Static Equilibrium - UPRMacademic.uprm.edu/pcaceres/Courses/INME4011/MD-2.pdf · Static...
Transcript of Static Equilibrium - UPRMacademic.uprm.edu/pcaceres/Courses/INME4011/MD-2.pdf · Static...
Static EquilibriumStatic equilibrium requires a balance of forces and a balance of moments.
0 0 0
0 0 0
=Σ=Σ=Σ
=Σ=Σ=Σ
zyx
zyx
MMM
FFF
Example 1:
A painter stands on a ladder that leans against the wall of a house at an angle of 20o. Assume that the painter is a midheight of the ladder. Calculate the minimum coefficient of friction for static equilibrium.
0)( 0
21
1221
=++−=Σ==−=Σ
PmmPFPPPPF
personladdery
x
μμμ
μμ
μ +
+=
+
+=
1)(
1
)(21
gmmP
gmmP plpl
0)20(2
)()20()20( 21 =+−− SinLgmmLCosPLSinP pl
1763.0=μ
Static equilibrium: Sum of vertical and horizontal forces must be zero.
Taking the moment about point O and setting it equals to zero:
Load Analysis – 3-D
Newton First Law: A body at rest tend to remain at rest and a body in motion at constant velocity will tend to maintain that velocity unless acted upon by an external force.
Newton Second Law: The time rate of change of momentum of a body is equal to the magnitude of applied force and acts in the direction of the force.
Newton Third Law: when two particles interact, a pair of equal and opposite reaction forces will exist at their contact point. This force pair will have the same magnitude and act along the same direction line but have opposite sense.
Newton’s second law can be written for a rigid body in two forms, one for linear forces and one for moments or torques
GG HMmaF&=Σ
=Σ F=Force; m=mass, a= acceleration
MG=moment about the center of gravity
GH& =time rate of change of the moment or the angular momentum about the CG
zzyyxx maFmaFmaF =Σ=Σ=Σ
kIjIiIH zzyyxxGˆˆˆ ωωω ++=
yxyxzzz
xzxzyyy
zyzyxxx
IIIM
IIIM
IIIM
ωωα
ωωα
ωωα
)(
)(
)(
−−=Σ
−−=Σ
−−=Σ
If the x, y and z axes are chosen to coincide with the principal axes of inertia of the body. Where Ix, Iy and Iz are the principal centroidal mass moments of inertia (second moments of mass). In 3-D
Euler’s Equations
For 2-D:
yyxx maFmaF =Σ=Σ zzz IM α=ΣWhere αz is the angular acceleration
Moment of a Force about an axis
3-D Equilibrium Example:Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The radius at A = 2.5” and at C = 2”. The axle rotates at constant speed. Find T and the reaction forces at B, D. Assume that bearing at D exerts no axial thrust and neglect weights of sheaves and axle.
8in
6in6in
x
y
A B C D33.75lb 33.75lb67.5lb
33.75lb
33.75lb
202.5lb-in
x
z
A B C D70lb 28lb42lb
28lb
42lb
336lb-in
DefinitionsCentroid of Area (Center of Gravity of an area): Point that defines the geometric center of the area
( )yx,
A
Axx
A
Ayy AA
∫∫==
δδ
Q axis-y therespect towith
Areaan ofMoment First
Q axis- x therespect towith
Areaan ofMoment First
y
x
AxAx
AyAy
A
A
⋅===
⋅===
∫
∫
δ
δ
IbVQ
=τ
Example 2:
Calculate the center of gravity of the rectangle:
(a)Without a hole
(b)With a hole of dimensions cand d
a) Without a hole
22
22
1
1
aba
baa
xA
Axx
A
Axx
bba
bab
yA
Ayy
A
Ayy
n
iiA
n
iiA
=⋅
⋅⋅⎟⎠⎞
⎜⎝⎛
===
=⋅
⋅⋅⎟⎠⎞
⎜⎝⎛
===
∑∫
∑∫
δ
δ
b) With a hole
dcba
dccebaa
xA
Axx
A
Axx
dcba
dcdfbab
yA
Ayy
A
Ayy
n
iiA
n
iiA
⋅−⋅
⋅⋅−−⋅⋅⎟⎠⎞
⎜⎝⎛
===
⋅−⋅
⋅⋅+−⋅⋅⎟⎠⎞
⎜⎝⎛
===
∑∫
∑∫
)2
(2
)2
(2
1
1
δ
δ
Second Moment or Moment of Inertia of an Area
∫∫ ==A
yA
x AxIAyI δδ 22
∫ +==A
yxO IIAJ δρ 2
Rectangular moments of inertia
Polar moments of inertia
Radius of Gyration
2222
2
2
yxOOO
yy
xx
rrrArJ
ArI
ArI
+==
=
=
Example 3:Find the moment of inertia of the circular area about the x and y axes, the polar moment of inertia and the radius of gyration about the x and y axes. φφ CosrxSinry ⋅=⋅= yCosrArea δφ ⋅⋅⋅= 2
δφφδ ⋅⋅= Cosry
2
4
42
44
2
2
4224
rIIJrI
rCosSinrI
yxzy
x
⋅=+=
⋅=
⋅=⋅⋅⋅⋅= ∫−
ππ
πδφφφπ
π
Parallel-Axis Theorem
2' yxx dAII ⋅+=
Example 4:
( )4
'
224
2'
4
2516
r4 4
4
rπ.I
rπ rπdAII
rI
x
yxx
x
⋅⋅=
⋅⋅⋅+⋅
=⋅+=
⋅=π
IMy
−=σ
Mass Moment of Inertia
It is the product of the element’s mass and the square of the element’s distance from the axis.
( )( )( )∫∫∫
⋅+=
⋅+=
⋅+=
amz
amy
amx
myxI
mzxI
mzyI
δ
δ
δ
22
22
22
Classification with respect to the method of application:
(a)Normal tensile
(b)Normal compressive
(c)Shear
(d)Bending
(e)Torsion
(f) Combined
Load Classification and Sign Convention
The sign convention that will be used here is as follows:
deflectionx
=> υδυδ 02
2
Distributed loads, Shear Force and
Bending Moment in Beams
4
4
3
3
2
2
2
2
xEI-qq
xV
xEIVV
xM
xEIM
EIM
x
δυδ
δδ
δυδ
δδ
δυδ
δυδ
=⇒−=
=⇒=
=⇒=
Successive Integration Method
1
2
1
2600)(
600)(
)()(
CxxV
CxxxV
xxqxV
qxV
+−=
+−=
−=
−=
∫∫
δ
δδδ
2/3m
=650N1050N=
10 << x
NVthenmxFor
xxV
CxFor
CxxV
xFor
350)1(_____1__
6502
600)(
650__0__2
600)(
10__
21
1
2
==
+−=
==
+−=
<<
NVthenmxFor
xxV
CVxFor
CxxV
xFor
1050)2(_____2__
1502
600)(
150__150500350)1(__1__2
600)(
21__
22
2
2
==
+−=
=−=−==
+−=
<<
550)1(____1__0__0)0(____0__
650100)(
6502
600)()(
6502
600)(
10__
3
33
2
2
=====
++−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−==
+−=
<<
∫∫
MthenxForCMthenxFor
CxxxM
xxxxVxM
xxV
xFor
δδ
0)2(____2__500__550)1(____1__
150100)(
1502
600)()(
1502
600)(
21__
4
43
2
2
=====
++−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−==
+−=
<<
∫∫
MthenxForCMthenxFor
CxxxM
xxxxVxM
xxV
xFor
δδ
MMax=550N-m
Normal Stress and Strain:Where F: force, normal to the cross-sectional area,
A0: original cross-sectional area 0AF
=σ
1Pa = 1 N.m-2; 1MPa = 106Pa; 1GPa=109Pa
Normal strain or tensile strain or Axial strain
00
0
ll
lll Δ=
−=ε
Hooke’s LawWhen strains are small, most of materials are linear elastic.
σ
ε
E
Normal: σ = Ε ε
Young’s modulus
EAlFl
o
o
⋅⋅
=Δ
Springs: the spring rate
o
o
lEA
lFk ⋅
=Δ
=
Torsion Loading resulting from the twist of a shaft.
lr
zrZ
θδδθγθ
⋅≅⋅=,
lrG
zrGG zz
θδδθγτ θθ
⋅⋅≅⋅⋅=⋅= ,,
G = Shear Modulus of Elasticity
Shear strain
Twist Moment or Torque
∫∫ ⋅⋅
=⋅⋅=A zA z Ar
lGArT δθδτ θθ
2,,
∫ ⋅=A
ArJ δ2Area Polar Moment of Inertia
JGlT
lJGT
⋅⋅
=⋅⋅
= θθ or
JrT
z⋅
≅,θτThus:JrT o
Max⋅
≅τAngular spring rate:
lGJTka⋅
==θ
We can obtain the normal and shear stresses from flexure and shear formulas
Where:σ is the normal stress acting on the cross section, M is the bending moment, y is the distance from the neutral axis and I is the moment of inertia of the cross sectional area with respect to
the neutral axis.τ is the shear stress at any point in the cross section, V is the shear force,Q is the first moment of the cross sectional area outside of the point in the cross section where the stress is being found, and b is the width of the cross section.
IMy
−=σ
Maximum Stresses in Beams
IbVQ
=τ
The normal stresses obtained from the flexure formula have their maximum values at the farthest distance from the neutral axis. The normal stresses are calculated at the cross section of maximum bending moment.
The shear stress obtained from the shear formula usually have their highest value at the neutral axis. The shear stresses are calculated at the cross section of maximum shear force. In most circumstances, these are the only stresses that are needed for design purposes. However to obtain a more complete picture of the stresses, we will need to determine the principal stresses and maximum shear stresses at various points in the beam.
Consider a simple rectangular beam below, and a cross section to the left of the load. Points A and E are at the top and bottom of the beam. Point C is in the midheight of the beam and points B and Dare in between.If Hooke’s law applies (linear elasticity), the normal and shear stresses at each of these five points can be readily calculated from the flexure and shear formulas.All the elements of vertical and horizontal faces, are in plane stress, because there is no stresses acting perpendicular to the plane of the figure.
Beams of Rectangular Cross Section
Stresses in a beam of rectangular cross section:
(a) simple beam with points A, B, C, D, and E on the side of the beam;
(b) normal and shear stresses acting on stress elements at points A, B, C, D, and E;
(c) principal stress; and (d) maximum shear stresses.
Points A and E elements are in uniaxialcompressive and tensile stresses respectively. Point C (neutral axis) element is in pure shear. Points B and Delements have both normal and shear stresses.
We may use either the transformation equations of plane stress or the Mohr’s circle to find the stresses at each point along the height of the beam or to describe how the principal stresses changes as we go from to top to the bottom of the beam.By investigating the stresses at many cross sections of the beam, we can determine how the principal stresses vary throughout the beam. Stress trajectory: Gives the directions of the principal stresses.Stress Contours: Curves connecting points of equal principal stress
Principal-stress trajectories for beams of rectangular cross section: (a) cantilever beam, and (b) simple beam. (Solid lines represent tensile principal stresses and dashed lines
represent compressive principal stresses.)
Example 8-3: A simple beam AB with a span length L = 6ft supports a concentrated load P = 10800lb acting a distance c = 2ft from the right-hand support (see figure below). The beam is made of steel and has a rectangular cross section (width b=2in and height h = 6in).Investigate the principal stresses and maximum shear stresses at cross section mnlocated at a distance x = 9in from the end A of the beam. (Consider only the in-plane stresses) Point y (in)
A 3B 2C 1D 0E -1F -2G -3
The reaction of the beam at support A is RA = P/3 = 3600lb, and therefore the bending moment and shear force at the section mn are
M = RA x = (3600lb)(9in) = 32400lb-inV = RA = 3600lb
Normal stress on cross section mn
Where y has units in inches and σx has units in psi. The stresses calculated are positive when in tension. Note that a positive value of y(upper half of the beam) gives a negative stress, as expected.
( )( )( )( )
yinin
yinlbbh
MyI
MyX 900
623240012
1233 −=
−−=−=−=σ
Solution
Shear stresses on cross section mnThe shear stresses are given by the shear formula
in which the first moment Q for a rectangular cross section is
IbVQ
=τ
( )( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟
⎠⎞
⎜⎝⎛==
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −+⎟
⎠⎞
⎜⎝⎛ −=
22
32
2
3
22
46
4212
becomes formulashear the thus,422
22
yhbh
Vyhbbbh
VIb
VQτ
yhbyhyyhbQ
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⋅⋅== ∫∫ 2
22
42 yh bδyby AyQ
h/
yA
x δ
AV
Max 23
=τ
The shear stresses τxy acting on the x face of the stress element are positive upwards, whereas the actual shear stresses τ act downward. Therefore
Substituting the numerical values into this equation gives
In which y has units of inches and τxy has units of psi
⎟⎟⎠
⎞⎜⎜⎝
⎛−−== 2
2
3 46 yhbh
VIb
VQτ
( )( )( )
( ) ( )222
3 9504
662
36006 yyinininlbτ XY −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
Note: The maximum shear stress that occurs at the neutral axis in a rectangular section can be simplified by using the following equation:
AV
Max 23
=τ
Point y (in) σx (psi) τxy (psi)A 3 -2700 0B 2 -1800 -250C 1 -900 -400D 0 0 -450E -1 900 -400F -2 1800 -250G -3 2700 0
The normal stresses vary linearly from a compressive stress of -2700psi at the top of the beam (point A) to a tensile stress of 2700psi at the bottom of the beam (point G). The shear stresses have a parabolic distribution with a maximum stress at the neutral axis (point D).
Calculation of stresses on cross section mnWe divide the height of the beam into six equal intervals and label the corresponding points from A to G.
Principal Stresses and Maximum Shear StressesThe principal stresses and maximum shear stresses at each of the seven points A through G may be determined from the following equations
( )22
2,1 22 xyyxyx τ
σσσσσ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
( )22
2 xyyx
MAX τσσ
τ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
Point y (in) σx (psi) τxy (psi) σ1 (psi) σ2 (psi) τmax (psi)-2700 1350
934602450602934
1350
-1834-1052-450-152-340
034152450
105218342700
A 3 -2700 0B 2 -1800 -250C 1 -900 -400D 0 0 -450E -1 900 -400F -2 1800 -250G -3 2700 0
( )22
2,1 22 xyxx τ
σσσ +⎟
⎠⎞
⎜⎝⎛±⎟
⎠⎞
⎜⎝⎛=( )2
2
2 xyx
MAX τσ
τ +⎟⎠⎞
⎜⎝⎛=
Since there is no normal stress in the y direction, this equation simplifies to:
Thus, by substituting the values of σx and τxy, we can calculate the principal stresses σ1 and σ2 and the maximum shear stress τmax.
The largest tensile stress anywhere in the beam is the normal stress at the bottom of the beam at the cross section of maximum bending moment (σtens)max = 14400psi).The largest shear stress occurs to the right of the load P (V = RB = 7200lb). Therefore, the largest value that occurs at the neutral axis is (τxy)max = 900psiThe largest shear stress anywhere in the beam occurs at 45o planes at either the top or bottom (τxy)max = 14400 / 2 = 7200psi