed to support tensile loads. It can generally be stated that most materials exhibit both ductile and brittle behavior. For example, steel has brittle behavior when it contains a high carbon content, and it is ductile when the carbon content is reduced. Also, at low temperatures materials become harder and more brittle, whereas when the temperature rises they become softer and more ductile.

The new stress–strain diagram, now has a higher yield point a consequence of strain-hardening. In other words, the material now has a greater elastic region; however, it has less ductility, a smaller plastic region, than when it was in its original state.

Strain Energy

Example

Example

Homework Prob.

Answers

Homework Prob.

Homework Prob.

Objectives of the Lecture Describe concept of Poisson’s Ratio

Solve Problems related to Poisson’s Ratio

Study the Shear Stress-Strain Diagram

Able to solve problems associated with Shear Stress-Strain Diagram

Describe concepts of Creep & Fatigue

Poisson’s Ratio

It has a numerical value that is unique for a particular material that is both homogeneous and isotropic

Homework Prob. A cylindrical bar, made up of a special material,

has a diameter of 40 mm and a length of 50 cm.

When this bar is subjected to a tensile loading of

100 N, it causes the bar to elongate. As a result

of elongation, the new length of the bar is 55 cm.

Determine the Poisson’s ratio for this material?

How much strain is required in the bar, if the bar

is required to have a radius of 15 mm.

Example

Homework Prob.

Shear Stress-Strain Diagram

Modulus of Rigidity

(Modulus of Elasticity) for Normal Stress-Strain (Modulus of Rigidity) for Shear Stress-Strain

Relation between E, G & v

Example

Example

Creep & Fatigue When a material has to support a load for a very long period of time, it may continue to deform until a sudden fracture occurs or its usefulness is impaired. This time-dependent permanent deformation is known as creep. In general, the creep strength will decrease for higher temperatures or for higher applied stresses

When a metal is subjected to repeated cycles of stress or strain, it causes its structure to break down, ultimately leading to fracture. This behavior is called fatigue. In order to specify a safe strength for a metallic material under repeated loading, it is necessary to determine a limit below which no evidence of failure can be detected after applying a load for a specified number of cycles. This limiting stress is called the endurance or fatigue limit.

Homework Prob.

Objectives of the Lecture Describe concept of Axial Loads

Define the Saint-Venant’s Principle

Study the Elastic Deformation of an Axially Loaded Member

Able to solve problems associated with Elastic Deformation of an Axially Loaded Member

Axial Load

An axial load is the kind of loading that occurs when the load acts along the axis of an object.

An Axial load that tends to elongate an object is known as tensile load.

An Axial load that tends to compress an object is known as compressive load.

Saint-Venant’s Principle

This principle states that the stress and strain produced at points in a body sufficiently removed from the region of load application will be the same as the stress and strain produced by any applied loadings that have the same statically equivalent resultant, and are applied to the body within the same region.

Saint-Venant’s Principle

Elastic Deformation of an Axially Loaded Member

Elastic Deformation of an Axially Loaded Member

Elastic Deformation of an Axially Loaded Member

Example

Example

Example

Example

Homework Prob.

Homework Prob.

Objectives of the Lecture Describe concept of Principle of Superposition

Define Statically Indeterminate Axially Loaded Members

Able to solve problems associated with Statically Indeterminate Axially Loaded Members

Study the Thermal Stresses.

Able to solve problems associated with Thermal Stresses.

Principle of Superposition

The following two conditions must be satisfied if the principle of superposition is to be applied.

The loading must be linearly related to the stress or displacement that is to be determined.

The loading must not significantly change the original geometry or configuration of the member.

Statically Indeterminate Axially Loaded Members

Statically Indeterminate Axially Loaded Members

Example

Example

Homework Prob.

Homework Prob.

Thermal Stresses

Example

Example

Homework Prob.

Objectives of the Lecture Describe concept of Torsion

Formulate torsion for circular shafts

Define the concept of Polar Moment of Inertia

Able to solve problems associated with the concept of torsion.

Torsional Deformation of a Circular Shaft

Torque is a moment that tends to twist a member about its longitudinal axis. Its effect is of primary concern in the design of axles or drive shafts used in vehicles and machinery.

Torsional Deformation of a Circular Shaft

we can assume that if the angle of twist is small, the length of the shaft and its radius will remain unchanged

Torsion Formulation

𝑆 = 𝑟𝜃 Here S = 𝐵𝐵′ 𝑟 = 𝜌 𝜃 = ∆∅

Torsion Formulation

Torsion Formulation

Torsion Formulation

Torsion Formulation At outer boundary with radius equal to “c”, shear stress will be maximum.

Using the above two equations, we can derive the following relation

Polar Moment of Inertia (J) Polar moment of area is a quantity used to predict an

object's ability to resist torsion, in objects (or segments of objects) with an invariant circular cross section and no significant warping or out-of-plane deformation.

It is used to calculate the angular displacement of an object subjected to a torque.

The larger the polar moment of area, the less the beam will twist, when subjected to a given torque.

The polar moment of area cannot be used to analyze shafts with non-circular cross-sections.

The SI unit for polar moment of area, like the area moment of inertia, is metre to the fourth power (m4).

Polar Moment of Inertia for a Solid Shaft

Polar Moment of Inertia for Tubular Shaft

Homework:- Prove that for a tubular shaft, the polar moment of inertia is as follows.

Sign Convention:-

Example

Example

Example

Section a-a

Homework Prob.

Homework Prob.

Homework Prob.

Objectives of the Lecture Describe concept of Power Transmission

Able to solve problems associated with Power Transmission

Define Angle of twist

Formulate angle of twist

Able to solve problems associated with the concept of angle of twist.

Power Transmission

Units of Power Transmission

Calculations for a Machinery

Note:-

Example

Example

Homework Prob.

Angle of Twist

Formulating Angle of Twist

Formulating Angle of Twist

Note the similarity in equations between Axially loaded and torque applied members

Torsion Testing Machine

Formulating Angle of Twist

Note the similarity between Axially loaded and torque applied members

Sign Convention

Example

Subscript Notation

Example

Homework Prob.

Homework Prob.

Objectives of the Lecture Describe concept of Bending

Define Beams and its types

Describe concept of Shear & Moment Diagrams

Study the base case solutions

Able to solve problems associated with the base case solution – Single Concentrated Load

Able to solve problems associated with the base case solution – Single Moment

Bending of Beams

Bending (also known as flexure) characterizes the behavior of a slender structural element subjected to an external load applied perpendicularly to a longitudinal axis of the element.

Beams & its types Members that are slender and support loadings that are applied perpendicular to their longitudinal axis are called beams. In general, beams are long, straight bars having a constant cross-sectional area. Often they are classified as to how they are supported. For example, a simply supported beam is pinned at one end and roller supported at the other, a cantilevered beam is fixed at one end and free at the other, and an overhanging beam has one or both of its ends freely extended over the supports.

Shear & Moment Diagrams

Shear & Moment Diagrams

Shear & Moment Diagrams

Beam Sign Conventions The positive directions are as follows: the distributed load acts downward on the beam; the internal shear force causes a clockwise rotation of the beam segment on which it acts; and the internal moment causes compression in the top fibers of the segment such that it bends the segment so that it holds water. Loadings that are opposite to these are considered negative.

Base Case Solutions We have a total of four base cases for analyzing the shear force and bending moment on beams. These base cases are as follows:-

1. Single Concentrated load

2. Single Moment

3. Uniformly Distributed Load

4. Uniformly Varying Load

Other than the base cases, there could be a number of complex cases that are a combination of various bases cases.

Example – Single Concentrated Load

‘X’ ‘V’ Eq. ‘M’ Eq.

0 P/2 1 0 2

L/4 P/2 1 PL/8 2

L/2 P/2 & -P/2 1 & 3 PL/4 2 & 4

3L/4 -P/2 3 PL/8 4

L -P/2 3 0 4

The table must contain all action points

S.F.D.

B.M.D.

Shear Force Diagram

Bending Moment Diagram

Example – Single Moment

(1)

(2)

(3)

(4)

‘X’ ‘V’ Eq. ‘M’ Eq.

0 -Mo/L 1 0 2

L/4 -Mo/L 1 -Mo/4 2

L/2 -Mo/L 1 & 3 -Mo/2 & Mo/2 2 & 4

3L/4 -Mo/L 3 Mo/4 4

L -Mo/L 3 0 4

The table must contain all action points

S.F.D.

B.M.D.

Shear Force Diagram

Bending Moment Diagram

Objectives of the Lecture Study the base case solutions

Able to solve problems associated with the base case solution – Uniformly Distributed Load

Able to solve problems associated with the base case solution – Uniformly Varying Load

Able to solve problems associated with the Complex cases

Example – Uniformly Distributed Load

Also determine the position of zero shear and evaluate

maximum bending moment on the beam.

‘X’ ‘V’ Eq. ‘M’ Eq.

0 wL/2 1 0 2

L/2 0 1 wL²/8 2

L -wL/2 1 0 2

The table must contain all action points

S.F.D.

B.M.D.

Shear Force Diagram

Bending Moment Diagram

Example – Uniformly Varying Load

‘X’ ‘V’ Eq. ‘M’ Eq.

0 wₒL/2 1 -wₒL²/3 2

L/2 3wₒL/8 1 -5wₒL²/48 2

L 0 1 0 2

The table must contain all action points

S.F.D.

B.M.D.

Shear Force Diagram

Bending Moment Diagram

Example – Complex Case

Also determine the position of zero shear and evaluate

maximum bending moment on the beam.

‘X’ ‘V’ Eq. ‘M’ Eq.

0 1 2

L/2 1 2

L 1 2

The table must contain all action points

S.F.D.

B.M.D.

Shear Force Diagram

Bending Moment Diagram

Example – Complex Case

(1)

(2)

(3)

(4)

‘X’ ‘V’ Eq. ‘M’ Eq.

0 1 2

2.5 1 2

5 1 & 3 2 & 4

7.5 3 4

10 3 4

The table must contain all action points

S.F.D.

B.M.D.

Shear Force Diagram

Bending Moment Diagram

Objectives of the Lecture Study Bending Deformation of a Straight Member

Formulate Flexural Formula

Perform Calculation for Moment of Inertia

Solve Problems relative to Flexural Formula

Bending Deformation of a Straight Member

Bending Deformation of a Straight Member

Formulating Flexural Formula (Bending Formula)

Formulating Flexural Formula (Bending Formula)

Formulating Flexural Formula (Bending Formula)

Interpreting the Normal Strain Distribution

Formulating Flexural Formula (Bending Formula)

Formulating Flexural Formula (Bending Formula)

Formulating Flexural Formula (Bending Formula)

Interpreting the Normal Stress Distribution

Formulating Flexural Formula (Bending Formula)

Formulating Flexural Formula (Bending Formula)

Formulating Flexural Formula (Bending Formula)

Interpreting the Flexural Formula

Calculation for Moment of Inertia

Calculation for Moment of Inertia

Example

Also draw the stress distribution across the cross-section of this beam

Example

Locating Neutral Axis

Calculating Moment of Inertia

Calculating Internal Moment

Example