Solution Manual For
Mechanics of Materials 6th Edition by James M. Gere 재료역학 솔루션
Chapter 1-12EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries,
including engineering, finance, data science, and technology
1
Tension, Compression,
and Shear
Normal Stress and Strain P1
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 dAB
uniformly distributed around the shelf at B. The diameters of P2
the upper and lower parts of the post are dAB = 1.25 in. and
dBC = 2.25 in., respectively.
B
(a) Calculate the normal stress δAB in the upper part of the post.
(b) If it is desired that the lower part of the post have the dBC
same compressive stress as the upper part, what should be
the magnitude of the load P2?
C
Solution 1.2-1 Circular post in compression
P1 = 2500 lb ALTERNATE SOLUTION
dAB = 1.25 in. s = P1 + P2 = P1 + P2
d = 2.25 in. BC w 2
d
ABC
BC 4 BC
P1 P1
(a) NORMAL STRESS IN PART AB sAB = =w sBC = sAB
A d2
AB 4 AB 2
P dBC
P1 2500 lb P1 + P2 1
s = = = or P = P B ¢ ≤ — 1R
AB A w 2 = 2040 psi — d2 d2 2 1 dAB
AB 4 (1.25 in.) BC AB
dBC
= 1.8
(b) LOAD P2 FOR EQUAL STRESSES dAB
P +P 2500 lb + P P1 P2 = 2.24 P1 = 5600 lb —
s = 1 2
= 2
BC w 2 A
ABC 4 (2.25 in.)
P2
B
C
, =δAB = 2040 psi
Solve for P2: P2 = 5600 lb —
EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science,
and technology
Problem 1.2-2 Calculate the compressive stress δc in the circular
50 mm
piston rod (see figure) when a force P = 40 N is applied to the
brake pedal. 5 mm
Assume that the line of action of the force P is parallel to the 225 mm
piston rod, which has diameter 5 mm. Also, the other dimensions
shown in the figure (50 mm and 225 mm) are measured perpen-
P = 40 N
dicular to the line of action of the force P.
Piston rod
Solution 1.2-2 Free-body diagram of brake pedal
50 mm EQUILIBRIUM OF BRAKE PEDAL
A ©MA = 0 ` “
225 mm F(50 mm) — P(275 mm) = 0
F
275 mm 275
= 40 N F = P¢ ≤ = (40 N) ¢ ≤ = 220 N
P
50 mm 50
COMPRESSIVE STRESS IN PISTON ROD (d = 5 mm)
F = compressive force in piston rod F 220 N
d = diameter of piston rod sc = = w = 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, calcu-
late the maximum normal stress δmax 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
, Solution 1.2-3 Long steel rod in tension
P = 200 lb W+P P
s = = gL +
max A
d A
L = 110 ft
1 ft2
3
L
d = 1⁄4 in. gL = (490 lb/ft )(110 ft)¢ ≤
144 in.2
Weight density: g = 490 lb/ft3 = 374.3 psi
W = Weight of rod P 200 lb = 4074 psi
=
w
A 4 (0.25 in.)2
= g(Volume)
δmax= 374 psi + 4074 psi = 4448 psi
= gAL
Rounding, we get
P = 200 lb
P = 200 lb δmax = 4450 psi —
EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science,
and technology
Problem 1.2-4 A circular aluminum tube of length L = 400 Strain gage
mm is loaded in compression by forces P (see figure). The out-
P P
side 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. L = 400 mm
(a) If the measured strain is s = 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?
Solution 1.2-4 Aluminum tube in compression
Strain gage
P P
e = 550 × 10—6 (b) COMPRESSIVE LOAD P
L = 400 mm δ = 4 0 MPa
w w
d = 60 mm A = [ d 2 — d 2] = [ (60 mm)2 — (50 mm)2 ]
2
4 2 1
4
d1 = 50 mm = 863.9 mm2
(a) SHORTENING OF THE BAR
P = δA = (40 MPa)(863.9 mm2)
= eL = (550 × 10—6)(400 mm) = 34.6 kN —
= 0.220 mm —
, Problem 1.2-5 The cross section of a concrete pier that is loaded y
uniformly in compression is shown in the figure.
20 in.
(a) Determine the average compressive stress δc in
the concrete if the load is equal to 2500 k.
(b) Determine the coordinates x¯ and ¯y of the point where the 16 in.
resultant load must act in order to produce uniform normal
stress. 48 in. 16 in.
16 in.
O 20 in. 16 in. x
EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science,
and technology
Solution 1.2-5 Concrete pier in compression
(a) AVERAGE COMPRESSIVE STRESS δc
y P = 2500 k
16 in. P 2500 k
x 2 s = = = 1.70 ksi —
c
48 in. C 16 in. 1 A 1472 in.2
3
y 4 (b) COORDINATES OF CENTROID c
16 in. 1
x
From symmetry, y = (48 in.) = 24 in.
O 20 in. 16 in. 2
© xi Ai
USE THE FOLLOWING AREAS: x= (see Chapter 12, Eq. 12 — 7a)
A
A = (48 in.)(20 in.) = 960 in.2 x=
1
(x A + 2x A + x A )
1 1 1 2 2 3 3
1 A
A2 = A4 = (16 in.) (16 in.) = 128 in.
2
1
2 = [(10 in.)(960 in.2)
1472 in.2
A3 = (16 in.)(16 in.) = 256 in.2
+ 2(25.333 in.)(128 in.2)
A = A1 + A2 + A3 + A4
+(28 in.)(256 in.2)]
= (960 + 128 + 256 + 128) in.2
=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 Cable
(see figure). The cable has an effective cross-sectional area
of 490 mm2, and the angle α of the incline is 30°.
Calculate the tensile stress δt in the cable.
α
Mechanics of Materials 6th Edition by James M. Gere 재료역학 솔루션
Chapter 1-12EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries,
including engineering, finance, data science, and technology
1
Tension, Compression,
and Shear
Normal Stress and Strain P1
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 dAB
uniformly distributed around the shelf at B. The diameters of P2
the upper and lower parts of the post are dAB = 1.25 in. and
dBC = 2.25 in., respectively.
B
(a) Calculate the normal stress δAB in the upper part of the post.
(b) If it is desired that the lower part of the post have the dBC
same compressive stress as the upper part, what should be
the magnitude of the load P2?
C
Solution 1.2-1 Circular post in compression
P1 = 2500 lb ALTERNATE SOLUTION
dAB = 1.25 in. s = P1 + P2 = P1 + P2
d = 2.25 in. BC w 2
d
ABC
BC 4 BC
P1 P1
(a) NORMAL STRESS IN PART AB sAB = =w sBC = sAB
A d2
AB 4 AB 2
P dBC
P1 2500 lb P1 + P2 1
s = = = or P = P B ¢ ≤ — 1R
AB A w 2 = 2040 psi — d2 d2 2 1 dAB
AB 4 (1.25 in.) BC AB
dBC
= 1.8
(b) LOAD P2 FOR EQUAL STRESSES dAB
P +P 2500 lb + P P1 P2 = 2.24 P1 = 5600 lb —
s = 1 2
= 2
BC w 2 A
ABC 4 (2.25 in.)
P2
B
C
, =δAB = 2040 psi
Solve for P2: P2 = 5600 lb —
EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science,
and technology
Problem 1.2-2 Calculate the compressive stress δc in the circular
50 mm
piston rod (see figure) when a force P = 40 N is applied to the
brake pedal. 5 mm
Assume that the line of action of the force P is parallel to the 225 mm
piston rod, which has diameter 5 mm. Also, the other dimensions
shown in the figure (50 mm and 225 mm) are measured perpen-
P = 40 N
dicular to the line of action of the force P.
Piston rod
Solution 1.2-2 Free-body diagram of brake pedal
50 mm EQUILIBRIUM OF BRAKE PEDAL
A ©MA = 0 ` “
225 mm F(50 mm) — P(275 mm) = 0
F
275 mm 275
= 40 N F = P¢ ≤ = (40 N) ¢ ≤ = 220 N
P
50 mm 50
COMPRESSIVE STRESS IN PISTON ROD (d = 5 mm)
F = compressive force in piston rod F 220 N
d = diameter of piston rod sc = = w = 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, calcu-
late the maximum normal stress δmax 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
, Solution 1.2-3 Long steel rod in tension
P = 200 lb W+P P
s = = gL +
max A
d A
L = 110 ft
1 ft2
3
L
d = 1⁄4 in. gL = (490 lb/ft )(110 ft)¢ ≤
144 in.2
Weight density: g = 490 lb/ft3 = 374.3 psi
W = Weight of rod P 200 lb = 4074 psi
=
w
A 4 (0.25 in.)2
= g(Volume)
δmax= 374 psi + 4074 psi = 4448 psi
= gAL
Rounding, we get
P = 200 lb
P = 200 lb δmax = 4450 psi —
EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science,
and technology
Problem 1.2-4 A circular aluminum tube of length L = 400 Strain gage
mm is loaded in compression by forces P (see figure). The out-
P P
side 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. L = 400 mm
(a) If the measured strain is s = 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?
Solution 1.2-4 Aluminum tube in compression
Strain gage
P P
e = 550 × 10—6 (b) COMPRESSIVE LOAD P
L = 400 mm δ = 4 0 MPa
w w
d = 60 mm A = [ d 2 — d 2] = [ (60 mm)2 — (50 mm)2 ]
2
4 2 1
4
d1 = 50 mm = 863.9 mm2
(a) SHORTENING OF THE BAR
P = δA = (40 MPa)(863.9 mm2)
= eL = (550 × 10—6)(400 mm) = 34.6 kN —
= 0.220 mm —
, Problem 1.2-5 The cross section of a concrete pier that is loaded y
uniformly in compression is shown in the figure.
20 in.
(a) Determine the average compressive stress δc in
the concrete if the load is equal to 2500 k.
(b) Determine the coordinates x¯ and ¯y of the point where the 16 in.
resultant load must act in order to produce uniform normal
stress. 48 in. 16 in.
16 in.
O 20 in. 16 in. x
EducationMathematics education spans a wide range of topics, from basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science,
and technology
Solution 1.2-5 Concrete pier in compression
(a) AVERAGE COMPRESSIVE STRESS δc
y P = 2500 k
16 in. P 2500 k
x 2 s = = = 1.70 ksi —
c
48 in. C 16 in. 1 A 1472 in.2
3
y 4 (b) COORDINATES OF CENTROID c
16 in. 1
x
From symmetry, y = (48 in.) = 24 in.
O 20 in. 16 in. 2
© xi Ai
USE THE FOLLOWING AREAS: x= (see Chapter 12, Eq. 12 — 7a)
A
A = (48 in.)(20 in.) = 960 in.2 x=
1
(x A + 2x A + x A )
1 1 1 2 2 3 3
1 A
A2 = A4 = (16 in.) (16 in.) = 128 in.
2
1
2 = [(10 in.)(960 in.2)
1472 in.2
A3 = (16 in.)(16 in.) = 256 in.2
+ 2(25.333 in.)(128 in.2)
A = A1 + A2 + A3 + A4
+(28 in.)(256 in.2)]
= (960 + 128 + 256 + 128) in.2
=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 Cable
(see figure). The cable has an effective cross-sectional area
of 490 mm2, and the angle α of the incline is 30°.
Calculate the tensile stress δt in the cable.
α
