TMEE3642: SOLID MECHANICS ICOURSE INTRODUCTION

Details of Lecturer Course Lecturer/s: Dr. A. Ogunmokun & Mr. Izaaks

Room Number: 211 Mechanical Building

Email: gizaaks@unam.na Tel. No. : 065 232 4114

Office Hours: 9 a.m. to 1 pm. (Tue, Wed and Friday)

COURSE GOALS

This course has two specific goals:

(i) To introduce students to basic concepts of stress, strains, bending and torsion of rigid bodies in two and three dimensions.

(ii) To develop analytical skills relevant to the areas mentioned in (i) above.

COURSE OBJECTIVES Upon successful completion of this course,

students should be able to: 

(i) Analyze equilibrium of rigid bodies subjected to 2 & 3 dimensional force systems.

(ii) Describe the principal of rigid equilibrium to trusses, frames and machines.

(iii) Apply the method of virtual work for equilibrium and stability analysis.

(iv) Apply properties of areas in solving mechanics problems.

COURSE OBJECTIVES CONTD.(v) Analyze statically determinate and statically

indeterminate problems.

(vi) Analyze thermal and assembly stresses and incorporate them in stress analysis.

(vii) Analyze stresses and strains under torsion, bending and combined torsion and bending.

(viii) Apply the principle of transformation of stresses and analyze stresses and strains using Mohr’s circle.

   Course Content

Statics:1.Properties of 3D force systems.2.Equilibrium of rigid bodies.3.Application of principals of rigid bodies

equilibrium to trusses, frames and machines.

4.Introduction to the method of virtual work for equilibrium and stability analysis of interconnected system.

Course Content Cont. Solid Mechanics:1. Second moment of area.2. Normal and shear stresses and strains.3. Statically indeterminate problems.4. Geometric compatibility.5. Thermal and assembly stresses.6. Torsion of shafts.7. Bending of beams.8. Combined bending and direct stresses.9. Bending and torsion stresses.10.Transformation of stresses and strains.11.Mohr’s circle.

Teaching Strategies

The course will be taught via Lectures and Tutorial Sessions, the tutorial being designed to complement and enhance both the lectures and the students appreciation of the subject.

Course work assignments will be reviewed

with the students.

Course Textbook & Lecture Times

To be communicated to you.

Lectures: Monday, 15h30 to 17h25

Tutorials: Wednesday, 09h00 to 09h55

Attendance at Lectures and Tutorials is Compulsory

Course Assessment

(i) One (1) mid-semester test, 1 hour duration counting for 25% of the total course.

(ii) Sum of the tutorials marks counting for 25% of the total course.

(iii) One (1) End-of-semester examination, 2 hours duration counting for 50% of the total course marks.

TMEE3642: Solids MechanicsINTRODUCTION

1.1 MECHANICS

Body of Knowledge which Deals with the Study and Prediction of the State of Rest or Motion of Particles and Bodies under the action of Forces

PARTS OF MECHANICS

1.2 STATICS

Statics deals with the equilibrium of bodies, that is those that are either at rest or move with a constant velocity.

Dynamics Is concerned with the accelerated motion of bodies and will be dealt in another course.

TMEE3642: SOLID MECHANICS

STATICS OF PARTICLES

2.1 PARTICLE

A particle has a mass but a size that can be neglected.

When a body is idealised as a particle, the principles of mechanics reduce to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem.

PARTICLE CONTINUED

All the forces acting on a body will be assumed to be applied at the same point, that is the forces are assumed concurrent.

2.2 FORCE ON A PARTICLE

A Force is a Vector quantity and must have Magnitude, Direction and Point of action.

F

P

Force on a Particle Contd.

Note: Point P is the point of action of force and and are directions. To notify that F is a vector, it is printed in bold as in text books.

Its magnitude is denoted as |F| or simply F.

Force on a Particle Contd.There can be many forces acting on a

particle.

The resultant of a system of forces on a particle is the single force which has the same effect as the system of forces.

The resultant of two forces can be found using the parallelogram law.

2.2.VECTOR OPERATIONS 2.2.1 EQUAL VECTORS

Two vectors are equal if they are equal in magnitude and act in the same direction.

pP

Q

Equal Vectors Contd.

Forces equal in Magnitude can act in opposite Directions

S

R

Q

P

R

2.2.2 Vector Addition

Using the Parallelogram Law, Construct a Parallelogram. with two Forces as Parts. The resultant of the forces is the diagonal.

Vector Addition Contd.

Triangle Rule: Draw the first Vector. Join the tail of the Second to the head of the First and then join the head of the third to the tail of the first force to get the resultant force, R

Q

P

R = Q + P

Triangle Rule Contd.

Also:

P

Q

R = P + Q

Q + P = P + Q.

This is the commutative law of vector addition

Polygon Rule

Can be used for the addition of more than two vectors.

Two vectors are actually summed and added to the third.

Polygon Rule contd.

P

QS

P

Q

S

R

R = P + Q + S

(P + Q)

Polygon Rule Contd. P + Q = (P + Q) ………. Triangle Rule

i.e. P + Q + S = (P + Q) + S = R The method of drawing the vectors is

immaterial . The following method can be used.

Polygon Rule contd.

P

QS

P

Q

S

R

R = P + Q + S(Q + S)

Polygon Rule Concluded Q + S = (Q + S) ……. Triangle Rule

P + Q + S = P + (Q + S) = R

i.e. P + Q + S = (P + Q) + S = P + (Q + S)

This is the associative Law of Vector Addition

2.2.3. Vector Subtraction

P - Q = P + (- Q) P

Q

P

-Q

P -QQ

PP - Q

Parallelogram Rule Triangle Rule

2.3 Resolution of ForcesIt has been shown that the

resultant of forces acting at the same point (concurrent forces) can be found.

In the same way, a given force, F can be resolved into components.

There are two major cases.

Resolution of Forces: Case 1

(a) When one of the two components, P is known: The second component Q is obtained using the triangle rule. Join the tip of P to the tip of F. The magnitude and direction of Q are determined graphically or by trigonometry.

F

P Q

i.e. F = P + Q

Resolution of Forces: Case 2(b) When the line of action of each component is known: The force, F can be

resolved into two components having lines of action along lines ‘a’ and ‘b’ using the

paralleogram law. From the head of F, extend a line parallel to ‘a’ until it intersects ‘b’.

Likewise, a line parallel to ‘b’ is drawn from the head of F to the point of intersection with

‘a’. The two components P and Q are then drawn such that they extend from the tail of

F to points of intersection.

a

Q F

P b

Example

Determine graphically, the magnitude and direction of the resultant of the two forces using (a) Parallelogram law and (b) the triangle rule.

900 N

600 N

30o45o

SolutionSolution: A parm. with sides equal to 900 N and 600 N is drawn to scale as shown.

The magnitude and direction of the resultant can be found by drawing to scale.

600 N R

15o 900 N

45o 30o

The triangle rule may also be used. Join the forces in a tip to tail fashion and

measure the magnitude and direction of the resultant.

600 N

R 45o

135o C

B 30o 900 N

900N600N

30o45o

Trignometric Solution U s i n g t h e c o s i n e l a w :

R 2 = 9 0 0 2 + 6 0 0 2 - 2 x 9 0 0 x 6 0 0 c o s 1 3 5 0

R = 1 3 9 0 . 6 = 1 3 9 1 N

U s i n g t h e s i n e l a w :

R

Bi e B

T h e a n g l e o f t h e r e s u l t

s i n s i n. . s i n

s i n

.

t a n . .

1 3 5

6 0 0 6 0 0 1 3 5

1 3 9 1

1 7 8

3 0 1 7 8 4 7 8

1

i e . R = 1 3 9 N

4 7 . 8 o

R

900 N

600N

135o

30oB

ExampleTwo structural members B and C are bolted to

bracket A. Knowing that both members are in tension and that P = 30 kN and Q = 20 kN, determine the magnitude and direction of the resultant force exerted on the bracket.

Q

P

25o

50o

Solution

R

P

25o105o

θ20kNQ

30kN75o

2.4 RECTANGULAR COMPONENTS OF FORCE

x

F

j

i Fx = Fx i

Fy = Fy j

y

RECTANGULAR COMPONENTS OF FORCE CONTD.

In many problems, it is desirable to resolve force F into two perpendicular components in the x and y directions.

Fx and Fy are called rectangular vector components.

In two-dimensions, the Cartesian unit vectors i and j are used to designate the directions of x and y axes.

Fx = Fx i and Fy = Fy j i.e. F = Fx i + Fy j Fx and Fy are scalar components of F

RECTANGULAR COMPONENTS OF FORCE CONTD.

While the scalars Fx and Fy may be positive or negative, depending on the sense of Fx and Fy, their absolute values are respectively equal to the magnitude of the components forces Fx and Fy. Scalar components of F have magnitudes

Fx = F cos θ and Fy = F Sin θ

Where F is the magnitude of the force F

Example

Determine the resultant of the three forces below.

25o45o

350 N

800 N600 N

60o

y

x

Solution F x = 3 5 0 c o s 2 5 o + 8 0 0 c o s 7 0 o - 6 0 0 c o s 6 0 o

= 3 1 7 . 2 + 2 7 3 . 6 - 3 0 0 = 2 9 0 . 8 N

F y = 3 5 0 s i n 2 5 o + 8 0 0 s i n 7 0 o + 6 0 0 s i n 6 0 o

= 1 4 7 . 9 + 7 5 1 + 5 1 9 . 6 = 1 4 1 9 . 3 N

i . e . F = 2 9 0 . 8 N i + 1 4 1 9 . 3 N j

R e s u l t a n t , F

F N

2 9 0 8 1 4 1 9 3 1 4 4 9

1 4 1 9 3

2 9 0 87 8 4

2 2

1 0

. .

t a n.

..

F = 1 4 4 9 N 7 8 . 4 o

25o

45o

350 N

800 N600 N

60o

y

x

ExampleA hoist trolley is subjected to the three forces

shown. Knowing that = 40o , determine (a) the magnitude of force, P for which the resultant of the three forces is vertical (b) the corresponding magnitude of the resultant.

1000 N

P

2000 N

Solution (a) The resultant being vertical means that the

horizontal component is zero.

F x = 1000 sin 40o + P - 2000 cos 40o = 0

P = 2000 cos 40o - 1000 sin 40o =

1532.1 - 642.8 = 889.3 = 889 kN

(b) Fy = - 2000 sin 40o - 1000 cos 40o =

- 1285.6 - 766 = - 2052 N = 2052 N

1000 N

P

2000 N40o

40o

2.5. EQUILIBRIUM OF A PARTICLEA particle is said to be at equilibrium when the resultant of all the forces acting on it is

zero. It two forces are involved on a body in equilibrium, then the forces are equal and

opposite.

.. 150 N 150 N

If there are three forces, when resolving, the triangle of forces will close, if they are in

equilibrium.

F2 F1 F2

F3

F1

F3

EQUILIBRIUM OF A PARTICLE CONTD.

If there are more than three forces, the polygon of forces will be closed if the particle is

in equilibrium.

F3

F2 F2

F3 F1 F4

F1

F4

The closed polygon provides a graphical expression of the equilibrium of forces.

Mathematically: For equilibrium:

R = F = 0

i.e. ( Fx i + Fy j) = 0 or (Fx) i + (Fy) j

EQUILIBRIUM OF A PARTICLE CONCLUDED For equilibrium: Fx = 0 and F y = 0. Note: Considering Newton’s first law of

motion, equilibrium can mean that the particle is either at rest or moving in a straight line at constant speed.

FREE BODY DIAGRAMS: Space diagram represents the sketch of the

physical problem. The free body diagram selects the significant particle or points and draws the force system on that particle or point.

Steps: 1. Imagine the particle to be isolated or cut free

from its surroundings. Draw or sketch its outlined shape.

Free Body Diagrams Contd.

2. Indicate on this sketch all the forces that act on the particle.

These include active forces - tend to set the particle in motion e.g. from cables and weights and reactive forces caused by constraints or supports that prevent motion.

Free Body Diagrams Contd.

3. Label known forces with their magnitudes and directions. use letters to represent magnitudes and directions of unknown forces.

Assume direction of force which may be corrected later.

ExampleThe crate below has a weight of 50 kg. Draw

a free body diagram of the crate, the cord BD and the ring at B.

CRATE

B ring C

A

D

45o

Solution(a) Crate

FD ( force of cord acting on crate)

50 kg (wt. of crate)

(b) Cord BD

FB (force of ring acting on cord)

FD (force of crate acting on cord)

CRATE

C45o

B

A

D

Solution Contd.

(c) Ring

FA (Force of cord BA acting along ring)

FC (force of cord BC acting on ring)

FB (force of cord BD acting on ring)

Example

Solution Contd.

FF

FB CA C

o

o A C s i n

c o s. . . . . . . . . . . . . . ( )

7 5

7 53 7 3 1

F y = 0 i . e . F B C s i n 7 5 o - F A C c o s 7 5 o - 1 9 6 2 = 0

FF

FB CA C

A C

1 9 6 2 0 2 6

0 9 6 62 0 3 1 2 0 2 7 2

.

.. . . . . . . . ( )

F r o m E q u a t i o n s ( 1 ) a n d ( 2 ) , 3 . 7 3 F A C = 2 0 3 1 . 2 + 0 . 2 7 F A C

F A C = 5 8 7 N

F r o m ( 1 ) , F B C = 3 . 7 3 x 5 8 7 = 2 1 9 0 N

RECTANGULAR COMPONENTS OF FORCE (REVISITED)

x

j

iFx = Fx i

Fy = Fy j

y

F = Fx + Fy

F = |Fx| . i + |Fy| . j

|F|2 = |Fx|2 + |Fy|2

F | | | | | |F Fx Fy 2 2

2.6 Forces in Space

Rectangular Components

Fy

Fx

Fz

j

i

k

F

Rectangular Components of a Force in Space

F = Fx + Fy + Fz

F = |Fx| . i + |Fy| . j + |Fz| . k

|F|2 = |Fx|2 + |Fy|2 + |Fz|2

| | | | | | | |F Fx Fy Fz 2 2 2

| | | | cos | | | | cos | | | |cos

, cos

,

Fx F Fy F Fz F

Cos Cos and Cos are called direction ines of

angles and

x y z

x y z

x y z

Forces in Space Contd.

i.e. F = F ( cos x i + cos y j + cos z k) = F

F can therefore be expressed as the product of scalar, F

and the unit vector where: = cos x i + cos y j + cos z k.

is a unit vector of magnitude 1 and of the same direction as F.

is a unit vector along the line of action of F.

Forces in Space Contd.Also:

x = cos x, y = cos y and z = cos z - Scalar vectors

i.e. magnitudes.

x2 + y

2 + z2 = 1 = 2

i.e. cos2 x, + cos2 y + cos2 z = 1

Note: If components, Fx, Fy, and Fz of a Force, F are known,

the magnitude of F, F = Fx2 + Fy

2 + Fz2

Direction cosines are: cos x = Fx/F , cos y = Fy/F and cos2 z = Fz/F

Force Defined by Magnitude and two Points on its Line of Action Contd.Unit vector, along the line of action of F = MN/MN

MN is the distance, d from M to N.

= MN/MN = 1/d ( dx i + dy j + dz k )

Recall that: F = F

F = F = F/d ( dx i + dy j + dz k )

FFd

dF

Fd

dF

Fd

dd x x d y y d z z

d d d d

d

d

d

d

d

d

xx

yy

zz

x y z

x y z

xx

yy

zz

, ,

, ,

cos , cos , cos

2 1 2 1 2 1

2 2 2

2.8.3 Addition of Concurrent Forces in Space

The resultant, R of two or more forces in space is obtained by

summing their rectangular components i.e.

R = F

i.e. Rx i + Ry j + Rz k = ( Fx i + Fy j + Fz k )

= ( Fx) i + ( Fy)j + ( Fz )k

R x = Fx, Ry = Fy , Rz = Fz

R = Rx2 + Ry

2 + Rz2

cos x = Rx/R cos y = Ry/R cos z = Rz/R

Solution

S o lu t io n :

P o s i t io n v e c t o r o f B H = 0 . 6 m i + 1 . 2 m j - 1 . 2 m k

M a g n i t u d e , B H = 0 6 1 2 1 2 1 82 2 2. . . . m

B H

B H B H B H B H

B H

x y z

B H

B Hm i m j m k

T T TB H

B H

N

mm i m j m k

T N i N j N k

F N F N F N

| | .( . . . )

| | . | || | .

. . .

( ) ( 5 0 0 ) ( 5 0 0 )

, ,

1

1 80 6 1 2 1 2

7 5 0

1 80 6 1 2 1 2

2 5 0

2 5 0 5 0 0 5 0 0

2.9 EQUILIBRIUM OF A PARTICLE IN SPACE

For equilibrium: Fx = 0, Fy = 0 and Fz =0.

The equations may be used to solve problems dealing with the equilibrium of a particle involving no more than three unknowns.