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Department of Mechanical Engineering Lecture 6-4 3-dimensional problems
Department of Mechanical Engineering
Lecture 6-4 3-dimensional problems
Department of Mechanical Engineering
Σ F = 0 Σ MO = Σ (r x F)
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
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.
Equilibrium in three-dimensions
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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.
∑ =∑ =∑ =∑ =∑ =∑ =
000000
zyx
zyxMMMFFF
• 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,
( )∑ ∑ =∑ ×== 00 FrMF O
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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.
Equilibrium of a Rigid Body in Three Dimensions
Department of Mechanical Engineering
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.
Equilibrium of a Rigid Body in Three Dimensions
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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 ProblemA 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
• 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.
( )
( )kjiT
kjiT
rrrrTT
kjiT
kjiT
rrrrTT
EC
EC
EC
ECECEC
BD
BD
BD
BDBDBD
72
73
76
32
31
32
7236
12848
++−=
++−=
−−
=
−+−=
−+−=
−−
=
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Sample Problem
• Apply the conditions for static equilibrium to develop equations for the unknown reactions.
( )
( ) ( )
0lb1080571.2667.2:
0714.1333.5:
0lb 270ft 4
0:
0lb 270:
0:
0lb 270
72
32
73
31
76
32
=−+
=−
=−×+×+×=∑
=+−
=−++
=−−
=∑ −++=
ECBD
ECBD
ECEBDBA
ECBDz
ECBDy
ECBDx
ECBD
TTk
TTj
jiTrTrM
TTAk
TTAj
TTAi
jTTAF
( ) ( ) ( )kjiA
TT ECBD
lb 22.5lb 101.2lb 338
lb 315lb 3.101
−+=
==Solve the 5 equations for the 5 unknowns,
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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
8 - 12
• 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 Fincreases as well until it reaches a maximum value Fm.
NF sm µ=• Further increase in P causes the block to begin
to move as F drops to a smaller kinetic-friction force Fk. NF kk µ=
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The Laws of Dry Friction. Coefficients of Friction
• Maximum static-friction force:NF sm µ=
• Kinetic-friction force:
sk
kk NFµµ
µ75.0≅
=
• Maximum static-friction force and kinetic-friction 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 Nand friction force F by their resultant R:
• No friction • Motion impending• No motion
ss
sms N
NN
F
µφ
µφ
=
==
tan
tan
• Motion
kk
kkk N
NNF
µφ
µφ
=
==
tan
tan
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Angles of Friction
8 - 16
• Consider block of weight W resting on board with variable inclination angle θ.
• No friction • No motion • Motion impending
• Motion
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Problems Involving Dry Friction
• All applied forces known
• Coefficient of static friction is known
• Determine whether body will remain at rest or slide
• All applied forces known
• Motion is impending
• Determine value of coefficient of static friction.
• Coefficient of static friction is known
• Motion is impending
• Determine magnitude or direction of one of the applied forces
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Sample Problem
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.
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.
• 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
8 - 19
SOLUTION:• Determine values of friction force and normal
reaction force from plane required to maintain equilibrium.
:0=∑ xF ( ) 0lb 300 - lb 100 53 =− F
lb 80−=F
:0=∑ yF ( ) 0lb 300 - 54 =N
lb 240=N
• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide.
( ) lb 48lb 24025.0 === msm FNF µ
The block will slide down the plane.
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Sample Problem
8 - 20
• If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force.
( )lb 24020.0=== NFF kkactual µ
lb 48=actualF
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Sample Problem
8 - 21
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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Sample ProblemSOLUTION:• 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.
BBsB
AAsANNFNNF
25.025.0
====
µµ
• Apply conditions for static equilibrium to find minimum x.:0=∑ xF 0=− AB NN AB NN =
:0=∑ yF
WNWNN
WFF
A
BA
BA
==−+
=−+
5.0025.025.0
0
WNN BA 2==
:0=∑ BM ( ) ( ) ( )( ) ( )
( ) ( ) ( ) 05.1275.02605.125.036
0in.5.1in.3in.6
=−−−=−−−
=−−−
xWWWxWNN
xWFN
AA
AA
in.12=x
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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.
Design: Example 1
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Free-Body diagram of the tie rod CB
Solution
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Solution
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Each member of the truss of Fig. P6-148 has a circular cross section and is made of structural steel. Iffailure is by yielding, and the factor of safety is 2.5, determine the minimum permissible diameter ofmember EJ.
Design: Example 2
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Design: Example 2
