Engineering Mechanics: Statics

671 Pages·2022·112.381 MB·English

by  Russell Hibbeler

This is a special edition of an established title widely used by colleges and
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Engineering Mechanics
Engineering Mechanics: Statics features Russell Hibbeler’s hallmark approach to teaching the
subject—a clear, thorough, and student-friendly presentation of theory alongside examples and E
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practice problems for the application of concepts. Now in its fifteenth edition, the text continues
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STATICS
to empower students with i
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• a variety of problems for practice. Sections end with Fundamental Problems and Conceptual S e
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Problems, and chapters conclude with Review Problems. i
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• an emphasis on free-body diagrams.Specific sections, examples, and homework problems T g Fifteenth Edition in SI Units

are devoted to the drawing of free-body diagrams, essential to solving problems. I M
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• the Procedure for Analysis feature. Introduced in the first chapter and customized in later S e
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ones, this procedure is a versatile approach to analyzing a variety of mechanics problems. h
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NEW TO THIS EDITION
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• New Fundamental Problems have been added, with partial solutions at the back of the book. s
• Over 300 new problems involve applications to many different fields of engineering.
• New or updated photos throughout the book illustrate how principles apply to real-world R. C. Hibbeler
situations and how materials behave under load.
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Available separately for purchase is Mastering Engineering for Engineering Mechanics: Statics, the ine
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teaching and learning platform that empowers instructors to personalize learning for every student.
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This optional suite helps deliver the desired learning outcomes when combined with Pearson’s
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trusted educational content and features like the following: itsdit
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• Video Solutions: Developed by the author, they summarize key concepts discussed in the text,
demonstrate how to solve problems, and model the best way to reach a solution.
• Enhanced feedback: Tutorials and many end-of-section problems provide enhanced feedback,
specific to student errors, and optional hints, which break problems down into simpler steps.
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• GeoGebra 3D Interactive Figures: These figures enable students to interact directly with ib
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the graph in a manner that replicates how they would graph on paper.
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SI Prefixes
Multiple Exponential Form Prefix SI Symbol
1 000 000 000 109 giga G
1 000 000 106 mega M
1 000 103 kilo k
Submultiple
0.001 10-3 milli m
0.000 001 10-6 micro μ
0.000 000 001 10-9 nano n
Conversion Factors (SI) to (FPS)
Unit of Unit of
Quantity Measurement (SI) Equals Measurement (FPS)
Force N 0.2248 lb
Mass kg 0.06852 slug
Length m 3.281 ft
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Fundamental Equations of Statics
Cartesian Vector Equilibrium
A = Axi + Ayj + Azk
Particle
Magnitude
ΣFx = 0, ΣFy = 0, ΣFz = 0
A = A2x + A2y + A2z
Rigid Body-Two Dimensions
2
Directions
A Ax Ay Az ΣFx = 0, ΣFy = 0, ΣMO = 0
uA = A = A i + A j + A k Rigid Body-Three Dimensions
= cos ai + cos bj + cos gk ΣFx = 0, ΣFy = 0, ΣFz = 0
cos2 a + cos2 b + cos2 g = 1 ΣMx = 0, ΣMy = 0, ΣMz = 0
Dot Product
A · B = AB cos u Friction
= AxBx + AyBy + AzBz Static (maximum) Fs = μsN
Cross Product
Kinetic Fk = μkN
i j k
Center of Gravity
C = A : B = Ax Ay Az
B B B Particles or Discrete Parts
x y z
Cartesian Position Vector Σ ∼r W
r =
r = (x2 - x1)i + (y2 -3 y1)j + (z2 - z13)k ΣW
Body
Cartesian Force Vector
∼
r r dW
F = Fu = F L
r r =
a b dW
Moment of a Force
L
MO = Fd
i j k Area and Mass Moments of Inertia
MO = r : F = rx ry rz I = r2dA I = r2dm
Fx Fy Fz L L
Moment of a Force about a Specified Axis Parallel-Axis Theorem
3 u u 3u I = I + Ad2 I = I + md2
x y z
#
M = u r : F = rx ry rz Radius of Gyration
F F F
x y z I I
k = k =
Simplification of a Force and Couple System A m
A A
FR = Σ3 F 3 Virtual Work
(MR)O = ΣM + ΣMO dU = 0
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Geometric Properties of Line and Area Elements Center of Gravity and Mass Moment of Inertia of Homogeneous Solids
z z
Centroid Location Centroid Location Area Moment of Inertia
y y
L 5 2ur A 5 ur2 r
ur uC x u r uC x Ix = 14 r4 (u - 12 sin 2u) V54–3pr3 r G V5pr2h G h–2
r suin u 23 r suin u Iy = 14 r4 (u + 12 sin 2u) y h–2 y
x
Circular arc segment Circular sector area x
Cylinder
1
L5p–2r L 5 pr y r y9A 5 14 pr3—4pr2x9 IIxy == 11166 ππrr44 Ixx = IyyS =ph Iezzr e= 25 mr2 Ixx = Iyy = 112 m(3r2z + h2) Izz = 12 mr2
r C —2pr C r C x Ix′ = 1 π 6 - 94 π r4 z
Quarter and semicircle arcs Quart3—4eprr circle area Iy′ = a1 π 6 - 94 π b r4 V 5–2pr3 V513– r2h G h–4 h
3
a b G y
a A5 –12 h (a 1 b) y A 5—pr2 1 r y x r
h C x 2—4r Ix = 8 πr4 x 3–r Cone
r 3p 8
b –31 2aa 11 bb h C x Iy = 18 πr4 Hemisphere Ixx = Iyy = 830 m(4r2 + h2) Izz = 130 mr2
Trapezoidal area Semicircular area Ixx = Iyy = 0.259 mr2 Izz = 25 mr2 z
y
b A52–ab A 5 pr2 z
3 z9
1
a C 35–a r C x Ix = 4πr4 G G y
1 r a
38–b Iy = 4πr4 y x b
Semiparabolic area Circular area x Thin plate
b C A1—350b—13ab h yC A 5x bh Ix = 112bh3 Ixx = Iyy = 14 mTrh2 i n cIizrzc =u la12r m dris2k Iz′z′ = 32 mr2 Ixx = 112 mb2 Iyy = 1z12 ma2 Izz = 112 m(a2 + b2)
–3a z
4 a b Iy = 112hb3 2l
Exparabolic area Rectangular area G
a y9 rr y
x l
GG 2
A 5 — 1 bh y
b 2
1
C A5—43ab h—h3 C—b3 x9 IIxy′′ == 33166bhhb33 x Thin ring x9 Slender rod y9
—2a b 1 1 1
5 Parabolic area Triangular area Ixx = Iyy = 2 mr2 Izz = mr2 Ixx = Iyy = 12 ml 2 Ix′x′ = Iy′y′ = 3 ml 2 Iz′z′ = 0
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Center of Gravity and Mass Moment of Inertia of Homogeneous Solids
z z
r
V5pr2h
V54–3pr3 r G h–
G 2
h y
y –
2
x
x
Cylinder
Sphere
1 1
Ixx = Iyy = Izz = 25 mr2 Ixx = Iyy = 12 m(3r2 + h2) Izz = 2 mr2
z
z
V51– r2h h
3 –
V 5–2pr3 G 4 h
3
G y
r y r
x
x 3–r Cone
8
3 3
Hemisphere Ixx = Iyy = 80 m(4r2 + h2) Izz = 10 mr2
Ixx = Iyy = 0.259 mr2 Izz = 25 mr2 z
z
z9
G
G
y
r a
b
y
x
x Thin plate
1 Thin circula1r disk 3 Ixx = 112 mb2 Iyy = 112 ma2 Izz = 112 m(a2 + b2)
Ixx = Iyy = 4 mr2 Izz = 2 mr2 Iz′z′ = 2 mr2 z
z
l
2
G
rr y
x l
GG 2
y
y9
x x9
Thin ring Slender rod
1 1 1
Ixx = Iyy = 2 mr2 Izz = mr2 Ixx = Iyy = 12 ml 2 Ix′x′ = Iy′y′ = 3 ml 2 Iz′z′ = 0
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ENGINEERING MECHANICS
STATICS
FIFTEENTH EDITION IN SI UNITS
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ENGINEERING MECHANICS
STATICS
FIFTEENTH EDITION IN SI UNITS
R. C. HIBBELER
SI Conversion by
Jun Hwa Lee
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© 2023 by R. C. Hibbeler
The right of R. C. Hibbeler to be identified as the author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988.
Authorized adaptation from the United States edition, entitled Engineering Mechanics: Statics, Fifteenth Edition, ISBN 978-0-13-481497-1, by Russell C. Hibbeler,
published by Pearson Education, Inc. © 2022.
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