Download - ENGINEERING MECHANICS
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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
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.