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At 3.1 the proportional limit, a 12-inch gage length of a 0.75-in.-diameter alloy bar has elongated 0.0325
in. and the diameter has been reduced 0.0006 in. The total tension force on the bar was 17.5 kips.
Determine the following properties of the material:
(a) the modulus of elasticity.
(b) Poissons ratio.
(c) the proportional limit.
Solution
(a) The bar cross-sectional area is
D2 =
(0.75 in.) 2 = 0.441786 in.2
4
4
and thus, the normal stress corresponding to the 17.5-kip force is
17.5 kips
=
= 39.611897 ksi
0.441786 in.2
The strain in the bar is
e 0.0325 in.
= =
= 0.002708 in./in.
L
12 in.
The modulus of elasticity is therefore
39.611897 ksi
= 14, 626 ksi = 14,630 ksi
E= =
0.002708 in./in.
A=
Ans.
(b) The longitudinal strain in the bar was calculated previously as
long = 0.002708 in./in.
The lateral strain can be determined from the reduction of the diameter:
D 0.0006 in.
=
= 0.000800 in./in.
lat =
0.75 in.
D
Poissons ratio for this specimen is therefore
0.000800 in./in.
= lat =
= 0.295421 = 0.295
long
0.002708 in./in.
Ans.
(c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,
PL = 39.6 ksi
Ans.
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3.2 At the proportional limit, a 20 mm thick 75 mm wide bar elongates 6.8 mm under an axial load of
480 kN. The bar is 1.6-m long. If Poissons ratio is 0.32 for the material, determine:
(a) the modulus of elasticity.
(b) the proportional limit
(c) the change in each lateral dimension.
Solution
(a) The bar cross-sectional area is
A = (20 mm)(75 mm) = 1,500 mm 2
and thus, the normal stress corresponding to the 480-kN axial load is
(480 kN)(1,000 N/kN)
=
= 320 MPa
1,500 mm 2
The strain in the bar is
e
6.8 mm
= =
= 0.004250 mm/mm
L (1.6 m)(1,000 mm/m)
The modulus of elasticity is therefore
320 MPa
E= =
= 75, 294 MPa = 75.3 GPa
0.004250 mm/mm
Ans.
(b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,
PL = 320 MPa
Ans.
(c) Poissont ratio is given as = 0.32. The longitudinal strain in the bar was calculated previously as
long = 0.004250 mm/mm
The corresponding lateral strain can be determined from Poissons ratio:
lat = long = (0.32)(0.004250 mm/mm) = 0.001360 mm/mm
Using this lateral strain, the change in bar width is
width = lat (width) = ( 0.001360 mm/mm)(75 mm) = 0.1020 mm
and the change in bar thickness is
thickness = lat (thickness) = (0.001360 mm/mm)(20 mm) = 0.0272 mm
Ans.
Ans.
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3.3 At an axial load of 22 kN, a 15 mm thick 45 mm wide polyimide polymer bar elongates 3.0 mm
while the bar width contracts 0.25 mm. The bar is 200 mm long. At the 22-kN load, the stress in the
polymer bar is less than its proportional limit. Determine:
(a) the modulus of elasticity.
(b) Poissons ratio
(c) the change in the bar thickness.
Solution
(a) The bar cross-sectional area is
A = (15 mm)(45 mm) = 675 mm 2
and thus, the normal stress corresponding to the 22-kN axial load is
(22 kN)(1,000 N/kN)
=
= 32.592593 MPa
675 mm 2
The strain in the bar is
e 3.0 mm
= =
= 0.0150 mm/mm
L 200 mm
The modulus of elasticity is therefore
32.592593 MPa
E= =
= 2,173 MPa = 2.17 GPa
0.0150 mm/mm
(b) The longitudinal strain in the bar was calculated previously as
long = 0.0150 mm/mm
The lateral strain can be determined from the reduction of the bar width:
width 0.25 mm
=
= 0.005556 mm/mm
lat =
width
45 mm
Poissons ratio for this specimen is therefore
0.005556 mm/mm
= 0.370370 = 0.370
= lat =
0.0150 mm/mm
long
(c) The change in bar thickness can be found from the lateral strain:
thickness = lat (thickness) = (0.005556 mm/mm)(15 mm) = 0.0833 mm
Ans.
Ans.
Ans.
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3.4 A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width
of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in the
longitudinal (x) and transverse (y) directions: x = 2,136 and y = 673 .
(a) Determine Poissons ratio for this
specimen.
(b) If the measured strains were produced
by an axial load of P = 50 kips, what is the
modulus of elasticity for this specimen?
Fig. P3.4
Solution
(a) Poissons ratio for this specimen is
673
= 0.315
= lat = y =
2,136
long
x
(b) The bar area is
A = (3.0 in.)(0.75 in.) = 2.25 in.2
and so the normal stress for an axial load of P = 50 kips is
50 kips
=
= 22.222222 ksi
2.25 in.2
The modulus of elasticity is thus
22.222222 ksi
= 10, 404 ksi = 10, 400 ksi
E= =
1 in./in.
(2,136 )
1,000,000
Ans.
Ans.
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3.5 A 6-mm-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width
of the bar is w = 30 mm. Strain gages bonded to the specimen measure the following strains in the
longitudinal (x) and transverse (y) directions: x = 1,525 and y = 540 .
(a) Determine Poissons ratio for this
specimen.
(b) If the measured strains were produced
by an axial load of P = 27.5 kN, what is
the modulus of elasticity for this
specimen?
Fig. P3.5
Solution
(a) Poissons ratio for this specimen is
540
= 0.354
= lat = y =
1,525
long
x
(b) The bar area is
A = (30 mm)(6 mm) = 180 mm 2
and so the normal stress for an axial load of P = 50 kips is
(27.5 kN)(1,000 N/kN)
=
= 152.777778 MPa
180 mm 2
The modulus of elasticity is thus
152.777778 MPa
= 100,182 MPa = 100.2 GPa
E= =
1 in./in.
(1,525 )
1,000,000
Ans.
Ans.
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3.6 A copper rod [E = 110 GPa] originally 600-mm long is pulled in tension with a normal stress of 275
MPa. If the deformation is entirely elastic, what is the resulting elongation?
Solution
Since the deformation is elastic, the strain in the rod can be determined from Hookes Law,
275 MPa
= 0.002500 mm/mm
= =
E 110, 000 MPa
The elongation in the rod is thus
e = L = (0.002500 mm/mm)(600 mm) = 1.500 mm
Ans.
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3.7 A 6061-T6 aluminum tube [E = 10,000 ksi; = 0.33] has an outside diameter of 4.000 in. and a wall
thickness of 0.065 in.
(a) Determine the tension force that must be applied to the tube to cause its outside diameter to contract
by 0.005 in.
(b) If the tube is 84-in. long, what is the overall elongation?
Solution
(a) The strain associated with the given diameter contraction is
0.005 in.
lat =
= 0.001250 in./in.
4.000 in.
From the given Poissons ratio, the longitudinal strain in the tube must be
0.001250 in./in.
long = lat =
= 0.003788 in./in.
0.33
and from Hookes Law, the normal stress can be calculated as
= E = (10, 000 ksi)(0.003788 in./in.) = 37.878788 ksi
The area of the tube is needed to determine the tension force. Given that the outside diameter of the
tube is 4.000 in. and the wall thickness is 0.065 in., the inside diameter of the tube is 3.870 in. The tube
cross-sectional area is thus
(4.000 in.) 2 (3.870 in.) 2 = 0.803541 in.2
4
and the force applied to the tube is
F = A = (37.878788 ksi)(0.803541 in.2 ) = 30.437154 kips = 30.4 kips
Ans.
(b) The strain was calculated previously. Use the longitudinal strain to determine the overall
elongation:
e = L = (0.003788 in./in.)(84 in.) = 0.318 in.
Ans.
A=
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3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in
tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the
fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent
reduction in area.
Solution
Percent elongation is simply the longitudinal strain at fracture:
e (3.08 in. 2.000 in.) 1.08 in.
=
= 0.54 in./in.
= =
L
2.000 in.
2.000 in.
percent elongation = 54%
Ans.
The initial cross-sectional area of the specimen is
D2 =
(0.500 in.) 2 = 0.196350 in.2
4
4
The final cross-sectional area at the fracture location is
A0 =
D2 =
(0.260 in.) 2 = 0.053093 in.2
4
4
The percent reduction in area is
A0 Af
(0.196350 in.2 0.053093 in.2 )
percent reduction of area =
(100%) =
(100%) = 73.0% Ans.
A0
0.196350 in.2
Af =
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3.9 A portion of the stress-strain curve for a
stainless steel alloy is shown in Fig. P3.9. A 350mm-long bar is loaded in tension until it elongates
2.0 mm and then the load is removed.
(a) What is the permanent set in the bar?
(b) What is the length of the unloaded bar?
(c) If the bar is reloaded, what will be the
proportional limit?
Fig. P3.9
Solution
(a) The normal strain in the specimen is
e 2.0 mm
= =
= 0.005714 mm/mm
L 350 mm
Construct a line parallel to the elastic modulus line that passes through the data curve at a strain of =
0.005714 mm/mm. The strain value at which this modulus line intersects the strain axis is the
permanent set:
permanent set = 0.0035 mm/mm
Ans.
(b) The length of the unloaded bar is therefore:
e = L = (0.0035 mm/mm)(350 mm) = 1.225 mm
L f = 350 mm + 2.225 mm = 352.225 mm
(c) From the stress-strain curve, the reload proportional limit is 444 MPa .
Ans.
Ans.
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3.10 The 16 22 25 mm rubber blocks
shown in Fig. P3.10 are used in a double U
shear mount to isolate the vibration of a
machine from its supports. An applied load of
P = 285 N causes the upper frame to be
deflected downward by 5 mm. Determine the
shear modulus G of the rubber blocks.
Fig. P3.10
Solution
Consider a FBD of the upper U frame. The
downward force P is resisted by two upward shear
forces V; therefore, V = 285 N / 2 = 142.5 N.
Next, consider a FBD of one of the rubber blocks.
The shear force acting on one rubber block is V =
142.5 N. The area of the rubber block that is
parallel to the direction of V is
AV = (22 mm)(25 mm) = 550 mm 2
Consequently, the shear stress in one rubber block is
V
142.5 N
=
=
= 0.259091 MPa
AV 550 mm 2
The shear strain associated with the 5-mm downward displacement of the rubber blocks is given by:
5 mm
= 0.312500
= 0.302885 rad
tan =
16 mm
From Hookes Law for shear stress and shear strain, the shear modulus G can be computed:
0.259091 MPa
= G
G = =
= 0.855411 MPa = 0.855 MPa
0.302885 rad
Ans.
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3.11 Two hard rubber blocks are used in an anti-vibration
mount to support a small machine, as shown in Fig. P3.11.
An applied load of P = 150 lb causes a downward deflection
of 0.25 in. Determine the shear modulus of the rubber blocks.
Assume a = 0.5 in., b = 1.0 in., and c = 2.5 in.
Fig. P3.11
Solution
Determine the shear strain from the angle formed by the
downward deflection and the block thickness a:
0.250 in.
tan =
= 0.500
= 0.463648 rad
0.50 in.
Note: The small angle approximation tan is not applicable in this instance.
Determine the shear stress from the applied load P and the block area.
Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block
is half of the total load: V = P/2 = 75 lb.
To determine the area needed here, consider the surface that is bonded to the plate. This area has
dimensions of bc. The shear stress acting on a single block is therefore:
V
75 lb
= =
= 30 psi
A (1.0 in.)(2.5 in.)
The shear modulus G can be calculated from Hookes law for shear stress and strain:
30 psi
G = =
= 64.704 psi = 64.7 psi
= G
0.463648 rad
Ans.
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3.12 Two hard rubber blocks [G = 350 kPa] are used in an antivibration mount to support a small machine, as shown in Fig.
P3.12. Determine the downward deflection that will occur for an
applied load of P = 900 N. Assume a = 20 mm, b = 50 mm, and
c = 80 mm.
Fig. P3.12
Solution
Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block
is half of the total load: V = P/2 = 450 N.
Determine the shear stress from the shear force V and the block area. To determine the area needed
here, consider the surface that is bonded to the plate. This area has dimensions of bc. The shear stress
acting on a single block is therefore:
V
450 N
= =
= 0.112500 MPa
A (50 mm)(80 mm)
Since the shear modulus G is given, the shear strain can be calculated from Hookes law for shear stress
and shear strain:
0.112500 MPa
= G
= =
= 0.321429 rad
G
0.350 MPa
From the angle and the block thickness a, the downward deflection of the block can be determined
from:
tan =
a
= a tan = (20 mm) tan(0.321429 rad) = 6.659512 mm = 6.66 mm
Ans.
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3.13 A load test on a 6 mm diameter by 150 mm long magnesium alloy rod found that a tension load of
780 N caused an elastic elongation of 0.55 mm in the rod. Using this result, determine the elastic
elongation that would be expected for a 19-mm-diameter rod of the same material if the rod were 1.2 m
long and subjected to a tension force of 2.6 kN.
Solution
The area of the 6-mm-diameter rod is
D2 =
(6 mm) 2 = 28.274334 mm 2
4
4
Thus, the normal stress in the rod due to a 780-N load is
F
780 N
= =
= 27.586857 MPa
A 28.274334 mm 2
The strain in the 150-mm long rod associated with a 0.55-mm elongation is
e 0.55 mm
= =
= 0.003667 mm/mm
L 150 mm
Therefore, the elastic modulus of the magnesium alloy is
27.586857 MPa
E= =
= 7,523.688217 MPa
0.003667 mm/mm
A=
The area of the 19-mm-diameter rod is
A=
D2 =
(19 mm) 2 = 283.528737 mm 2
4
4
Thus, the normal stress in the 19-mm-diameter rod due to a 2.6-kN load is
F (2.6 kN)(1,000 N/kN)
= =
= 9.170146 MPa
A
283.528737 mm 2
From Hookes Law, the strain that would be expected is
9.170146 MPa
= =
= 0.001219 mm/mm
E 7,523.688217 MPa
Since the 19-mm-diameter rod is 1.2-m long, the expected elongation is
e = L = (0.001219 mm/mm)(1.2 m)(1,000 mm/m) = 1.462604 mm = 1.463 mm
Ans.
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3.14 The stress-strain diagram for a
particular stainless steel alloy is shown in
Fig. P3.14. A rod made from this material is
initially 800 mm long at a temperature of
20C. After a tension force is applied to the
rod and the temperature is increased by
200C, the length of the rod is 804 mm.
Determine the stress in the rod and state
whether the elongation in the rod is elastic
or inelastic. Assume the coefficient of
thermal expansion for this material is 18
106/C.
Fig. P3.14
Solution
The 4-mm total elongation of the rod is due to a combination of load and temperature increase. The
200C temperature increase causes a normal strain of:
T = T = (18 106 / C )(200C ) = 0.003600 mm/mm
which means that the rod elongates
eT = T L = (0.003600 mm/mm)(800 mm) = 2.8800 mm
The portion of the 4-mm total elongation due to load is therefore
e = e eT = 4 mm 2.8800 mm = 1.1200 mm
The strain corresponding to this elongation is
e
1.1200 mm
= =
= 0.001400 mm/mm
L
800 mm
By inspection of the stress-strain curve, this strain is clearly in the linear region. Therefore, the rod is
elastic in this instance.
For the linear region, the elastic modulus can be determined from the lower scale plot:
(400 MPa 0)
E=
=
= 200,000 MPa
(0.002 mm/mm 0)
Using Hookes Law (or directly from the - diagram), the stress corresponding to the 0.001400
mm/mm strain is
= E = (200, 000 MPa)(0.001400 mm/mm) = 280 MPa
Ans.
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3.15 In Fig. P3.15, rigid bar ABC is supported by
axial member (1), which has a cross-sectional area
of 400 mm2, an elastic modulus of E = 70 GPa, and
a coefficient of thermal expansion of = 22.5
106 /C. After load P is applied to the rigid bar
and the temperature rises 40C, a strain gage
affixed to member (1) measures a strain increase of
1,650 . Determine:
(a) the normal stress in member (1).
(b) the magnitude of applied load P.
(c) the deflection of the rigid bar at C.
Fig. P3.15
Solution
(a) The strain measured in member (1) is due to both the internal force in the member and the
temperature change. The strain caused by the temperature change is
T = T = (22.5 106 / C )(40C ) = 0.000900 mm/mm
Since the total strain is = 1,650 = 0.001650 mm/mm, the strain caused by the internal force in
member (1) must be
= T = 0.001650 mm/mm 0.000900 mm/mm = 0.000750 mm/mm
The elastic modulus of member (1) is E = 70 GPa; thus, from Hookes Law, the stress in the member is:
1 = E = (70, 000 MPa)(0.000750 mm/mm) = 52.5 MPa
Ans.
(b) If the normal stress in member (1) is 52.5 MPa, the axial force in the member is
F1 = 1 A1 = (52.5 N/mm 2 )(400 mm 2 ) = 21, 000 N
Consider moment equilibrium of rigid bar ABC about joint
A to determine the magnitude of P:
M A = (1.4 m)(21,000 N) (2.4 m)P = 0
P = 12, 250 N = 12.25 kN
Ans.
(c) The strain in member (1) was measured as = 1,650 = 0.001650 mm/mm; therefore, the total
elongation of member (1) is
e1 = 1 L1 = (0.001650 mm/mm)(2,500 mm) = 4.125 mm
The deflection of the rigid bar at B is equal to this elongation; therefore, vB = e1 = 4.125 mm
(downward). By similar triangles, the deflection of the rigid bar at C is given by:
v
vB
=C
1.4 m 2.4 m
2.4 m
2.4
Ans.
vC = vB
= (4.125 mm)
= 7.071429 mm = 7.07 mm
1.4 m
1.4
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3.16 A tensile test specimen of 1045 hot-rolled
steel having a diameter of 0.505 in. and a gage
length of 2.00 in. was tested to fracture. Stress
and strain data obtained during the test are
shown in Fig. P3.16. Determine
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.20% offset).
(e) the fracture stress.
(f) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
0.392 in.
Fig. P3.16
Solution
From the stress-strain curve, the proportional limit will be taken as = 60 ksi at a strain of = 0.0019.
(Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.)
(a) The modulus of elasticity is
60 ksi
E= =
= 31, 600 ksi
0.0019 in./in.
(b) From the diagram, the proportional limit is taken as
PL = 60 ksi
(c) The ultimate strength is
ult = 105 ksi
(d) The yield strength is
Y = 68 ksi
(e) The fracture stress is
fracture = 98 ksi
Ans.
Ans.
Ans.
Ans.
Ans.
(f) The original cross-sectional area of the specimen is
Af =
D2 =
(0.505 in.) 2 = 0.200296 in.2
4
4
The area of the specimen at the fracture location is
A0 =
D2 =
(0.392 in.) 2 = 0.120687 in.2
4
4
The true fracture stress is therefore
0.200296 in.2
true fracture = (98 ksi)
= 162.6 ksi
0.120687 in.2
Ans.
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3.17 A tensile test specimen of stainless
steel alloy having a diameter of 0.495 in.
and a gage length 2.00 of in. was tested to
fracture. Stress and strain data obtained
during the test are shown in Fig. P3.17.
Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.20% offset).
(e) the fracture stress.
(f) the true fracture stress if the final
diameter of the specimen at the location of
the fracture was 0.350 in.
Fig. P3.17
Solution
From the stress-strain curve, the proportional limit will be taken as = 60 ksi at a strain of = 0.002.
(Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.)
(a) The modulus of elasticity is
60 ksi
E= =
= 30, 000 ksi
0.002 in./in.
(b) From the diagram, the proportional limit is taken as
PL = 60 ksi
(c) The ultimate strength is
ult = 159 ksi
(d) The yield strength is
Y = 80 ksi
(e) The fracture stress is
fracture = 135 ksi
Ans.
Ans.
Ans.
Ans.
Ans.
(f) The original cross-sectional area of the specimen is
D2 =
(0.495 in.) 2 = 0.192442 in.2
4
4
The area of the specimen at the fracture location is
A0 =
D2 =
(0.350 in.) 2 = 0.096211 in.2
4
4
The true fracture stress is therefore
0.192442 in.2
true fracture = (135 ksi)
= 270 ksi
0.096211 in.2
Af =
Ans.
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3.18 A bronze alloy specimen having a
diameter of 12.8 mm and a gage length
of 50 mm was tested to fracture. Stress
and strain data obtained during the test
are shown in Fig. P3.18. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.20% offset).
(e) the fracture stress.
(f) the true fracture stress if the final
diameter of the specimen at the location
of the fracture was 10.5 mm.
Fig. P3.18
Solution
From the stress-strain curve, the proportional limit will be taken as = 210 MPa at a strain of = 0.002.
(Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.)
(a) The modulus of elasticity is
210 MPa
E= =
= 105, 000 MPa
0.002 in./in.
(b) From the diagram, the proportional limit is taken as
PL = 210 MPa
(c) The ultimate strength is
ult = 380 MPa
(d) The yield strength is
Y = 290 MPa
(e) The fracture stress is
fracture = 320 MPa
Ans.
Ans.
Ans.
Ans.
Ans.
(f) The original cross-sectional area of the specimen is
A0 =
D2 =
(12.8 mm) 2 = 128.679635 mm 2
4
4
The area of the specimen at the fracture location is
D2 =
(10.5 mm) 2 = 86.590148 mm 2
4
4
The true fracture stress is therefore
128.679635 mm 2
true fracture = (320 MPa)
= 476 MPa
86.590148 mm 2
Af =
Ans.
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3.19 An alloy specimen having a diameter of 12.8
mm and a gage length of 50 mm was tested to
fracture. Load and deformation data obtained
during the test are given. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
11.3 mm.
(kN)
Change in
Length
(mm)
(kN)
Change in
Length
(mm)
0
7.6
14.9
22.2
28.5
29.9
30.6
32.0
33.0
33.3
36.8
41.0
0
0.02
0.04
0.06
0.08
0.10
0.12
0.16
0.20
0.24
0.50
1.00
43.8
45.8
48.3
49.7
50.4
50.7
50.4
50.0
49.7
47.9
45.1
1.50
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
fracture
Load
Load
Solution
The plot of the stress-strain data is shown below.
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(a) The modulus of elasticity is
224.976 MPa
E= =
= 140, 600 MPa
0.0016 mm/mm
Ans.
(b) From the diagram, the proportional limit is taken as
PL = 234 MPa
Ans.
(c) The ultimate strength is
ult = 400 MPa
Ans.
(d) The yield strength by the 0.05% offset method is
Y = 239 MPa
Ans.
(e) The yield strength by the 0.2% offset method is
Y = 259 MPa
Ans.
(f) The fracture stress is
fracture = 356 MPa
Ans.
(f) The original cross-sectional area of the specimen is
D2 =
(12.8 mm) 2 = 128.679635 mm 2
4
4
The area of the specimen at the fracture location is
A0 =
D2 =
(11.3 mm) 2 = 100.287492 mm 2
4
4
The true fracture stress is therefore
128.679635 mm 2
true fracture = (356 MPa)
= 457 MPa
100.287492 mm 2
Af =
Ans.
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3.20 A 1035 hot-rolled steel specimen with a
diameter of 0.500 in. and a 2.0-in. gage length was
tested to fracture. Load and deformation data
obtained during the test are given. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
0.387 in.
(lb)
Change in
Length
(in.)
0
2,690
5,670
8,360
11,050
12,540
13,150
13,140
12,530
12,540
12,840
12,840
0
0.0009
0.0018
0.0028
0.0037
0.0042
0.0046
0.0060
0.0079
0.0098
0.0121
0.0139
Load
(lb)
Change
in Length
(in.)
12,540
12,540
14,930
17,020
18,220
18,820
19,110
19,110
18,520
17,620
16,730
16,130
15,900
0.0209
0.0255
0.0487
0.0835
0.1252
0.1809
0.2551
0.2968
0.3107
0.3246
0.3339
0.3385
fracture
Load
Solution
The plot of the stress-strain data is shown below.
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(a) The modulus of elasticity is
63.849 ksi
E= =
= 30, 400 ksi
0.0021 in./in.
Ans.
(b) From the diagram, the proportional limit is taken as
PL = 63.8 ksi
Ans.
(c) The ultimate strength is
ult = 97.3 ksi
Ans.
(d) The yield strength using the 0.05% offset method is
Y = 65.4 ksi
Ans.
(e) The yield strength using the 0.2% offset method is
Y = 63.8 ksi
Ans.
(f) The fracture stress is
fracture = 82.1 ksi
Ans.
(g) The original cross-sectional area of the specimen is
D2 =
(0.500 in.) 2 = 0.196350 in.2
4
4
The area of the specimen at the fracture location is
A0 =
D2 =
(0.387 in.) 2 = 0.117628 in.2
4
4
The true fracture stress is therefore
0.196350 in.2
true fracture = (82.1 ksi)
= 137.0 ksi
0.117628 in.2
Af =
Ans.
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3.21 A 2024-T4 aluminum test specimen with a
diameter of 0.505 in. and a 2.0-in. gage length was
tested to fracture. Load and deformation data
obtained during the test are given. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
0.452 in.
Load
(lb)
0
1,300
2,390
3,470
4,560
5,640
6,720
7,380
8,240
8,890
9,330
9,980
10,200
10,630
Change in
Length
(in.)
0.0000
0.0014
0.0023
0.0032
0.0042
0.0051
0.0060
0.0070
0.0079
0.0088
0.0097
0.0107
0.0116
0.0125
Load
(lb)
11,060
11,500
12,360
12,580
12,800
13,020
13,230
13,450
13,670
13,880
14,100
14,100
14,100
14,100
14,100
Change in
Length
(in.)
0.0139
0.0162
0.0278
0.0394
0.0603
0.0788
0.0974
0.1159
0.1391
0.1623
0.1994
0.2551
0.3200
0.3246
fracture
Solution
The plot of the stress-strain data is shown below.
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(a) The modulus of elasticity is
33.55 ksi
E= =
= 11,180 ksi
0.003 in./in.
Ans.
(b) From the diagram, the proportional limit is taken as
PL = 33.6 ksi
Ans.
(c) The ultimate strength is
ult = 70.4 ksi
Ans.
(d) The yield strength using the 0.05% offset method is
Y = 44.4 ksi
Ans.
(e) The yield strength using the 0.2% offset method is
Y = 54.5 ksi
Ans.
(f) The fracture stress is
fracture = 70.4 ksi
Ans.
(g) The original cross-sectional area of the specimen is
A0 =
D2 =
(0.505 in.) 2 = 0.200296 in.2
4
4
The area of the specimen at the fracture location is
Af =
D2 =
(0.452 in.) 2 = 0.160460 in.2
4
4
The true fracture stress is therefore
0.200296 in.2
true fracture = (70.4 ksi)
= 87.9 ksi
0.160460 in.2
Ans.
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3.22 A 1045 hot-rolled steel tension test specimen
has a diameter of 6.00 mm and a gage length of 25
mm. In a test to fracture, the stress and strain data
below were obtained. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
4.65 mm.
(kN)
Change in
Length
(mm)
0.00
2.94
5.58
8.52
11.16
12.63
13.02
13.16
13.22
13.22
13.25
13.22
0.00
0.01
0.02
0.03
0.05
0.05
0.06
0.08
0.08
0.10
0.14
0.17
Load
(kN)
Change in
Length
(mm)
13.22
16.15
18.50
20.27
20.56
20.67
20.72
20.61
20.27
19.97
19.68
19.09
18.72
0.29
0.61
1.04
1.80
2.26
2.78
3.36
3.83
3.94
4.00
4.06
4.12
fracture
Load
Solution
The plot of the stress-strain data is shown below.
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(a) The modulus of elasticity is
394.765 MPa
E= =
= 247,000 MPa
0.0016 mm/mm
Ans.
(b) From the diagram, the proportional limit is taken as
PL = 400 MPa
Ans.
(c) The ultimate strength is
ult = 732 MPa
Ans.
(d) The yield strength by the 0.05% offset method is
Y = 465 MPa
Ans.
(e) The yield strength by the 0.2% offset method is
Y = 465 MPa
Ans.
(f) The fracture stress is
fracture = 675 MPa
Ans.
(f) The original cross-sectional area of the specimen is
D2 =
(6 mm) 2 = 28.274334 mm 2
4
4
The area of the specimen at the fracture location is
A0 =
D2 =
(4.65 mm) 2 = 16.982272 mm 2
4
4
The true fracture stress is therefore
28.274334 mm 2
true fracture = (675 MPa)
= 1,124 MPa
16.982272 mm 2
Af =
Ans.
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3.23 Rigid bar BCD in Fig. P3.23 is
supported by a pin at C and by aluminum
rod (1). A concentrated load P is applied to
the lower end of aluminum rod (2), which is
attached to the rigid bar at D. The crosssectional area of each rod is A = 0.20 in.2
and the elastic modulus of the aluminum
material is E = 10,000 ksi. After the load P
is applied at E, the strain in rod (1) is
measured as 900 (tension).
(a) Determine the magnitude of load P.
(b) Determine the total deflection of point E
relative to its initial position.
Fig. P3.23
Solution
(a) From the measured strain, the stress in rod (1) is
1 = E11 = (10, 000 ksi)(900 106 in./in.) = 9 ksi
and thus, the force in rod (1) is
F1 = 1 A1 = (9 ksi)(0.20 in.2 ) = 1.8 kips (T)
Consider the equilibrium of the rigid bar, and write a
moment equilibrium equation about C to determine the
magnitude of load P:
M C = (20 in.)(1.8 kips) (30 in.)P = 0
P = 1.2 kips
Ans.
(b) From the measured strain, the elongation of rod (1) is
e1 = 1 L1 = (900 106 in./in.)(50 in.) = 0.0450 in.
From similar triangles, the deflection of the rigid bar at D can
be expressed in terms of the deflection at B:
vB
v
=D
20 in. 30 in.
30 in.
30 in.
v D = vB
= (0.045 in.)
= 0.0675 in.
20 in.
20 in.
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The elongation of rod (2) due to the 1.2-kip load must be determined. The stress in rod (2) is
F 1.2 kips
2 = 2 =
= 6 ksi
A2 0.2 in.2
and consequently, the strain in rod (2) is
6 ksi
2 = 2 =
= 0.000600 in./in.
E2 10,000 ksi
From the strain, the elongation in rod (2) can be computed:
e2 = 2 L2 = (0.000600 in./in.)(100 in.) = 0.06 in.
The deflection of joint E is the sum of the rigid bar deflection at D and the elongation in rod (2):
vE = vD + e2 = 0.0675 in. + 0.06 in. = 0.1275 in.
Ans.
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3.24 The rigid bar AC in Fig. P3.24 is
supported by two axial bars (1) and (2).
Both axial bars are made of bronze [E =
100 GPa; = 18 106 mm/mm/C]. The
cross-sectional area of bar (1) is A1 = 240
mm2 and the cross-sectional area of bar (2)
is A2 = 360 mm2. After load P has been
applied and the temperature of the entire
assembly has increased by 20C, the total
strain in bar (2) is measured as 800
(elongation). Determine:
(a) the magnitude of load P.
(b) the vertical displacement of pin A.
Fig. P3.24
Solution
(a) The total strain in bar (2) is caused partly by the axial force in the bar and partly by the increase in
temperature. The strain caused by the 20C temperature increase is:
T = T = (18 106 mm/mm/ C)(20C) = 0.000360 mm/mm
The strain caused by the axial force in the bar is thus:
2, = 2 2,T = 0.000800 mm/mm 0.000360 mm/mm = 0.000440 mm/mm
The stress in bar (2) is
2 = E2 2, = (100, 000 MPa)(0.000440 mm/mm) = 44 MPa
and the force in bar (2) is
F2 = 2 A2 = (44 N/mm 2 )(360 mm 2 ) = 15,840 N
Next, consider a FBD of the rigid bar AC. Equilibrium
equations for this FBD are:
Fy = F1 + F2 P = 0
M A = (1, 400 mm)F2 (500 mm)P = 0
which can be solve simultaneously to give:
500 mm
F2 =
P = 0.357143P
1,400 mm
and
F1 = 0.642857 P
The applied load P can be expressed in terms of F2 as
1
P=
F2 = 2.8 F2
0.357143
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and so the magnitude of load P is
P = 2.8F2 = 2.8(15,840 N) = 44,352 N = 44.4 kN
Ans.
(b) The force in bar (1) is
F1 = 0.642857 P = (0.642857)(44,352 N) = 28,512 N
Thus, the stress in bar (1) is
F 28,512 N
1 = 1 =
= 118.800 MPa
A1 240 mm 2
The normal strain due to the axial force in bar (1) is
118.800 MPa
1, = 1 =
= 0.001188 mm/mm
E1 100, 000 MPa
The normal strain caused by the 20C temperature increase is:
1,T = T = (18 106 mm/mm/ C)(20C) = 0.000360 mm/mm
Therefore, the total strain in bar (1) is
1 = 1, + 1,T = 0.001188 mm/mm + 0.000360 mm/mm = 0.001548 mm/mm
and the elongation in bar (1) is
e1 = 1, L1 = (0.001548 mm/mm)(1,300 mm) = 2.012400 mm
Since rigid bar ABC is connected to bar (1) (with a perfect connection), joint A displaces downward by
an amount equal to the elongation of bar (1); therefore,
vA = e1 = 2.01 mm
Ans.
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3.25 The rigid bar in Fig. P3.25 is supported
by axial bar (1) and by a pin connection at
C. Axial bar (1) has a cross-sectional area of
A1 = 275 mm2, an elastic modulus of E =
200 GPa, and a coefficient of thermal
expansion of = 11.9 106 mm/mm/C.
The pin at C has a diameter of 25 mm. After
load P has been applied and the temperature
of the entire assembly has been increased by
20C, the total strain in bar (1) is measured
as 675 (elongation). Determine:
(a) the magnitude of load P.
(b) the shear stress in pin C.
Fig. P3.25
Solution
The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:
= + T
The normal strain due to the increase in temperature is:
T = T = (11.9 106 mm/mm/C)(20C) = 0.000238 mm/mm
Therefore, the normal stress in bar (1) causes a normal strain of:
= T = 0.000675 mm/mm 0.000238 mm/mm = 0.000437 mm/mm
From Hookes law, the normal stress in bar (1) can be calculated as:
1 = E = (200, 000 MPa)(0.000437 mm/mm) = 87.4 MPa
and thus the axial force in bar (1) must be:
F1 = 1 A1 = (87.4 N/mm 2 )(275 mm 2 ) = 24, 035 N
Next, consider a free-body diagram of the rigid bar. Write
a moment equilibrium equation about pin C:
M C = (200 mm)F1 (380 mm)P
= (200 mm)(24, 035 N) (380 mm)P = 0
P = 12, 650 N = 12.65 kN
Ans.
Now that P is known, the horizontal and vertical reactions
at C can be calculated:
Fx = Cx F1 = 0
Cx = F1 = 24, 035 N
Fy = C y P = 0
C y = P = 12, 650 N
The resultant force acting on pin C is:
2
C = Cx2 + C y = (24, 035 N) 2 + (12, 650 N) 2 = 27,160.702 N
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Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the
resultant force: V = 13,580.351 N. The area of one shear plane of the 25-mm-diameter pin at C (in
other words, the cross-sectional area of the pin) is:
Apin =
(25 mm)2 = 490.874 mm 2
4
and thus the shear stress in pin C is:
V
13,580.351 N
C =
=
= 27.666 N/mm 2 = 27.7 MPa
2
AV 490.874 mm
Ans.
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3.26 The rigid bar in Fig. P3.26 is
supported by axial bar (1) and by a pin
connection at C. Axial bar (1) has a crosssectional area of A1 = 275 mm2, an elastic
modulus of E = 200 GPa, and a coefficient
of thermal expansion of = 11.9 106
mm/mm/C. The pin at C has a diameter
of 25 mm. After load P has been applied
and the temperature of the entire assembly
has been decreased by 30C, the total
strain in bar (1) is measured as 675
(elongation). Determine:
(a) the magnitude of load P.
(b) the shear stress in pin C.
Fig. P3.26
Solution
The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:
= + T
The normal strain due to the decrease in temperature is:
T = T = (11.9 106 mm/mm/ C)( 30C) = 0.000357 mm/mm
Therefore, the normal stress in bar (1) causes a normal strain of:
= T = 0.000675 mm/mm (0.000357 mm/mm) = 0.001032 mm/mm
From Hookes law, the normal stress in bar (1) can be calculated as:
1 = E = (200, 000 MPa)(0.001032 mm/mm) = 206.4 MPa
and thus the axial force in bar (1) must be:
F1 = 1 A1 = (206.4 N/mm 2 )(275 mm 2 ) = 56, 760 N
Next, consider a free-body diagram of the rigid bar. Write
a moment equilibrium equation about pin C:
M C = (200 mm)F1 (380 mm)P
= (200 mm)(56, 760 N) (380 mm)P = 0
P = 29,874 N = 29.9 kN
Ans.
Now that P is known, the horizontal and vertical reactions
at C can be calculated:
Fx = Cx F1 = 0
Cx = F1 = 56, 760 N
Fy = C y P = 0
C y = P = 29,874 N
The resultant force acting on pin C is:
2
C = Cx2 + C y = (56, 760 N) 2 + (29,874 N) 2 = 64,142 N
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Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the
resultant force: V = 32,071 N. The area of one shear plane of the 25-mm-diameter pin at C (in other
words, the cross-sectional area of the pin) is:
Apin =
(25 mm)2 = 490.874 mm 2
4
and thus the shear stress in pin C is:
V
32, 071 N
C =
=
= 65.3345 N/mm 2 = 65.3 MPa
2
AV 490.874 mm
Ans.
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U. Memphis - CIVL - 3322

4.1 A stainless steel alloy bar 25 mm wide by16 mm thick is subjected to an axial load ofP = 145 kN. Using the stress-strain diagramgiven in Fig. P4.1, determine:(a) the factor of safety with respect to theyield strength defined by the 0.20% offsetm

U. Memphis - CIVL - 3322

5.1 A steel [E = 200 GPa] rod with a circular cross section is 6-m long. Determine the minimumdiameter D required if the rod must transmit a tensile force of 30 kN without exceeding an allowablestress of 180 MPa or stretching more than 5 mm.SolutionIf

U. Memphis - CIVL - 3322

5.14 Rigid bar ABCD is loaded and supported asshown in Fig. P5.14. Steel [E = 30,000 ksi] bars(1) and (2) are unstressed before the load P isapplied. Bar (1) has a cross-sectional area of0.625 in.2 and bar (2) has a cross-sectional areaof 1.25 in.2.

U. Memphis - CIVL - 3322

5.23 The 200 200 1,200-mm oak [E = 12 GPa] block (2)shown in Fig. P5.23 was reinforced by bolting two 6 200 1,200mm steel [E = 200 GPa] plates (1) to opposite sides of the block. Aconcentrated load of 360 kN is applied to a rigid cap. Determine:(a) th

U. Memphis - CIVL - 3322

5.44 A 20-mm-diameter by 3.5-m-long steel rod (1)is stress free after being attached to rigid supports,as shown in Fig. P5.44. At A, a 25-mm-diameterbolt is used to connect the rod to the support.Determine the normal stress in steel rod (1) and thesh

U. Memphis - CIVL - 3322

5.62 The machine part shown in Fig. P5.62is in. thick and is made of SAE 4340 heattreated steel (see Appendix D for properties).Determine the maximum safe load P if afactor of safety of 2 with respect to failure byyield is specified.Fig. P5.62Soluti

U. Memphis - CIVL - 3322

6.1 A solid circular steel shaft having an outside diameter of D = 1.25 in. is subjected to a pure torque ofT = 1,900 lb-in. Determine the maximum shear stress in the shaft.SolutionThe polar moment of inertia for the shaft isD4 =(1.25 in.) 4 = 0.2396

U. Memphis - CIVL - 3322

6.26 A torque of TA = 525 lb-ft is applied to gear Aof the gear train shown in Fig. P6.26. The bearingsshown allow the shafts to rotate freely.(a) Determine the torque TD required for equilibriumof the system.(b) Assume shafts (1) and (2) are solid 1

U. Memphis - CIVL - 3322

6.42 The driveshaft of an automobile is being designed to transmit 280 hp at 3,500 rpm. Determine theminimum diameter required for a solid steel shaft if the allowable shear stress in the shaft is not toexceed 4,000 psi.SolutionThe torque in the drive

U. Memphis - CIVL - 3322

6.67 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 2.00 in.and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainlesssteel [G2 = 12,500 ksi] core (2) as shown in

U. Memphis - CIVL - 3322

6.79 The torsional assembly of Fig. P6.79 consistsof a cold-rolled stainless steel tube connected to asolid cold-rolled brass segment at flange C. Theassembly is securely fastened to rigid supports at Aand D. Stainless steel tube (1) and (2) has anou

U. Memphis - CIVL - 3322

6.94 A fillet with a radius of 0.15 in. is used at the junction in a stepped shaft where the diameter isreduced from 4.00 in. to 3.00 in. Determine the maximum shear stress in the fillet when the shaft istransmitting a torque of 4,000 lb-ft.SolutionFi

U. Memphis - CIVL - 3322

6.101 A torque of magnitude T = 1.5kip-in. is applied to each of the barsshown in Fig. P6.101. If the allowableshear stress is specified as allow = 8 ksi,determine the minimum requireddimension b for each bar.Fig. P6.101Solution(a) Circular Sectio

U. Memphis - CIVL - 3322

7.1 For the cantilever beam and loading shown,(a) Derive equations for the shear force V and thebending moment M for any location in the beam.(Place the origin at point A.)(b) Plot the shear-force and bending-momentdiagrams for the beam using the der

U. Memphis - CIVL - 3322

7.7 For the simply supported beam subjected to theloading shown,(a) Derive equations for the shear force V and thebending moment M for any location in the beam.(Place the origin at point A.)(b) Plot the shear-force and bending-momentdiagrams for the

U. Memphis - CIVL - 3322

7.24 Use the graphical method to construct theshear-force and bending-moment diagrams for thebeams shown. Label all significant points on eachdiagram and identify the maximum moments (bothpositive and negative) along with their respectivelocations. C

U. Memphis - CIVL - 3322

7.31 Draw the shear-force and bending-momentdiagram for the beam shown in Fig. P7.31. Assumethe upward reaction provided by the ground to beuniformly distributed. Label all significant pointson each diagram. Determine the maximum value of(a) the inte

U. Memphis - CIVL - 3322

8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E =1,900 ksi] planks is bent to a radius of curvature of 12 ft. Determine the maximum bending stressdeveloped in the plank.SolutionFrom Eq. (8.3):E

U. Memphis - CIVL - 3322

8.19 A WT230 26 standard steel shape is used to support the loads shown on the beam in Fig. P8.19a.The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch ofthe cross section (Fig. P8.19b). Consider the entire

U. Memphis - CIVL - 3322

8.31 A solid steel shaft supports loadsPA = 200 lb and PD = 300 lb as shownin Fig. P8.31. Assume L1 = 6 in.,L2 = 20 in., and L3 = 10 in. The bearingat B can be idealized as a roller supportand the bearing at C can be idealizedas a pin support. If th

U. Memphis - CIVL - 3322

8.42 A composite beam is fabricated by bolting two 3 in. wide 12 in. deep timber planks to the sidesof a 0.50 in. 12 in. steel plate (Fig. P8.42b). The moduli of elasticity of the timber and the steel are1,800 ksi and 30,000 ksi, respectively. The simpl

U. Memphis - CIVL - 3322

8.52 A steel pipe assembly supports aconcentrated load of 17 kN as shown in Fig.P8.52. The outside diameter of the pipe is 142mm and the wall thickness is 6.5 mm.Determine the normal stresses produced atpoints H and K.Fig. P8.52SolutionSection pro

U. Memphis - CIVL - 3322

8.65 A beam with a box cross section is subjected toa resultant moment magnitude of 2,100 N-m actingat the angle shown in Fig. P8.65. Determine:(a) the maximum tension and the maximumcompression bending stresses in the beam.(b) the orientation of the

U. Memphis - CIVL - 3322

9.1 For the following problems, a beam segment subjected to internal bending moments at sections Aand B is shown along with a sketch of the cross-sectional dimensions. For each problem:(a) Sketch a side view of the beam segment and plot the distribution

U. Memphis - CIVL - 3322

9.11 A 1.6-m long cantilever beam supports a concentrated load of 7.2 kN, as shown below. The beamis made of a rectangular timber having a width of 120 mm and a depth of 280 mm. Calculate themaximum horizontal shear stresses at points located 35 mm, 70

U. Memphis - CIVL - 3322

9.18 A 50-mm-diameter solid steel shaftsupports loads PA = 1.5 kN and PC = 3.0 kN,as shown in Fig. P9.18. Assume L1 = 150mm, L2 = 300 mm, and L3 = 225 mm. Thebearing at B can be idealized as a rollersupport and the bearing at D can be idealizedas a

U. Memphis - CIVL - 3322

9.38 A wooden beam is fabricated from one 2 8 and two 2 4 piecesof dimension lumber to form the I-beam cross section shown in Fig.P9.38. The flanges of the beam are fastened to the web with nails thatcan safely transmit a force of 100 lb in direct shea

U. Memphis - CIVL - 3322

10.1 For the loading shown, use the doubleintegration method to determine (a) theequation of the elastic curve for thecantilever beam, (b) the deflection at the freeend, and (c) the slope at the free end.Assume that EI is constant for each beam.Fig.

U. Memphis - CIVL - 3322

10.21 For the beam and loading shown inFig. P10.21, integrate the load distribution todetermine (a) the equation of the elasticcurve for the beam, and (b) the maximumdeflection for the beam. Assume that EI isconstant for the beam.Fig. P10.21Solutio

U. Memphis - CIVL - 3322

10.29a For the beams and loadings shownbelow, determine the beam deflection atpoint H. Assume that EI = 8 104 kN-m2 isconstant for each beam.Fig. P10.29aSolutionDetermine beam slope at A.[Appendix C, SS beam with concentrated moment.]Relevant equa

U. Memphis - CIVL - 3322

10.47 The simply supported beam shown inFig. P10.47 consists of a W410 60structural steel wide-flange shape [E = 200GPa; I = 216 106 mm4]. For the loadingshown, determine:(a) the beam deflection at point B.(b) the beam deflection at point C.(c) the

U. Memphis - CIVL - 3322

11.1 A beam is loaded and supported asshown in Fig. P11.1. Use the doubleintegration method to determine themagnitude of the moment M0 required tomake the slope at the left end of the beamzero.Fig. P11.1SolutionMoment equation:xM a a = M ( x) + w

U. Memphis - CIVL - 3322

11.18a For the beams and loadings shownbelow, assume that EI = 3.0 104 kN-m2 isconstant for each beam.(a) For the beam in Fig. P11.18a, determinethe concentrated upward force P required tomake the total beam deflection at B equal tozero (i.e., vB =

U. Memphis - CIVL - 3322

11.26 For the beam and loading shownbelow, derive an expression for the reactionsat supports A and B. Assume that EI isconstant for the beam.Fig. P11.26SolutionChoose the reaction force at A as the redundant; therefore, the released beam is a cantil

U. Memphis - CIVL - 3322

11.35 The beam shown in Fig. P11.35consists of a W360 79 structural steelwide-flange shape [E = 200 GPa; I = 225 106 mm4]. For the loading shown,determine:(a) the reactions at A, B, and C.(b) the magnitude of the maximum bendingstress in the beam.

U. Memphis - CIVL - 3322

11.47 A W530 92 structural steel wideflange shape [E = 200 GPa; I = 554 106mm4] is loaded and supported as shown inFig. P11.47. Determine:(a) the force and moment reactions atsupports A and C.(b) the maximum bending stress in thebeam.(c) the deflec

U. Memphis - CIVL - 3322

12.1 The stresses shown in the figure act at a point in a stressedbody. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.Fig. P12.1SolutionFn = n dA (215 MPa) cos 25(dA cos 25)

U. Memphis - CIVL - 3322

12.9 The stresses shown in the figure act at a point in a stressed body.Determine the normal and shear stresses at this point on the inclinedplane shown.Fig. P12.9SolutionThe given stress values are: x = 4, 200 psi, y = 1,800 psi, xy = 0 psi, = +50

U. Memphis - CIVL - 3322

12.25 Consider a point in a structural member that is subjected to planestress. Normal and shear stresses acting on horizontal and vertical planesat the point are shown.(a) Determine the principal stresses and the maximum in-plane shear stressacting a

U. Memphis - CIVL - 3322

12.47 Mohrs circle is shown for a point in aphysical object that is subjected to plane stress.(a) Determine the stresses x, y, and xy and showthem on a stress element.(b) Determine the principal stresses and themaximum in-plane shear stress acting at

U. Memphis - CIVL - 3322

12.75 At a point in a stressed body, the known stresses are x = 40 MPa (T), y = 20 MPa (C), z = 20MPa (T), xy = +40 MPa, yz = 0, and zx = +30 MPa. Determine:(a) the normal and shear stresses on a plane whose outward normal is oriented at angles of 40, 7

U. Memphis - CIVL - 3322

13.1 The thin rectangular plate shown in Fig. P13.1 isuniformly deformed such that x = +890 , y = 510 ,and xy = +680 rad.(a) Determine the normal strain AC along diagonal AC ofthe plate.(b) Determine the normal strain BD along diagonal BD ofthe plat

U. Memphis - CIVL - 3322

In Problems 13.23 through 13.26, the principal strains are given for a point in a body subjected to planestrain. Construct Mohrs circle and use it to(a) determine the strains x, y, and xy. (Assume x > y)(b) determine the maximum in-plane shear strain a

U. Memphis - CIVL - 3322

13.39 The strain rosette shown in the figure was used to obtain normalstrain data at a point on the free surface of a machine part.(a) Determine the strain components x, y, and xy at the point.(b) Determine the principal strains and the maximum in-plan

U. Memphis - CIVL - 3322

13.49 A 10-mm-thick aluminum [E = 70 GPa; = 0.33]plate is subjected to biaxial stress with x = 120 MPa andy = 60 MPa. The plate dimensions are b = 100 mm and h= 50 mm (see Fig. P13.49).(a) Determine the change in length of edges AB and AD.(b) Determi

U. Memphis - CIVL - 3322

14.1 Determine the normal stress in a ball, which has an outside diameterof 220 mm and a wall thickness of 3 mm, when the ball is inflated to agage pressure of 110 kPa.Fig. P14.1SolutionD = 220 mmt = 3 mmd = 220 mm 2(3 mm) = 214 mma =pd (0.110 MP

U. Memphis - CIVL - 3322

15.1 A 3-in.-diameter solid shaft is subjected toboth a torque of T = 25 kip-in. and an axialtension load of P = 40 kips, as shown in Fig.P15.1.(a) Determine the principal stresses and themaximum shear stress at point H on thesurface of the shaft.(

U. Memphis - CIVL - 3322

15.19 A tee-shaped flexural member (Fig. P15.19b) is subjected to an internal axial force of P = 1,000lb, an internal shear force of V = 600 lb, and an internal bending moment of M = 1,500 lb-ft, as shown inFig. P15.19a. Determine the principal stresses

U. Memphis - CIVL - 3322

15.29 For the vertical flexural member shown,determine the principal stresses and the maximumshear stress acting at points H and K, as shown onFigs. P15-29a and P15-29b. Show these stresses onan appropriate sketch for each point.Fig. P15.29aFig. P15

U. Memphis - CIVL - 3322

15.37 A short rectangular post supports a compressive load of P = 120 kN as shown in Fig. P15.37a. Atop view of the post showing the location where load P is applied to the top of the post is shown in Fig.P15.37b. Determine the vertical normal stresses

U. Memphis - CIVL - 3322

15.43 A 2.5-in.-diameter solid aluminum post issubjected to a horizontal force of V = 3 kips, a verticalforce of P = 7 kips, and a concentrated torque of T = 11kip-in., acting in the directions shown in Fig. P15.43.Assume L = 3.5 in. Determine the nor

U. Memphis - CIVL - 3322

15.53 A steel shaft with an outside diameter of1.25 in. is supported in flexible bearings at itsends. Two pulleys are keyed to the shaft, andthe pulleys carry belt tensions as shown in Fig.P15.53.(a) Determine the normal and shear stresses onthe top

FSU - ENC - 1101

How To Write A Paper (3rd draft) By Colton BatehamMe: Heyy. Carl: Yo dude Me: Whatcha up to bro? Carl: nm, just ate. U? Me: Dangg. Tryin to procrastinate on this paper I gotta write. Carl: oh, well, I already told some friends I was down to longboard man

FSU - ENC - 1101

Crots Paper Process Memo Colton BatehamNothing directly inspired me to write about procrastinating in my crots paper. I started texting my friends, and I noticed I was putting off my work by texting. Anything was allowed to be in the essay, so, I put my

FSU - ENC - 1101

Colton Bateham p.61-p.81 - = New slide - = big dash * = smaller (sub) dash-btw Rachael, make a title page from the chapter title and the list on page 62. -Google vs. the Library -Google: *Convenient *Only about 50% of results were considered "good" -the

FSU - ENC - 1101

Research Paper Process Memo Colton BatehamThe research paper was the least enjoyable of the three essays for me. The topic idea came to me easily though. After thinking about what interests me, my paper topic became an obvious choice. I am obsessed with

FSU - ENC - 1101

Colton Bateham Professor Burnett ENC 1105 27 September 2010 Florida State on the Upswing In College Football there are a handful of programs known as the elite. These teams dominate their opponents year after year and create a dynasty for their school. Fl

FSU - ENC - 1101

Radical Remediation Process Memo Colton BatehamFor my radical remediation or, radical revision, I chose to create a post secret postcard. The postcard was modeled as if it were created by the Narrator in my 3rd paper, the short story paper. My Narrator i

FSU - AMH - 2097

Exam 3 (Final Exam) Study GuideItalians I. Stereotypes of Italians A. Organized Crime B. Food Pasta & Pizza C. Guido D. Family-Oriented II. Italy A. No such thing as "Italy" A.1. City-states are controlled by families. A.2. 1859-1919 Unification A.2.a. S

FSU - AMH - 2097

Notes for Race and Ethnicity in the United States Test 1 Terms Race the assumption of differences based on real or imagined physical characteristics Racism assuming that someone is inferior or superior to you based on race Ethnicity similar cultural

FSU - AMH - 2097

Notes for test 2 V. Forced Labor Movement Atlantic/ Triangle Slave Trade Involvement in the trade: 1500-1880 (this class focuses on the English involvement) Types of slaves traded: o 75% male o 75% are adults (18-26 years old) Passage to the New World (Fo