Review Lecture Slides

VDB1023

Engineering Mechanics

Lecturer

Dr Zubair Imam Syed

Email: [email protected]

Ph: 05 368 7313

Room: 14.03.13

Week 14 Review Lecture

mailto:[email protected]

Force

Resultant of Forces

Components of a force

Position vector

Force in Cartesian vector form

Can you find the magnitude and direction of the resultant of the force system given

below

Force and Moment

30m

6

12

8

A man is pulling the cord with a 300 N force. Represent this force as a Cartesian

vector and determine its direction.

Exercise If F3 = 15 kN, =35, and = 40, determine the magnitude and coordinate angle of the resultant force acting on the ball-and-socket joint.

Moment of a Force Vector Formulation

zyx

zyxO

FFF

rrr

kji

FrM

Determine the moment produced by each of the forces

Equilibrium Conditions and applications

Fx = 0

Fy = 0

Fz = 0

These equations are the three scalar equations of equilibrium.

Reactions on a Rigid body

Two force member

Reactions at hinge and fixed supports

Example

Truss Analysis

Problem

Find the member forces in members KJ, CD and CJ

for the forces shown in the figure below.

Centre of Gravity, Centroid

A y dA

A dA

y =

~

A x dA

A dA

x =

~

Centroid of Complex Shapes

i i

i

x Ax=

A

i i

i

y Ay=

A

x

y

i i

i

z Az=

A

Centroid Location of Complex Shapes

x

y

Centroid for an area is determined by the first moment of an area about an axis

Second moment of an area is referred as the moment of inertia

Moments of Inertia

Consider area A lying in the x-y plane

Be definition, moments of inertia of the differential plane area dA about the x and y axes

For entire area, moments of

inertia are given by

Ay

Ax

yx

dAxI

dAyI

dAxdIdAydI

2

2

22

y

y

x

C

x

50

35

10

50

35 20 20

10

10 10

Calculate the moment of inertia about the centroid

axis,

Problem

x

oo

oo

o

ssavv

attvss

atvv

2

2

1

22

2

Rectilinear Kinematics

dt

dsv

dt

dva dvvdsa

Curvilinear Motion

nt uua nt aa

2va

vdvdsava

n

tt

Equations of Motion

maF

dt

dvmmaF tt

2vmmaF nn

Solution

Conservation of Linear Momentum

smv

v

vmmvmvm BABBAA

/5.0

)27000()75.0)(12000()5.1)(15000(

)()()(

2

2

211( )

Momentum and Impulse

Plan:

1) Determine the speed of the crate just before the collision using projectile

motion or an energy method.

2) Analyze the collision as a central impact

problem.

Problem A 2 kg crate B is released from rest, falls a distance h = 0.5 m, and

strikes plate P (3 kg mass). The coefficient of restitution between B and

P is e = 0.6, and the spring stiffness is k = 30 N/m. The velocity of crate

B just after the collision.

0 + 0 = 0.5(2)(v2)2 + (2)(9.81)(-0.5)

T1 + V1 = T2 + V2

0.5m(v1)2 + mgh1 = 0.5m(v2)

2 + mgh2

v2 = 3.132 m/s

Apply conservation of momentum to the system in the

vertical direction:

Analyze the collision as a central impact problem.

+ mB(vB)1 + mP(vP)1 = mB(vB)2 + mP(vP)2

(2)(-3.132) + 0 = (2)(vB)2 + (3)(vP)2

(vB)2

(vP)2 (vP)1 = 0

(vB)1 = 3.132 m/s

B

P

Using the coefficient of restitution:

+ e = [(vP)2 (vB)2]/[(vB)1 (vP)1]

0.6 = [(vP)2 (vB)2]/[-3.132 0]

-1.879 = (vP)2 (vB)2

Solving the two equations simultaneously yields

(vB)2 = -0.125 m/s and (vP)2 = -2.00 m/s

Both the block and plate will travel down after the collision.

(vB)2

(vP)2 (vP)1 = 0

(vB)1 = 3.132 m/s

B

P