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Unformatted text preview: 1.21 A hollow circular post ABC (see figure) supports a
load P. = l?00lb acting at the top. A second load P2 is
uniformly distributed around the cap plate at B. The diam—
eters and thicknesses of the upper and lower parts of the
post are aim = 1.25 111.. L15 = 0.5 in.. dm = 2.25 in.. and
rm — 0.375 in.. respectively. ta) Calculate the normal stress 0",”; 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 P3? (c) If PI remains at 1700 lb and P3 is now set at 2260 lb,
what new thickness of HC will result in the same compres
sive stress in both part s‘? PRDB. 1.21 CHAPTER 1 Problems 1.22 A force F of 70 N is applied by a rider to the
front hand brake of a bicycle (P is the resultant of an
evenly distributed pressure]. As the hand brake pivots at A,
a tension Tdevelops in the 460—1nm long brake cable (Ar =
[.025 Inma) which elongates by 5 = 0.2 [4 mm. Find normal
stress or and strain .9 in the brake cable. Brake cable. L = 460 mm Hand brake pivot A P (Resultant
of distributed
pressure)  . H I ‘
“700 “HR /¢ Uniform hand “he.
0303.. .r brake pressure PHDE. 1.2—2 CHAPTER 1 Tension, Compression, and Shear 1.23 A bicycle rider would like to compare the effective—
ness of cantilever hand brakes [see figure part (all versus V
brakes [figure part (btl. (a) Calculate the braking force RH at the wheel rims for
each of the bicycle brake systems shown. Assume that all
forces act in the plane of the figure and that cable tension
1" = 45 lbs. Also. what is the average compressive normal
stress or. on the brake pad (A — 0.035 ml)? anchored to frame ta) Cantilever brakes FHCIB. 1.2v3 1.24 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 E = 550 X IO '5. what is
the shortenng 5 of the bar? {b} If the compressive stress in the bar is intended to
he 40 MPa. what should be the load I"? (b) For each braking system. what is the stress in the
brake cable (assume effective cross—sectional area of
0.00 l 6? if}? (HINT: Because of symmetry. you only need to use the
right halfof each figure in your analysis.) — PIVOII porntS—__ anchored to frame  [b] V brakes Strain gage
/. u— — ——L=4(imm —’i PRIDE. 1.2—4 1.25 The cross section of a concrete corner column that
is loaded uniformly in compression is shown in the
figure. CHAPTER 1 Problems 3:25 (a) Determine the average compressive stress rrf. in the ,8 = 48°. Both wires have a diameter of 30 mils. (Wire (lia
concrete if the load is equal to 3200 k. meters are often expressed in mils: one mil equals 0.001 in.)
(b) Determine the coordinates xr and v. of the point Determine the tensile stresses or. and 0'; in the two wires. where the resultant load must act in order to produce
unilorrn normal stress in the column. y
4— 24 in.—»lv 20 lit—v PRIJB. 1.25
PRDB. 1.27 1.26 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 crosssectional area of 490 mmz. 1.28 A long retaining wall is braced by wood shores set at and the angle a of the incline is 30°. an angle of 30° and supported by concrete thrust blocks. as
Calculate the tensile stress Uf in the cable. shown in the ﬁrst part of the figure. The shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are ideal
ized 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 X 150 mm square cross
section. what is the compressive stress (rt in the shores? Soil Retaining
wall Concrete
Shore thrust
/ PRIJB. 1.26 K block 1.2? Two steel wires support a moveablc overhead camera
weighing W = '25 ll) (see figure) used for closeup viewing
of field action at sporting events. At some instant. wire 1 is at
an angle a = 20° to the horizontal and wire 2 is at an angle PRDB. 1.28 51} CHAPTER 1 Tension, Compression, and Shear 1.29 A pickup truck tailgate supports a crate (WC = 1501b),
as shown in the figure. The tailgate weighs IMF=60 lb
and is supported by two cables (only one is shown in
the figure). Each cable has an effective cross—sectional area
A. = 0.017 inz. (a) Find the tensile force T and normal stress 0' in each
cable. (b) If each cable clongales 5: 0.0] in. due to the
weight of both the crate and the tailgate, what is the
average strain in the cable'? PROBS.1.29 and 1210 1.210 Solve the preceding problem if the mass of the tail gate
is MT = 27 kg and that ofthe crate is MC = 68 kg. Use dimen
sions H = 305 mm, L = 406 mm. dc = 460 mm, and (if:
350 mm. The cable crosssectional area is AC, = l l .0 mmz. (a) Find the tensile force T and normal stress 0' in each
cable. (b) If each cable elongates 5 = 0.25 mm due to the
weight of both the crate and the tailgate. what is the
average strain in the cable? L = 406 mm PRDB. 1.240 1"1.2" An Lshaped reinforced concrete slab [2 ft X 12
ft (but with a 6 ft X 6 ft cutout) and thickness t = 9.0 in. is
lifted by three cables attached at 0, B and D, as shown in
the figure. The cables are combined at point Q, which is
7.0 ft above the top of the slab and directly above the center
of mass at C. Each cable has an effective cross~sectional
area 01°14.e = 0.12 inz. (a) Find the tensile force T, (i = l, 2, 3) in each cable due
to the weight W of the concrete slab [ignore weight of cables). (b) Find the average stress o, in each cable. (See Table
11—1 in Appendix H for the weight density of reinforced
concrete.) (0‘ U, ) _ l‘ 6 ft L B(IZ,0,0) lb Concrete slab ’y = 150 E Thickness t, cg at (5 ft, 5 ft. 0)
PHOB. 1.211 52? CHAPTER 1 Tension, Compression, and Shear 1.33 Three different materials. designated A, B, and C, are
tested in tension using test specimens having diameters of
0.505 in. and gage lengths of2.0 in. (see figure). At failure.
the distances between the gage marks are found to be 2.13,
2.48. and 2.78 in.. respectively. Also. at the failure cross
sections the diameters are found to be 0.484. 0.398. and
0.253 in.. respectively. Determine the percent elongation and percent reduction
in area of each specimen. and then. using your Own judg
ment, classify each material as hrittle or ductile. Gage PRIJB. 1.33 1.34 The .s'tr'engrliton‘cight ratio of a structural material
is defined as its loadcarrying capacity divided by
its weight. For materials in tension. we may use a charac—
teristic tensile stress (as obtained from a stressstrain
curve) as a measure of strength. For instance. either the
yield stress or the ultimate stress could he used. depending
upon the particular application. Titus. the strengthto—
weight ratio REM for a material in tension is defined as 0'
Rs'xw : _
1’ in which 0' is the characteristic stress and 'y is the weight
density. Note that the ratio has units of length. Using the ultimate stress or“ as the strength parameter.
calculate the strength—to—weight ratio (in units of meters)
for each of the following materials: alumintlm alloy
60bl—T6. Douglas fir (in bending). nylon, structural steel
ASTMA572. and a titanium alloy. (Obtain the material
properties from Tables H—1 and 113 of Appendix H. When
a range of values is given in a table. use the average value.) 1.35 A symmetrical framework consisting of three pin—
connected bars is loaded by a force P (see figure). The angle
between the inclined bars and the horizontal is a = 48°. The
axial strain in the middle bar is measured as 0.0? [3. Determine the tensile stress in the outer bars if they are
constructed of aluminum alloy having the stressstrain
diagram shown in Fig. 113. (Express the stress in USCS
units.) PRDB. 1.35 1.36 A specimen of a methacrylate plastic is tested in tension
at room temperature (see ﬁgure), producing the stressstrain
data listed in the accompanying table (see the next page). Plot the stress—strain curve and determine the propor
tional limit. modulus of elasticity tie. the slope of the
initial part of the stress—strain curve), and yield stress at
0.2% offset. Is the material ductile or brittle? P Md.th
( gteTHi
PROB. 1.35 STRESSSTRAIN DATA FOR PROBLEM 1.36 Stress (Ml’21) Strain
8.0 0.0032
17.5 0.0073
25.6 0.01 1 1
31.1 0.0129
39.8 0.0163
44.0 0.0184
48.2 0.0209
53.9 0.0260
58.1 0.0331
62.0 0.0429
62.1 Fracture its 7 The data shown in the accompanying table were
_i"'ll'i from a tensile test of highstrength steel. The test
gs en had a diameter of 0.505 in. and a gage length of in. {see ﬁgure for Prob. 1.33). At fracture. the elonga—
_a between the gage marks was 0. [2 in. and the minimum
in was 0.42 in. Plot the conventional stressstrain curve for the steel determine the proportional limit, modulus of elasticity
., the slope of the initial part of the stress—strain curve),
"_f.I stress 310.1% offset. ultimate stress, percent elonga—
_:: in 2.00 in., and percent reduction in area. TENSILETEST DATA FOR PROBLEM 1.37 Load (lb) Elongation (in)
l ,000 0.0002
2,000 0.0006
6,000 0.00 19 10.000 0.0033
12.000 00039
12,900 0.0043
I 3 ,400 0.004?
I 3 .600 0.0054
1 3 .800 0.0063
I 4,000 0.0090
14.400 0.0102
1 5 .200 0.0 I 30
16.8 00 0.02 30
18,400 0.0336
20,000 0.050?
22,400 0.1 108
22.600 Fracture  'city and Plasticity . 'I A bar made of structural steel having the stressstrain
am shown in the ﬁgure has a length of 48 in. The yield
5 of the steel is 42 ksi and the slope of the initial linear
_ of the stress—strain curve (modulus of elasticity) is
X 103 ksi. The bar is loaded axially until it elongates
.320 in., and then the load is removed. CHAPTER 1 Problems 63 How does the ﬁnal length of the bar compare with
its original length? (Him: Use the concepts illustrated in
Fig. l—le.) 0(ksi] 60 ‘7 40 20 0 . 0 0.002 0.004 0.006
e
PHUB. 1.41 1.42 A bar of length 2.0 m is made of a structural steel
having the stressstrain diagram shown in the ﬁgure. The
yield stress of the steel is 250 MPa and the slope of the
initial linear pan of the stressstrain curve (modulus of
elasticity) is 200 (Spa. The bar is loaded axially until it
clongates 6.5 mm. and then the load is removed. How does the ﬁnal length of the bar compare with
its original length? (Hint: Use the concepts illustrated in Fig. 118b.) 0'( MPa)
300 200 100 0 0.002 0.004 0.006 PRUB. 1.4~2 64 CHAPTER 1 Tension, Compression, and Shear 1.43 An aluminum bar has length l. = 5 ft and diameter
d = 1.25 in. The stress—strain curve for the aluminum is
shown in Fig. 1—13 of Section 1.3. The initial straightw
line part of the curve has a slope (modulus of elasticity) of
10 X 10‘I psi. The bar is loaded by tensile forces P = 39 k
and then unloaded. (a) What is the permanent set of the bar? (b) If the bar is reloaded. what is the proportional
limit? (Him: Use the concepts illustrated in Figs. 1—18b and
1—19.) 1.44 A circular bar of magnesium alloy is 750 mm long.
The stress—strain diagram for the material is shown in the
ﬁgure. The bar is loaded in tension to an elongation of
6.0 mm. and then the load is removed. (a) What is the permanent set ofthe bar? (b) lfthe bar is reloaded, what is the proportional limit?
(Him: Use the concepts illustrated in Figs. 118b and 119.] 200 rI(MPa) 100 l1
0 0.005 0.010 P9035. 1.43 and 1.44 in1.45 A wire of length L = 4 ft and diameter d = 0.125 in.
is stretched by tensile forces P = 600 lb. The wire is made of
a copper alloy having a stressstrain relationship that may be
described mathematically by the following equation: 18.0005 0 003 k _
= — E S I = .
‘7 1 + 3005 E (a 51) in which 6 is nondimensional and a" has units of kips per
square inch {ksi}.
(a) Construct a stress—strain diagram for the material.
(b) Determine the elongation of the wire due to the
forces P. (c) If the forces are removed, what is the permanent set
of the bar? (d) If the forces are applied again, what is the propor—
tional limit? Roche’s Law and Poisson‘s Basia When solving the problems for Secrirm l .5. assume that the
material behaves linearly elastically. 1.51 A highstrength steel bar used in a large crane has
diameter d = 2.00 in. {see figure). The steel has modulus of
elasticity E = 29 X 10" psi and Poisson's ratio V = 0.29.
Because of clearance requirements. the diameter of the bar
is limited to 2.001 in. when it is compressed by axial
forces. What is the largest compressive load Pm.“1x that is
permitted? PRUB. 1.5—1 1.52 A round bar of 10 mm diameter is made of
aluminum alloy 7075—T6 (see figure}. When the bar is
stretched by axial forces P, its diameter decreases by
0.016 mm. Find the magnitude of the load P. (Obtain the material
properties from Appendix H.) TOYS—T6 PROB. 1.52 1.53 A polyethylene bar having diameter d1 = 4.0 in. is
placed inside a steel tube having inner diameter d: = 4.01 in.
(see figure). The polyethylene bar is then compressed by an
axial force P. At what value of the force P will the space between
'the polyethylene bar and the steel tube be closed? (For
polyethylene, assume E = 200 ksi and v = 04.) Steel
tube Polyethylene
bar PHDB. 1.53 1.54 A prismatic bar with a circular cross section is
loaded by tensile forces P = 65 kN (see ﬁgure). The bar
has length L = 1.75 m and diameter d = 32 mm. It is made
of aluminum alloy with modulus of elasticity E = 75 GPa
and Poisson’s ratio U = U3. Find the increase in length of the bar and the percent
decrease in its cross—sectional area. P8083. 1.54 and 1.55 1.55 A bar of monel metal as in the ﬁgure (length L = 9 in.,
diameter d = 0.225 in.) is loaded axially by a tensile force P.
If the bar elongates by 0.0195 in., what is the decrease in
diameter a"? What is the magnitude of the load P? Use the
data in Table 112, Appendix H. 1.55 A tensile test is peformed on a brass specimen 10 mm
in diameter using a gage length of 50 mm (see ﬁgure]. When
the tensile load P reaches a value of 20 kN, the distance
between the gage marks has increased by 0.122 mm. CHAPTER 1 Problems 55 (a) What is the modulus of elasticity E of the brass?
(b) If the diameter decreases by 0.00830 mm, what is
Poisson’s ratio? PRDB. 1.56 1.57 A hollow, brass circular pipe ABC (see figure) supports
a load P. = 26.5 kips acting at the top. A second load P2 =
22.0 kips is unifonnly distributed around the cap plate at B.
The diameters and thicknesses of the upper and lower parts of
the pipe are LEM = 1.25 in., .5”, = 0.5 in.. (lac = 2.25 in., and
rm; = 0.375 in., respectively. The modulus of elasticity is
[4,000 ksi. When both loads are fully applied, the wall thick
ness of pipe BC increases by 200 X 10 h in. (a) Find the increase in the inner diameter of pipe
segment BC. (b) Find Poisson‘s ratio for the brass. (c) Find the increase in the wall thickness of
pipe segment AB and the increase in the inner diameter
ofAB. PRUB. 1.5? 1.21 1.22
1.23 1.24
1 .25
1.26
1.2?
1 .28
1.29
1.210 1.2“ 1.2“!
1.213 1.214
1.31
1.32
1.33 1.34 1.35
1.36 1.3? CHAPTER 1 (1003”, = 1443 psi; (b) Pg = 1487.5 lbs; (c) rm; = 0.5 in. (11):)" = 65 MPa; (b) a: = 4.652 X 10 4 (a) Rh. = 127.3 lb {cantilever}, 191.3 lb (V—brakes);
t. = 204 psi (cantilever), 306 psi (V—brakes); (b) (b) (rum.e = 26,946 psi (both) (a15 = 0.220 mm; (b) P = 34.6 kN (a) 05 = 2.128 ksi: IC = 19.22 in., ya = 19.22 in.
or, = 133 MPa or. = 25.5 ksi: U2 = 35.8 ksi: cr,_. = 5.21 MPa (3) T: 184 1b, 0' = 10.8 ksi; (b) em“, = 5 x 10'4
(a) T = 819 N, 0' = 74.5 MPa;
(b) em“, = 4.923 >< 10‘4
(a) T. = 5877 lb, T2 = 46791b. T3 = 7159 lb;
(610. = 49 k51,0'2 = 39 ksi. 03 = 60 ksi
(a) (er = yw2(L2  x2)7'2g: (b) Um“ = yw2L272g
((1)7113 = 16201b, TM: = 1536 lb, To) = 164016
(b) (IN, = 13,501 psi, 17m; = 12.799 psi,
(rtp = 13.667 psi
(a) TAQ = TBQ = 50.5 kN; (b} 0' = 166 MPa
(a) Lmax = 11,800 ft; (6)12”, = 13,500 ft
(a) Lmax = 7900 m; (b) Lnm = 8330 m
% elongation = 6.5, 24.0, 39.0;
% reduction = 8.1, 37.9, 74.9:
Brittle. ductile, ductile
11.9 X103m;12.7 ><103m;6.1><103 m;
6.5 X 103 m; 23.9 X 103 m
rr H”: 31 ksi
Up. = 47 MPa, Slope % 2.4 GPa. 01' = 53 MPa;
Brittle
up. = 65,000 psi, Slope 4: 30 x 10“ psi,
cry ‘= 69,000 psi, 0'” 2 113,000 psi;
Elongation = 6%, Reduction = 31%
0.13 in. longer Answers to Problems 1.42
1.43
1.44
1.45
1.51
1.52
1.53
1.54 1.55
1.56
1.57 1 .53
1.61
1.62
1.63 1.6—4 1.65
1.66 1.87 1.68
1.69
1.6"!
1.6“
1.812
1.613 4.0 mm longer
(a) 2.809 in.; (b) 31.8 ksi
(a) 2.966 mm; (b) 180 MPa
(1)) 0.71 in.: (1:10.58 ir1.; {(1)49 ksi
P11m = 157 k
P = 27.4 kN (tension)
P =  15.708 kips
AL = 1.886 mm; 92: decrease
in .r—sec area = 0.072%
Ad 2 —1.56 x 10“1 in., P = 2.154 kips
(a) E = 104 GPa; (b) v = 0.34
(a) AdﬂCinner : 8 X 10—4 in
16) 9W, 2 0.34
{0)AL13 = 2.?32 x 10416, Admmr = 1.366 >< 10‘4 in.
M = 9789 mm3
0;, = 7.04 ksi, “rave = 10.756 ksi
01:, = 139.86 MPa; Pun = 144.45 kN
(a) 1' = 12.732 ksi; (b) any: = 20 ksi: 0b,, = 26.667 ksi (a) Ax = 254.6 N, A}. = 1072 N, 3,, = —254.6 N
(b) Aresullam 2 11018 N
(c) T = 5.48 MPa, 0!, = 6.886 MPa
{3) “rm, = 2979 psi; 1b) orbmx = 936 psi
T] = 13.176 kN, T3 = 10.772 RN,
TIM, = 25.888 MPa, 72m = 21.166 MPa,
orb. = 7.32 MPa, 0b; = 5.985 MPa
(6) Resultant = 1097 lb;
(b) 0;», = 4999 psi
(1:) Tm“ = 2793 psi, 7p] 2 609 psi
G = 2.5 MPa
(at) ‘yaver = 0.004; (b) V = 89.6 k
(a) yam, = 0.50; (b) 5 = 4.50 mm
(a) rave, = 6050 psi; (b) 0;, = 9500 psi
7;..." = 429 MPa
(a) Ax = 0, 14).: 17016114,, = 4585 in.—lb
1MB,r = 253.6 lb, B). = 1601b, Bras = 299.8 lb, ct = _B.'( 995 ...
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This note was uploaded on 09/18/2011 for the course EM 319 taught by Professor Staff during the Spring '11 term at Oklahoma State.
 Spring '11
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