Engineering Mechanics: Statics, SI Units (15th Edition, 2023, eBook) by Hibbeler

E-TEXTBOOK

, GLOBAL
EDITION


Engineering Mechanics
STATICS
Fifteenth Edition in SI Units




R. C. Hibbeler




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,SI Prefixes

Multiple Exponential Form Prefix SI Symbol
1 000 000 000 109 giga G
6
1 000 000 10 mega M
3
1 000 10 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
A = 2A2x + A2y + A2z
ΣFx = 0, ΣFy = 0, ΣFz = 0
Rigid Body-Two Dimensions
Directions
ΣFx = 0, ΣFy = 0, ΣMO = 0
A Ax Ay Az
uA = = i + j + k Rigid Body-Three Dimensions
A A A A
= 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 F k = μkN

3 Az 3
i j k
C = A : B = Ax Ay Center of Gravity
Bx By Bz Particles or Discrete Parts
Cartesian Position Vector Σ∼r W
r =
r = (x2 - x1)i + (y2 - y1)j + (z2 - z1)k ΣW
Body
Cartesian Force Vector

L
∼
F = Fu = Fa b
r r dW
r r =

L
Moment of a Force dW

MO = Fd
Area and Mass Moments of Inertia
= r : F = 3 rx rz 3
i j k


L L
MO ry
I = r 2dA I = r 2dm
Fx Fy Fz

Moment of a Force about a Specified Axis Parallel-Axis Theorem
I = I + Ad 2 I = I + md 2
M = u # r : F = 3 rx rz 3
ux uy uz
ry Radius of Gyration
Fx Fy Fz
AA Am
I I
k = k =
Simplification of a Force and Couple System
FR = ΣF Virtual Work
( MR ) O = ΣM + ΣMO dU = 0

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,Geometric Properties of Line and Area Elements
Centroid Location Centroid Location Area Moment of Inertia
y y
L 5 2ur A 5 ur 2
r r
u u 1 4 1
x x Ix = r (u - sin 2u)
C C 4 2
u u
r sin u 2 r sin u 1 4 1
u Iy = r (u + sin 2u)
3 u 4 2
Circular arc segment Circular sector area

y y9 1 4
Ix = πr
16
A 5 14 pr 2 1 4
L 5 p–2 r L 5 pr 4r Iy = πr
— 16
r 3p
x9
r C
Ix′ = a - b r4
2r
— C r x π 4
C p
4r
—
16 9 π
3p
Iy′ = a - b r4
π 4
Quarter and semicircle arcs Quarter circle area 16 9 π

a A 5 –12 h (a 1 b) y
pr 2
A5—
2 1 4
C Ix = πr
h x 4r 8
—
3p
r
–1 2 a 1 b h C
b 3 a 1b x 1 4
Iy = πr
8
Trapezoidal area Semicircular area

y
b A5 23– ab A 5 pr2
1
r Ix = πr4
a C 3– 4
5a x
C
1
3–
Iy = πr4
8b
4

Semiparabolic area Circular area


1 y A 5 bh
A5 —
3
ab 1 3
b Ix = bh
C 3
— b h x 12
10 C
–3 a
4 1 3
a b Iy = hb
12
Exparabolic area Rectangular area
a y9
1
A 5— bh
b 2
1 3
C h Ix′ = bh
h C x9 36
—
3
4 ab 1
A 5— b
3 —
3 Iy′ = hb3
b 36
2
—
5
a
Parabolic area Triangular area
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,Center of Gravity and Mass Moment of Inertia of Homogeneous Solids
z z


r
V 5 pr 2 h
V5 4–3 pr 3 r h
–
G G 2

y h
– y
2
x
x
Cylinder
Sphere
1 1
2 Ixx = Iyy = m(3r2 + h2) Izz = mr2
Ixx = Iyy = Izz = mr2 12 2
5
z
z

1
V 5 – r 2h h
3 –
G 4 h
V 5 –23 pr 3
G y
r y r
x
x 3
–r Cone
8
3 3
Hemisphere Ixx = Iyy = m(4r2 + h2) Izz = mr2
80 10
2 z
Ixx = Iyy = 0.259 mr2 Izz = mr2
5

z
z9

G
G y
r a
b
y
x
x Thin plate
Thin circular disk 1 1 1
Ixx = mb2 Iyy = ma2 Izz = m(a2 + b2)
1 1 3 12 12 12
Ixx = Iyy = mr2 Izz = mr2 Iz′z′ = mr2
4 2 2 z

z
l
2
G
r y
x l
G 2
y
y9
x x9

Thin ring Slender rod
1 1 1
Ixx = Iyy = mr2 Izz = mr2 Ixx = Iyy = ml 2 Ix′x′ = Iy′y′ = ml 2 Iz′z′ = 0
2 12 3
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,ENGINEERING MECHANICS




STATICS
FIFTEENTH EDITION IN SI UNITS




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