[Solution Manual] Mechanics of Material, 7th Edition - James M. Gere y Barry J. Goodno PDF

Title	[Solution Manual] Mechanics of Material, 7th Edition - James M. Gere y Barry J. Goodno
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1 Tension, Compression, and Shear

P1

Normal Stress and Strain A

Problem 1.2-1 A solid circular post ABC (see figure) supports a load P1 2500 lb acting at the top. A second load P2 is uniformly distributed around the shelf at B. The diameters of the upper and lower parts of the post are dAB 1.25 in. and dBC 2.25 in., respectively.

dAB P2 B

(a) Calculate the normal stress AB in the upper part of the post. (b) If it is desired that the lower part of the post have the same compressive stress as the upper part, what should be the magnitude of the load P2?

dBC

C

Solution 1.2-1 P1 2500 lb

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Circular post in compression

ALTERNATE SOLUTION FOR PART (b) P1 P2 P1 P2 2 ABC 4 dBC P1 P1 sAB 2 sBC sAB AAB 4 dAB

dAB 1.25 in.

sBC

dBC 2.25 in. (a) NORMAL STRESS IN PART AB P1 2500 lb sAB 2040 psi AAB 4 (1.25 in.) 2

dBC 2 P1 P2 P1 2 or P2 P1 B ¢ ≤ 1R 2 dAB dBC dAB dBC 1.8 dAB ∴ P2 2.24 P1 5600 lb

(b) LOAD P2 FOR EQUAL STRESSES sBC

P1 P2 2500 lb P2 2 ABC 4 (2.25 in.)

P1

AB 2040 psi

A

Solve for P2: P2 5600 lb P2 B

C

1

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2

CHAPTER 1

Tension, Compression, and Shear

Problem 1.2-2 Calculate the compressive stress c in the circular piston rod (see figure) when a force P 40 N is applied to the brake pedal. Assume that the line of action of the force P is parallel to the piston rod, which has diameter 5 mm. Also, the other dimensions shown in the figure (50 mm and 225 mm) are measured perpendicular to the line of action of the force P.

50 mm 5 mm 225 mm P = 40 N Piston rod

Solution 1.2-2

Free-body diagram of brake pedal EQUILIBRIUM OF BRAKE PEDAL

50 mm

©MA 0

A F

F(50 mm) P(275 mm) 0

225 mm P = 40 N

F P¢

275 mm 275 ≤ (40 N) ¢ ≤ 220 N 50 mm 50

COMPRESSIVE STRESS IN PISTON ROD (d 5 mm)

F compressive force in piston rod

sc

d diameter of piston rod

F 220 N 11.2 MPa A 4 (5 mm) 2

5 mm

Problem 1.2-3 A steel rod 110 ft long hangs inside a tall tower and holds a 200-pound weight at its lower end (see figure). If the diameter of the circular rod is 1⁄4 inch, calculate the maximum normal stress max in the rod, taking into account the weight of the rod itself. (Obtain the weight density of steel from Table H-1, Appendix H.)

110 ft 1 — in. 4

200 lb

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www.elsolucionario.org SECTION 1.2

Solution 1.2-3

3

Normal Stress and Strain

Long steel rod in tension P 200 lb

smax

L 110 ft

d

gL (490 lbft3 )(110 ft) ¢

d 1⁄4 in.

L

WP P gL A A 1 ft2 ≤ 144 in.2

374.3 psi P 200 lb 4074 psi A 4 (0.25 in.) 2

Weight density: 490 lb/ft3 W Weight of rod (Volume)

max 374 psi 4074 psi 4448 psi

AL

Rounding, we get max 4450 psi P = 200 lb

Problem 1.2-4 A circular aluminum tube of length L 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction.

Strain gage P

P L = 400 mm

(a) If the measured strain is 550 106, what is the shortening of the bar? (b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?

Solution 1.2-4

Aluminum tube in compression Strain gage P

P

e 550 106

(b) COMPRESSIVE LOAD P

L 400 mm d1 50 mm

40 MPa A [d22 d12] [ (60 mm) 2 (50 mm) 2 ] 4 4 863.9 mm2

(a) SHORTENING OF THE BAR

P A (40 MPa)(863.9 mm2)

d2 60 mm

eL (550 106)(400 mm)

34.6 kN

0.220 mm

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4

CHAPTER 1

Tension, Compression, and Shear

y

Problem 1.2-5 The cross section of a concrete pier that is loaded uniformly in compression is shown in the figure.

20 in.

(a) Determine the average compressive stress c in the concrete if the load is equal to 2500 k. (b) Determine the coordinates x and y of the point where the resultant load must act in order to produce uniform normal stress.

16 in. 16 in.

48 in.

16 in. O

Solution 1.2-5

(a) AVERAGE COMPRESSIVE STRESS c P 2500 k

16 in. x

C

16 in.

2 1

y 16 in. O

x

16 in.

Concrete pier in compression

y

48 in.

20 in.

20 in.

16 in.

3 4

x

(b) COORDINATES OF CENTROID C 1 From symmetry, y (48 in.) 24 in. 2 © xi Ai (see Chapter 12, Eq. 12-7a) A 1 x (x1 A1 2x2 A2 x3 A3 ) A x

USE THE FOLLOWING AREAS: A1 (48 in.)(20 in.) 960 in.2 1 A2 A4 (16 in.)(16 in.) 128 in.2 2 A3 (16 in.)(16 in.) 256 in.2

1 [ (10 in.)(960 in.2 ) 1472 in.2

2(25.333 in.)(128 in.2)

A A1 A2 A3 A4 (960 128 256 128)

P 2500 k 1.70 ksi A 1472 in.2

sc

(28 in.)(256 in.2)] in.2

1472 in.2

15.8 in.

Problem 1.2-6 A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). The cable has an effective cross-sectional area of 490 mm2, and the angle of the incline is 30°. Calculate the tensile stress t in the cable.

Cable

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SECTION 1.2

Solution 1.2-6

5

Normal Stress and Strain

Car on inclined track

FREE-BODY DIAGRAM OF CAR W

TENSILE STRESS IN THE CABLE W Weight of car

R2 R1

st

T W sin A A

T Tensile force in cable

SUBSTITUTE NUMERICAL VALUES:

Angle of incline

W 130 kN 30

A Effective area of cable

A 490 mm2 st

R1, R2 Wheel reactions (no friction force between wheels and rails)

(130 kN)(sin 30) 490 mm2

133 MPa

EQUILIBRIUM IN THE INCLINED DIRECTION ©FT 0Q b T W sin 0 T W sin Problem 1.2-7 Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle 34° to the horizontal and wire BC is at an angle 48°. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) Determine the tensile stresses AB and BC in the two wires.

C A

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B

Solution 1.2-7 Two steel wires supporting a lamp FREE-BODY DIAGRAM OF POINT B

SUBSTITUTE NUMERICAL VALUES: TAB(0.82904) TBC(0.66913) 0

TBC

TAB

34

d 30 mils 0.030 in.

y

A

W = 18 lb 0

48

d 2 706.9 10 6 in.2 4

x

TAB(0.55919) TBC(0.74314) 18 0 SOLVE THE EQUATIONS: TAB 12.163 lb TBC 15.069 lb TENSILE STRESSES IN THE WIRES TAB 17,200 psi A TBC sBC 21,300 psi A sAB

EQUATIONS OF EQUILIBRIUM Fx 0 Fy 0

TAB cos TBC cos 0 TAB sin TBC sin W 0

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www.elsolucionario.org 6

CHAPTER 1

Tension, Compression, and Shear

Problem 1.2-8 A long retaining wall is braced by wood shores set at an angle of 30° and supported by concrete thrust blocks, as shown in the first part of the figure. The shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are idealized as shown in the second part of the figure. Note that the base of the wall and both ends of the shores are assumed to be pinned. The pressure of the soil against the wall is assumed to be triangularly distributed, and the resultant force acting on a 3-meter length of the wall is F 190 kN. If each shore has a 150 mm 150 mm square cross section, what is the compressive stress c in the shores?

Solution 1.2-8

30°

B F 30°

1.5 m A

C 0.5 m

4.0 m

F 190 kN

Wall

A area of one shore

Shore

F 30°

1.5 m A

A (150 mm)(150 mm)

C

22,500 mm2

0.5 m

0.0225 m2

4.0 m

FREE-BODY DIAGRAM OF WALL AND SHORE

SUMMATION OF MOMENTS ABOUT POINT A ©MA 0

B

F(1.5 m)CV (4.0 m)CH (0.5 m) 0 30°

A AH

CH CV

30°

C

AV

C compressive force in wood shore CH horizontal component of C CV vertical component of C CH C cos 30

Retaining wall Concrete Shore thrust block

Retaining wall braced by wood shores

B

F 1.5 m

Soil

or (190 kN)(1.5 m) C(sin 30)(4.0 m) C(cos 30)(0.5 m) 0 ∴ C 117.14 kN COMPRESSIVE STRESS IN THE SHORES sc

C 117.14 kN A 0.0225 m2 5.21 MPa

CV C sin 30

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SECTION 1.2

Problem 1.2-9 A loading crane consisting of a steel girder ABC supported by a cable BD is subjected to a load P (see figure). The cable has an effective cross-sectional area A 0.471 in2. The dimensions of the crane are H 9 ft, L1 12 ft, and L2 4 ft.

7

Normal Stress and Strain

D

Cable H

(a) If the load P 9000 lb, what is the average tensile stress in the cable? (b) If the cable stretches by 0.382 in., what is the average strain?

Girder B

A L1

C L2 P

Solution 1.2-9

Loading crane with girder and cable EQUILIBRIUM

D

©MA 0 TV (12 ft) (9000 lb)(16 ft) 0 TV 12,000 lb TH L1 12 ft TV H 9 ft 12 ∴ TH TV ¢ ≤ 9

H

B

A L1

H 9 ft

C L2 P = 9000 lb

L1 12 ft

L2 4 ft

TH (12,000 lb) ¢

12 ≤ 9

16,000 lb

A effective area of cable

A 0.471

TENSILE FORCE IN CABLE

in.2

T TH2 TV2 (16,000 lb) 2 (12,000 lb) 2

P 9000 lb

20,000 lb

FREE-BODY DIAGRAM OF GIRDER T

(a) AVERAGE TENSILE STRESS IN CABLE

TV

s TH

A 12 ft

B

C

(b) AVERAGE STRAIN IN CABLE L length of cableL H 2 L21 15 ft

4 ft P = 9000 lb

T tensile force in cable

T 20,000 lb 42,500 psi A 0.471 in.2

stretch of cable e

P 9000 lb

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0.382 in.

0.382 in. 2120 10 6 L (15 ft)(12 in.ft)

8

CHAPTER 1

Tension, Compression, and Shear

Problem 1.2-10 Solve the preceding problem if the load P 32 kN; the cable has effective cross-sectional area A 481 mm2; the dimensions of the crane are H 1.6 m, L1 3.0 m, and L2 1.5 m; and the cable stretches by 5.1 mm. Figure is with Prob. 1.2-9. Solution 1.2-10

Loading crane with girder and cable

D

H

B

A L1

H 1.6 m

L1 3.0 m

L2 1.5 m

A effective area of cable

A 481 mm2

P 32 kN

C L2 P = 32 kN

TENSILE FORCE IN CABLE

FREE-BODY DIAGRAM OF GIRDER T

TH

A 3.0 m

TV

T = tensile force in cable

B

C

T TH2 TV2 (90 kN) 2 (48 kN) 2 102 kN (a) AVERAGE TENSILE STRESS IN CABLE

1.5 m

P = 32 kN

EQUILIBRIUM ©MA 0

T 102 kN 212 MPa A 481 mm2

s

(b) AVERAGE STRAIN IN CABLE

TV (3.0 m) (32 kN)(4.5 m) 0

L length of cable

TV 48 kN TH L1 3.0 m TV H 1.6 m 3.0 ∴ TH TV ¢ ≤ 1.6 3.0 TH (48 kN) ¢ ≤ 1.6

L H 2 L21 3.4 m stretch of cable 5.1 mm e

5.1 mm 1500 10 6 L 3.4 m

90 kN Problem 1.2-11 A reinforced concrete slab 8.0 ft square and 9.0 in. thick is lifted by four cables attached to the corners, as shown in the figure. The cables are attached to a hook at a point 5.0 ft above the top of the slab. Each cable has an effective cross-sectional area A 0.12 in2. Determine the tensile stress t in the cables due to the weight of the concrete slab. (See Table H-1, Appendix H, for the weight density of reinforced concrete.)

Cables

Reinforced concrete slab

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www.elsolucionario.org SECTION 1.2

Solution 1.2-11

9

Normal Stress and Strain

Reinforced concrete slab supported by four cables

W

T tensile force in a cable Cable AB:

A

TV H T LAB

H Cable

TV T ¢ t

H H L22 2

≤

(Eq. 1)

EQUILIBRIUM

B

Fvert 0 ↑ ↓

L

L

W 4TV 0 H height of hook above slab

TV

L length of side of square slab t thickness of slab

W 4

(Eq. 2)

COMBINE EQS. (1) & (2):

weight density of reinforced concrete T¢

W weight of slab L2t D length of diagonal of slab L2

H H L 2

T

2

2

≤

W 4

W H 2 L22 W 1 L22H 2 4 H 4

DIMENSIONS OF CABLE AB TENSILE STRESS IN A CABLE

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LAB H B

LAB length of cable L2 H2 B 2

D= L 2 2

FREE-BODY DIAGRAM OF HOOK AT POINT A

A effective cross-sectional area of a cable st

T W 1 L22H2 A 4A

SUBSTITUTE NUMERICAL VALUES AND OBTAIN t : H 5.0 ft

L 8.0 ft

150

A 0.12

lb/ft3

t 9.0 in. 0.75 ft

in.2

W L2t 7200 lb W

TH

st A

T

T

W 1 L22H2 22,600 psi 4A

TV T

T

T

Problem 1.2-12 A round bar ACB of length 2L (see figure) rotates about an axis through the midpoint C with constant angular speed (radians per second). The material of the bar has weight density . (a) Derive a formula for the tensile stress x in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress max?

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A

C L

B

x L

10

CHAPTER 1

Tension, Compression, and Shear

Solution 1.2-12 D

Rotating Bar dM

Consider an element of mass dM at distance from the midpoint C. The variable ranges from x to L.

B

C x

g dM g A dj

d

L

dF Inertia force (centrifugal force) of element of mass dM g dF (dM)(j 2 ) g A 2jdj

angular speed (rad/s) A cross-sectional area

Fx

weight density

B

dF

D

g g mass density

L

x

g gA 2 2 A 2jdj (L x 2) g 2g

(a) TENSILE STRESS IN BAR AT DISTANCE x Fx g 2 2 (L x 2) — A 2g

sx We wish to find the axial force Fx in the bar at Section D, distance x from the midpoint C. The force Fx equals the inertia force of the part of the rotating bar from D to B.

(b) MAXIMUM TENSILE STRESS x 0smax

g 2L2 — 2g

Mechanical Properties of Materials Problem 1.3-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon. (a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea water from Table H-1, Appendix H.)

Solution 1.3-1

Hanging wire of length L W total weight of steel wire S weight density of steel

L

Lmax

11,800 ft

490 lb/ft3 W weight density of sea water

63.8 lb/ft3

A cross-sectional area of wire max 40 ksi (yield strength)

(b) WIRE HANGING IN SEA WATER F tensile force at top of wire F (gS gW ) ALsmax Lmax

(a) WIRE HANGING IN AIR W S AL W smax gSL A

smax 40,000 psi (144 in.2ft2 ) gS 490 lbft3

F (gS gW )L A

smax gS gW 40,000 psi (144 in.2ft2 ) (490 63.8)lbft3

13,500 ft

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SECTION 1.3

Mechanical Properties of Materials

11

Problem 1.3-2 Imagine that a long wire of tungsten hangs vertically from a high-altitude balloon. (a) What is the greatest length (meters) it can have without breaking if the ultimate strength (or breaking strength) is 1500 MPa? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of tungsten and sea water from Table H-1, Appendix H.) Solution 1.3-2

Hanging wire of length L W total weight of tungsten wire T weight density of tungsten 190

L

kN/m3

F tensile force at top of wire F (TW)AL

W weight density of sea water 10.0

(b) WIRE HANGING IN SEA WATER

kN/m3

A cross-sectional area of wire max 1500 MPa (breaking strength)

F (gT gW )L A smax Lmax gT gW

smax

(a) WIRE HANGING IN AIR

8300 m

W T AL smax

W gTL A

Lmax

smax 1500 MPa gT 190 kNm3

1500 MPa (190 10.0) kNm3

7900 m

Problem 1.3-3 Three different materials, designated A, B, and C, are tested in tension using test specimens having diameters of 0.505 in. and gage lengths of 2.0 in. (see figure). At failure, the distances between the gage marks are found to be 2.13, 2.48, and 2.78 in., respectively. Also, at the failure cross sections the diameters are found to be 0.484, 0.398, and 0.253 in., respectively. ...