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ENGINEERING MECHANICS

Part Mechanics of MaterialsⅡ or Strength of Materials

Mechanics of deformable bodies

Theoretical Mechanics------particles or rigid bodies

Mechanics of Materials-----deformable solid bodies

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Introduction Mechanics of Materials is a branch of applied mechanics, it deals with the behaviors of deformable solid bodies subjected to various types of loading.

Behaviors:

stress, strain, deformation

Abilities:

(a) Strength — the ability to prevent failure

(b) Rigidity — the ability to resist deformation

(c) Stability — the ability to keep the original equilibrium state

(d) Toughness — the ability to resist fracture

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IntroductionSolid bodies(1D) and various types of loading

axially loaded members in tension or compression

shafts in torsion

beams in bending

columns in compression

The principle of superposition

The resultant response in a system due to several forces is the algebraic sum of their effects when separately applied only if each effect is linearly related to the force causing it.

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Tension and Compression

Torsion

Bending

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Torsion:Loaded by pairs of forces——torque, twisting couples, or twisting moments.There will be a rotation about the longitudinal axis of one end of the bar with respect to the other.

Forces acting transverse to the axis of structural member

Bending

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The resistance forces that set up within a body to balance the effect of the externally applied forces.

Chapter 4 inner forces

4.2 Method of Section for Internal Force

Make an imaginary cut at cross section perpendicular to longitudinal axis.

Using for forces representing action of the moved part upon remained one, the forces are continuously distributed over the cross section.

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L

PP

PP

Resultant of forces N

For equilibrium ： x=0 N - P=0

Axial internal force ： N=P

N

A

4.3 Tension and compression

Prismatic bar loaded by axial force at the centroid of ends (Axially loaded members).

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Example determine the axial forces and draw the axial-force diagram.

x

C P2=15kN

P1=5kN

B P3=10kNA

P1=5kN

A1

1

N1

N1-P1=0 N1=P1=5kN

N2 = -10kNN

x

5kN

10kN

P3 =10kNB2

2N2

Axial Force Diagram

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Example P1=15kN, P2=10kN, P3=5kN, determine the axial forces on the cross sections 1-1, 2-2, and draw the axial-force diagram.Solution:

X=0 N1 +P1=0 N1=-P1=-15kN

X=0 N2 +P1 -P2=0 N2=-5kN

-10-5

xN(kN)

(-) (-)

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4.4 diagram of torque4.4.1 Transmission of power by circular shafts

Work:

Power:

W T

Motor driven shaftAngular speed ω(rad/s)Transmitted torque T(N·m)

7024 ( )N

T N mn

609549 ( )

2

N NT N m

n n

Φ—angular rotation

22 ( , / , / )

60

dW d nTN T T f T Watt N m s J s

dt dt

f—frequency of revolution, Hz=s-1

n—number of revolutions per minute (rpm)

T

If N is expresses in kilowatt or horsepower, 1 hp=735.5 W

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4.4.2 Diagram of Torsion Moment

Method of Section for Internal Force

nmx

nSign convention for torque

——right hand law.

T

Tn m

n

m m

n

nEquilibrium mx = 0

T - m = 0

T = m

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T1

10 20 15

a b c d10 20 15

I b c d

T2

20 15

II c d 15

T3III d

x = 0

15-20+10+T1 = 0

T1 = -5kN-m

15-20+T2 = 0

T2 = 5kN-m

15+T3 = 0

T3 = -15kN-m

15kN-m

5kN-m

5kN-mx

T

Determine the torsion moment of the shaft

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q

BA

P

Simple beam

Simply supported beam

Beam: the structural member to resist forces acting transverse to its axis

4.5 Shear Forces and Bending Moments4.5.1 Beams

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BA

P2P1

Beam with an overhang

P

A

B

Cantilever beam

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Plane Bending

•having symmetric cross sections

•Loads act in the symmetric plane•the bending deflections occur in this plane

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Plane bending and 3D bending

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Concentrated Load P

Distributed Load q

Couple M

Type of Loads

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Simple Beam

Beam with an

Overhang

Cantilever Beam

FAx

FAy

FBy

FAx

FAy FBy

FAx

FAy

MA

Reactions in various types of supports

How to find the reactions?

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Ax

P

4.5.2 Shear Forces and Bending Moments

M

Method of section

BM

Y=0 P - V =0 V = P

o

Mo=0 M - Px =0 M = Px

Sign Convention

M V

V M

M V

V M

VVV

V

MMM M

V

ABm

nx

P

V

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Bending Moment and Shear Force

RA

Find the Bending Moment and Shear Force on cross section of I-I

Reactions RA, RB P 2Pa

x

I

IA B

l

aMB=0

RAl - P(l-a) - 2Pa = 0

Pl

alRA

Y=0

RA + RB - 3P = 0

Pl

alRB

2RB

Pa

xRA

V

M

V

M

RB

2P

M0=0

Y=0

RA - P - V = 0

PRV A

RA x - P(x-a) - M = 0

Pal

xlaxPPx

l

alaxPxRM A

)()(

Pl

aPP

l

al

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Shear-Force and Bending-Moment Diagrams

RA RB

x

A

RA

M(x)

V(x)

Concentrated Loads

Reactions

l

PbRA

l

PaRB

Y = 0 RA - V(x) = 0l

PbRxV A )(

(0 < x < a)

M= 0 RA x - M(x) = 0l

PbxxRxM A )(

Equations for shear force and bending moment

Pa

x

A B

l

b

C

C

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M(x)

V(x)xA

RA

Pa

Y = 0 RA - V(x) - P = 0

l

PaPRxV A )(

M = 0 Rax - P(x-a) - M(x) = 0

)()( axPxRxM A

(a < x < l)

xll

Pa

Pa

x

A B

l

b

C

C

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l

PaxV )(

l

Pa

V l

Pb

Ml

Pab

xll

PaxM )(

l

PbxV )(

l

PbxxM )(

Pa

x

A B

l

b

C

(a < x < l)

(0 < x < a)

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A beam loaded by several concentrated forces

RA RB

Reactions Ra and RB

V(x) = RA M(x)=RAx

(0 < x < a1)

(a1< x < a2)

V(x) = RA- P1 M(x)=RA x -P1(x-a1)M(x)

V(x)l

b

P3

x

B

RB

V(x) = RA- P1 – P2

M(x)= RA x -P1(x-a1)- P2(x-a2)

(a2< x < a3)

V(x) = - RB M(x)=RB (l - x)

(a3< x < l)

V P1P2

P3 RB

RA

MM1

M3

M2

P1

P3

a1

x

A B

l

P2

a2

a3

b

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RA RB

M(x)

V(x)

Distributed Loads

2

qlRA

2

qlRB

BA

lx

Y=0 RA - V(x) - qx = 0

M = 0

2)(

2qxxRxM A

(0 < x < l)

V

M

8

2ql

q

02

xMx

qxxRA2

ql

2

ql

q

x

A

RA

qxql

qxRxV A 2

)(

22

2qxqlx

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Cantilever beam loaded by uniformly distributed forces

M(x)

V(x)

A

B

x

l

q

B

l-x

Y=0

V(x) - q (l-x) = 0

V(x) = q (l-x)

M = 0

02

xl

xlqxM

22

2xlq

xlxlqxM

V

M

2

2ql

ql

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Beam with an overhang q=3kN/m ， m=3kNm

4m 2m2m

BA

q

x

mC

D

RBRA

V

M

+8.5kN

6kN 3.5kN

+

6kNm

6.04kNm

4kNm

7kNm

4.83m

V(x) = - qx M(x) = - qx2 / 2

V(x) = RA- qx=14.5 - 3x

M(x) =RA(x-2) - qx2 / 2

= 14.5(x-2)-1.5x2

035.14)(

xdx

xdM x = 4.83m

M (x = 4.83m )=14.5(4.83-2)-1.5×4.832 = 6.04kNm(6m<x<8m)

M(x) = RB(8-x) = 3.5(8-x)

RA=14.5kN RB=3.5kN(0<x<2m)

(2m<x<6m)

V(x) = - RB= -3.5 kN

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4.5.3 Relationships between Loads, Shear Forces and Bending Moments

0)]()([)()( xdVxVdxxqxV

dxx

xo

y

q(x)

V(x) V(x)+dV

M(x)M(x)+dMdx

q(x)

C

Y = 0

02

)()()()]()([ dx

dxxqdxxVxMxdMxM

)()(

xVdx

xdM

)()()(

2

2

xqdx

xdV

dx

xMd

M = 0

)()(

xqdx

xdV

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Shear force and bending moment diagrams

RB

RA

B C

D EAF

1.5m 1.5m1.5m

20kN

10kN-m

Reactions

RA= 8.9kN RB= 11.1kN

Shear force and bending moment at special cross sections

C

RA

10kN-m

1.5mV M

B

RA 1.5mV M

Special cross sections of Special cross sections of beam are where loads beam are where loads occur abrupt changeoccur abrupt change

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Techniques of how to draw shear force and bending moment diagram

4m 2m2m

BA

q=3

x

m=3C

D

RBRA

V

M

+8.5kN

6kN 3.5kN

+

6kNm

6.04kNm

4kNm

7kNm

4.83m

q is positive when it acts downwardq=0, V=c, M=f(x)q=c, V=f(x), M=f(x2 )q=f(x), V =f(x2 ) , M= f(x3 )V has a sudden increase where P acts upward.M has a sudden increase where m acts clockwise.

)()(

xVdx

xdM )(

)()(2

2

xqdx

xdV

dx

xMd)(

)(xq

dx

xdV

VV

M M

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Examples

AB

a a

P Pa CBA

aa

PPaC

RARB

RA= 0 RB= P

PV

PaM

P

V

PaM

MA

RA

RA= P MA= 0

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Examplesimple beam

RB

RA

B C

D EAF

1.5m 1.5m1.5m

20kN

10kN-m

RA= 8.9kN RB= 11.1kN

A:V= -8.9kN M=0

B:V= -8.9kN M= -13.35kN-m

C:V= -8.9kN M= -3.35kN-m

D:V= -8.9kN M= -16.65kN-m

E:V= 11.11kN M= -16.65kM-m

C

RA

10kN-m

1.5mV M

F:V= 11.11kN M= 0 8.9

V

x

11.1

M

x3.35

13.35 16.65

RA V M

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example

RA=10kN RB=5kN

RA RB

0.6m 0.6m 1.2m

P=3kNm=3.6kN-m

q=10kN/mA

CD B

Vx

3kN

7kN

5kN

Mx

1.8kN-m1.2kN-m

2.4kN-m 1.25kN-m

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example

qa2qa

aa a

qa

qq

qa

V

x

qa

M

x

0. 5qa2

0. 5qa2

Find mistakes in shear force and bending moment diagram Correct

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Example

L/2 L/2

BA Cq

RA=3qL/8 RB=qL/8

3qL/8

qL/83L/8

V

M9qL2/128

qL2/16

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V

V

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V

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How to get the inner force in following bent shaft

1. Establish coordinate;2. Find reactions;3. Treat loads;4. Find inner force components in different segments.