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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
Third EditionLECTURE
101, 2, and 3
Chapters
REVIEW FOR EXAM #1
byDr. Ibrahim A. Assakkaf
SPRING 2003ENES 220 – Mechanics of Materials
Department of Civil and Environmental EngineeringUniversity of Maryland, College Park
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 1ENES 220 ©Assakkaf
Review: Statics
Equations of Equilibrium– Rigid Body
F1
F2F2
x
yz
i
jk
2
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 2ENES 220 ©Assakkaf
Review: Statics
Equations of Equilibrium– For a rigid body to be in equilibrium, both
the resultant force R and a resultant moments (couples) C must vanish.
– These two conditions can be expressed mathematically in vector form as
0kji
0kjirr
rr
∑∑ ∑∑∑ ∑
=++=
=++=
zyx
zyx
MMMC
FFFR
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 3ENES 220 ©Assakkaf
Review: Statics
Equations of Equilibrium– The two conditions can also be expressed
in scalar form as
∑ ∑ ∑∑ ∑ ∑
===
===
0 0 0
0 0 0
zyx
zyx
MMM
FFF
3
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 4ENES 220 ©Assakkaf
Review: Statics
Equilibrium in Two Dimensions– The term “two dimensional” is used to
describe problems in which the forces under consideration are contained in a plane (say the xy-plane)
x
y
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 5ENES 220 ©Assakkaf
Review: Statics
Equilibrium in Two Dimensions– For two-dimensional problems, since a
force in the xy-plane has no z-component and produces no moments about the x- or y-axes, hence
0k
0jirr
rr
==
=+=
∑∑ ∑
z
yx
MC
FFR
4
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 6ENES 220 ©Assakkaf
Review: Statics
Equilibrium in Two Dimensions– In scalar form, these conditions can be
expressed as
∑∑ ∑
=
==
0
0 0
A
yx
M
FF
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 7ENES 220 ©Assakkaf
Review: Statics
Cartesian Vector Representation of A Force
F
x
y
z
Fx
Fy
Fzkji zyx FFFF ++=
5
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 8ENES 220 ©Assakkaf
Review: Statics
Cartesian Vector Representation of A Force in Two Dimensions
θ
F
F cos θ
F sin θ jsinicos
ji
θθ FF
FFF yx
+=
+=r
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 9ENES 220 ©Assakkaf
Review: Vector Operations
Dot or Scalar Product– The dot or scalar product or two
intersecting vectors is defined as the product of the magnitudes of the vectors and the cosine of the angle between them.
A • B = AB cos θ
B
Aθ
6
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 10ENES 220 ©Assakkaf
Review: Vector Operations
Cross or Vector Product
A
B
C
θ
C = A × B = (AB sin θ) eC
A × B = - B × A
CBBBAAABA
zyx
zyx
rrr==×
kji
eC
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 11ENES 220 ©Assakkaf
Review: Vector Operations
Example 3– If A = -3.75i – 2.50j + 1.50k and
B = 32i + 44j + 64 kdetermine the magnitude and direction of the vector C = A × B
6444325.15.275.3
kji−−=×= BAC
rrr
7
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 12ENES 220 ©Assakkaf
Review: Vector Operations
Example 3 (cont’d)
[ ] [ ][ ]
k85j288i226 (-2.5)32-3.75(44)-
j1.5(32)-3.75(64)- - i1.5(44)-2.5(64)- 644432
5.15.275.3kji
−+−=+
=
−−=×= BACrrr
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 13ENES 220 ©Assakkaf
Review: Vector Operations
Example 3 (cont’d)( ) 376)85(288)226( 222 =−++−== CC
r
011
011
011
1.103376
0.85coscos
0.40376288coscos
9.126376226coscos
=−
==
===
=−
==
−−
−−
−−
CC
C
C
CC
zzx
yyx
xx
r
r
r
θ
θ
θ
8
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 14ENES 220 ©Assakkaf
Introduction
Objectives
Mechanics of Materials answers two questions:
Is the material strong enough?
Is the material stiff enough?
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 15ENES 220 ©Assakkaf
Introduction
Objectives– If the material is not strong enough, your
design will break.– If the material isn’t stiff enough, your
design probably won’t function the way it’s intended to.
9
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 16ENES 220 ©Assakkaf
Internal Forces for Axially Loaded Members
Analysis of Internal Forces
Assume that F1 = 2 k, F3 = 5 k, and F4 = 8 k
F1F2
F3 F4
k 1528or ;0
Then
2
31424321
=−−=
−−=⇒=+−−−+→ ∑F
FFFFFFFF
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 17ENES 220 ©Assakkaf
Internal Forces for Axially Loaded Members
Analysis of Internal Forces– What is the internal force developed on
plane a-a and b-b?a
a
b
b
2k 1k 5k 8k
2k 1kRa-a
k3=−aaR
10
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 18ENES 220 ©Assakkaf
Internal Forces for Axially Loaded Members
Analysis of Internal Forcesa
a
b
b
2k 1k 5k 8k
k8=−bbR2k 1k
Rb-b5k
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 19ENES 220 ©Assakkaf
Axial Loading: Normal Stress
Stress– Stress is the intensity of internal force.– It can also be defined as force per unit
area, or intensity of the forces distributed over a given section.
AreaForce Stress = (1)
11
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 20ENES 220 ©Assakkaf
Axial Loading: Normal Stress
Normal StressF = magnitude of the force FA = area of the cross sectional area of the
eye bar.Units of Stress
lb/in2 = psiKip/in2 = ksi = 1000 psi
1 kPa = 103 Pa = 103 N/m2
1 MPa = 106 Pa = 106 N/m2
1 GPa = 109 Pa = 109 N/m2
U.S. Customary UnitsSI System
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 21ENES 220 ©Assakkaf
Shearing Stress
Illustration of Shearing StressP
P
P
P
ss AP
AV
==avgτ VP
12
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 22ENES 220 ©AssakkafStresses on an Inclined Plane in
an Axially Loaded Member
Illustration
θ
P
PF
Original Area, AInclined Area, An
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 23ENES 220 ©Assakkaf
Design Loads, Working Stresses, and Factor of Safety (FS)
Factor of Safety– The factor of safety (FS) can be defined as
the ratio of the ultimate stress of the material to the allowable stress
stress allowablestress ultimateFS =
13
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 24ENES 220 ©Assakkaf
Displacement, Deformation, and Strain
Strain– Two general types of strain:
• Axial (normal) Strain• Shearing Strain
δL
θ
φ = γxy
δs
Normal Strain Shearing Strain
L
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 25ENES 220 ©Assakkaf
Displacement, Deformation, and Strain
Average Axial Strain
True Axial Strain
Lnδε =avg
( )dLd
Lp nn
L
δδε =∆∆
=→∆ 0
lim
δnL
14
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 26ENES 220 ©Assakkaf
Displacement, Deformation, and Strain
Average Shearing Strain
Since δs is vary small,
sin φ = tan φ = φ, therefore,
φγ tanavg =
Lsδγ =avg
θ
φ = γxy
δs
L
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 27ENES 220 ©Assakkaf
ExampleA rigid steel plate A is supported by three rods as shown. There is no strain in the rods before the load P is applied. After Pis applied, the axial strain in rod C is 900µin/in. Determine(a) The axial strain in rods B.(b) The axial strain in rods B if there is a 0.006-in
clearance in the connections between A and B before the load is applied.
Displacement, Deformation, and Strain
15
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 28ENES 220 ©Assakkaf
Displacement, Deformation, and Strain
Example (cont’d)
42"72"
B C B
AP
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 29ENES 220 ©Assakkaf
Displacement, Deformation, and Strain
Example (cont’d)42"
72"B C B
A
µδε
µδε
εδδε
1400001400.042
006.00648.0 a)
1543001543.0420648.0 a)
in 0648.0)72(10900 6
==−
==
====
=×==⇒= −
B
BB
B
BB
CCCC
CC
L
L
LL
16
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 30ENES 220 ©Assakkaf
Stress-Strain-Temperature Relationships
General Stress-Strain Diagram
Lδε =
AP
=σ
P
δ
LRupture
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 31ENES 220 ©Assakkaf
Stress-Strain-Temperature Relationships
Modulus of Elasticity, E– The initial portion of the stress-strain curve
(diagram) is a straight line. The equation for this straight line is called the modulus of elasticity or Young’s Modulus E
εσ E=σ
σ=Eε
ε
17
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 32ENES 220 ©Assakkaf
Stress-Strain-Temperature Relationships
Shear Modulus of Elasticity, G– The shear modulus is similar to the
modulus of elasticity. However it is applied to shear stress-strain.
γτ G=
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 33ENES 220 ©Assakkaf
Stress-Strain-Temperature Relationships
ττ=Gγ
γ
γτ
==G Slope
Shear Modulus of Elasticity, G
18
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 34ENES 220 ©Assakkaf
Stress-Strain-Temperature Relationships
Poisson’s Ratio– A material loaded in one direction will
undergo strains perpendicular to the direction of the load in addition to those parallel to the load. The ratio of the lateral or perpendicular strain to the longitudinal or axial strain is called Poisson’s ratio.
( )GEa
l νεε
εεν +=== 12
long
lat
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 35ENES 220 ©Assakkaf
Stress-Strain-Temperature Relationships
Thermal Strain– The thermal strain due a temperature
change of ∆T degrees is given by
TT ∆=αε
19
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 36ENES 220 ©Assakkaf
Stress-Strain-Temperature Relationships
Total Strain– The sum of the normal strain caused by
the loads and the thermal strain is called the total strain, and it is given by
TET ∆+=+= total ασεεε σ
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 37ENES 220 ©Assakkaf
Rods: Stress Concentrations
Fig. 1. Stress distribution near circular hole in flat barunder axial loading
20
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 38ENES 220 ©Assakkaf
Rods: Stress Concentrations
Fig. 2. Stress distribution near fillets in flat bar underaxial loading
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 39ENES 220 ©Assakkaf
Rods: Stress ConcentrationsHole
Discontinuities of cross section may result in high localized or concentrated stresses. ave
maxσσ
=K
Fig. 3
21
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 40ENES 220 ©Assakkaf
Rods: Stress ConcentrationsFillet Fig. 4
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 41ENES 220 ©Assakkaf
Rods: Stress Concentrations
To determine the maximum stress occurring near discontinuity in a given member subjected to a given axial load P, it is only required that the average stress σave = P/A be computed in the critical section, and the result be multiplied by the appropriate value of the stress-concentration factor K.It is to be noted that this procedure is valid as long as σmax ≤ σy
22
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 42ENES 220 ©Assakkaf
Deformations of Members under Axial Loading
Uniform Member– The deflection (deformation),δ, of the
uniform member subjected to axial loading P is given by
EAPL
=δ (1)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 43ENES 220 ©Assakkaf
Deformations of Members under Axial Loading
Multiple Loads/Sizes– The deformation of of various parts of a rod
or uniform member can be given by
∑∑==
==n
i ii
iin
ii AE
LP11
δδ
E1 E2 E3
L1 L2 L3
(2)
23
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 44ENES 220 ©Assakkaf
Deformations of Members under Axial Loading
Relative Deformation– On the other hand, since both ends of bars
AB move, the deformation of AB is measured by the difference between the displacements δA and δB of points A and B.
– That is by relative displacement of B with respect to A, or
EAPL
ABAB =−= δδδ /(3)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 45ENES 220 ©Assakkaf
Statically Indeterminate Structures
Determinacy of Beams– For a coplanar (two-dimensional) structure,
there are at most three equilibrium equations for each part, so that if there is a total of n parts and r reactions, we have
ateindetermin statically ,3edeterminat statically ,3
⇒>⇒=
nrnr
(4)
24
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 46ENES 220 ©Assakkaf
Statically Indeterminate Structures
Determinacy of Trusses– For a coplanar (two-dimensional) truss,
there are at most two equilibrium equations for each joint j, so that if there is a total of bmembers and r reactions, we have
ateindetermin statically ,2edeterminat statically ,2
⇒>+⇒=+
jrbjrb
(5)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 47ENES 220 ©Assakkaf
Statically Indeterminate Axially Loaded Members
Example 4 (cont’d)
End plate
P
Tube (A2, E2)
Rod (A2, E2)
L
25
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 48ENES 220 ©Assakkaf
Statically Indeterminate Axially Loaded Members
Example 4 (cont’d)
PP
P
FT/2
FT/2FR
Tube (A2, E2)
Rod (A1, E1)
PFF
FFFPF
TR
RTT
x
=+
=−−−=+→ ∑
022
;0
FBD:
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 49ENES 220 ©Assakkaf
Torsional Loading
IntroductionCylindrical members
Fig. 1
26
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 50ENES 220 ©Assakkaf
Stresses in Circular Shaft due to Torsion
ρ
dF = τρ dA
T T
B C
Fig. 7
Torsional Loading
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 51ENES 220 ©Assakkaf
Shearing Strain
φρ
L
Fig. 8
c
ργ
Torsional Shearing Strain
27
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 52ENES 220 ©Assakkaf
Shearing StrainFor radius ρ, the shearing strain for circular shaft is
For radius c, the shearing strain for circular shaft is
Lρφγ ρ =
Lc
cφγ =
(6)
(7)
Torsional Shearing Strain
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 53ENES 220 ©Assakkaf
Polar Moment of InertiaThe integral of equation 12 is called the polar moment of inertia (polar second moment of area).It is given the symbol J. For a solid circular shaft, the polar moment of inertia is given by
( )2
d 2d4
0
22 cAJc πρπρρρ === ∫∫ (13)
Torsional Shearing Strain
28
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 54ENES 220 ©Assakkaf
Polar Moment of InertiaThe integral of equation 12 is called the polar moment of inertia (polar second moment of area).It is given the symbol J. For a solid circular shaft, the polar moment of inertia is given by
( )2
d 2d4
0
22 cAJc πρπρρρ === ∫∫ (13)
Torsional Shearing Strain
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 55ENES 220 ©Assakkaf
Shearing Stress in Terms of Torque and Polar Moment of Inertia
JT
JTc
ρτ
τ
ρ =
=max (17a)
(18a)
τ= shearing stress, T = applied torqueρ = radius, and J = polar moment on inertia
Torsional Shearing Strain
29
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 56ENES 220 ©Assakkaf
Torsional Displacements
Angle of Twist in the Elastic RangeThe angle of twist for a circular uniform shaft subjected to external torque T is given by
GJTL
=θ (22)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 57ENES 220 ©AssakkafTorsional Displacements
Multiple Torques/Sizes
E1 E2 E3
L1 L2 L3
∑∑==
==n
i ii
iin
ii JG
LT11
θθ
Circular ShaftsFig. 12
30
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 58ENES 220 ©AssakkafTorsional Displacements
Angle of Twist in the Elastic RangeThe angle of twist of various parts of a shaft of uniform member can be given by
∑∑==
==n
i ii
iin
ii JG
LT11
θθ (24)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 59ENES 220 ©Assakkaf
Torsional Displacements
Angle of Twist in the Elastic RangeIf the properties (T, G, or J) of the shaft are functions of the length of the shaft, then
∫=L
dxGJT
0
θ (25)
31
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 60ENES 220 ©Assakkaf
Stresses in Oblique Planes
Other Stresses Induced By Torsion
x
y AFig. 16
x
y
xyτxyτ
yxτ
yxτ
(a)
(b)x
yt
α
α
n
σn dA
τn t dA
α
τyx dA sin α
τ x y
dA
cosα
(c)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 61ENES 220 ©Assakkaf
Stresses in Oblique Planes
Other Stresses Induced By Torsionyt
α
n
σn dA
τn t dA
α
τyx dA sin α
τ x y
dA
cosα
( ) ( )
( ) ατααττ
αατααττ
2cossincos whichFrom
0sinsincoscos0
22xyxynt
yxxynt
t
dAdAdAF
=−=
=+−
=+ ∑
Fig. 16c
(29)
32
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 62ENES 220 ©Assakkaf
Stresses in Oblique Planes
Other Stresses Induced By Torsionyt
α
n
σn dA
τn t dA
α
τyx dA sin α
τ x y
dA
cosα
( ) ( )
αταατσ
ααταατσ
2sincossin2 whichFrom
0cossinsincos0
xyxyn
yxxyn
t
dAdAdAF
==
=−−
=+ ∑
Fig. 16c
(31)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 63ENES 220 ©Assakkaf
Stresses in Oblique Planes
Maximum Normal Stress due to Torsion on Circular Shaft
The maximum compressive normal stress σmax can be computed from
JcTmax
maxmax ==τσ (32)
33
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 64ENES 220 ©Assakkaf
Power Transmission
Power Transmission by Torsional Shaft– But ω = 2π f, where f = frequency. The unit
of frequency is 1/s and is called hertz (Hz).– If this is the case, then the power is given
by
fT
fT
π
π
2Power
or2Power
=
=
(37)
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 65ENES 220 ©Assakkaf
Power Transmission
Power Transmission by Torsional Shaft– Units of Power
hp (33,000 ft·lb/min)watt (1 N·m/s)
US CustomarySI
34
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 66ENES 220 ©Assakkaf
Power Transmission
Power Transmission by Torsional Shaft– Some useful relations
lb/sin 6600lb/sft 550hp 1
Hz601
601 rpm 1 1
⋅=⋅=
== −s
rpm = revolution per minute
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 67ENES 220 ©Assakkaf
Summary
Axially Loaded Versus TorsionallyLoaded Members
Deformation
Stress
TorsionAxial
EAPL
=δ
JTρτ ρ =
GJTL
=θ
AP
=σ
35
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3) Slide No. 68ENES 220 ©Assakkaf
Summary
Strain– Two general types of strain:
• Axial (normal) Strain• Shearing Strain
δL
θ
φ = γxy
δs
Normal Strain Shearing Strain
L