Solucionario solidos

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1
Tension, Compression, and Shear

Normal Stress and Strain
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. (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 FOR PART (b) sBC sAB 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 P12500 lb sAB 2040 psi AAB 4 (1.25 in.) 2 (b) LOAD P FOR EQUAL STRESSES 2 sBC P1 P2 ABC
AB

P1 P2 2 dBC

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

P1 A

∴ P2

dBC 1.8 dAB 2.24 P1 5600 lb

P1 B ¢

sAB dBC 2 ≤ dAB

1R

Solve for P2: P2

P2 B

C

1

2

CHAPTER 1

Tension, Compression, and Shear

Problem 1.2-2 Calculate the compressive stress c in the circular pistonrod (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
50mm 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



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 (seefigure). 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.)
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 ft
1

smax gL 490 lb/ft3 P A
maxW A

P

gL

d L

L d

⁄4 in.

Weight density: W Weight of rod (Volume) AL

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 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 insidediameters 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 A
2 [d 2 d1 ] 4 2 863.9 mm2

400 mm 60 mm 50 mm
OF 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 isloaded 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 1.2-5
y

Concrete pier in compression (a)...
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