Lecture 6-4
3-dimensional problems
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Equilibrium in three-dimensions
When considering the equilibrium of a three-dimensional body, each
of the reactions exerted on the body by its supports can involve between
one and six unknowns, depending upon the type of support.
In the general case of the equilibrium of a three-dimensional body, the
six scalar equilibrium equations listed at the beginning of this review
should be used and solved for six unknowns. In most cases these
equations are more conveniently obtained if we first write
ΣF=0
Σ MO = Σ (r x F)
and express the forces F and position vectors r in terms of scalar
components and unit vectors. The vector product can then be
computed either directly or by means of determinants, and the
desired scalar equations obtained by equating to zero the coefficients
of the unit vectors.
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Equilibrium of a Rigid Body in Three
Dimensions
• Six scalar equations are required to express
the conditions for the equilibrium of a rigid
body in the general three dimensional case.
∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0
∑Mx = 0 ∑My = 0 ∑Mz = 0
• These equations can be solved for no more than 6
unknowns which generally represent reactions at
supports or connections.
• The scalar equations are conveniently obtained by
applying the vector forms of the conditions for
equilibrium,


 
∑ F = 0 ∑ M O = ∑ (r × F ) = 0
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Equilibrium of a Rigid Body in Three
Dimensions
As many as three unknown reaction components can be
eliminated from the computation of Σ MO through a
judicious choice of point O. Also, the reactions at two points
A and B can be eliminated from the solution of some
problems by writing the equation Σ MAB = 0, which
involves the computation of the moments of the forces about
an axis AB joining points A and B.
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Equilibrium of a Rigid Body in Three
Dimensions
If the reactions involve more than six unknowns, some of
the reactions are statically indeterminate; if they involve
fewer than six unknowns, the rigid body is only partially
constrained.
Even with six or more unknowns, the rigid body will be
improperly constrained if the reactions associated with the
given supports either are parallel or intersect the same line.
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Reactions at Supports and Connections
for a Three-Dimensional Structure
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Reactions at Supports and Connections
for a Three-Dimensional Structure
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Sample Problem
A sign of uniform density weighs 270 lb and is
supported by a ball-and-socket joint at A and
by two cables.
Determine the tension in each cable and the
reaction at A.
SOLUTION:
• Create a free-body diagram for the sign.
• Apply the conditions for static equilibrium to develop equations
for the unknown reactions.
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Sample Problem

TBD = TBD
= TBD
= TBD

TEC = TEC
• Create a free-body diagram for the sign.
Since there are only 5 unknowns, the sign
is partially constrain. It is free to rotate
about the x axis. It is, however, in
equilibrium for the given loading.
= TEC
= TEC


rD − rB


rD − rB



− 8i + 4 j − 8k
12
 1 2
2
− 3i + 3 j − 3k
 
rC − rE
 
rC − rE



− 6i + 3 j + 2 k
7
 3 2
6
−7i +7 j +7k
(
)
(
)
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Sample Problem

∑F =

i:

j:

k:

∑MA

j:

k:
• Apply the conditions for
static equilibrium to develop
equations for the unknown
reactions.
 


(
)
A + TBD + TEC − 270 lb j = 0
Ax − 23 TBD − 76 TEC = 0
Ay + 13 TBD + 73 TEC − 270 lb = 0
Az − 23 TBD + 72 TEC = 0



 

= rB × TBD + rE × TEC + (4 ft )i × (− 270 lb ) j = 0
5.333 TBD − 1.714 TEC = 0
2.667 TBD + 2.571TEC − 1080 lb = 0
Solve the 5 equations for the 5 unknowns,
TBD = 101.3 lb TEC = 315 lb




A = (338 lb )i + (101.2 lb ) j − (22.5 lb )k
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Friction
• 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.
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The Laws of Dry Friction.
Coefficients of Friction
• Block of weight W placed on horizontal
surface. Forces acting on block are its weight
and reaction of surface N.
• Small horizontal force P applied to block. For
block to remain stationary, 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 = µ s N
• Further increase in P causes the block to begin
to move as F drops to a smaller kinetic-friction
force Fk.
Fk = µ k N
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8 - 12
The Laws of Dry Friction.
Coefficients of Friction
• Maximum static-friction force:
Fm = µ s N
• Kinetic-friction force:
Fk = µ k N
µ k ≅ 0.75µ s
• Maximum static-friction force and kineticfriction force are:
- proportional to normal force
- dependent on type and condition of
contact surfaces
- independent of contact area
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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)
• No motion,
(Px < Fm)
• Motion impending,
(Px = Fm)
• Motion,
(Px > Fm)
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Angles of Friction
• It is sometimes convenient to replace normal force N
and friction force F by their resultant R:
• No friction
• No motion
tan φ s =
Fm µ s N
=
N
N
tan φ s = µ s
• Motion
• Motion impending
Fk µ k N
tan φ k =
=
N
N
tan φ k = µ k
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Angles of Friction
• Consider block of weight W resting on board
with variable inclination angle θ.
• No friction
• No motion
• Motion
impending
• Motion
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8 - 16
Problems Involving Dry Friction
• All applied forces known
• All applied forces known
• Coefficient of static
friction is known
• Coefficient of static friction
• Motion is impending
is known
• Motion is impending
•
Determine
value
of
coefficient
• Determine whether body
of static friction.
• Determine magnitude or
will remain at rest or slide
direction of one of the
applied forces
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Sample Problem
SOLUTION:
• 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.
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.
• If maximum friction force is less
than friction force required for
equilibrium, block will slide.
Calculate kinetic-friction force.
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Sample Problem
SOLUTION:
• 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
∑ Fy = 0 :
N - 54 (300 lb ) = 0
N = 240 lb
• Calculate maximum friction force and compare
with friction force required for equilibrium. If it is
greater, block will not slide.
Fm = µ s N
Fm = 0.25(240 lb ) = 48 lb
The block will slide down the plane.
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8 - 19
Sample Problem
• If maximum friction force is less than friction
force required for equilibrium, block will slide.
Calculate kinetic-friction force.
Factual = Fk = µ k N
= 0.20(240 lb )
Factual = 48 lb
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8 - 20
Sample Problem
The moveable bracket shown may be
placed at any height on the 3-in.
diameter pipe. If the coefficient of
friction between the pipe and bracket is
0.25, determine the minimum distance
x at which the load can be supported.
Neglect the weight of the bracket.
SOLUTION:
• When W is placed at minimum x, the bracket is
about to slip and friction forces in upper and lower
collars are at maximum value.
• Apply conditions for static equilibrium
to find minimum x.
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8 - 21
Sample Problem
SOLUTION:
• When W is placed at minimum x, the bracket is about to
slip and friction forces in upper and lower collars are at
maximum value.
FA = µ s N A = 0.25 N A
FB = µ s N B = 0.25 N B
• Apply conditions for static equilibrium to find minimum x.
∑ Fx = 0 : N B − N A = 0
∑ Fy = 0 : FA + FB − W = 0
0.25 N A + 0.25 N B − W = 0
0.5 N A = W
NB = N A
N A = N B = 2W
∑ M B = 0 : N A (6 in.) − FA (3 in.) − W ( x − 1.5 in.) = 0
6 N A − 3(0.25 N A ) − W ( x − 1.5) = 0
6(2W ) − 0.75(2W ) − W ( x − 1.5) = 0
x = 12 in.
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Design: Example 1
The structure shown in Fig. P6-146 consists of
a circular tie rod AB and a rigid member BC. If
the structure is to support a load P = 40 kN,
determine the required diameters of the pins at
A, B, and C, and the required diameter of the tie
rod.
The tie rod is made of structural steel and the
pins are made of 0.2% C hardened steel. All
pins are in double shear. The tie rod is
adequately reinforced around the pins so that
tensile failure does not occur at the pins. Failure
is by yielding, and the factor of safety is 1.3.
Take the yield strength in shear to be one-half
the yield strength in tension.
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Solution
Free-Body diagram of the tie rod CB
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Solution
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Design: Example 2
Each member of the truss of Fig. P6-148 has a circular
cross section and is made of structural steel. If
failure is by yielding, and the factor of safety is 2.5,
determine the minimum permissible diameter of
member EJ.
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Design: Example 2
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