# 01-01 chap gere

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1Tension, Compression, and Shear

Normal Stress and StrainProblem 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. (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?C A

P1

dAB P2 B dBC

Solution 1.2-1 P1 2500 lb dAB dBC 1.25 in. 2.25 in.

Circular post in compression ALTERNATE SOLUTION sBC sAB 2040 psi P1 P2 P1 P2 2 ABC 4 dBC P1 P1 sBC 2 AAB 4 dAB P1 2 or P2 dAB

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

P1 P2 2 dBC

(b) LOAD P FOR EQUAL STRESSES 2 sBC P1 P2 ABCAB

2500 lb P2 2 4 (2.25 in.) 2040 psi 5600 lb B

P1 A

P2

dBC 1.8 dAB 2.24 P1 5600 lb

P1 B

sAB dBC 2 dAB

1R

Solve for P2: P2

P2

C

1

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 pedal50 mm A

EQUILIBRIUM OF BRAKE PEDAL MA 0 275 mm 50 mm F(50 mm) (40 N) P(275 mm) 275 50 0 220 N 5 mm)

F

225 mm P = 40 N

F

COMPRESSIVE STRESS IN PISTON ROD (d F d compressive force in piston rod diameter of piston rod 5 mm sc F A 220 N (5 mm) 2 4 11.2 MPa

P

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 14 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.)1 in. 4

110 ft

200 lb

SECTION 1.2

Normal Stress and Strain

3

Solution 1.2-3

Long steel rod in tension P 200 lb 110 ft1

smax gL 490 lb/ft3 P Amax

W A

P

gL

d L

L d

4 in.

Weight density: W Weight of rod (Volume) AL P 200 lb

374.3 psi 200 lb 2 4 (0.25 in.) 374 psi

(490 lb ft3 )(110 ft) 4074 psi

P A

4074 psi

1 ft2 144 in.2 4448 psi

Rounding, we getmax

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. (a) If the measured strain is 550 10 6, 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?

Strain gage P L = 400 mm P

Solution 1.2-4

Aluminum tube in compressionStrain gage P P

e L d2 d1

550

10

6

(b) COMPRESSIVE LOAD P 40 MPa A2 [d 2 d1 ] 4 2 863.9 mm2

400 mm 60 mm 50 mmOF THE BAR

4

[ (60 mm) 2

(50 mm) 2 ]

(a) SHORTENING eL

P

A

(40 MPa)(863.9 mm2)

(550

10 6)(400 mm)

34.6 kN

0.220 mm

4

CHAPTER 1

Tension, Compression, and Shear

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

y 20 in.

16 in. 48 in. 16 in. 16 in. O 20 in. 16 in. x

Solution 1.2-5y

Concrete pier in compression (a) AVERAGE COMPRESSIVE STRESS16 in.c

P sc

2500 k P A 2500 k 1472 in.2 1.70 ksi

x 48 in.

C y

16 in. 1 16 in.

2 3 4 x

(b) COORDINATES OF CENTROID c From symmetry, y x x 1 (48 in.) 2 24 in. 7a)

O

20 in.

16 in.

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

xi Ai (see Chapter 12, Eq. 12 A 1 (x A 2x2 A2 x3 A3 ) A 1 1 1 [ (10 in.)(960 in.2 ) 1472 in.2 2(25.333 in.)(128 in.2) (28 in.)(256 in.2)]

1 (16 in.)(16 in.) 2

(16 in.)(16 in.) A1 A2 (960 A3

256 in.2 A4 256 128) in.2

128

15.8 in.

1472 in.2

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

SECTION 1.2

Normal Stress and Strain

5

Solution 1.2-6

Car on inclined track TENSILE STRESS IN THE CABLE W T Weight of car Tensile force in cable Angle of inclineR2

FREE-BODY DIAGRAM OF CARW

st

T A

W sin A

SUBSTITUTE NUMERICAL VALUES: W A st 130 kN 490 mm2 (130 kN)(sin 30 ) 490 mm2 133 MPa 30

A

R1

Effective area of cable

R1, R2

Wheel reactions (no friction force between wheels and rails)

EQUILIBRIUM IN THE INCLINED DIRECTION FT T 0 Q b T W sin 0

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

B

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

SUBSTITUTE NUMERICAL VALUES: TAB(0.82904) 34 d 30 mils d2 4 48 0.030 in. 106

TBC(0.66913) TBC(0.74314)

0 18 0

TAB(0.55919)

SOLVE THE EQUATIONS: in.2 TAB 12.163 lb TBC 15.069 lb

y W = 18 lb 0 x

A

706.9

TENSILE STRESSES IN THE WIRES sAB TAB A TBC A 17,200 psi 21,300 psi

EQUATIONS OF EQUILIBRIUM Fx Fy 0 0 TAB cos TAB sin TBC cos TBC sin 0 W 0 sBC

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?

Soil

Retaining wall Concrete Shore thrust block 30

B F 1.5 m A 30 0.5 m 4.0 m C

Solution 1.2-8

Retaining wall braced by wood shores

Wall B F 1.5 m A 4.0 m 30 0.5 m C Shore

F A A

190 kN area of one shore (150 mm)(150 mm) 22,500 mm2 0.0225 m2

FREE-BODY DIAGRAM OF WALL AND SHOREB F 1.5 m 30 A AH AV CV CH30

SUMMATION OF MOMENTS ABOUT POINT A MA 0 0

F(1.5 m) CV (4.0 m) CH (0.5 m) or (190 kN)(1.5 m) C 117.14 kN C(sin 30 )(4.0 m)C

C(cos 30 )(0.5 m)

0

C CH CV CH CV

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

COMPRESSIVE STRESS IN THE SHORES sc C A 117.14 kN 0.0225 m2 5.21 MPa

SECTION 1.2

Normal Stress and Strain

7

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. (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?

D

Cable H Girder A L1 B L2 P C

Solution 1.2-9

Loading crane with girder and cable EQUILIBRIUM

D

MA

0 (9000 lb)(16 ft) 0

TV (12 ft)H

TV TH TVA L1 B L2 P = 9000 lb C

TH

12,000 lb L1 12 ft H 9 ft 12 TH TV 9 16,000 lb

H L2 A P

9 ft 4 ft 0.471

L1 A in.2

12 ft effective area of cable

(12,000 lb) 2 TH 2 TV

12 9 (16,000 lb) 2 (12,000 lb) 2

TENSILE FORCE IN CABLE T

9000 lb

20,000 lb FREE-BODY DIAGRAM OF GIRDERT TV

(a) AVERAGE TENSILE STRESS IN CABLE s T A 20,000 lb 0.471 in.2 42,500 psi

A 12 ft

TH

B 4 ft

C

(b) AVERAGE STRAIN IN CABLE L length of cable stretch of cable e L 0.382 in. (15 ft)(12 in. ft) L H2 L2 1 15 ft

P = 9000 lb

0.382 in. 2120 106

T P

tensile force in cable 9000 lb

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 w