# Vector Mechanics for Engineers: Statics

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18 Ιουλ 2012 (πριν από 6 χρόνια)

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1
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 1
How to prepare for the final
• The final will be based on Chapters 6, 7, 8, and sections 10.1-10.5. It will be
three-hour, take-home, open-textbook and open-notes exam.
• Read “Review and Summary” after each Chapter. Brush up on topics that are
not familiar.
• Make sure you know how to solve HW problems and sample problems.
Additional review problems for the final will be posted on the web.
• Review important tables/formulae from the book (such as supports and their
reactions) so that you can use them easily.
• Remember, the correct reasoning and an error in computation will get you most
of the points. However, the right answer with no explanation will get you no
• Do not forget about the honor code. Carefully read the instructions on the front
page of the final. You cannot discuss anything about the final until after the due
date.
• The rest of this document is a brief summary of important topics we have learned
in the second half of the term.
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 2
Analysis of Trusses by the Method of Joints
• Dismember the truss and create a freebody
diagram for each member and pin.
• The two forces exerted on each member are
equal, have the same line of action, and
opposite sense.
• Forces exerted by a member on the pins or
joints at its ends are directed along the member
and equal and opposite.
• Conditions of equilibrium on the pins provide
2n equations for 2n unknowns. For a simple
truss, 2n = m + 3. May solve for m member
forces and 3 reaction forces at the supports.
• Conditions for equilibrium for the entire truss
provide 3 additional equations which are not
independent of the pin equations.
2
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 3
Analysis of Trusses by the Method of Sections
• When the force in only one member or the
forces in a very few members are desired, the
method of sections works well.
• To determine the force in member BD, pass a
section through the truss as shown and create
a free body diagram for the left side.
• With only three members cut by the section,
the equations for static equilibrium may be
applied to determine the unknown member
forces, including F
BD
.
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 4
Machines
• Machines are structures designed to transmit
and modify forces. Their main purpose is to
transform input forces into output forces.
• Given the magnitude of P, determine the
magnitude of Q.
• Create a free-body diagram of the complete
machine, including the reaction that the wire
exerts.
• The machine is a nonrigid structure. Use
one of the components as a free-body.
P
b
a
QbQaPM
A
=−==

0
3
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 5
Shear and Bending Moment Diagrams
• Variation of shear and bending
moment along beam may be
plotted.
• Determine reactions at
supports.
• Cut beam at C and consider
member AC,
22 PxMP
V
+
=
+
=
• Cut beam at E and consider
member EB,
( )
22 xLPMP
V

+
=

=
• For a beam subjected to
, shear is
and moment varies linearly.
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 6
Relations Among Load, Shear, and Bending Moment
• Relations between load and shear:
(
)
w
x
V
dx
dV
x
wVVV
x
−=
Δ
Δ
=
=
Δ

Δ
+

Δ
0
lim
0
( )

D
C
x
x
CD
dxwVV
• Relations between shear and bending moment:
( )
( )
VxwV
x
M
dx
dM
x
xwxVMMM
x
x
=Δ−=
Δ
Δ
=
=
Δ
Δ+Δ−−Δ+

Δ

Δ
2
1
00
limlim
0
2
( )
curveshear under area==−

D
C
x
x
CD
dxVMM
4
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 7
Relations Among Load, Shear, and Bending Moment
• Reactions at supports,
2
wL
RR
BA
==
• Shear curve,

−=−=−=
−=−=−

x
L
wwx
wL
wxVV
wxdxwVV
A
x
A
22
0
• Moment curve,
( )

===
−=

−=
=−

0at
8
22
2
max
2
0
0
V
dx
dM
M
wL
M
xxL
w
dxx
L
wM
VdxMM
x
x
A
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 8
Sample Problem 7.4
Draw the shear and bending-
moment diagrams for the beam
SOLUTION:
• Taking entire beam as a free-body, determine
reactions at supports.
shear variation is linear.
points, and shear is
constant.
0
=

=
wdxdV
points, The change
in moment between load application points is
equal to area under shear curve between
points.
.constant
=
=
VdxdM
• With a linear shear variation between D
and E, the bending moment diagram is a
parabola.
5
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 9
Sample Problem 7.4
points, The change
in moment between load application points is
equal to area under the shear curve between
points.
.constant
=
=
VdxdM
• With a linear shear variation between D
and E, the bending moment diagram is a
parabola.
048
ftkip 48140
ftkip 9216
ftkip 108108
=+=−
⋅−=−=−
⋅+=−=−

+
=
+
=

E
D
E
DCD
CBC
BAB
MMM
MMM
MMM
MMM
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 10
• Consider entire cable as free-body. Slopes of
cable at A and B are not known - two reaction
components required at each support.
• Four unknowns are involved and three
equations of equilibrium are not sufficient to
determine the reactions.
• For other points on cable,
2
yields0
2
yM
C
=

yxyx
θθFF, yield 0,0
=
=



=
=
=
x
x
ATT
θ
• Additional equation is obtained by
considering equilibrium of portion of cable
AD and assuming that coordinates of point D
on the cable are known. The additional
equation is
.0

=
D
M
6
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 11
• For cable carrying a distributed load:
a) cable hangs in shape of a curve
b) internal force is a tension force directed along
tangent to curve.
• Consider free-body for portion of cable extending
from lowest point C to given point D. Forces are
horizontal force T
0
at C and tangential force T at D.
• From force triangle:
0
22
0
0
tan
sincos
T
W
WTT
WTTT
=+=
==
θ
θθ
• Horizontal component of T is uniform over cable.
• Vertical component of T is equal to magnitude of W
measured from lowest point.
• Tension is minimum at lowest point and maximum
at A and B.
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 12
Parabolic Cable
• Consider a cable supporting a uniform, horizontally
distributed load, e.g., support cables for a
suspension bridge.
point D given by internal tension force
magnitude and direction are
,wxW
=
0
222
0
tan
T
wx
xwTT =+= θ
0
2
:0
0
=−=

yT
x
wxM
D
0
2
2T
wx
y =
or
The cable forms a parabolic curve.
7
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 13
Catenary
• Consider a cable uniformly loaded along the cable
itself, e.g., cables hanging under their own weight.
• With loading on the cable from lowest point C to a
point D given by the internal tension force
magnitude is
w
T
cscwswTT
0
22222
0
=+=+=
,
wsW
=
• To relate horizontal distance x to cable length s,
c
x
cs
c
s
c
csq
ds
x
csq
ds
T
T
dsdx
s
sinhandsinh
coscos
1
0
22
22
0
==
+
=
+
===

θθ
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 14
Catenary
• To relate x and y cable coordinates,
c
x
cy
c
c
x
cdx
c
x
cy
dx
c
x
dx
c
s
dx
T
W
dxdy
x
cosh
coshsinh
sinhtan
0
0
=
−==−
====

θ
which is the equation of a catenary.
8
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 15
The Laws of Dry Friction. Coefficients of Friction
• Four situations can occur when a rigid body is in contact with
a horizontal surface:
• No friction,
(P
x
= 0)
• No motion,
(P
x
< F
m
)
• Motion impending,
(P
x
= F
m
)
• Motion,
(P
x
> F
m
)
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 16
Wedges
• Wedges - simple
machines used to raise
• Friction prevents wedge
from sliding out.
• Want to find minimum
force P to raise block.
• Block as free-body
0
:0
0
:0
21
21
=+−−
=
=+−
=

NNW
F
NN
F
s
y
s
x
μ
μ
or
0
21
=++ WRR

( )
( )
06sin6cos
:0
0
6sin6cos
:0
32
32
=°−°+−
=
=+
°−°−−
=

s
y
ss
x
NN
F
P
NN
F
μ
μμ
• Wedge as free-body
or
0
32
=+− RRP

9
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 17
• Square-threaded screws frequently used in jacks, presses, etc.
Analysis similar to block on inclined plane. Recall friction
force does not depend on area of contact.
• Thread of base has been “unwrapped” and shown as straight
line. Slope is 2
π
r horizontally and lead L vertically.
• Moment of force Q is equal to moment of force P.
rPaQ =
• Impending motion
upwards. Solve for
Q
.
• Self-locking, solve
for
Q
,
θ
φ
>
s
• Non-locking, solve
,
θ
φ
>
s
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 18
Journal Bearings. Axle Friction
• Angle between R and
normal to bearing
surface is the angle of
kinetic friction ϕ
k
.
k
k
Rr
R
r
M
μ
φ

= sin
• May treat bearing
reaction as force-
couple system.
• For graphical solution,
R must be tangent to
circle of friction.
k
kf
r
rr
μ
φ

=
sin
10
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 19
Belt Friction
• Relate T
1
and T
2
when belt is about to slide to right.
• Draw free-body diagram for element of belt
( )
0
2
cos
2
cos:0 =Δ−
Δ

Δ
Δ+=

NTTTF
sx
μ
θ
θ
( )
0
2
sin
2
sin:0 =
Δ

Δ
Δ+−Δ=

θ
θ
TTTNF
y
• Combine to eliminate ΔN, divide through by Δθ,
(
)
2
2sin
22
cos
θ
θ
μ
θ
θ
Δ
Δ

Δ
+−
Δ
Δ
Δ
T
T
T
s
• In the limit as Δθgoes to zero,
0=− T
d
dT
s
μ
θ
• Separate variables and integrate from
β
θ
θ
== to0
βμ
βμ
s
e
T
T
T
T
s
==
1
2
1
2
orln
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 20
Principle of Virtual Work
• Imagine a small virtual displacement of a particle which
is acted upon by several forces.
• The corresponding virtual work,
(
)
r
R
rFFFrFrFrFU

δ
δδδδδ

=
⋅++=⋅+⋅+⋅=
321321
Principle of Virtual Work:
• A particle is in equilibrium if and only if the total virtual
work of forces acting on the particle is zero for any virtual
displacement.
• A rigid body is in equilibrium if and only if the total
virtual work of external forces acting on the body is
zero for any virtual displacement of the body.
• If a system of connected rigid bodies remains connected
during the virtual displacement, only the work of the
external forces need be considered.
11
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 21
Sample Problem 10.1
Determine the magnitude of the couple Mrequired to
maintain the equilibrium of the mechanism.
SOLUTION:
• Apply the principle of virtual work
D
PM
xPM
UUU
δδθ
δ
δ
δ
+=
+
=
=
0
0
θδθδ
θ
sin3
cos3
lx
lx
D
D
−=
=
(
)
θ
δ
θ
δθ
sin30 lP
M

+
=
θ
sin3Pl
M
=
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 22
Potential Energy and Equilibrium
(not covered in the final)
• When the potential energy of a system is known,
the principle of virtual work becomes
θ
δθ
θ
δδ
d
dV
d
dV
VU
=
−=−==
0
0
• For the structure shown,
( ) ( )
θθ cossin2
2
2
1
2
2
1
lWlk
WykxVVV
CBge
+=
+=+=
• At the position of equilibrium,
( )
Wkll
d
dV
−==
θθ
θ
cos4sin0
indicating two positions of equilibrium.
12
V
ector Mechanics for Engineers: Statics
Eighth
Edition
3 - 23
Stability of Equilibrium
(not covered in the final)
0
2
2
>
θ
d
Vd
2
2
0
d V

<
0=
θ
d
dV
Must examine higher
order derivatives.