Solucionario mecanica materiales gere 6ed
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Publicado: 11 de junio de 2011
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 thatthe lower part of the post have the
same compressive stress as the upper part, what should be
the magnitude of the load P2?
Solution 1.2-1 Circular post in compression
1
Tension, Compression,
and Shear
1
A
B
C
P1
dAB
dBC
P2
P1 2500 lb
dAB 1.25 in.
dBC 2.25 in.
(a) NORMAL STRESS IN PART AB
(b) LOAD P2 FOR EQUAL STRESSES
AB 2040 psi
Solve for P2: P2 5600 lb
sBCP1P2
ABC
2500 lbP2
4
(2.25 in.)2
sAB
P1
AAB
2500 lb
4
(1.25 in.)2 2040 psi
ALTERNATE SOLUTION FOR PART (b)
∴ P22.24 P15600 lb
dBC
dAB
1.8
P1P2
dBC
2
P1
dAB
2 or P2P1 B ¢
dBC
dAB
≤
2
1 R
sAB
P1
AAB
P1
4 dAB
2 sBCsAB
sBC
P1P2
ABC
P1P2
4 dBC
2
A
B
C
P1
P2
Piston rod
5 mm
50 mm
225 mm
P = 40 N
Problem 1.2-2 Calculatethe 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.
Solution 1.2-2 Free-body diagram of brakepedal
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.)
2 CHAPTER 1 Tension, Compression, and ShearF compressive force in piston rod
d diameter of piston rod
5 mm
EQUILIBRIUM OF BRAKE PEDAL
F(50 mm) P(275 mm) 0
COMPRESSIVE STRESS IN PISTON ROD (d 5 mm)
sc
F
A
220 N
4
(5 mm)2 11.2 MPa
FP¢
275 mm
50 mm
≤ (40 N) ¢
275
50
≤220 N
©MA0
50 mm
225 mm
P = 40 N
F
A
in.
200 lb
110 ft
1
4 —
Solution 1.2-3 Long steel rod in tension
Problem 1.2-4 Acircular 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 550 106, what is the
shortening of the bar?
(b) If the compressive stress in the baris intended to be 40
MPa, what should be the load P?
Solution 1.2-4 Aluminum tube in compression
SECTION 1.2 Normal Stress and Strain 3
374.3 psi
max 374 psi 4074 psi 4448 psi
Rounding, we get
max 4450 psi
P
A
200 lb
4
(0.25 in.)2 4074 psi
gL (490 lbft3)(110 ft)¢
1
144
ft2
in.2 ≤
smax
WP
A
gL
P
A
d
L
P = 200 lb
P 200 lb
L 110 ft
d 1⁄4 in.Weight density: 490 lb/ft3
W Weight of rod
(Volume)
AL
Strain gage
L = 400 mm
P P
e 550 106
L 400 mm
d2 60 mm
d1 50 mm
(a) SHORTENING OF THE BAR
eL (550 106)(400 mm)
0.220 mm
(b) COMPRESSIVE LOAD P
40 MPa
863.9mm2
P A (40 MPa)(863.9 mm2)
34.6 kN
A
4
[d2
2d1
2]
4
[ (60 mm)2 (50 mm)2 ]
Strain gage
P P
Problem 1.2-5The 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.
Solution 1.2-5 Concrete pier in compression
4 CHAPTER 1...
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 thatthe lower part of the post have the
same compressive stress as the upper part, what should be
the magnitude of the load P2?
Solution 1.2-1 Circular post in compression
1
Tension, Compression,
and Shear
1
A
B
C
P1
dAB
dBC
P2
P1 2500 lb
dAB 1.25 in.
dBC 2.25 in.
(a) NORMAL STRESS IN PART AB
(b) LOAD P2 FOR EQUAL STRESSES
AB 2040 psi
Solve for P2: P2 5600 lb
sBCP1P2
ABC
2500 lbP2
4
(2.25 in.)2
sAB
P1
AAB
2500 lb
4
(1.25 in.)2 2040 psi
ALTERNATE SOLUTION FOR PART (b)
∴ P22.24 P15600 lb
dBC
dAB
1.8
P1P2
dBC
2
P1
dAB
2 or P2P1 B ¢
dBC
dAB
≤
2
1 R
sAB
P1
AAB
P1
4 dAB
2 sBCsAB
sBC
P1P2
ABC
P1P2
4 dBC
2
A
B
C
P1
P2
Piston rod
5 mm
50 mm
225 mm
P = 40 N
Problem 1.2-2 Calculatethe 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.
Solution 1.2-2 Free-body diagram of brakepedal
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.)
2 CHAPTER 1 Tension, Compression, and ShearF compressive force in piston rod
d diameter of piston rod
5 mm
EQUILIBRIUM OF BRAKE PEDAL
F(50 mm) P(275 mm) 0
COMPRESSIVE STRESS IN PISTON ROD (d 5 mm)
sc
F
A
220 N
4
(5 mm)2 11.2 MPa
FP¢
275 mm
50 mm
≤ (40 N) ¢
275
50
≤220 N
©MA0
50 mm
225 mm
P = 40 N
F
A
in.
200 lb
110 ft
1
4 —
Solution 1.2-3 Long steel rod in tension
Problem 1.2-4 Acircular 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 550 106, what is the
shortening of the bar?
(b) If the compressive stress in the baris intended to be 40
MPa, what should be the load P?
Solution 1.2-4 Aluminum tube in compression
SECTION 1.2 Normal Stress and Strain 3
374.3 psi
max 374 psi 4074 psi 4448 psi
Rounding, we get
max 4450 psi
P
A
200 lb
4
(0.25 in.)2 4074 psi
gL (490 lbft3)(110 ft)¢
1
144
ft2
in.2 ≤
smax
WP
A
gL
P
A
d
L
P = 200 lb
P 200 lb
L 110 ft
d 1⁄4 in.Weight density: 490 lb/ft3
W Weight of rod
(Volume)
AL
Strain gage
L = 400 mm
P P
e 550 106
L 400 mm
d2 60 mm
d1 50 mm
(a) SHORTENING OF THE BAR
eL (550 106)(400 mm)
0.220 mm
(b) COMPRESSIVE LOAD P
40 MPa
863.9mm2
P A (40 MPa)(863.9 mm2)
34.6 kN
A
4
[d2
2d1
2]
4
[ (60 mm)2 (50 mm)2 ]
Strain gage
P P
Problem 1.2-5The 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.
Solution 1.2-5 Concrete pier in compression
4 CHAPTER 1...
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