Hibbeler R.C. Structural Analysis

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370 CHAPTER 9DEFLECTIONS USING ENERGY M ETHODS

Fig. 9–19

(a)

10 ft 10 ft 15 ft

80 kft

6 k

0.75 k

1.75 k

1 k

1x

1.75 k

1 k

 15

 0.75x

15

0.75 k

0.75x

(b)

virtual loads

Determine the displacement at D of the steel beam in Fig. 9–19a. Take

I = 800 in

.E = 29110

2 ksi,

EXAMPLE 9.9

SOLUTION

Virtual Moments

. The beam is subjected to a virtual unit load at

D as shown in Fig. 9–19b. By inspection, three coordinates, such as

and must be used to cover all the regions of the beam.

Notice that these coordinates cover regions where no discontinuities

in either real or virtual load occur. The internal moments m have been

computed in Fig. 9–19b using the method of sections.

9.7 METHOD OF VIRTUAL WORK: BEAMS AND FRAMES 371

Real Moments

. The reactions on the beam are computed first;

then, using the same x coordinates as those used for m, the internal

moments M are determined as shown in Fig. 9–19c.

Virtual-Work Equation. Applying the equation of virtual work to

the beam using the data in Figs. 9–19b and 9–19c, we have

Ans.

The negative sign indicates the displacement is upward, opposite to

the downward unit load, Fig. 9–19b. Also note that did not actually

have to be calculated since M

= 0.

=-0.466 in.

-6250 k

1122

>ft

29110

2 k>in

1800 in

3500

2750

6250 k

1-0.75x

2180 - 1x

2 dx

1-1x

2102 dx

10.75x

- 15217x

2 dx

1 k

7 k

80 kft

6 k

 0

7 k

 7x

1 k

 80  1x

80 kft

real loads

(c)

372 CHAPTER 9DEFLECTIONS USING ENERGY M ETHODS

1 k

1.25 k

8 ft

10 ft

1.25 k

1 k

virtual loadings

(b)

 1x

1.25 k

1 k

 1.25x

1.25 k

1 k

Determine the horizontal displacement of point C on the frame

shown in Fig. 9–20a. Take and for both

members.

I = 600 in

E = 29110

2 ksi

EXAMPLE 9.10

8 ft

4 k/ft

10 ft

(a)

Fig. 9–20

SOLUTION

Virtual Moments

. For convenience, the coordinates and in

Fig. 9–20a will be used. A horizontal unit load is applied at C,

Fig. 9–20b. Why? The support reactions and internal virtual moments

are computed as shown.

9.7 METHOD OF VIRTUAL WORK: BEAMS AND FRAMES 373

25 k

8 ft

5 ft

25 k

40 k

real loadings

(c)

 40x

 2x

25 k

 25x

40 k

25 k

40 k

10 kft

8 ft

10 ft

(

)

200 kft

8 ft

10 ft

(e)

Real Moments

. In a similar manner the support reactions and

real moments are computed as shown in Fig. 9–20c.

Virtual-Work Equation. Using the data in Figs. 9–20b and 9–20c,

we have

dx =

11x

40x

- 2x

11.25x

2125x

2dx

(1)

If desired, the integrals can also be evaluated graphically

using the table on the inside front cover. The moment diagrams for the

frame in Figs. 9–20b and 9–20c are shown in Figs. 9–20d and 9–20e,

respectively.Thus, using the formulas for similar shapes in the table yields

This is the same as that calculated in Eq. 1. Thus

Ans.= 0.113 ft = 1.36 in.

13 666.7 k

[29110

2 k>in

11122

>ft

2][600 in

1ft

>1122

= 8333.3 + 5333.3 = 13 666.7 k

mM dx =

1102120021102+

110212002182

mM>dx

8333.3

5333.3

13 666.7 k

374 CHAPTER 9DEFLECTIONS USING ENERGY M ETHODS

5 kN

7.5 kNm

 7.5

real loads

(c)

7.5 kNm

5 kN

30

5 kN 5 kN

7.5 kNm

5 kN

 2.5x

1.5 m

Determine the tangential rotation at point C of the frame shown in

Fig. 9–21a. Take I = 15110

2 mm

.E = 200 GPa,

EXAMPLE 9.11

Fig. 9–21

2 m

3 m

5 kN

60

(a)

1 kNm

 1

1 kNm

 1

virtual loads

(b)

SOLUTION

Virtual Moments . The coordinates and shown in Fig. 9–21a

will be used. A unit couple moment is applied at C and the internal

moments are calculated, Fig. 9–21b.

Real Moments

. In a similar manner, the real moments M are

calculated as shown in Fig. 9–21c.

Virtual-Work Equation. Using the data in Figs. 9–21b and 9–21c,

we have

Ans. = 0.00875 rad

26.25 kN

200110

2 kN>m

[15110

2 mm

]110

-12

>mm

11.25

26.25 kN

dx =

1-121-2.5x

2 dx

11217.52 dx

9.8 VIRTUAL STRAIN ENERGY CAUSED BY AXIAL LOAD, SHEAR, TORSION, AND TEMPERATURE 375

9.8 Virtual Strain Energy Caused by Axial

Load, Shear, Torsion, and Temperature

Although deflections of beams and frames are caused primarily by

bending strain energy, in some structures the additional strain energy

of axial load, shear, torsion, and perhaps temperature may become

important. Each of these effects will now be considered.

Axial Load. Frame members can be subjected to axial loads, and the

virtual strain energy caused by these loadings has been established in

Sec. 9–4. For members having a constant cross-sectional area, we have

(9–24)

where

internal virtual axial load caused by the external virtual unit load.

internal axial force in the member caused by the real loads.

modulus of elasticity for the material.

cross-sectional area of the member.

member’s length.

Shear. In order to determine the virtual strain energy in a beam due

to shear, we will consider the beam element dx shown in Fig. 9–22. The

shearing distortion dy of the element as caused by the real loads is

If the shearing strain is caused by linear elastic material

response, then Hooke’s law applies, Therefore,

We can express the shear stress as where K is a form factor

that depends upon the shape of the beam’s cross-sectional area A.

Hence, we can write The internal virtual work done

by a virtual shear force , acting on the element while it is deformed dy,

is therefore For the entire beam, the

virtual strain energy is determined by integration.

(9–25)

where

internal virtual shear in the member, expressed as a function of x

and caused by the external virtual unit load.

internal shear in the member, expressed as a function of x and

caused by the real loads.

cross-sectional area of the member.

form factor for the cross-sectional area:

for rectangular cross sections.

for circular cross sections.

for wide-flange and I-beams, where A is the area of the web.

shear modulus of elasticity for the material. G =

K L 1

K = 10>9

K = 1.2

K =

A =

V =

v =

b dx

= v dy = v1KV>GA2 dx.

dy = K1V>GA2 dx.

t = K1V>A2,

dy = 1t>G2 dx.g = t>G.

gdy = g dx.

L =

A =

E =

N =

n =

nNL

Fig. 9–22



376 CHAPTER 9DEFLECTIONS USING ENERGY M ETHODS

Torsion. Often three-dimensional frameworks are subjected to

torsional loadings. If the member has a circular cross-sectional area, no

warping of its cross section will occur when it is loaded. As a result, the

virtual strain energy in the member can easily be derived.To do so consider

an element dx of the member that is subjected to an applied torque T,

Fig. 9–23. This torque causes a shear strain of Provided

linear elastic material response occurs, then where

Thus, the angle of twist If a

virtual unit load is applied to the structure that causes an internal virtual

torque t in the member, then after applying the real loads, the virtual strain

energy in the member of length dx will be

Integrating over the length L of the member yields

(9–26)

where

internal virtual torque caused by the external virtual unit load.

internal torque in the member caused by the real loads.

shear modulus of elasticity for the material.

polar moment of inertia for the cross section, where c

is the radius of the cross-sectional area.

member’s length.

The virtual strain energy due to torsion for members having noncircular

cross-sectional areas is determined using a more rigorous analysis than

that presented here.

Temperature. In Sec. 9–4 we considered the effect of a uniform

temperature change on a truss member and indicated that the

member will elongate or shorten by an amount In some

cases, however, a structural member can be subjected to a temperature

difference across its depth, as in the case of the beam shown in Fig. 9–24a.

If this occurs, it is possible to determine the displacement of points along

the elastic curve of the beam by using the principle of virtual work.To do

so we must first compute the amount of rotation of a differential element

dx of the beam as caused by the thermal gradient that acts over the

beam’s cross section. For the sake of discussion we will choose the most

common case of a beam having a neutral axis located at the mid-depth

noted that the mean temperature is If the

temperature difference at the top of the element causes strain

elongation, while that at the bottom causes strain contraction. In both

cases the difference in temperature is ¢T

= T

- T

= T

- T

7 T

= 1T

+ T

2>2.

¢L = a ¢TL.

¢T

L =

J = pc

>2, J =

G =

T =

t =

tTL

= t du = tT dx>GJ.

du = 1g dx2>c = 1t>Gc2 dx = 1T>GJ2 dx.

t = Tc>J.g = t>G,

g = 1cdu2>dx.

Fig. 9–23

9.8 VIRTUAL STRAIN ENERGY CAUSED BY AXIAL LOAD, SHEAR, TORSION, AND TEMPERATURE 377

Since the thermal change of length at the top and bottom is

Fig. 9–24c, then the rotation of the element is

If we apply a virtual unit load at a point on the beam where a displacement

is to be determined, or apply a virtual unit couple moment at a point

where a rotational displacement of the tangent is to be determined, then

this loading creates a virtual moment m in the beam at the point where

the element dx is located.When the temperature gradient is imposed, the

virtual strain energy in the beam is then

(9–27)

where

internal virtual moment in the beam expressed as a function

of x and caused by the external virtual unit load or unit

couple moment.

coefficient of thermal expansion.

temperature difference between the mean temperature and

the temperature at the top or bottom of the beam.

mid-depth of the beam.

Unless otherwise stated, this text will consider only beam and frame

deflections due to bending. In general, though, beam and frame members

may be subjected to several of the other loadings discussed in this

section. However, as previously mentioned, the additional deflections

caused by shear and axial force alter the deflection of beams by only a

few percent and are therefore generally ignored for even “small”

two- or three-member frame analysis of one-story height. If these and

the other effects of torsion and temperature are to be considered for the

analysis, then one simply adds their virtual strain energy as defined by

Eqs. 9–24 through 9–27 to the equation of virtual work defined by

Eq. 9–22 or Eq. 9–23. The following examples illustrate application of

these equations.

c =

¢T

a =

m =

temp

ma ¢T

du =

a ¢T

dx = a ¢T

dx,

 T

(a)

 T



_______

T

temperature profile

(b)

(c)

positive rotation

Fig. 9–24

Determine the horizontal displacement of point C on the frame

shown in Fig. 9–25a. Take

and for both members. The cross-sectional

area is rectangular. Include the internal strain energy due to axial load

and shear.

A = 80 in

I = 600 in

G = 12110

2 ksi,E = 29110

2 ksi,

EXAMPLE 9.12

378 CHAPTER 9DEFLECTIONS USING ENERGY M ETHODS

SOLUTION

Here we must apply a horizontal unit load at C. The necessary

free-body diagrams for the real and virtual loadings are shown in

Figs. 9–25b and 9–25c.

8 ft

4 k/ft

10 ft

(a)

1 k

1.25 k

8 ft

10 ft

1.25 k

1 k

virtual loadings

(b)

 1x

1.25 k

1 k

 1.25x

 1.25

1.25 k

1 k

 1

 1.25

25 k

8 ft

5 ft

25 k

40 k

real loadings

(c)

 40x

 2x

25 k

 25x

 0

 25

40 k

25 k

40 k

 25

 40  4x

Fig. 9–25

9.8 VIRTUAL STRAIN ENERGY CAUSED BY AXIAL LOAD, SHEAR, TORSION, AND TEMPERATURE 379

Bending. The virtual strain energy due to bending has been deter-

mined in Example 9–10. There it was shown that

Axial load. From the data in Fig. 9–25b and 9–25c, we have

Shear. Applying Eq. 9–25 with for rectangular cross

sections, and using the shear functions shown in Fig. 9–25b and 9–25c,

we have

Applying the equation of virtual work, we have

Ans.

Including the effects of shear and axial load contributed only a 0.6%

increase in the answer to that determined only from bending.

= 1.37 in.

1 k

= 1.357 in.

k + 0.001616 in.

k + 0.00675 in.

540 k

ft112 in.>ft2

[12110

2 k>in

]180 in

= 0.00675 in.

1.2112140 - 4x

2dx

1.21-1.2521-252dx

bdx

K = 1.2

= 0.001616 in.

1.25 k125 k21120 in.2

80 in

[29110

2 k>in

]

1 k102196 in.2

80 in

[29110

2 k>in

]

nNL

mM dx

13 666.7 k

112

>1 ft

[29110

2 k>in

]1600 in

= 1.357 in.