ENGINEERING MECHANICS
Part Ⅱ Mechanics of Materials
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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Introduction
Solid 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.
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Bending
Forces acting transverse to the axis of structural member
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Chapter 4 inner forces
4.2 Method of Section for Internal Force
The resistance forces that set up within a body to balance the effect
of the externally applied forces.
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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4.3 Tension and compression
Prismatic bar loaded by axial force at the
centroid of ends (Axially loaded
members).
P
P
L
P
P
N
s
A
Resultant of forces N
For equilibrium：S x=0
N - P=0
Axial internal force ：
N=P
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MOM-1-9
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Axial Force Diagram
Example determine the axial forces
and draw the axial-force diagram.
A
C P2=15kN B
P3=10kN
x
P1=5kN
1
N1-P1=0
A
N1
P1=5kN
1
N
2
B
N2 2
5kN
N1=P1=5kN
P3 =10kN
N2 = -10kN
x
10kN
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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
N(kN)
(-)
x
(-)
-10
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-5
MOM-1-12
4.4 diagram of torque
4.4.1 Transmission of power by circular shafts
Motor driven shaft
T Angular speed ω(rad/s)
Transmitted torque T(N·m)
Φ—angular rotation
W  T 
Power: N  dW  T d   T   2 f T  2 nT (Watt , N  m / s , J / s )
dt
dt
60
Work:
f—frequency of revolution, Hz=s-1
n—number of revolutions per minute (rpm)
If N is expresses in kilowatt or horsepower, 1 hp=735.5 W
60 N
N
T
 9549 ( N  m )
2 n
n
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N
T  7024 ( N  m)
n
MOM-1-13
4.4.2 Diagram of Torsion Moment
Method of Section for Internal Force
Equilibrium Smx = 0
n
m
T-m=0
T=m
m
n
n
m
x
T
Sign convention for torque
——right hand law.
n
n
m
T
n
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MOM-1-14
10
a
20
b
T1
T
I
15
c
10
20
b
T2
20
Determine the torsion
moment of the shaft
SMx = 0
d
15
c
15
15-20+10+T1 = 0
T1 = -5kN-m
d
15-20+T2 = 0
T2 = 5kN-m
II
c
d
15+T3 = 0
T3 = -15kN-m
15
5kN-m
T3
III d
x
5kN-m
15kN-m
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4.5 Shear Forces and Bending Moments
4.5.1 Beams
Beam: the structural member to resist
forces acting transverse to its axis A
q
B
P
Simple beam
Simply supported beam
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P1
P2
A
B
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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Type of Loads
Concentrated Load P
Distributed Load q
Couple M
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Reactions in various types of supports
FAx
FBy
FAy
FAx
FAy
FBy
FAx
MA
Simple
Beam
Beam with
an
Overhang
Cantilever
Beam
FAy
How to find the reactions?
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4.5.2 Shear Forces and Bending Moments
Method of section
P
SY=0
SMo=0
P - V =0
V=P
M - Px =0 M = Px
A
P
VM
M
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M
o
A
M
V
V
n
x
Sign Convention
M
V
m B
x
V
VM
B
V
V
M
V
V
M
M
M
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MOM-1-23
Bending Moment and Shear Force
Find the Bending Moment and Shear Force on cross section of I-I
Reactions RA, RB
S MB=0 RAl - P(l-a) - 2Pa = 0 A
S Y=0
I
2l  a
RB 
P
l
B
I
x
RA + RB - 3P = 0
la
RA 
P
l
a
2P
a
P
S Y=0
RA - P - V = 0
RA
l a
a
V  RA  P 
PP P
l
l
RA x - P(x-a) - M = 0
S M0=0
a
P
M
x
V
2P
M
V
la
lx
M  RA x  P ( x  a ) 
Px  P( x  a ) 
Pa
l
l
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RB
l
RA
RB
MOM-1-24
Shear-Force and Bending-Moment Diagrams
Concentrated Loads
a
C P
b
B
A
Reactions
x
Pb
RA 
l
Pa
RB 
l
l
C
RA
RB
M(x)
A
Equations for shear force and
bending moment
RA
x
V(x)
(0 < x < a)
SY = 0
S M= 0
RA - V(x) = 0
Pb
V ( x)  RA 
l
RA x - M(x) = 0
Pbx
M ( x )  RA x 
l
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MOM-1-25
(a < x < l)
SY = 0
C P
a
RA - V(x) - P = 0
b
B
A
x
Pa
V ( x )  RA  P  
l
l
C
P
a
M(x)
S M = 0 Rax - P(x-a) - M(x) = 0 A
M ( x )  RA x  P( x  a )
RA
x
V(x)
Pa
l  x 

l
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MOM-1-26
(0 < x < a)
Pb
V ( x) 
l
a
P
b
B
A
x
Pbx
M ( x) 
l
l
C
Pb
l
V
Pa
l
(a < x < l)
Pa
V ( x)  
l
M
Pab
l
Pa
l  x 
M ( x) 
l
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MOM-1-27
A beam loaded by several
concentrated forces
Reactions Ra and RB
a1
P1
a2
P3
P2
A
B
(0 < x < a1)
x
l
RA
V(x) = RA M(x)=RAx
RB
b
(a1< x < a2)
P3
M(x)
V(x) = RA- P1
M(x)=RA x -P1(x-a1)
V RA
V(x) = RA- P1 – P2
M(x)= RA x -P1(x-a1)- P2(x-a2)
(a3< x < l)
M
M(x)=RB (l - x)
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RB
P1
P2
M2
M1
B
V(x)
x
l
(a2< x < a3)
V(x) = - RB
b
a3
P3
RB
M3
MOM-1-28
q
Distributed Loads
ql
RA 
2
SY=0
A
ql
RB 
2
RA
RA - V(x) - qx = 0
B
x
RB
q
M(x)
A
ql
x
V ( x )  RA  qx   qx
RA
2
ql
x
RA x  qx   M x   0 V 2
SM=0
2
2
qx
M ( x )  RA x 
2 2
qlx qx


(0 < x < l)
M
2
2
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l
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V(x)
ql
8
ql
2
2
MOM-1-29
Cantilever beam loaded by uniformly distributed forces
B
q
A
SY=0
x
l
V(x) - q (l-x) = 0
M(x)
V(x) = q (l-x)
V(x)
SM=0
lx
M  x   ql  x  
0
2
2

lx
l  x
M  x    ql  x  
 q
2
2
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B
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V
l-x
ql
M
ql 2
2
MOM-1-30
q q=3kN/m，m=3kNm
Beam with an overhang
m
C
RA=14.5kN
RB=3.5kN
A
V
M(x) = - qx2 / 2
2m
4m
RA
(0<x<2m)
V(x) = - qx
D
x
2m
B
RB
8.5kN
+
(2m<x<6m)
V(x) = RA- qx=14.5 - 3x
M(x) =RA(x-2) - qx2 / 2
4.83m
M
6.04kNm
+
= 14.5(x-2)-1.5x2
dM ( x )
 14.5  3x  0
dx
6kN
7kNm
4kNm
6kNm
x = 4.83m
M (x = 4.83m )=14.5(4.83-2)-1.5×4.832 = 6.04kNm
(6m<x<8m)
V(x) = - RB= -3.5 kN
M(x) = RB(8-x) = 3.5(8-x)
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3.5k
N
4.5.3 Relationships between Loads, Shear Forces and Bending Moments
SY = 0
V ( x )  q( x )dx  [V ( x )  dV ( x )]  0 o
SM = 0
y
x
q(x)
x
dx
dx
[ M ( x )  dM ( x )]  M ( x )  V ( x )dx  q( x )dx 
0
2
dV ( x )
 q( x )
dx
M(x)
dM ( x )
 V ( x)
dx
C
V(x)
d 2 M ( x ) dV ( x )

 q( x )
2
dx
dx
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dx M(x)+dM
V(x)+dV
q(x)
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MOM-1-32
Shear force and bending moment diagrams
Reactions
RA= 8.9kN
RB= 11.1kN
10kN-m
B C
A
Shear force and bending
moment at special cross
sections
RA
Special cross sections of
beam are where loads
occur abrupt change
1.5m
F
D E
1.5m
1.5m
RB
20kN
B
RA 1.5m
V M
10kN-m
C
RA
1.5m
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V M
MOM-1-33
Techniques of how to draw shear force and bending moment
diagram
V
V
d 2 M ( x ) dV ( x )

 q( x )
2
dx
dx
dM ( x )
 V ( x)
dx
dV ( x )
 q( x )
dx
M
q=3
M
C
q is positive when it acts downward
q=0, V=c, M=f(x)
q=c, V=f(x), M=f(x2 )
V
2
3
q=f(x), V =f(x ) , M= f(x )
V has a sudden increase where P acts
upward.
M has a sudden increase where m acts M
clockwise.
m=3
A
D
x
2m
2m
4m
RA
Kylinsoft
RB
8.5kN
+
4.83m
3.5kN
6kN
6.04kNm
+
7kNm
4kNm
6kNm
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B
MOM-1-34
Examples
P
C
A
P
Pa
B
MAA
Pa
C
B
a
a
RA
RA= 0
a
RB
RA
RA= P
RB= P
a
MA= 0
V
V
P
P
M
Pa
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M
Kylinsoft
Pa
MOM-1-35
Example
simple beam
RA= 8.9kN
A:V= -8.9kN
10kN-m
B C
A
RB= 11.1kN
1.5m
RA
M=0
20kN
RB
10kN-m
C:V= -8.9kN M= -3.35kN-m R
A
C
VM
RA
1.5m
11.1
V
E:V= 11.11kN M= -16.65kM-m
F:V= 11.11kN
1.5m
1.5m
B:V= -8.9kN M= -13.35kN-m
D:V= -8.9kN M= -16.65kN-m
F
D E
V M
x
M= 0
8.9
M
3.35
13.35
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x
16.65
MOM-1-36
example
RA=10kN
P=3kN
RB=5kN
C
m=3.6kN-m
q=10kN/m
A
D
B
0.6m 0.6m
RA
V
1.2m
RB
7kN
x
3kN
M
2.4kN-m
5kN
1.25kN-m
x
1.2kN-m
1.8kN-m
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MOM-1-37
example
Find mistakes in shear force
and bending moment diagram
Correct
q
a
q
qa
a
q
a
a
2qa
V
q
a
x
q
a
M
0. 5qa2
x
0. 5qa2
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MOM-1-38
Example
q
C
A
L/2
B
L/2
RA=3qL/8
RB=qL/8
V 3qL/8
3L/8
9qL2/128
M
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qL/8
qL2/16
MOM-1-39
V
V
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MOM-1-40
V
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MOM-1-41
How to get the inner force in following bent shaft
1.
2.
3.
4.
2016/3/17
Establish coordinate;
Find reactions;
Treat loads;
Find inner force components in different segments.
Kylinsoft
Statics-3-42