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Sample Problem 7.2
Strategy:
• Taking entire beam as a free-body,
calculate reactions at B and D.
• Find equivalent internal force-couple
systems for free bodies formed by
cutting beam on either side of load
application points.
Draw the shear and bending moment
diagrams for the beam and loading
shown.
• Plot results.
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Modeling and Analysis:
• Taking entire beam as a free body, calculate
reactions at B and D.
• Find equivalent internal force-couple systems
at sections on either side of load application
points. For stub to left of point 1,
F
y
M
1
= 0:
−20 kN − V1 = 0
= 0:
( 20 kN )( 0 m ) + M 1 = 0
Similarly,
V2 = −20 kN M 2 = −50 kN m
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V3 = 26 kN
M 3 = −50 kN m
V4 = 26 kN
M 4 = +28 kN m
V5 = −14 kN
M 5 = +28 kN m
V6 = −14 kN
M6 = 0
V1 = −20 kN
M1 = 0
• Plot results.
Note that shear is of constant value
between concentrated loads and
bending moment varies linearly.
© 2019 McGraw-Hill Education.
Sample Problem 7.2
3
• Plot results.
Note that shear is of constant value
between concentrated loads and
bending moment varies linearly.
Reflect and Think:
The calculations are pretty similar
for each new choice of free body.
However, moving along the beam,
the shear changes magnitude
whenever you pass a transverse
force and the graph of the bending
moment changes slope at these
points.
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Sample Problem 7.3
1
Strategy:
• Taking entire beam as free body,
calculate reactions at A and B.
Draw the shear and bending moment
diagrams for the beam AB. The
distributed load of 40 lb/in. extends
over 12 in. of the beam, from A to C,
and the 400 lb load is applied at E.
• Determine equivalent internal
force-couple systems at
sections cut within segments
AC, CD, and DB.
• Plot results.
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Sample Problem 7.3
2
Modeling and Analysis:
• Taking entire beam as a free-body, calculate
reactions at A and B.
M = 0 :
By ( 32in.) − ( 480 lb )( 6in.) − ( 400 lb )( 22in.) = 0
A
By = 365lb
M = 0 :
( 480 lb )( 26in.) + ( 400 lb )(10in.) − A (32in.) = 0
B
A = 515lb
F
x
= 0:
Bx = 0
• Note: The 400 lb load at E may be replaced
by a 400 lb force and 1600 lb·in. couple at D.
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• Evaluate equivalent internal force-couple systems
at sections cut within segments AC, CD, and DB.
From A to C:
F
y
M
1
= 0:
515 − 40 x − V = 0
= 0:
−515 x − 40 x ( 12 x ) + M = 0
V = 515 − 40 x
M = 515 x − 20 x 2
From C to D:
 Fy = 0 :
40*12=480
515 − 480 − V = 0
V = 35 lb
M2 = 0 :
− 515x + 480( x − 6) + M = 0
M = (2880 + 35 x ) lb  in.
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From D to B:
 Fy = 0 :
515 − 480 − 400 − V = 0
V = −365 lb
M2 = 0 :
− 515 x + 480( x − 6) − 1600 + 400( x − 18) + M = 0
M = (11,680 − 365 x ) lb  in.
© 2019 McGraw-Hill Education.
• Plot results.
From A to C:
X=12
V = 515 − 40 x
M = 515 x − 20 x 2
3510+1600
From C to D:
V = 35 lb
X=18
M = (2880 + 35 x ) lb  in.
From D to B:
V = −365 lb
X=32
M = (11,680 − 365 x ) lb  in.
© 2019 McGraw-Hill Education.
Sample Problem 7.3
6
Reflect and Think:
Shear and bending-moment diagrams typically feature various kinds of curves
and discontinuities. In such cases, it is often useful to express V and M as
functions of location x as well as to determine certain numerical values.
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Procedure for Analysis
12
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Procedure for Analysis
13
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7.3 Relations Among Load, Shear, and Bending
Moment
• Relations between load and shear:
V − (V + V ) − wx = 0
dV
V
= lim
= −w
dx x →0 x
VD − VC = −  w dx = −(area under load curve )
xD
xC
• Relations between shear and bending moment:
(M + M ) − M − Vx + wx x = 0
(
2
)
dM
M
= lim
= lim V − 12 wx = V
dx x →0 x x →0
M D − M C =  V dx = (area under shear curve )
xD
xC
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Relations Among Load, Shear, and Bending Moment
R A = RB =
• Reactions at supports,
wL
2
• Shear curve,
Shear at any point
x
V − VA = −  w dx = − wx
0
V = VA − wx =
wL
L

− wx = w − x 
2
2

• Moment curve,
x
M − M A =  Vdx
0
(
)
x L
w

M =  w − x dx = L x − x 2
0
2
2

wL2 
dM

M max =
= V = 0
 M at
8
dx


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Sample Problem 7.4
1
Strategy:
• Taking entire beam as a free body,
determine reactions at supports.
• Between concentrated load application
points, dV = − w = 0 and shear is constant.
dx
• With uniform loading between D and E,
the shear variation is linear.
Draw the shear and bendingmoment diagrams for the beam
and loading shown.
• Between concentrated load application
points, dM = v = constant. The change
dx
in moment between load application points
is equal to area under shear curve between
points.
• With a linear shear variation between D
and E, the bending moment diagram is a
parabola.
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Sample Problem 7.4 2
Modeling and Analysis:
• Taking entire beam as a free-body, determine
reactions at supports.
MA = 0:
D(24 ft ) − (20 kips )(6 ft ) − (12 kips )(14 ft )
− (12 kips )(28 ft ) = 0
 F y =0 :
D = 26 kips
Ay − 20 kips − 12 kips + 26 kips − 12 kips = 0
Ay = 18 kips
• Between concentrated loads,
dV
= −w = 0
dx
and shear is constant and determined by
appropriate section cut and solution.
• With uniform loading between D and E, the shear
variation is linear.
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Sample Problem 7.4
3
Between concentrated load application
points, dM = V = constant. Thus, the slope
dx
of the bending moment diagram is constant
in these regions. The change in moment
between load application points is equal to
area under the shear curve between points.
18X6
1/2X8X12
8x2
M B − M A = +108
M B = +108 kip  ft
M C − M B = −16
M C = +92 kip  ft
M D − M C = −140 M D = −48 kip  ft
M E − M D = +48
ME = 0
14X10
• With a linear shear variation between D
and E, the bending moment diagram is a
parabola.
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Sample Problem 7.4
4
Reflect and Think:
As expected, the values of shear and slopes of the bending-moment
curves show abrupt changes at the points where concentrated loads
act. Useful for design, these diagrams make it easier to determine the
maximum values of shear and bending moment for a beam and its
loading.
© 2019 McGraw-Hill Education.
Sample Problem 7.6
Strategy:
• The change in shear between A and B is
equal to the negative of the area under the
load curve between these points. The linear
load curve results in a parabolic shear curve.
• With zero load, the change in shear between
B and C is zero.
Sketch the shear and bendingmoment diagrams for the
cantilever beam and loading
shown.
• The change in moment between A and B is
equal to the area under the shear curve
between these points. The parabolic shear
curve results in a cubic moment curve.
• The change in moment between B and C is
equal to area under the shear curve between
these points. The constant shear curve
results in a linear moment curve.
© 2019 McGraw-Hill Education.
Sample Problem 7.6
Modeling and Analysis:
• The change in shear between A and B is equal to the
negative of the area under the load curve between
these points.
• At the free end of the beam, we find VA = 0. Betwe A
and B, the area under the load curve is
dV
= − w = − w0
dx
VB − V A = − 12 w0 a
at A, V A = 0,
1/3*X*y
VB = − 12 w0 a
Between B and C, the beam is not loaded; thus
VC = VB.
Loading decreases
linearly and the
shear curve is
parabolic
At, A W=wo, and the slope dv/dx = -wo.
At, B W=0, and the slope dv/dx = 0
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• The change in moment between A and B is equal to
the area under the shear curve between these
points. The parabolic shear curve results in a cubic
moment curve.
dM
at A, M A = 0,
=V = 0
dx
M B − M A = − 13 w0 a 2
M B = − 13 w0 a 2
M C − M B = − 12 w0 a( L − a ) M C = − 16 w0 a(3L − a )
•
The change in moment between B and C is equal to area
under shear curve between points. The constant shear curve
results in a linear moment curve.
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• Vector Mechanics For Engineers:
• Twelfth
Statics
Edition
Chapter 8
Friction
©Renato Bordoni/Alamy
© 2019
Education. Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill
• McGraw-Hill
© 2019 McGraw-Hill
Application
© 2019 McGraw-Hill Education.
Introduction
• In preceding chapters, it was assumed that surfaces in contact were either
frictionless (surfaces could move freely with respect to each other) or rough
(tangential forces prevent relative motion between surfaces).
• Actually, no perfectly frictionless surface exists. For two surfaces in
contact, tangential forces, called friction forces, will develop if one attempts
to move one relative to the other.
• However, the friction forces are limited in magnitude and will not prevent
motion if sufficiently large forces are applied.
• The distinction between frictionless and rough is, therefore, a matter of
degree.
• There are two types of friction: dry or Coulomb friction and fluid friction.
Fluid friction applies to lubricated mechanisms. The present discussion is
limited to dry friction between nonlubricated surfaces.
© 2019 McGraw-Hill Education.
8.1A The Laws of Dry Friction – Coefficients of Friction
• Block of weight W is placed on a horizontal
surface. Forces acting on the block are its weight
and reaction of surface N.
• Small horizontal force P is applied. For the block
to remain stationary (i.e., in equilibrium) a
horizontal component F of the surface reaction is
required. F is a static-friction force.
• As P increases, the static-friction force F
increases as well until it reaches a maximum
value Fm.
Fm = µs N
• Further increase in P causes the block to begin to
move, while F drops to a smaller kinetic-friction
force Fk.
Fk = µk N
© 2019 McGraw-Hill Education.
8.1A The Laws of Dry Friction – Coefficients of Friction 2
Table 8.1 Approximate Values of
Coefficient of Static Friction for
Dry Surfaces
Maximum static-friction force:
Fm = µs N
Kinetic-friction force:
Metal on metal
0.15-0.60
Metal on wood
0.20-0.60
Metal on stone
0.30-0.70
Metal on leather
0.30-0.60
Wood on wood
0.25-0.50
Maximum static-friction force and
kinetic-friction force are:
Wood on leather
0.25-0.50
• proportional to normal force.
Stone on stone
0.40-0.70
Earth on earth
0.20-1.00
• dependent on type and condition
of contact surfaces.
Rubber on concrete
0.60-0.90
• independent of contact area.
© 2019 McGraw-Hill Education.
Fk = µk N
k  0.75 s
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,
(Px = 0)
© 2019 McGraw-Hill Education.
•
No motion,
(Px < F)
Since there is no
evidence that F has
reached its
maximum value,
the equation
• Motion impending,
(Px = Fm)
The applied forces are such that
the body is just about to slide.
We say that motion is impending.
•
Motion,
(Px > Fm)
The body is sliding under
the action of the applied
forces, and the equations
of equilibrium do not
apply any more
8.1BAngles of Friction 1
• It is sometimes convenient to replace the normal force N and friction force
F with their resultant R:
•
No friction
•
No motion
• Motion impending
if the applied force P has a
horizontal component Px
which tends to move the
block, the force R will have
a horizontal component F
and, thus, will form an
angle f with the normal to
the surface,
© 2019 McGraw-Hill Education.
tan  s =
Fm  s N
=
N
N
tan  s =  s
If Px is increased until motion
becomes impending, the angle
between R and the vertical grows
and reaches a maximum
valueThis value is called the angle
of static friction
•
Motion
tan  k =
Fk k N
=
N
N
tan  k = k
If motion actually takes place,
the magnitude of the friction
force drops to Fk; similarly, the
angle f between R and N drops
to a lower value ⱷk, called the
angle of kinetic friction
8.1B– Angles of Friction
•
•
Consider a block of weight W resting on a board of variable inclination angle q.
If the board is given a small angle of inclination θ, the force R will deviate from perpendicular
and have normal and tangential components
• The value of the angle of inclination corresponding to impending motion is called
the angle of repose.(figure c)
angle of kinetic friction
• No friction
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• No motion
• Motion
impending
• Motion
8.4 Problems Involving Dry Friction
A number of common machines and mechanisms can be analyzed by applying the laws of dry
friction. These include wedges, screws, journal and thrust bearings, and belt transmissions.
If a problem involves only a motion of translation, with no possible rotation, the body
under consideration can usually be treated as a particle, and the methods of Chap. 2.
If involve rotation -----chapter 4
Problems----- in three groups
•
•
•
All applied forces
known
•
Coefficient of static
friction is known
All applied forces
known
•
Motion is impending
•
Determine value of
coefficient of static
friction.
Determine whether
body will remain at
rest or slide.
© 2019 McGraw-Hill Education.
•
Coefficient of static
friction is known
•
Motion is impending
•
Determine magnitude or
direction of one of the
applied forces.
Sample Problem 8.1
Strategy:
• Draw the free body diagram for the
block. Remember that the friction
force is opposite the direction of
impending motion.
A 100 lb force acts as shown on a 300
lb block placed on an inclined plane.
The coefficients of friction between
the block and plane are µs = 0.25 and
µk = 0.20. Determine whether the
block is in equilibrium and find the
value of the friction force.
© 2019 McGraw-Hill Education.
•
Determine values of friction force
and normal reaction force from plane
required to maintain equilibrium.
•
Calculate maximum friction force and
compare with friction force required
for equilibrium. If it is greater, block
will not slide.
•
If maximum friction force is less than
friction force required for equilibrium,
block will slide. Calculate kineticfriction force.
Sample Problem 8.1
2
Modeling and Analysis:
• Determine values of friction force and normal
reaction force from plane required to maintain
equilibrium.
 Fx = 0 : 100 lb − 53 ( 300 lb ) − F = 0
F = −80 lb
F
y
What does the sign tell
you about the assumed
direction of impending
motion
•
= 0: N −
4
5
( 300 lb ) = 0
N = 240 lb
Calculate maximum friction force and compare
with friction force required for equilibrium.
Fm =  s N
Fm = 0.25 ( 240 lb ) = 60 lb
What does this solution imply about the block?.
The block will slide down
the plane.
© 2019 McGraw-Hill Education.
the value of the force required to maintain equilibrium
(80 lb) is larger than the maximum value which may be
obtained (60 lb), equilibrium will not be maintained
• If maximum friction force is less than friction
force required for equilibrium, block will slide.
Calculate kinetic-friction force.
Factual = Fk = µk N
=0.20(240 lb)
Factual = 48 lb
Reflect and Think:
This is a typical friction problem of the first type. Note that you used the
coefficient of static friction to determine if the crate moves, but once you
found that it does move, you needed the coefficient of kinetic friction to
determine the friction force.
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