Beer F.P., Johnston E.R., DeWolf J.T., Mazurek D.F. Mechanics of Materials

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PROBLEMS

727

11.42 The cylindrical block E has a speed v

5 16 ft/s when it strikes

squarely the yoke BD that is attached to the

-in.-diameter rods

AB and CD. Knowing that the rods are made of a steel for which

5 50 ksi and E 5 29 3 10

psi, determine the weight of block

E for which the factor of safety is five with respect to permanent

deformation of the rods.

11.4 3 The 18-lb cylindrical block E has a horizontal velocity v

when it

strikes squarely the yoke BD that is attached to the

-in.-diameter

rods AB and CD. Knowing that the rods are made of a steel for which

5 50 ksi and E 5 29 3 10

psi, determine the maximum allow-

able speed v

if the rods are not to be permanently deformed.

11.4 4 Collar D is released from rest in the position shown and is stopped

by a small plate attached at end C of the vertical rod ABC. Deter-

mine the mass of the collar for which the maximum normal stress

in portion BC is 125 MPa.

3.5 ft

Fig. P11.42 and P11.43

Bronze

E  105 GPa

12-mm diamete

Aluminum

E  70 GPa

9-mm diameter

0.6 m

2.5 m

4 m

Fig. P11.44

2.5 m

AEC

Fig. P11.46

11.45 Solve Prob. 11.44, assuming that both portions of rod ABC are

made of aluminum.

11. 46 The 48-kg collar G is released from rest in the position shown and

is stopped by plate BDF that is attached to the 20-mm-diameter

steel rod CD and to the 15-mm-diameter steel rods AB and EF.

Knowing that for the grade of steel used s

all

5 180 MPa and E 5

200 GPa, determine the largest allowable distance h.

11. 47 Solve Prob. 11.46, assuming that the 20-mm-diameter steel rod CD

is replaced by a 20-mm-diameter rod made of an aluminum alloy

for which s

all

5 150 MPa and E 5 75 GPa.

11.48 The steel beam AB is struck squarely at its midpoint C by a

45-kg block moving horizontally with a speed v

5 2 m/s. Using

E 5 200 GPa, determine (a) the equivalent static load, (b) the

maximum normal stress in the beam, (c) the maximum deflection

of the midpoint C of the beam.

11.49 Solve Prob. 11.48, assuming that the W150 3 13.5 rolled-steel beam

is rotated by 908 about its longitudinal axis so that its web is vertical.

1.5 m

W150  13.5

Fig. P11.48

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728

Energy Methods

11.50 A 25-lb block C moving horizontally with at velocity v

hits the

post AB squarely as shown. Using E 5 29 3 10

psi, determine

the largest speed v

for which the maximum normal stress in the

pipe does not exceed 18 ksi.

11. 51 Solve Prob. 11.50, assuming that the post AB has been rotated 908

about its longitudinal axis.

11.52 and 11. 53 The 2-kg block D is dropped from the position

shown onto the end of a 16-mm-diameter rod. Knowing that E 5

200 GPa, determine (a) the maximum deflection of end A, (b) the

maximum bending moment in the rod, (c) the maximum normal

stress in the rod.

7.5 ft

W5  16

Fig. P11.50

0.6 m

2 kg

40 mm

Fig. P11.52

2 kg

40 mm

0.6 m 0.6 m

Fig. P11.53

11.54 The 45-lb block D is dropped from a height h 5 0.6 ft onto the

steel beam AB. Knowing that E 5 29 3 10

psi, determine (a) the

maximum deflection at point E, (b) the maximum normal stress in

the beam.

11. 55 Solve Prob. 11.54, assuming that a W4 3 13 rolled-steel shape is

used for beam AB.

11.56 A block of weight W is dropped from a height h onto the horizontal

beam AB and hits it at point D. (a) Show that the maximum deflec-

tion y

at point D can be expressed as

5 y

a1 1

1 1

where y

represents the deflection at D caused by a static load W

applied at that point and where the quantity in parenthesis is

referred to as the impact factor. (b) Compute the impact factor for

the beam and the impact of Prob. 11.52.

11.57 A block of weight W is dropped from a height h onto the horizontal

beam AB and hits point D. (a) Denoting by y

the exact value of

the maximum deflection at D and by y9

the value obtained by

neglecting the effect of this deflection on the change in potential

energy of the block, show that the absolute value of the relative

error is (y9

2 y

)yy

, never exceeding y9

y2h. (b) Check the result

obtained in part a by solving part a of Prob. 11.52 without taking

into account when determining the change in potential energy

of the load, and comparing the answer obtained in this way with

the exact answer to that problem.

S5  10

2 ft 4 ft

Fig. P11.54

Fig. P11.56 and P11.57

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729

Problems

11.58 a n d 11.59 Using the method of work and energy, determine

the deflection at point D caused by the load P.

Fig. P11.58

Fig. P11.59

11.60 and 11.61 Using the method of work and energy, determine

the slope at point D caused by the couple M

Fig. P11.60

Fig. P11.61

11.62 and 11.63 Using the method of work and energy, determine

the deflection at point C caused by the load P.

2EI

EI EI

aaaa

Fig. P11.62

L/2

2EI EI

L/2

Fig. P11.63

11.64 Using the method of work and energy, determine the slope at point

A caused by the couple M

11.65 Using the method of work and energy, determine the slope at point

D caused by the couple M

2EI

L/2L/2

Fig. P11.64

2EI

L/2L/2

Fig. P11.65

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730

Energy Methods

11.6 6 Torques of the same magnitude T are applied to the steel shafts

AB and CD. Using the method of work and energy, determine the

length L of the hollow portion of shaft CD for which the angle of

twist at C is equal to 1.25 times the angle of twist at A.

60 in.

2 in.

1.5 in.

Fig. P11.66

11.67 The 20-mm diameter steel rod BC is attached to the lever AB and

to the fixed support C. The uniform steel lever is 10 mm thick and

30 mm deep. Using the method of work and energy, determine

the deflection of point A when L 5 600 mm. Use E 5 200 GPa

and G 5 77.2 GPa.

11.68 The 20-mm diameter steel rod BC is attached to the lever AB and

to the fixed support C. The uniform steel lever is 10 mm thick and

30 mm deep. Using the method of work and energy, determine

the length L of the rod BC for which the deflection at point A is

40 mm. Use E 5 200 GPa and G 5 77.2 GPa.

11.69 Two solid steel shafts are connected by the gears shown. Using the

method of work and energy, determine the angle through which

end D rotates when T 5 820 N ? m. Use G 5 77.2 GPa.

450 N

500 mm

Fig. P11.67 and P11.68

60 mm

100 mm

50 mm

0.60 m

0.40 m

40 mm

Fig. P11.69

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731

Problems

11.70 The thin-walled hollow cylindrical member AB has a noncircular

cross section of nonuniform thickness. Using the expression given

in Eq. (3.53) of Sec. 3.13, and the expression for the strain-energy

density given in Eq. (11.19), show that the angle of twist of mem-

ber AB is

f 5

where ds is an element of the center line of the wall cross section

and A is the area enclosed by that center line.

11.71 Each member of the truss shown has a uniform cross-sectional area A.

Using the method of work and energy, determine the vertical

deflection of the point of application of the load P.

Fig. P11.70

Fig. P11.71

11.72 Each member of the truss shown is made of steel and has a cross-

sectional area of 400 mm

. Using E 5 200 GPa, determine the

deflection of point D caused by the 16-kN load.

11.73 Each member of the truss shown is made of steel and has a cross-

sectional area of 5 in

. Using E 5 29 3 10

psi, determine the

vertical deflection of point B caused by the 20-kip load.

16 kN

1.5 m

0.8 m

Fig. P11.72

6 ft 6 ft

20 kips

2.5 ft

Fig. P11.73

6 ft 6 ft

2.5 ft

15 kips

Fig. P11.74

11.74 Each member of the truss shown is made of steel and has a uni-

form cross-sectional area of 5 in

. Using E 5 29 3 10

psi, deter-

mine the vertical deflection of joint C caused by the application of

the 15-kip load.

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11.75 Each member of the truss shown is made of steel; the cross-

sectional area of member BC is 800 mm

and for all other members

the cross-sectional area is 400 mm

. Using E 5 200 GPa, determine

the deflection of point D caused by the 60-kN load.

11.76 The steel rod BC has a 24-mm diameter and the steel cable

ABDCA has a 12-mm diameter. Using E 5 200 GPa, determine

the deflection of joint D caused by the 12-kN load.

12 kN

360 mm

480 mm 480 mm

Fig. P11.76

1.2 m

0.5 m

1.2 m

60 kN

Fig. P11.75

*11.11 WORK AND ENERGY UNDER SEVERAL LOADS

In this section, the strain energy of a structure subjected to several

loads will be considered and will be expressed in terms of the loads

and the resulting deflections.

Consider an elastic beam AB subjected to two concentrated loads

and P

. The strain energy of the beam is equal to the work of P

and P

as they are slowly applied to the beam at C

and C

, respectively

(Fig. 11.33). However, in order to evaluate this work, we must first

express the deflections x

and x

in terms of the loads P

and P

Fig. 11.33 Beam with multiple loads.

Fig. 11.34

Fig. 11.35

Let us assume that only P

is applied to the beam (Fig. 11.34).

We note that both C

and C

are deflected and that their deflections

are proportional to the load P

. Denoting these deflections by x

and

, respectively, we write

5 a

(11.54)

where a

and a

are constants called influence coefficients. These

constants represent the deflections of C

and C

, respectively, when

a unit load is applied at C

and are characteristics of the beam AB.

Let us now assume that only P

is applied to the beam

(Fig. 11.35). Denoting by x

and x

, respectively, the resulting

deflections of C

and C

, we write

5 a

(11.55)

732

Energy Methods

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733

where a

and a

are the influence coefficients representing the

deflections of C

and C

, respectively, when a unit load is applied at

. Applying the principle of superposition, we express the deflections

and x

of C

and C

when both loads are applied (Fig. 11.33) as

5 x

1 x

5 a

1 a

(11.56)

5 x

1 x

5 a

1 a

(11.57)

To compute the work done by P

and P

, and thus the strain

energy of the beam, it is convenient to assume that P

is first applied

slowly at C

(Fig. 11.36a). Recalling the first of Eqs. (11.54), we

express the work of P

(11.58)

and note that P

does no work while C

moves through x

, since it

has not yet been applied to the beam.

Now we slowly apply P

at C

(Fig. 11.36b); recalling the sec-

ond of Eqs. (11.55), we express the work of P

(11.59)

But, as P

is slowly applied at C

, the point of application of P

moves

through x

from C9

to C

, and the load P

does work. Since P

fully applied during this displacement (Fig. 11.37), its work is equal

to P

or, recalling the first of Eqs. (11.55),

5 P

5 a

(11.60)

11.11 Work and Energy under Several Loads

(b)

(a)

Fig. 11.36

(a) Load-displacement

diagram for C

(b) Load-displacement

diagram for C

Fig. 11.37 Load-displacement diagrams.

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734

Energy Methods

*11.12 CASTIGLIANO’S THEOREM

We recall the expression obtained in the preceding section for the strain

energy of an elastic structure subjected to two loads P

and P

U 5

1 2a

1 a

(11.61)

where a

, a

, and a

are the influence coefficients associated with

the points of application C

and C

of the two loads. Differentiating

both members of Eq. (11.61) with respect to P

and recalling Eq.

(11.56), we write

5 a

1 a

5 x

(11.63)

(a) Load-displacement

diagram for C

(b) Load-displacement

diagram for C

Fig. 11.39 Alternative load-displacement diagrams.

Adding the expressions obtained in (11.58), (11.59), and (11.60), we

express the strain energy of the beam under the loads P

and P

U 5

1 2a

1 a

(11.61)

If the load P

had first been applied to the beam (Fig. 11.38a),

and then the load P

(Fig. 11.38b), the work done by each load would

have been as shown in Fig. 11.39. Calculations similar to those we

have just carried out would lead to the following alternative expres-

sion for the strain energy of the beam:

U 5

1 2a

1 a

(11.62)

Equating the right-hand members of Eqs. (11.61) and (11.62), we find

that a

5 a

, and thus conclude that the deflection produced at C

a unit load applied at C

is equal to the deflection produced at C

by a

unit load applied at C

. This is known as Maxwell’s reciprocal theorem,

after the British physicist James Clerk Maxwell (1831–1879).

While we are now able to express the strain energy U of a

structure subjected to several loads as a function of these loads, we

cannot use the method of Sec. 11.10 to determine the deflection of

such a structure. Indeed, computing the strain energy U by integrat-

ing the strain-energy density u over the structure and substituting

the expression obtained into (11.61) would yield only one equation,

which clearly could not be solved for the various coefficients a.

(b)

(a)

Fig. 11.38

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735

Differentiating both members of Eq. (11.61) with respect to P

recalling Eq. (11.57), and keeping in mind that a

5 a

, we have

5 a

1 a

5 x

(11.64)

More generally, if an elastic structure is subjected to n loads

, P

, . . ., P

, the deflection x

of the point of application of P

measured along the line of action of P

, can be expressed as the

partial derivative of the strain energy of the structure with respect

to the load P

. We write

(11.65)

This is Castigliano’s theorem, named after the Italian engineer

Alberto Castigliano (1847–1884) who first stated it.†

Recalling that the work of a couple M is

Mu, where u is the

angle of rotation at the point where the couple is slowly applied, we

note that Castigliano’s theorem may be used to determine the slope

of a beam at the point of application of a couple M

. We have

(11.68)

Similarly, the angle of twist f

in a section of a shaft where a torque

is slowly applied is obtained by differentiating the strain energy

of the shaft with respect to T

(11.69)

†In the case of an elastic structure subjected to n loads P

, P

, . . ., P

, the deflection of

the point of application of P

, measured along the line of action of P

, can be expressed as

(11.66)

and the strain energy of the structure is found to be

U 5

(11.67)

Differentiating U with respect to P

, and observing that P

is found in terms corresponding

to either i 5 j or k 5 j, we write

or, since a

5 a

Recalling Eq. (11.66), we verify that

(11.65)

11.12 Castigliano’s Theorem

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736

Energy Methods

*11.13 DEFLECTIONS BY CASTIGLIANO’S THEOREM

We saw in the preceding section that the deflection x

of a structure

at the point of application of a load P

can be determined by comput-

ing the partial derivative 0U

of the strain energy U of the struc-

ture. As we recall from Secs. 11.4 and 11.5, the strain energy U is

obtained by integrating or summing over the structure the strain

energy of each element of the structure. The calculation by Castigli-

ano’s theorem of the deflection x

is simplified if the differentiation

with respect to the load P

is carried out before the integration or

summation.

In the case of a beam, for example, we recall from Sec. 11.4

that

U 5

2EI

(11.17)

and determine the deflection x

of the point of application of the load

by writing

(11.70)

In the case of a truss consisting of n uniform members of length

, cross-sectional area A

, and internal force F

, we recall Eq. (11.14)

and express the strain energy U of the truss as

U 5

i51

(11.71)

The deflection x

of the point of application of the load P

is obtained

by differentiating with respect to P

each term of the sum. We

write

i51

(11.72)

EXAMPLE 11.12

The cantilever beam AB supports a uniformly distributed load w and

a concentrated load P as shown (Fig. 11.40). Knowing that L 5 2 m,

w 5 4 kN/m, P 5 6 kN, and EI 5 5 MN ? m

, determine the deflec-

tion at A.

Fig. 11.40

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