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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