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Lying by Approximation by Paul D. Gessler, Christopher Papadopoulos, Vincent C. Prantil

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2.3. IDEALIZED STRUCTURAL RESPONSES 17
applied to yield the well-known formula to predict the bending stress at a given distance from
the neutral axis:
bending
D
My
I
;
where M refers to the resultant moment, I represents the cross-sectional property known as the
area moment of inertia about an axis passing through the centroid of the cross section, and y
represents the distance from the neutral axis toward the outer edge of the cross section where the
stress is being evaluated. e displacement of a beam due to a transverse loading can be determined
/FVUSBM
"YJT
$PNQSFTTJWF 4USFTT
5FOTJMF 4USFTT
Figure 2.4: Stress distribution due to bending loads varies linearly through the cross section.
by integrating the fourth-order differential equation
d
4
v
d x
4
D
w
EI
;
where E is the modulus of elasticity and w is the load per unit length applied transversely to the
beam. is basic theory of beam bending is often referred to as Euler-Bernoulli beam theory.
2.3.4 TORSIONAL RESPONSE
Shear stresses can also develop when a torque is applied to a shaft. If the shaft is circular or
annular in cross section, it can be assumed that cross sections remain parallel and circular. From
this assumption, the shear stress due to torsion can be predicted at a point at a given radial distance,
, away from the center by the well-known formula
torsion
D
T
J
;
where T is the total torque carried by the section and J is the polar moment of inertia of the cross
section. ese stress components are illustrated in Fig. 2.5. Under these conditions, the axial twist
(sometimes referred to as angular displacement) along such a shaft of length L can be calculated
from the formula
D
TL
GJ
;
where G is the modulus of rigidity.
18 2. LET’S GET STARTED
Figure 2.5: Internal stresses due to torsion loads are distributed as shear tractions.
Example 2.1: Simple Truss Analysis
A weight is suspended by three bars as shown in Fig. 2.6. All three bars are made of
steel, a D 16 in, b D 12 in, c D 12 in, the diameter of each bar is 0:5 in, and W D 5000 lbf.
Determine the force carried by each bar.
y
x
a b
c
Q R S
W
(0; 0)
P
Figure 2.6: A three-bar structure supporting a weight forms an indeterminate truss.
A Free Body Diagram (FBD) of point P reveals that there are three unknown forces,
as shown in Fig. 2.7.
I
2.3. IDEALIZED STRUCTURAL RESPONSES 19
Example 2.1: Simple Truss Analysis (continued)
P
F
PQ
F
PR
F
PS
W
x
P
PQ
PR
PS
Figure 2.7: Free body diagram of point P with bar angle conventions.
However, there are only two equations of static equilibrium:
X
F
x
W F
PQ
cos
PQ
C F
PR
cos
PR
C F
PS
cos
PS
D 0;
X
F
y
W F
PQ
sin
PQ
C F
PR
sin
PR
C F
PS
sin
PS
D W;
where the angle for each bar is measured in the counterclockwise direction from the pos-
itive x-axis. Such a system is called statically indeterminate because the equations of static
equilibrium are insufficient to determine the forces in the structural elements. Analysis of a
statically indeterminate system requires additional equations that account for the structural
deformation, i. e., how the bars deform under their applied load.
Inverting the force-displacement relation from Section 2.3.1, F D .EA=L/ı, where
E is the modulus of elasticity, A is the cross-sectional area of the bar, and L is the (initial)
length of the bar, allows us to interpret each bar as a spring with equivalent stiffness
k D
EA
L
:
Denoting the stiffness of each bar by k
PQ
, k
PR
, and k
PS
, and the deformation of each
bar by ı
PQ
, ı
PR
, and ı
PS
, we can rewrite the equilibrium equations as follows:
X
F
x
W k
PQ
ı
PQ
cos
PQ
C k
PR
ı
PR
cos
PR
C k
PS
ı
PS
cos
PS
D 0;
X
F
y
W k
PQ
ı
PQ
sin
PQ
C k
PR
ı
PR
sin
PR
C k
PS
ı
PS
sin
PS
D W:
I
20 2. LET’S GET STARTED
Example 2.1: Simple Truss Analysis (continued)
After the load is applied, the point P , which is initially located at .0; 0/, will move to a new
location P
0
. We use u and v to denote, respectively, the horizontal and vertical components of
the displacement from point P to point P
0
, as illustrated in Fig. 2.8. Note that by convention,
we have illustrated the case such that u > 0 and v > 0, but in general, one or both of these
values could be negative.
y
x
a b
c
Q R S
W
(0; 0)
P
P
0
Figure 2.8: e structure deforms and point P displaces as the load is applied.
As suggested by Fig. 2.8, both the length and direction of each bar change after the load
is applied. However, under many common circumstances, the displacements are small enough
such that the change in direction is negligible. erefore we will assume, as an approximation,
that each deformed bar is parallel to its original position. is is illustrated in Fig. 2.9 which
shows initial and deformed positions of the bar PS near point P , and how the deformation
ı
PS
is geometrically related to the displacements u and v.
I

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