2. Introduction

Mechanics of Materials

Dr. Rami Zakaria

References:

1. Engineering Mechanics: Statics, R.C. Hibbeler, 12th ed, Pearson

2. Mechanics of Materials: R.C. Hibbeler, 9th ed, Pearson

3. Mechanics of Materials: J.M. Gere & B.J. Goodno, 8th ed, Cengage Learning

Terminology

2

Term Definition Notes

Stress the intensity of the internal force acting on an area

Deformation Changing of the shape of a subject

Load/Loading Forces/force curve

Resultant Force Overall force

Strain The intensity of deformation at a point in an object

Member A structure unit (such as a bar, a beam, etc.)

Compression Applying two forces towards each others

Tension Applying two forces away from each others

Translation Moving along a straight line

Rotation Moving in a circular way

3

Mechanics of materials is a branch of applied mechanics that studies the

behavior of solid bodies (such as bars and beams) subjected to external loading.

This involves finding the internal effects (stress and strain ) in a solid body,

which is important for the safe design of structures (in bridges, buildings,

airplanes, etc…).

Stress is associated with the strength of the material from which the body is made,

while strain is a measure of the deformation of the body.

What is Mechanics of Materials?

This steel framework is subjected to stress.

How does this affect the design of the framework and its connections?

• We will review some of the important principles of

statics and how to determine the internal resultant

loadings in a body.

• Afterwards we will learn the concepts of normal

stress (σ) and shear stress (τ) and study members

subjected to an axial load or direct shear.

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A SHORT REVIEW OF

STATICS

5

External Loads

Reactive forces

Body forces: They are developed when one body exerts a

force on another, without direct physical

contact between them.

For examples: the effects caused by the

earth’s gravitation (weight!), or its

electromagnetic field. Although body forces

affect all the particles of the body, we usually

represent them by a single concentrated force

(for example centre of Gravity).

Surface forces: caused by the direct contact of one

body with another. They are

distributed over the area of contact

between the bodies.

-If it is applied on a small area (in

comparison with the total surface

area of the body), then the

surface force can be idealized as

a single concentrated force (at

one point on the body).

-If it is applied along a narrow

strip of area, the loading can be

idealized as a distributed line of

forces.

Active Forces

Support reactions 6

Why do we need “statics” here?!

Because in Mechanics of Materials we need to know all forces acting on the body.

When in equilibrium, the net force and net moment acting on a body are zero.

Equations of equilibrium :

0

0

M

F

0,0,0

0,0,0

zyx

zyx

MMM

FFF

0

,0,0

o

yx

M

FF

Equilibrium of a Rigid Body

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Steps for Solving Free Body Equilibrium problems:

1. If not given, establish a suitable coordinate system (x,y or x,y,z).

2. Draw a free body diagram (FBD) of the object.

3. Apply the three equations of equilibrium (E-of-E) to solve for the unknowns.

The FBD will show us all the external loads (known & unknown forces &

moments) and the support reactions.

How can I find the support reactions ?

As a general rule, if a support prevents translation of a body in a given direction,

then a force is developed on the body in the opposite direction.

Similarly, if a support prevents rotation, then a couple moment is exerted on the

body in the opposite direction.

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Prevents translation in one

direction

Prevents translation in

all directions

Prevents translation in one

direction, and prevents rotation

Pre

ven

ts tra

nsl

atio

n in

all

dir

ecti

ons,

an

d

pre

ven

ts r

ota

tio

n

Support Reaction

• q is a line load (N/m or lb/ft)

The resultant force FR of q , acts through the

centroid C (or centre of area) and it is

equivalent to the area under the distributed

loading curve.

Overall FBD Separate Parts FBD

Distributed Loads

Ref [3]

Notice the internal forces and moments:

11 Ref [3]

•p is a surface pressure (N/m2 or lb/ft2)

• w is a body force (N/m3 or lb/m3)

In case of 3D systems

Remember:

When we have a simple structure, members can be

subjected to either compression or tension forces.

We can calculate these forces using the concepts of

statics.

When we know the forces at the ends of each of the members, we can determine the

internal forces in these members, by looking at a cross-section at an arbitrary point.

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In such cases, w(x) is the load intensity

distribution function and it has units of

force per length.

We will analyze the most common case of

a distributed pressure loading. This is a

uniform load along one axis of a flat

rectangular body.

dxxwdF )(

The net force on the beam is given by

+

Here A is the area under the loading curve w(x).

LL

R AdxxwdFF )(

Distributed Loading Review

13 Ref [1]

Location of the Resultant Force

The total moment about point O is given as

Assuming that FR acts at , it will produce the

moment about point O as

The force dF will produce a moment of (x)(dF)

about point O.

x

LL

Ro dxxxwxdFM )(

L

RRo dxxwxFxM )()()(

14 Ref [1]

Comparing the last two equations, we get

FR acts through a point “C”, which is called the

geometric center or centroid of the area under the

loading curve w(x).

15 Ref [1]

Example: Determine the magnitude

and location of the equivalent resultent

force (FR) acting on the shaft

Ref [1]

Note: for simples shapes we can find the shift easily using the geometric centroid

Area Cx Cy

b.h b/2 h/2

b.h/2 (a +b)/3 h/3

π.a2 a a

π.a2 /2 a 4.a /3.π

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The rectangular load:

ftx

lbFR

52

10

400010400

The triangular loading:

mx

NFR

43

66

180066002

1

18

Examples:

Example about support reaction calculations

Given: The 4kN load at B of the beam

is supported by pins at A and C .

Find: The support reactions at A and C.

FBD of the beam:

AX

AY

A

1.5 m

C B

4 kN

FC

45°

1.5 m Solution:

+ FX = AX + 11.31 cos 45 = 0; AX = – 8.00 kN

+ FY = AY + 11.31 sin 45 – 4 = 0; AY = – 4.00 kN

MA = (FC sin 45) (1.5) – (4) (3) = 0

Fc = 11.31 kN

Ref [1]

Simple Truss

20 Ref [1]

The solution has finished until here! But let’s see the

reaction components at joint (A)

All reaction forces together 21 Ref [1]

22

Statics is used to determine the resultant loadings that act within a body.

Consider a body in equilibrium by four external forces (neglect the weight for now). In

order to obtain the internal loadings acting on a specific region, let’s pass an imaginary

section (cut) through the region of interest. Separate the two parts and draw the free-

body diagram of one of them.

There is a distribution of internal forces acting on the area of the cross section.

We can represent these forces by a resultant force FR and moment MRo, at any point

O (usually the centroid). Here, we are interested in the normal and perpendicular

components of FR and MRo.

Internal Resultant Loadings

Therefore, we can define 4 types of loading:

Normal force, N: This force acts perpendicular to the area. It

is developed when the external loads tend to push or pull on

the two segments of the body.

Shear force, V: The shear force lies in the plane of the area. It

is developed when the external loads tend to cause the two

segments to slide over one another.

Bending moment, M: It is developed when the external loads tend to bend the body

about an axis lying on the area.

Torsional moment (torque) , T: It is developed when the external loads tend to twist

one segment of the body about an axis perpendicular to the area.

This exists only in a 3D system

23 Ref [2]

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

external loads applied to a body and the stress and strain caused by

the internal loads within the body.

• A support produces a force in a particular direction on its attached

member if it prevents translation of the member in that direction,

and it produces a couple moment on the member if it prevents

rotation.

• In general the internal resultant loadings acting on a body are :a

normal force, shear force, torsional moment, and bending moment.

We can determine them by making a cross-section through the body.

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Important Points:

(Calculating The resultant internal loadings at a point located on the section of a body )

1. Determine the Support Reactions.

2. Draw the Free-Body Diagram.

( indicate the unknown resultants N, V, M, and T )

3. Write and solve the Equations of Equilibrium.

Note: If the solution is negative, the assumed directional sense of the resultant is opposite to

that shown on the free-body diagram

only in a 3D system

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Procedure for Analysis:

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Examples:

Ref [2]

27 Ref [2]

EXAMPLE 1.2

Ref [2] 28

29 Ref [2]

30 Ref [2]

EXAMPLE 1.3 (CONTINUED) lbFFF

lbFFF

BDBDy

BABAx

465005

377500

775005

462000

Ref [2] 31

32 Ref [2]

33 Ref [2]

Problems:

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Student’s solution:

35

36

37

Student’s solution:

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39

Student’s solution:

40

42

Student’s solution: