Solutions Manual for Introduction to Continuum Mechanics (4th Edition) by W. Michael Lai, David Rubin & Erhard Krempl | Mechanics PDF

ALL CHAPTERS COVERED

, 1


Chapter 1: INTRODUCTION

1.1 Newton’s second law can be expressed as
F = ma (1)
where F is the net force acting on the body, m mass of the body, and a the acceleration
of the body in the direction of the net force. Use Eq. (1) to determine the governing
equation of a free-falling body. Consider only the forces due to gravity and the air
resistance, which is assumed to be proportional to the square of the velocity of the
falling body.
Solution: From the free-body-diagram shown in Fig.P1.1 it follows that
dv
m = Fg − F d , Fg = mg, Fd = cv 2
dt
where v is the downward velocity (m/s) of the body, Fg is the downward force (N or
kg m/s2 ) due to gravity, Fd is the upward drag force, m is the mass (kg) of the body,
g the acceleration (m/s2 ) due to gravity, and c is the proportionality constant (drag
coefficient, kg/s). The equation of motion is
dv c
+ αv 2 = g, α=
dt m


Fd cv2




v
Fg mg


Fig. P1.1


1.2 Consider steady-state heat transfer through a cylindrical bar of nonuniform cross sec-
tion. The bar is subject to a known temperature T0 (◦ C) at the left end and exposed,
both on the surface and at the right end, to a medium (such as cooling fluid or air)
at temperature T∞ . Assume that temperature is uniform at any section of the bar,
T = T (x), and neglect thermal expansion of the bar (that is, assume rigid). Use the
principle of conservation of energy (which requires that the rate of change (increase)
of internal energy is equal to the sum of heat gained by conduction, convection, and
internal heat generation) to a typical element of the bar (see Fig. P1.2) to derive the
governing equations of the problem.
Solution: If q denotes the heat flux (heat flow per unit area, W/m2 ), then [Aq]x is
the net heat flow into the volume element at x, [Aq]x+∆x is the net heat flow out
of the volume element at x + ∆x. If h denotes the film conductance [W/(m2 ·◦ C)],
βP ∆x(T∞ − T ) is the heat flow through the surface of the rod into the body, where T∞
is the temperature of the surrounding medium and P is the perimeter (m). Suppose
that there is a heat source within the rod generating energy at a rate of g (W/m3 ).
Then the energy balance gives
[Aq]x − [Aq]x+∆x + βP ∆x(T∞ − T ) + gA ∆x = 0 (1)
or, dividing throughout by ∆x,
[Aq]x+∆x − [Aq]x
− + βP (T∞ − T ) + Ag = 0
∆x

,2 SOLUTIONS MANUAL


and taking the limit ∆x → 0, we obtain
d
− (Aq) + βP (T∞ − T ) + Ag = 0 (2)
dx


Convection from lateral
g(x), internal heat generation
surface
Maintained at
temperature, T0
Exposed to ambient
temperature, T∞
L


x

∆x


g(x)
heat flow in, heat flow out,
(Aq)x (Aq)x+∆x
∆x


Fig. P1.2


1.3 The Euler–Bernoulli hypothesis concerning the kinematics of bending deformation of
a beam assumes that straight lines perpendicular to the beam axis before deformation
remain (1) straight, (2) perpendicular to the tangent line to the beam axis, and (3)
inextensible during deformation. These assumptions lead to the following displacement
field:
dv
u1 (x, y) = −y , u2 = v(x), u3 = 0, (1)
dx
where (u1 , u2 , u3 ) are the displacements of a point (x, y, z) along the x, y, and z coor-
dinates, respectively, and v is the vertical displacement of the beam at point (x, 0, 0).
Suppose that the beam is subjected to a distributed transverse load q(x). Determine
the governing equation by summing the forces and moments on an element of the beam
(see Fig. P1.3). Note that the sign conventions for the moment and shear force are
based on the definitions
Z Z
V = σxy dA, M = y σxx dA,
A A

and may not agree with the sign conventions used in some mechanics of materials books.
Solution: Summation of the forces in the transverse direction on the element of the
beam gives
(V + ∆V ) − V + q(x)∆x = 0.
Dividing throughout with ∆x and taking the limit ∆x → 0 gives
dV
+ q = 0. (2)
dx
Taking the moment of forces about the right end of the element, we obtain
X
Mz = 0 : −V ∆x − M + (M + ∆M ) + q∆x · α∆x = 0,

where α is a number 0 ≤ α ≤ 1. Again, dividing throughout with ∆x and taking the
limit ∆x → 0 gives
dM
− V = 0. (3)
dx

, CHAPTER 1: INTRODUCTION 3

y, v q(x)
y



• x z •

L Beam
cross section


y q(x) q(x)

σ xy + dσ xy

σ xx σ xx + dσ xx
x +
• •
+
M M + dM
V V + dV σ xy
dx dx

M= ò y⋅s xx dA, V = òs xy dA
A A


Fig. P1.3


Note that V and M denote the shear force and bending moment on the entire cross
section, and they have the meaning
Z
dM
M (x) = σxx y dA, V (x) = .
A dx
Here A denotes the area of cross section. The stress resultants (V, M ) can be related to
the deflection v. Using the linear elastic constitutive relation for an isotropic material

d2 v
 
σxx = Eεxx = E −y 2 .
dx
Substituting into the definition of M , we obtain

d2 v d2 v
Z Z  
M (x) = σxx y dA = E −y 2 y dA = −EI 2 , (4)
A A dx dx

where I is the moment of inertia about the axis of bending (z−axis). Then

d2 v
 
d
V =− EI 2 . (5)
dx dx

Equations (2)-(5) can be combined to obtain the following fourth-order equation for v:

d2 d2 v
 
EI = q(x). (6)
dx2 dx2

1.4 A cylindrical storage tank of diameter D contains a liquid column height h(x, t). Liquid
is supplied to the tank at a rate of qi (m3 /day) and drained at a rate of q0 (m3 /day).
Assume that the fluid is incompressible (that is, constant mass density ρ) and use the
principle of conservation of mass to obtain a differential equation governing h(x, t).
Solution: The conservation of mass requires

time rate of change in mass = mass inflow - mass outflow.