Third Edition
LECTURE
REVIEW FOR EXAM #1
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
10
by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park
Chapters
1, 2, and 3
Slide No. 1
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Review: Statics
„
Equations of Equilibrium
– Rigid Body
F2
F2
F1
z
k
j
i
y
x
1
Slide No. 2
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 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
r
r
R = ∑ Fx i + ∑ Fy j + ∑ Fz k =0
r
r
C = ∑ M x i + ∑ M y j + ∑ M z k =0
Slide No. 3
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Statics
„
ENES 220 ©Assakkaf
Equations of Equilibrium
– The two conditions can also be expressed
in scalar form as
∑F = 0
∑M = 0
x
x
∑F
∑M
=0
y
y
=0
∑F = 0
∑M = 0
z
z
2
Slide No. 4
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Statics
„
ENES 220 ©Assakkaf
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)
y
x
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Statics
„
Slide No. 5
ENES 220 ©Assakkaf
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
r
r
R = ∑ Fx i + ∑ Fy j = 0
r
r
C = ∑ M zk = 0
3
Slide No. 6
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Statics
„
ENES 220 ©Assakkaf
Equilibrium in Two Dimensions
– In scalar form, these conditions can be
expressed as
∑F
y
∑M
A
=0
=0
Slide No. 7
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Statics
„
∑F
=0
x
ENES 220 ©Assakkaf
Cartesian Vector Representation of A
Force
z
Fz
F
y
F = Fx i + Fy j + Fz k
Fy
Fx
x
4
Slide No. 8
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Statics
„
ENES 220 ©Assakkaf
Cartesian Vector Representation of A
Force in Two Dimensions
r
F = Fx i + Fy j
F sin θ
θ
F
= F cosθi + F sin θj
F cos θ
Slide No. 9
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Vector Operations
„
ENES 220 ©Assakkaf
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
A
θ
B
5
Slide No. 10
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Vector Operations
„
Cross or Vector Product
C
C = A × B = (AB sin θ) e
eC
C
B
θ
A×B=-B×A
A
i
r r
A × B = Ax
j
k
Ay
r
Az = C
Bx
By
Bz
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Vector Operations
„
ENES 220 ©Assakkaf
Slide No. 11
ENES 220 ©Assakkaf
Example 3
– If A = -3.75i – 2.50j + 1.50k and
B = 32i + 44j + 64 k
determine the magnitude and direction of
the vector C = A × B
i
j
k
r r r
C = A × B = − 3.75 − 2.5 1.5
32
44 64
6
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Vector Operations
„
Slide No. 12
ENES 220 ©Assakkaf
Example 3 (cont’d)
i
j
k
r r r
C = A × B = − 3.75 − 2.5 1.5
32
44 64
= [- 2.5(64) - 1.5(44)]i - [- 3.75(64) - 1.5(32)]j
+ [- 3.75(44) - (-2.5)32]
= −226i + 288 j − 85k
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Review: Vector Operations
„
Slide No. 13
ENES 220 ©Assakkaf
Example 3 (cont’d)
r
2
C = C = (−226) 2 + (288) + (−85) 2 = 376
C
θ x = cos −1 rx = cos −1
C
C
− 226
= 126.9 0
376
θ yx = cos −1 ry = cos −1
C
C
θ zx = cos −1 rz = cos −1
C
288
= 40.0 0
376
− 85.0
= 103.10
376
7
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Introduction
„
Slide No. 14
ENES 220 ©Assakkaf
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)
Introduction
„
Slide No. 15
ENES 220 ©Assakkaf
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.
8
Slide No. 16
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Internal Forces for Axially
Loaded Members
„
F1
ENES 220 ©Assakkaf
Analysis of Internal Forces
F3
F2
F4
Assume that F1 = 2 k, F3 = 5 k, and F4 = 8 k
Then
→ + ∑ − F1 − F2 − F3 + F4 = 0; ⇒ F2 = F4 − F1 − F3
or F2 = 8 − 2 − 5 = 1 k
Slide No. 17
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Internal Forces for Axially
Loaded Members
„
ENES 220 ©Assakkaf
Analysis of Internal Forces
– What is the internal force developed on
plane a-a and b-b?
a
2k
b
5k
1k
a
2k
1k
8k
b
Ra-a
Ra −a = 3k
9
Slide No. 18
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Internal Forces for Axially
Loaded Members
„
ENES 220 ©Assakkaf
Analysis of Internal Forces
a
2k
b
5k
1k
a
8k
b
Rb −b = 8k
2k
1k
5k
Rb-b
Slide No. 19
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Axial Loading: Normal Stress
„
ENES 220 ©Assakkaf
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.
Stress =
Force
Area
(1)
10
Slide No. 20
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Axial Loading: Normal Stress
„
Normal Stress
F = magnitude of the force F
A = area of the cross sectional area of the
eye bar.
„
Units of Stress
SI System
103
1 kPa =
Pa =
1 MPa = 106 Pa = 106 N/m2
1 GPa = 109 Pa = 109 N/m2
103
N/m2
U.S. Customary Units
lb/in2 = psi
Kip/in2 = ksi = 1000 psi
Slide No. 21
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Shearing Stress
„
Illustration of Shearing Stress P
P
P
τ avg =
P
V
P
=
As As
V
P
11
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Stresses on an Inclined Plane in
an Axially Loaded Member
„
Slide No. 22
ENES 220 ©Assakkaf
Illustration
P
Original Area, A
Inclined Area, An
θ
P
F
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Slide No. 23
Design Loads, Working Stresses,
and Factor of Safety (FS)
ENES 220 ©Assakkaf
„
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
FS =
ultimate stress
allowable stress
12
Slide No. 24
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Displacement, Deformation, and
Strain
„
Strain
– Two general types of strain:
φ = γxy
• Axial (normal) Strain
• Shearing Strain
L
δs
δ
L
θ
Shearing Strain
Normal Strain
Slide No. 25
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Displacement, Deformation, and
Strain
ENES 220 ©Assakkaf
„
Average Axial Strain
ε avg =
„
δn
L
δn
L
True Axial Strain
∆δ n dδ n
=
∆L →0 ∆L
dL
ε ( p ) = lim
13
Slide No. 26
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Displacement, Deformation, and
Strain
„
Average Shearing Strain
γ avg = tan φ
Since δs is vary small,
sin φ = tan φ = φ, therefore,
γ avg =
δs
L
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
φ = γxy
δs
L
θ
Slide No. 27
ENES 220 ©Assakkaf
Displacement, Deformation, and
Strain
„
Example
A 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 P
is applied, the axial strain in rod C is 900µ
in/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.
14
Slide No. 28
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Displacement, Deformation, and
Strain
ENES 220 ©Assakkaf
„
Example (cont’d)
72"
C
B
42"
B
A
P
Slide No. 29
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Displacement, Deformation, and
Strain
ENES 220 ©Assakkaf
„
Example (cont’d)
C
B
72
"
B
42"
A
εC =
δC
LC
a)
εB =
a)
εB =
⇒ δ C = ε C LC = 900 ×10 −6 (72) = 0.0648 in
δB
LB
δB
LB
=
0.0648
= 0.001543 = 1543µ
42
=
0.0648 − 0.006
= 0.001400 = 1400µ
42
15
Slide No. 30
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Stress-Strain-Temperature
Relationships
„
σ=
ENES 220 ©Assakkaf
General Stress-Strain Diagram
P
A
Rupture
L
δ
ε=
δ
P
L
Slide No. 31
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Stress-Strain-Temperature
Relationships
„
ENES 220 ©Assakkaf
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
σ
σ = Eε
σ=Eε
ε
16
Slide No. 32
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Stress-Strain-Temperature
Relationships
„
ENES 220 ©Assakkaf
Shear Modulus of Elasticity, G
– The shear modulus is similar to the
modulus of elasticity. However it is applied
to shear stress-strain.
τ = Gγ
Slide No. 33
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Stress-Strain-Temperature
Relationships
Shear Modulus of Elasticity, G
Slope = G =
τ
τ=Gγ
τ
γ
γ
17
Slide No. 34
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Stress-Strain-Temperature
Relationships
„
ENES 220 ©Assakkaf
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.
ν=
ε lat ε l
=
ε long ε a
E = 2(1 + ν )G
Slide No. 35
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Stress-Strain-Temperature
Relationships
„
ENES 220 ©Assakkaf
Thermal Strain
– The thermal strain due a temperature
change of ∆T degrees is given by
ε T = α∆T
18
Slide No. 36
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Stress-Strain-Temperature
Relationships
„
ENES 220 ©Assakkaf
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
ε total = ε σ + ε T =
σ
E
+ α ∆T
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Rods: Stress Concentrations
Slide No. 37
ENES 220 ©Assakkaf
Fig. 1. Stress distribution near circular hole in flat bar
under axial loading
19
Slide No. 38
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Rods: Stress Concentrations
ENES 220 ©Assakkaf
Fig. 2. Stress distribution near fillets in flat bar under
axial loading
Slide No. 39
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Rods: Stress Concentrations
„
Hole
ENES 220 ©Assakkaf
Fig. 3
Discontinuities of cross section may result in
high localized or concentrated stresses.
σ
K = max
σ ave
20
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Rods: Stress Concentrations
„
Fillet
Fig. 4
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Rods: Stress Concentrations
„
„
Slide No. 40
ENES 220 ©Assakkaf
Slide No. 41
ENES 220 ©Assakkaf
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
21
Slide No. 42
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 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
PL
δ=
EA
(1)
Slide No. 43
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Multiple Loads/Sizes
– The deformation of of various parts of a rod
or uniform member can be given by
n
n
i =1
i =1
δ = ∑δi = ∑
Pi Li
Ei Ai
(2)
E1
E2
E3
L1
L2
L3
22
Slide No. 44
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
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
δ B/ A = δ B − δ A =
PL
EA
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Statically Indeterminate
Structures
„
(3)
Slide No. 45
ENES 220 ©Assakkaf
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
r = 3n, ⇒ statically determinate
r > 3n, ⇒ statically indeterminate
(4)
23
Slide No. 46
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 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 b
members and r reactions, we have
b + r = 2 j , ⇒ statically determinate
(5)
b + r > 2 j , ⇒ statically indeterminate
Slide No. 47
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Statically Indeterminate Axially
Loaded Members
„
Example 4 (cont’d)
Tube (A2, E2)
Rod (A2, E2)
L
P
End plate
24
Slide No. 48
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Statically Indeterminate Axially
Loaded Members
„
Example 4 (cont’d)
P
Tube (A2, E2)
P
Rod (A1, E1)
FBD:
FT/2
P
FR
FT/2
FT FT
−
− FR = 0
2
2
FR + FT = P
→ + ∑ Fx = 0; P −
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Torsional Loading
„
Introduction
Slide No. 49
ENES 220 ©Assakkaf
Cylindrical members
Fig. 1
25
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Slide No. 50
Torsional Loading
ENES 220 ©Assakkaf
„
Stresses in Circular Shaft due to
Torsion
B
C
T
ρ
T
Fig. 7
dF = τρ dA
Slide No. 51
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Torsional Shearing Strain
„
Shearing Strain
c
γρ
φ
ρ
L
Fig. 8
26
Slide No. 52
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Torsional Shearing Strain
„
Shearing Strain
For radius ρ, the shearing strain for circular
shaft is
ρφ
γρ =
L
(6)
For radius c, the shearing strain for circular
shaft is
γc =
cφ
L
(7)
Slide No. 53
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Torsional Shearing Strain
„
Polar Moment of Inertia
The 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
c
4
J = ∫ ρ 2 dA = ∫ ρ 2 (2πρ dρ ) =
0
πc
2
(13)
27
Slide No. 54
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Torsional Shearing Strain
„
Polar Moment of Inertia
The 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
c
4
J = ∫ ρ 2 dA = ∫ ρ 2 (2πρ dρ ) =
0
πc
2
(13)
Slide No. 55
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Torsional Shearing Strain
„
Shearing Stress in Terms of Torque and
Polar Moment of Inertia
τ max =
τρ =
Tc
J
Tρ
J
(17a)
(18a)
τ= shearing stress, T = applied torque
ρ = radius, and J = polar moment on inertia
28
Slide No. 56
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Torsional Displacements
„
Angle of Twist in the Elastic Range
The angle of twist for a circular uniform
shaft subjected to external torque T is
given by
TL
θ=
GJ
(22)
Slide No. 57
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Torsional Displacements
„
Fig. 12
Multiple Torques/Sizes
ENES 220 ©Assakkaf
n
n
i =1
i =1
θ = ∑θ i = ∑
E1
E2
E3
L1
L2
L3
Ti Li
Gi J i
Circular Shafts
29
Slide No. 58
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Torsional Displacements
„
ENES 220 ©Assakkaf
Angle of Twist in the Elastic Range
The angle of twist of various parts of a
shaft of uniform member can be given by
n
n
θ = ∑θ i = ∑
i =1
i =1
Ti Li
Gi J i
(24)
Slide No. 59
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Torsional Displacements
„
Angle of Twist in the Elastic Range
If the properties (T, G, or J) of the shaft are
functions of the length of the shaft, then
L
T
dx
GJ
0
θ =∫
(25)
30
Slide No. 60
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Stresses in Oblique Planes
„
Other Stresses Induced By Torsion
Fig. 16
y
y
t
n
τn t dA
τx y dA cos α
A
(c)
α
α
x
y
σn dA
τ xy
τyx dA sin α
τ yx
(a)
τ xy
α
x
τ yx
(b)
x
Slide No. 61
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Stresses in Oblique Planes
„
Other Stresses Induced By Torsion
t
∑F
t
=0
τ nt dA − τ xy (dA cos α ) cos α + τ yx (dA sin α )sin α = 0
From which
τ nt = τ xy (cos 2 α − sin 2 α ) = τ xy cos 2α
(29)
n
τn t dA
τx y dA cos α
+
y
α
α
σn dA
τyx dA sin α
Fig. 16c
31
Slide No. 62
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Stresses in Oblique Planes
„
Other Stresses Induced By Torsion
t
∑F
t
=0
σ n dA − τ xy (dA cos α )sin α − τ yx (dA sin α ) cos α = 0
From which
σ n = 2τ xy sin α cos α = τ xy sin 2α
(31)
n
τn t dA
τx y dA cos α
+
y
α
α
σn dA
τyx dA sin α
Fig. 16c
Slide No. 63
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 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
σ max = τ max =
Tmax c
J
(32)
32
Slide No. 64
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 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
Power = 2πfT
or
T=
(37)
Power
2πf
Slide No. 65
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Power Transmission
„
ENES 220 ©Assakkaf
Power Transmission by Torsional Shaft
– Units of Power
SI
US Customary
watt (1 N·m/s)
hp (33,000 ft·lb/min)
33
Slide No. 66
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Power Transmission
„
Power Transmission by Torsional Shaft
– Some useful relations
1 −1 1
s =
Hz
60
60
1 hp = 550 ft ⋅ lb/s = 6600 in ⋅ lb/s
1 rpm =
rpm = revolution per minute
Slide No. 67
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
Summary
„
ENES 220 ©Assakkaf
Axially Loaded Versus Torsionally
Loaded Members
Stress
Deformation
Axial
Torsion
P
A
PL
δ=
EA
Tρ
J
TL
θ=
GJ
σ=
τρ =
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Slide No. 68
LECTURE 10. REVIEW FOR EXAM I (CH. 1, 2, AND 3)
ENES 220 ©Assakkaf
Summary
„
Strain
– Two general types of strain:
φ = γxy
• Axial (normal) Strain
• Shearing Strain
L
δ
δs
L
Normal Strain
θ
Shearing Strain
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