View
47
Download
0
Category
Preview:
Citation preview
11/16/2010
1
Simply Supported Beam witha Centered Load
Relationships:R1 = R2 = F / 2 VAB = R1VBC = -R2
Nomenclature:F = Loading force R = Reaction force V = Shear force M = Moment l = beam length x = location
2xFMAB
=
2)xl(FMBC =
Simply Supported Beam with an Intermediate Load
Relationships:R1 = F . b / lR2 = F . a / l VAB = R1VBC = -R2
lxbFMAB
=
l)xl(aFMBC =
11/16/2010
2
Simply Supported Beam with a Moment Load
Relationships:R1 = -R2 = MB / lV = MB / l
lxMM BAB
=
l)lx(MM BBC =
Simply Supported Beam with an Overhang Load
Relationships:R1 = -F * a / lR2 = F / l * (l + a)VAB = -F * a / lVBC = F
lxaFMAB
=
)alx(FMBC =
11/16/2010
3
Simply Supported Beam with an Uniform Load
Relationships:R1 = R2 = w . l / 2
xw2
lwV =
)xl(2
xwM =
Simply Supported Beam with Twin Loads
Relationships:R1 = R2 = F VAB = FVBC = 0VCD = -FMAB = F . xMBC = F . aMCD = F (l -x)
11/16/2010
4
Cantilever Beam with End Load
Relationships:R1 = V = F M1 = -F . lM = F (x - l)
Cantilever Beam with Intermediate Load
Relationships:R1 = V = F M1 = -F . aMAB = F (x - a)MBC = 0
11/16/2010
5
Cantilever Beam with Uniform Load
Relationships:R1 = w. l M1 = -(w. l 2 ) / 2V = w (l - x)
2)xl(2wM =
Cantilever Beam with Uniform Load
Relationships:R1 = w. l M1 = -(w. l 2 ) / 2V = w (l - x)
2)xl(2wM =
HW
45
P
P=2 angka terakhir (N)w = 5 N/ml= 10 m
A
B
11/16/2010
6
Cantilever Beam with a Moment Load
Relationships:R1 = 0 M1 = MBM = MBV = 0
Fundamental of
Mechanical DesignPart II Mechanics of Material
y
x
z
yzyx
zx
zy
xy
11/16/2010
7
TopicsPart I. Structural Statics1. Introduction2. Force System3. Moment System4. Centroid and Center of Gravity5. Moment of Inertia6. Equilibrium7. Structural Analysis8. Internal ForcesPart II. Mechanics of Material9. Bending Stress10. Transverse Shear Stress11. Torsion12. Stress Analysis13. Deflection14. Statically Indeterminate Beams15. Column
Bending Stress>>
11/16/2010
8
Bending StressStraight member x
xslim0x
=
r
yr
r)yr(lim0x
=
+
=
= rx+= )yr(s
r/cr/y
max
=
y sx
r
c
maxc
y=
maxc
y=
Bending Stress
Force equilibrium
====A
maxmax
AA AydA
cdA
c
ydAdP0
Moment
0ydAso,0c A
max=
===A
2max
A AdAy
cdAydPyM =
A
2dAyIif
IMc
max =
Stress at any intermediate distance y is: IMy
=
maxc
y=
11/16/2010
9
Bending StressExample:Aluminum beam with rectangular cross section (4 x 4 cm) experience force F = 100 N as shown in figure. If length of beam l = 2 m, find the maximum bending stress.
Solution:
MPa69.4m/N1069.4104
121
02,050I
Mc 2684
max ==
==
MB=50 Nm2cm
4cm
D
MD= ?
Bending StressExample:Stress distribution
11/16/2010
10
Physical Properties of Materials
Source: Machine Elements, BJ, Hamrock
Material Density Modulus ofElasticity
(GPa)Yield
Strength(MPa)
Ultimatestrength
(MPa)Ductility,%EL in
2 in
Poisson'sratio
Iron 7870 207 130 260 45 0.29
Low Carbonsteel (AISI 1020)
7860 207 295 395 37 0.30
Stainless steelsFerritic, type 446
7500 200 345 552 20 0.30
Alumunium(>99.5%)
2710 69 17 55 25 0.33
Copper (99.9%) 8940 110 69 220 45 0.35
Bending StressCurved member
=r
d)rR(
=
EdrR
r
o
p
d
r
e
ri
y
k
l
m
n
roR r
CentroidalNeutral axis
==r
d)rR(EEM
11/16/2010
11
Bending StressNeutral axisif Fn = 0
=A
0dA
E, R , , d = constan =
A0dA
r
d)rR(E
=
A0dA
r
)rR(Ed
0dAr
dAREdAA
=
=
Ar
dAAR
Bending StressMoment Mz = 0
==A A
2dA
r
d)rR(E)rR(dAM
=
=
A A
22 dA
r
d)rR(rR
rdA)rR(EdM
dAr
rRrRrRrR
rMA
22
+
=
=
AAAA
2 dArdARdARr
dARrR
rM
11/16/2010
12
Bending StressCont.
=
AAAAdArdARdA
r
dARRrR
rM
)RAAr(rR
rM
=
So,)Rr(Ar)rR(M
=
Bending Stress>>
)Rr(Ar)rR(M
=
If y positive from neutral axis and eRr =
)yR(AeMy
=
===
i
or
rA r
rln
h
rdrb
h.b
rdAAR
0
i
For rectangular cross section:
b
h
For circular cross section:
c 2crrR
22+
=
11/16/2010
13
Bending Stress
The curved beam with centroidal radius 25 cm experiencesmoment M=100 Nm. The beam has rectangular cross section asshown in figure. Determine bending stress for y = 1 cm.
Example:
4cm
6cm
MPa454.110)188.24(12.024
1011008
2
1y =
=
=
cm88.242412.06
2228ln
6R ==
=
Bending Stress
The curved beam with centroidal radius 25 cm experience moment M=100 Nm. The beam has rectangular cross section as shown in figure. Determine bending stress for :1. Top layer2. Centroidal layer3. Bottom layerPredict the stress distribution!
Problem:
4cm
6cm
11/16/2010
14
Bending StressSolution
4cm
6cm
MPa869.310))12.3(88.24(12.024
1012.31008
2
TOP =
=
cm88.242412.06
2228ln
6R ==
=
Bending StressCont.
MPa167.010))12.0(88.24(12.024
1012.01008
2
centroidal =
=
MPa545.410))88.288.24(12.024
1088.21008
2
bottom =
=
11/16/2010
15
Transverse Shear Stress
>>
Transverse Shear StressConcept:
How transverse shear is developed.
11/16/2010
16
Transverse Shear StressConcept:
Deformation due to Transverse Shear
Cantilevered bar made of highly deformable material and marked with horizontal and vertical grid lines to show deformation due to transverse shear. (a) Undeformed; (b) deformed.
Transverse Shear StressConcept:
Three-dimensional and profile views of moments and stresses associated with shaded top segment of element that has been sectioned at y about neutral axis. (a) Three-dimensional view; (b) profile view.
11/16/2010
17
Transverse Shear StressConcept:
ItyA
dxdM
tdxdF
==
yAI
dMdAyI
dMdFarea
==
VdxdM
=
ItyA.V
=
and
so
Transverse Shear StressFormula:
ItyVA
=
V is always the absolute value of the total shear (from the V diagram) on thebeam at the section being investigated (this is usually the point where the shearis greatest).A is always the area on the face of the structural section on either side of theplane at which shear is being investigated. For most beam applications this is thearea above (or below) a horizontal plane through the neutral axis of a beam.y is always the distance from the neutral axis to the centroid of the area A.I is always the moment of inertia of the entire cross sectional area of the beamwith respect to the neutral axis.
t is always the total width of the beam at the plane where the shear is beinginvestigated (this is usually the width at the neutral axis).
ItVQ
= If QyA =or
11/16/2010
18
Transverse Shear StressExample:Aluminum beam with rectangularcross section (4 x 4 cm)experience force F = 100 N asshown in figure. If length of beaml = 2 m, find the maximumtransverse shear stress atcentroidal line.
Solution:
24
84
6
max /10....04.0104
121
)10124)(50(mN
ItVQ
=
==
50 N
Transverse Shear StressProblem:Given:The beam with loading condition shown belowDetermine:The horizontal shearing stress at 2 cm increments from top to bottom of the beam. Plot these stresses - use maximum vertical shear on the beam
3kN 5kN 2kN
80cm 60cm 60cm 80cm
8 cm
8 cm
4 cm
11/16/2010
19
Transverse Shear StressSo
lu
tio
n
Transverse Shear Stress1st increment
2
83m/kN96.563
04.010164121
07.0)02.004.0(5.5It
yVA=
==
2nd increment
2
83m/kN79.966
04.010164121
06.0)04.004.0(5.5It
yVA=
==
4
8
8 7
11/16/2010
20
Transverse Shear Stress3rd increment
2
83m/kN49.1208
04.010164121
05.0)06.004.0(5.5It
yVA=
==
4th increment
2
83m/kN05.1289
04.010164121
04.0)08.004.0(5.5It
yVA=
==
Transverse Shear Stress
Stress distribution
8 cm
8 cm
4 cm
kPa96.563=kPa79.966=
kPa49.1208=kPa05.1289=
11/16/2010
21
Transverse Shear StressMaximum Shear Stress
Torsion>>
11/16/2010
22
TorsionConcept:
TdAcA
max =
TdAc A
2max=
oc
max
dA
TorsionCont.
pmax I
Tc=
TdAc A
2max=
pA
2 IdAif =
pIT
=For general relationship:
= area polar moment of inertia
So,
11/16/2010
23
TorsionPower TransferPower is the rate of doing work orhp = Force x Velocity
=
phT
Where : F = force, Nu = velocity, m/sT = torque, Nm = 2pif = rotational speed, rad/sf = frequency, hertz
hp = F . u
hp = T .
or
TorsionExample:A shaft with diameter 10 mm carries a torque of 30 Nm. Find the maximum shear stress due to torsion!
1044
0
4
0
32 1082.92
)005.0(24
22 ====== pipi
pipi cddAIcc
Ap
MPaITc
p
8.1521082.9005.030
10max =
==
Solution:
11/16/2010
24
TorsionAngle of twist
= dcdxmax
Hookes law:
Gmax
max
=GI
Tcp
max =
then: = dcdxGI
Tc
p
d
maxdx
cA
BSee linear deflection (segment de) :
e
TorsionAngle of twist
d
maxdx
cA
B
GITdxd
p=
==B
A px
xB
A GIdxTd
11/16/2010
25
TorsionExample:A 50 mm-long, circular, hollow shaft made of carbon steel must carry a torque of 5000 Nm at maximum shear stress of 70 MPa. The inside diameter is 0.5 of the outside diameter 72.94 mm. If the shear modulus rigidity (G) for carbon steel is 80 GPa, find the angle of twist !
GITLdx
GIT
GIdxT
p
L
0p
B
A px
x===
rad396 10799.1108010737.105.05000
=
=
4612412444
2 10737.12
10.235.182
10.47.3622
mrrdAI io
Ap
==== pipipipi
o103.0296.5700179.0 ==
Torsion
Problem:1. A shaft is used for electric motor with power 8 kilowatts at
frequency 40 Hz. Find the maximum shear stress if thediameter of shaft is 16 mm!
2. A 50 mm-long, circular, hollow shaft made of carbon steelmust carry a torque of 5000 Nm at maximum shear stress of70 MPa. The inside diameter is 0.5 of the outside diameter.Find the minimum outside diameter !
11/16/2010
26
TorsionSolution (1):
Nm83.31402
8000hT p =pi
=
=
4944
A
2p m10434.62
)008.0(2cdAI =pi=pi==
MPa58.3910434.6
008.083.31ITc
9p
max =
==
TorsionSolution (2):
pi=
pi
pi=pi==
4
o
i4
o4
i4
or
r
3
A
2p r
r12r
2r
2rd2dAI
o
i
5.0c
r
r
r
dd i
o
i
0
i===
366
max
p m1043.711070
5000Tc
I=
=
=
2d
rc oo ==
11/16/2010
27
TorsionSolution (2) cont.:
363o m1051.48r =
364
o
i3
o m1043.71r
r12r
=
pi
So, the minimum outside diameter is 72.94 mm
m03657.0ro =
PR:
Sebuah motor listrik dengan daya 100 kW, putaran 1500 rpm, menggerakan
Sebuah pompa centrifugal dihubungan dengan sebuah poros pejal (solid shaft). Bahan poros adalah stell st 40 dengan modulus elastisitas geser (G) =70 GPA, dan tegangan geser adalah 70 MPA. Panjang poros 40 cm. Tentukan
1. Diameter Poros
2. Sudut puntir poros.
11/16/2010
28
Stress Analysis
.
y
x
z
yzyx
zx
zy
xy
Stress AnalysisStress Element
A B
C
x
y
x
xy
xy
xy
yx
y
x
z
yzyx
zx
zy
xyx
y
yx
xyx
y
Plane Stress
11/16/2010
29
Plane Stress
+
++
= 2sin2cos22 xy
yxyx'x
= 0F 'x+++= cossindAsincosdAsinsindAcoscosdAdA xyxyyx'x
( ) ( ) +++= 2sin2
2cos12
2cos1xyyx'x
++= cossin2sincos xy2y2x'x
Plane Stress Transformation
A B
C
x dA cos
y dA sin
x dA
xy dA cos
xydA
xy dA sin
yx
Plane Stress
yx
xy1
22tan
=
= 0F 'y
Similary, the shear stress in oblique plane can be expressed as:
02cos22sin22d
dxy
yx'x=+
=
Plane of maximum and minimum normal stress:
+
= 2cos2sin2 xy
yx'y'x
So,
11/16/2010
30
Plane Stress
0'y'x =
In this condition,
Principle Normal StressMaximum and minimum normal stress:
( ) xy22
yxyx21min
max'x 22
, +
+==
Plane Strain22
21 221
2,
+
+
=yx
xyyx
=
yx
xy
1tan2
Maximum shear strain,
Principle Normal Strain
22
max 221
+
= yxxy
Orientation of principle axis,
11/16/2010
31
Plane Stress
xy
yx2 2
2tan
=
2'
yx +=
In this condition,
0d
d'y'x
=
xy2
2yx
minmax 2
+
=
Plane of maximum and mimimum shear stress:
Principle Shear Stress
Maximum and minimum shear stress:
Plane Stress
++= 2cos222121
= 2sin2yx
If the principle stress are known, the normal and shear stress at any oblique plane at angle can be determined from following equation:
Stress transformation
11/16/2010
32
Three-Dimensional Stresses
0)2()()(
222
22223
=+
+++++
xyzzxyyzxzxyzxyzyx
zxyzxyzyzxyxzyx
231
3/1
=
221
2/1
=
232
3/2
=
Principle shear stresses
Principle stresses can be determined by finding the three roots to the cubic equation:
Plane Strain
2cos222121 +
+=
2sin)( 21 =
If the principle strains are known, the strains at any oblique plane at angle can be determined from following equation:
11/16/2010
33
Mohrs Circle
xy
x
2
Plane Stress
2
2yx
c
+=
y
y
1
2
1
2
Deflection>>
11/16/2010
34
DeflectionFrom bending stress
r
y=
r
Ey=
m
dxx
dr
ds
dAr
EyP =
= y.ydArEM
r
EIM =EIM
r
1=
= drdsdsd
r
1 =
ds dx and tan = dy/dx
Mdx
ydEI 22
=
m1
x
y
Nomenclature:y = Deflection = Slope M = Moment x = location E = Modulus of elasticity I = Area moment of Inertia
The equation of the elastic curve (moment curvature relation) :Equation of the Elastic Curve
( )
( )
( ) 2100
10
2
21
CxCdxxMdxyEI
CdxxMdxdyEIEI
xMdx
ydEIEI
xx
x
++=
+=
==
Constants C1 & C2 are determined from boundary conditions
11/16/2010
35
Boundary condition:Equation of the Elastic Curve
Three cases for statically determinant beams,
Simply supported beam
0,0 == BA yy
Overhanging beam0,0 == BA yy
Cantilever beam0,0 == AAy
Equation of the Elastic CurveOverhanging beam Reactions at A and C Bending moment diagram
Curvature is zero at points where the bending moment is zero, i.e., at each end and at E.
EIxM )(1
= Beam is concave upwards where the bending
moment is positive and concave downwards where it is negative.
Maximum curvature occurs where the moment magnitude is a maximum.
An equation for the beam shape or elastic curve is required to determine maximum deflection and slope.
11/16/2010
36
A simply supported beam with length l experiences uniform load asshown in figure. Assume that the cross section is constant along thebeam and that the material is the same throughout, implying that thearea moment of inertia I and the modulus of elasticity E are constant.Find the deflection of any x.
Example 1:
q = wo N/m
lx
y
Example
2xw
2LxwM
2oo
=
)xLx2xL(EI24
wy 433o +=
0y,lx0y,0x
==
==
Moment:
2xw
2Lxw
dxydEI
2oo
2
2
=
1
3o
2o C
6xw
4Lxw
dxdyEI +=
21
4o
3o CxC
24xw
12LxwEIy ++=
Boundary condition :1. x = 0, y = 02. x = L, y = 0From boundary condition 1: C2 = 0From boundary condition 2: C1 = - wo L3 /24
ExampleSolution
11/16/2010
37
Example
ft 4ft15kips50psi1029in723 64
===
==
aLPEI
For portion AB of the overhanging beam,(a) derive the equation for the elastic curve,(b) determine the maximum deflection, (c) evaluate ymax.
ExampleSOLUTION:
Develop an expression for M(x) and derive differential equation for elastic curve.- Reactions:
+==LaPR
LPaR BA 1
- From the free-body diagram for section AD,
( )LxxLaPM
11/16/2010
38
Example
PaLCLCLLaPyLx
Cyx
61
610:0,at
0:0,0at
113
2
=+===
===
Integrate differential equation twice and apply boundary conditions to obtain elastic curve.
213
12
6121
CxCxLaPyEI
CxLaP
dxdyEI
++=
+=
xLaP
dxydEI =2
2
Example
=
32
6 Lx
Lx
EIPaLy
PaLxxLaPyEI
Lx
EIPaL
dxdyPaLx
LaP
dxdyEI
61
61
3166
121
3
22
+=
=+=
Substituting,
11/16/2010
39
Example Locate point of zero slope or
point of maximum deflection.
0dxdy
=
=
2
Lx31
EI6PaL
dxdy
=
2m
Lx31
EI6PaL0
L577.03
Lxm ==
Example
=
32
6 Lx
Lx
EIPaLy
Evaluate corresponding maximum deflection.
( )[ ]32max 577.0577.06 = EIPaLyEI
PaLy6
0642.02
max =
( )( )( )( )( )462
maxin723psi10296in180in48kips500642.0
=y
in238.0max =y
11/16/2010
40
Method of superposition
q = wo N/m
L/2x
yF
L/2
Principle of Superposition:Deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual loadings
A simply supported beam with length L experiences uniformload q and concentrated load F as shown in figure. Assume thatthe cross section is constant along the beam and that thematerial is the same throughout, implying that the area momentof inertia I and the modulus of elasticity E are constant.Determine the equation of the deflection curve!
Example:
Method of superposition
q = wo N/m
L/2x
yF
L/2
11/16/2010
41
If there are multiple loads on beam, you can compute deflections foreach load separately and add the results.Procedure is facilitated by tables of solutions for common types ofloadings and supports.
Solution:
Method of superposition
L/2
F
L/2
wo
L/2 L/2
wo
L/2
F
L/2
= +
y(x) = yI (x) yII (x)+
Moment-Area Theorems Geometric properties of the elastic curve
can be used to determine deflection and slope.
=
=
==
D
C
D
C
D
C
x
x
CD
x
x
dxEIM
dxEIMd
EIM
dxyd
dxd
2
2
Consider a beam subjected to arbitrary loading,
First Moment-Area Theorem:
area under (M/EI) diagram between C and D.
=CD
11/16/2010
42
Moment-Area Theorems
Second Moment-Area Theorem:The tangential deviation of C with respect to D is equal to the first moment with respect to a vertical axis through C of the area under the (M/EI) diagram between Cand D.
Tangents to the elastic curve at P and Pintercept a segment of length dt on the vertical through C.
=
==
D
C
x
x
DC dxEIM
xt
dxEIM
xdxdt
1
11
= tangential deviation of C with respect to D
Application of Moment-Area Theorems
Cantilever beam - Select tangent at A as the reference.
ADD
ADD
A
ty
=
=
=
,0with
Simply supported, symmetrically loaded beam - select tangent at C as the reference.
CBB
CBB
C
ty
=
=
=
,0with
11/16/2010
43
Bending Moment Diagrams by Parts
Determination of the change of slope and the tangential deviation is simplified if the effect of each load is evaluated separately.
Construct a separate (M/EI) diagram for each load.
- The change of slope, D/C, is obtained by adding the areas under the diagrams.
- The tangential deviation, tD/C is obtained by adding the first moments of the areas with respect to a vertical axis through D.
Bending moment diagram constructed from individual loads is said to be drawn by parts.
Moment-Area Theorems
For the prismatic beam shown below, determine the slope and deflection at E.
EXAMPLE:
11/16/2010
44
Moment-Area TheoremsSOLUTION:
Determine the reactions at supports.
waRR DB ==
Construct shear, bending moment and (M/EI) diagrams.
( )EI6
waa
EI2wa
31A
EI4Lwa
2L
EI2waA
32
2
22
1
=
=
=
=
Moment-Area Theorems Slope at E:
EI6wa
EI4LwaAA
32
21
CECECE
=+=
=+=
( )a2L3EI12
wa2
E +=
=
+
+=
=
EI16Lwa
EI8wa
EI16Lwa
EI4Lwa
4LA
4a3A
4L
aA
tty
224223
121
CDCEE
( )aL2EI8
way3
E +=
Deflection at E:
11/16/2010
45
Application of Moment-Area Theorems to Beams With Unsymmetrical Loadings
Define reference tangent at support A. Evaluate A by determining the tangential deviation at Bwith respect to A.
Lt AB
A =
The slope at other points is found with respect to reference tangent.
ADAD +=
The deflection at D is found from the tangential deviation at D.
ABADD
AB
tLx
tEFEDy
tLxEF
LHB
x
EF
==
==
Maximum Deflection Maximum deflection occurs at point
K where the tangent is horizontal.
AAK
AKAK
ABA L
t
=
+==
=
0
Point K may be determined by measuring an area under the (M/EI) diagram equal to -A .
Obtain ymax by computing the first moment with respect to the vertical axis through A of the area between Aand K.
Recommended