PROBLEM 1.1
Two solid cylindrical rods AB and BC are welded together at B and
loaded as shown. Knowing that the average normal stress must not
exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the
smallest allowable values of d1 and d2.
SOLUTION
(a)
Rod AB
P = 40 + 30 = 70 kN = 70 × 103 N
σ AB =
d1 =
(b)
P
=
AAB
P
4P
=
2
d
π d12
4 1
π
4P
=
πσ AB
(4)(70 × 103 )
= 22.6 × 10−3 m
π (175 × 106 )
d1 = 22.6 mm
Rod BC
P = 30 kN = 30 × 103 N
σ BC =
d2 =
P
=
ABC
4P
πσ BC
P
4P
=
2
d
π d 22
4 2
π
=
(4)(30 × 103 )
= 15.96 × 10−3 m
π (150 × 106 )
d 2 = 15.96 mm
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PROBLEM 1.2
Two solid cylindrical rods AB and BC are welded together at B and loaded
as shown. Knowing that d1 = 50 mm and d 2 = 30 mm, find the average
normal stress at the midsection of (a) rod AB, (b) rod BC.
SOLUTION
(a)
Rod AB
P = 40 + 30 = 70 kN = 70 × 103 N
A=
σ AB =
(b)
π
4
d12 =
π
4
(50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2
P
70 × 103
=
= 35.7 × 106 Pa
−3
A 1.9635 × 10
σ AB = 35.7 MPa
Rod BC
P = 30 kN = 30 × 103 N
A=
σ BC =
π
4
d 22 =
π
4
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
P
30 × 103
=
= 42.4 × 106 Pa
−6
A 706.86 × 10
σ BC = 42.4 MPa
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PROBLEM 1.3
Two solid cylindrical rods AB and BC are welded together
at B and loaded as shown. Determine the magnitude of the
force P for which the tensile stress in rod AB is twice the
magnitude of the compressive stress in rod BC.
SOLUTION
σ AB
π
(2) 2 = 3.1416 in 2
4
P
P
=
=
AAB
3.1416
AAB =
= 0.31831 P
ABC =
σ BC
π
(3)2 = 7.0686 in 2
4
(2)(30) − P
=
AAB
=
60 − P
= 8.4883 − 0.14147 P
7.0686
Equating σ AB to 2σ BC
0.31831 P = 2(8.4883 − 0.14147 P)
P = 28.2 kips
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PROBLEM 1.4
In Prob. 1.3, knowing that P = 40 kips, determine the
average normal stress at the midsection of (a) rod AB,
(b) rod BC.
PROBLEM 1.3 Two solid cylindrical rods AB and BC
are welded together at B and loaded as shown. Determine
the magnitude of the force P for which the tensile stress
in rod AB is twice the magnitude of the compressive
stress in rod BC.
SOLUTION
(a)
Rod AB
P = 40 kips (tension)
AAB =
σ AB
(b)
2
π d AB
=
π (2) 2
4
4
P
40
=
=
AAB
3.1416
= 3.1416 in 2
σ AB = 12.73 ksi
Rod BC
F = 40 − (2)(30) = −20 kips, i.e., 20 kips compression.
ABC =
σ BC
2
π d BC
=
π (3) 2
4
4
−20
F
=
=
ABC
7.0686
= 7.0686 in 2
σ BC = −2.83 ksi
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PROBLEM 1.5
Two steel plates are to be held together by means of 16-mmdiameter high-strength steel bolts fitting snugly inside cylindrical
brass spacers. Knowing that the average normal stress must not
exceed 200 MPa in the bolts and 130 MPa in the spacers,
determine the outer diameter of the spacers that yields the most
economical and safe design.
SOLUTION
At each bolt location the upper plate is pulled down by the tensile force Pb of the bolt. At the same time, the
spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium.
Pb = Ps
For the bolt,
σb =
Fb
4Pb
=
Ab
π db2
or
Pb =
For the spacer,
σs =
Ps
4Ps
=
As
π (d s2 − db2 )
or
Ps =
π
4
π
4
σ bdb2
σ s (d s2 − db2 )
Equating Pb and Ps ,
π
4
σ bdb2 =
ds =
π
4
σ s (d s2 − db2 )
1+
σb
d =
σs b
1+
200
(16)
130
d s = 25.2 mm
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PROBLEM 1.6
Two brass rods AB and BC, each of uniform diameter, will be brazed
together at B to form a nonuniform rod of total length 100 m, which will
be suspended from a support at A as shown. Knowing that the density of
brass is 8470 kg/m3, determine (a) the length of rod AB for which the
maximum normal stress in ABC is minimum, (b) the corresponding
value of the maximum normal stress.
SOLUTION
Areas:
AAB =
ABC =
From geometry,
Weights:
π
4
π
4
(15 mm)2 = 176.71 mm 2 = 176.71 × 10−6 m 2
(10 mm) 2 = 78.54 mm 2 = 78.54 × 10−6 m 2
b = 100 − a
WAB = ρ g AAB
AB
= (8470)(9.81)(176.71 × 10−6 ) a = 14.683 a
WBC = ρ g ABC
BC
= (8470)(9.81)(78.54 × 10−6 )(100 − a) = 652.59 − 6.526 a
Normal stresses:
At A,
PA = WAB + WBC = 652.59 + 8.157a
σA =
At B,
PA
= 3.6930 × 106 + 46.160 × 103 a
AAB
PB = WBC = 652.59 − 6.526a
σB =
(a)
(1)
(2)
PB
= 8.3090 × 106 − 83.090 × 103 a
ABC
Length of rod AB. The maximum stress in ABC is minimum when σ A = σ B or
4.6160 × 106 − 129.25 × 103 a = 0
a = 35.71 m
(b)
AB
= a = 35.7 m
Maximum normal stress.
σ A = 3.6930 × 106 + (46.160 × 103 )(35.71)
σ B = 8.3090 × 106 − (83.090 × 103 )(35.71)
σ A = σ B = 5.34 × 106 Pa
σ = 5.34 MPa
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PROBLEM 1.7
Each of the four vertical links has an 8 × 36-mm uniform rectangular
cross section and each of the four pins has a 16-mm diameter.
Determine the maximum value of the average normal stress in the
links connecting (a) points B and D, (b) points C and E.
SOLUTION
Use bar ABC as a free body.
ΣM C = 0 :
(0.040) FBD − (0.025 + 0.040)(20 × 103 ) = 0
FBD = 32.5 × 103 N
Link BD is in tension.
3
ΣM B = 0 : − (0.040) FCE − (0.025)(20 × 10 ) = 0
FCE = −12.5 × 103 N
Link CE is in compression.
Net area of one link for tension = (0.008)(0.036 − 0.016) = 160 × 10−6 m 2.
For two parallel links,
(a)
σ BD =
A net = 320 × 10−6 m 2
FBD
32.5 × 103
=
= 101.56 × 106
Anet
320 × 10−6
σ BD = 101.6 MPa
Area for one link in compression = (0.008)(0.036) = 288 × 10−6 m 2.
For two parallel links,
(b)
σ CE =
A = 576 × 10−6 m 2
FCE
−12.5 × 103
=
= −21.70 × 10−6
A
576 × 10−6
σ CE = −21.7 MPa
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PROBLEM 1.8
Knowing that the link DE is 18 in. thick and 1 in. wide, determine
the normal stress in the central portion of that link when
(a) θ = 0°, (b) θ = 90°.
SOLUTION
Use member CEF as a free body.
Σ M C = 0 : − 12 FDE − (8)(60 sin θ ) − (16)(60 cos θ ) = 0
FDE = −40 sin θ − 80 cos θ lb.
ADE = (1)
σ DE =
(a)
1
= 0.125 in.2
8
FDE
ADE
θ = 0: FDE = −80 lb.
σ DE =
(b)
−80
0.125
σ DE = −640 psi
θ = 90°: FDE = −40 lb.
σ DE =
−40
0.125
σ DE = −320 psi
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PROBLEM 1.9
Link AC has a uniform rectangular cross section 161 in.
thick and 14 in. wide. Determine the normal stress in the
central portion of the link.
SOLUTION
Free Body Diagram of Plate
Note that the two 240-lb forces form a couple of moment
(240 lb)(6 in.) = 1440 lb ⋅ in.
Σ M B = 0 : 1440 lb ⋅ in − ( FAC cos 30°)(10 in.) = 0
FAC = 166.277 lb.
Area of link:
AAC =
1
in.
16
1
in. = 0.015625 in.2
4
Stress:
σ AC =
FAC
166.277
=
= 10640 psi
AAC
0.015625
σ AC = 10.64 ksi
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PROBLEM 1.10
Three forces, each of magnitude P = 4 kN, are applied to
the mechanism shown. Determine the cross-sectional area
of the uniform portion of rod BE for which the normal
stress in that portion is +100 MPa.
SOLUTION
Draw free body diagrams of AC and CD.
Free Body CD:
ΣM D = 0: 0.150 P − 0.250C = 0
C = 0.6P
Free Body AC:
Required area of BE:
M A = 0: 0.150 FBE − 0.350P − 0.450 P − 0.450C = 0
FBE =
1.07
P = 7.1333 P = (7.133)(4 kN) = 28.533 kN
0.150
σ BE =
FBE
ABE
ABE =
FBE
σ BE
=
28.533 × 103
= 285.33 × 10−6 m 2
100 × 106
ABE = 285 mm 2
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PROBLEM 1.11
The frame shown consists of four wooden members, ABC,
DEF, BE, and CF. Knowing that each member has a 2 × 4-in.
rectangular cross section and that each pin has a 1/2-in.
diameter, determine the maximum value of the average
normal stress (a) in member BE, (b) in member CF.
SOLUTION
Add support reactions to figure as shown.
Using entire frame as free body,
ΣM A = 0: 40 Dx − (45 + 30)(480) = 0
Dx = 900 lb.
Use member DEF as free body.
Reaction at D must be parallel to FBE and FCF .
Dy =
4
Dx = 1200 lb.
3
ΣM F = 0: − (30)
4
FBE − (30 + 15) DY = 0
5
FBE = −2250 lb.
ΣM E = 0: (30)
4
FCE − (15) DY = 0
5
FCE = 750 lb.
Stress in compression member BE
Area:
(a)
A = 2 in × 4 in = 8 in 2
σ BE =
FBE
−2250
=
A
8
σ BE = −281 psi
Minimum section area occurs at pin.
Amin = (2)(4.0 − 0.5) = 7.0 in 2
Stress in tension member CF
(b)
σ CF =
FCF
750
=
Amin
7.0
σ CF =107.1 psi
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PROBLEM 1.12
For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the crosssectional area of that member is 5.87 in2.
SOLUTION
Use entire truss as free body.
ΣM H = 0: (9)(80) + (18)(80) + (27)(80) − 36 Ay = 0
Ay = 120 kips
Use portion of truss to the left of a section cutting members
BD, BE, and CE.
+ ↑ Σ Fy = 0: 120 − 80 −
σ BE =
12
FBE = 0
15
∴ FBE = 50 kips
FBE
50 kips
=
A
5.87 in 2
σ BE = 8.52 ksi
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PROBLEM 1.13
An aircraft tow bar is positioned by means of a single
hydraulic cylinder connected by a 25-mm-diameter steel
rod to two identical arm-and-wheel units DEF. The mass
of the entire tow bar is 200 kg, and its center of gravity
is located at G. For the position shown, determine the
normal stress in the rod.
SOLUTION
FREE BODY – ENTIRE TOW BAR:
W = (200 kg)(9.81 m/s 2 ) = 1962.00 N
Σ M A = 0 : 850 R − 1150(1962.00 N) = 0
R = 2654.5 N
FREE BODY – BOTH ARM & WHEEL UNITS:
tan α =
100
675
α = 8.4270°
Σ M E = 0 : ( FCD cos α )(550) − R(500) = 0
FCD =
500
(2654.5 N)
550 cos 8.4270°
= 2439.5 N (comp.)
σ CD = −
2439.5 N
FCD
=−
ACD
π (0.0125 m) 2
= −4.9697 × 106 Pa
σ CD = −4.97 MPa
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PROBLEM 1.14
A couple M of magnitude 1500 N ⋅ m is applied to the crank of an
engine. For the position shown, determine (a) the force P required to
hold the engine system in equilibrium, (b) the average normal stress in
the connecting rod BC, which has a 450-mm2 uniform cross section.
SOLUTION
Use piston, rod, and crank together as free body. Add wall reaction H
and bearing reactions Ax and Ay.
Σ M A = 0 : (0.280 m) H − 1500 N ⋅ m = 0
H = 5.3571 × 103 N
Use piston alone as free body. Note that rod is a two-force member;
hence the direction of force FBC is known. Draw the force triangle
and solve for P and FBE by proportions.
l =
2002 + 602 = 208.81 mm
P
200
=
H
60
∴
P = 17.86 × 103 N
(a)
P = 17.86 kN
FBC
208.81
=
∴ FBC = 18.643 × 103 N
H
60
Rod BC is a compression member. Its area is
450 mm 2 = 450 × 10−6 m 2
Stress,
σ BC =
− FBC
−18.643 × 103
=
= −41.4 × 106 Pa
A
450 × 10−6
(b)
σ BC = −41.4 MPa
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PROBLEM 1.15
When the force P reached 8 kN, the wooden specimen shown failed in
shear along the surface indicated by the dashed line. Determine the
average shearing stress along that surface at the time of failure.
SOLUTION
Area being sheared:
A = 90 mm × 15 mm = 1350 mm 2 = 1350 × 10−6 m2
Force:
P = 8 × 103 N
Shearing stress:
τ =
P
8 × 103
−
= 5.93 × 106 Pa
A 1350 × 10−6
τ = 5.93 MPa
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PROBLEM 1.16
The wooden members A and B are to be joined by plywood splice
plates, that will be fully glued on the surfaces in contact. As part of
the design of the joint, and knowing that the clearance between the
ends of the members is to be 14 in., determine the smallest allowable
length L if the average shearing stress in the glue is not to exceed
120 psi.
SOLUTION
There are four separate areas that are glued. Each of these areas transmits one half the 5.8 kip force. Thus
F =
1
1
P = (5.8) = 2.9 kips = 2900 lb.
2
2
Let l = length of one glued area and w = 4 in. be its width.
For each glued area,
A = lw
Average shearing stress:
τ =
F
F
=
A lw
The allowable shearing stress is τ = 120 psi
F
2900
=
= 6.0417 in.
τ w (120)(4)
Solving for l,
l =
Total length L:
L = l + (gap) + l = 6.0417 +
1
+ 6.0417
4
L = 12.33 in.
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PROBLEM 1.17
A load P is applied to a steel rod supported as shown by an
aluminum plate into which a 0.6-in.-diameter hole has been
drilled. Knowing that the shearing stress must not exceed
18 ksi in the steel rod and 10 ksi in the aluminum plate,
determine the largest load P that can be applied to the rod.
SOLUTION
A1 = π dt = π (0.6)(0.4)
For steel:
= 0.7540 in 2
τ1 =
P
∴ P = A1τ1 = (0.7540)(18)
A
= 13.57 kips
A2 = π dt = π (1.6)(0.25) = 1.2566 in 2
For aluminum:
τ2 =
P
∴ P = A2τ 2 = (1.2566)(10) = 12.57 kips
A2
Limiting value of P is the smaller value, so
P = 12.57 kips
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PROBLEM 1.18
Two wooden planks, each 22 mm thick and 160 mm
wide, are joined by the glued mortise joint shown.
Knowing that the joint will fail when the average
shearing stress in the glue reaches 820 kPa, determine
the smallest allowable length d of the cuts if the joint is
to withstand an axial load of magnitude P = 7.6 kN.
SOLUTION
Seven surfaces carry the total load P = 7.6 kN = 7.6 × 103.
Let t = 22 mm.
Each glue area is A = dt
τ =
P
7A
A=
P
7.6 × 103
=
= 1.32404 × 10−3 m 2
7τ
(7)(820 × 103 )
= 1.32404 × 103 mm 2
d =
A 1.32404 × 103
=
= 60.2
t
22
d = 60.2 mm
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PROBLEM 1.19
The load P applied to a steel rod is distributed to a timber support by an annular
washer. The diameter of the rod is 22 mm and the inner diameter of the washer
is 25 mm, which is slightly larger than the diameter of the hole. Determine the
smallest allowable outer diameter d of the washer, knowing that the axial normal
stress in the steel rod is 35 MPa and that the average bearing stress between the
washer and the timber must not exceed 5 MPa.
SOLUTION
Steel rod: A =
π
4
(0.022) 2 = 380.13 × 10−6 m 2
σ = 35 × 106 Pa
P = σ A = (35 × 106 )(380.13 × 10−6 )
= 13.305 × 103 N
Washer: σ b = 5 × 106 Pa
Required bearing area:
Ab =
But, Ab =
π
4
P
σb
=
13.305 × 103
= 2.6609 × 10−3 m 2
5 × 106
(d 2 − di2 )
d 2 = di2 +
4 Ab
π
= (0.025)2 +
(4)(2.6609 × 10−3 )
π
−3
= 4.013 × 10 m
d = 63.3 × 10−3 m
2
d = 63.3 mm
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PROBLEM 1.20
The axial force in the column supporting the timber beam shown is
P = 20 kips. Determine the smallest allowable length L of the bearing
plate if the bearing stress in the timber is not to exceed 400 psi.
SOLUTION
Bearing area: Ab = Lw
σb =
L=
P
P
=
Ab
Lw
20 × 103
P
=
= 8.33 in.
σ b w (400)(6)
L = 8.33 in.
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PROBLEM 1.21
An axial load P is supported by a short W8 × 40 column of crosssectional area A = 11.7 in.2 and is distributed to a concrete foundation
by a square plate as shown. Knowing that the average normal stress in
the column must not exceed 30 ksi and that the bearing stress on the
concrete foundation must not exceed 3.0 ksi, determine the side a of
the plate that will provide the most economical and safe design.
SOLUTION
For the column σ =
P
or
A
P = σ A = (30)(11.7) = 351 kips
For the a × a plate, σ = 3.0 ksi
A=
P
σ
=
351
= 117 in 2
3.0
Since the plate is square, A = a 2
a =
A = 117
a = 10.82 in.
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PROBLEM 1.23
A 58 -in.-diameter steel rod AB is fitted to a round hole near end C of
the wooden member CD. For the loading shown, determine (a) the
maximum average normal stress in the wood, (b) the distance b for
which the average shearing stress is 100 psi on the surfaces indicated
by the dashed lines, (c) the average bearing stress on the wood.
SOLUTION
(a)
Maximum normal stress in the wood
Anet = (1) 4 −
σ =
(b)
5
= 3.375 in.2
8
1500
P
=
= 444 psi
3.375
Anet
σ = 444 psi
Distance b for τ = 100 psi
For sheared area see dotted lines.
P
P
=
A 2bt
1500
P
b=
=
= 7.50 in.
2tτ
(2)(1)(100)
τ =
(c)
b = 7.50 in.
Average bearing stress on the wood
σb =
P
P
1500
=
=
= 2400 psi
5
Ab
dt
(1)
8
σ b = 2400 psi
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PROBLEM 1.24
Knowing that θ = 40° and P = 9 kN, determine (a) the smallest
allowable diameter of the pin at B if the average shearing stress in
the pin is not to exceed 120 MPa, (b) the corresponding average
bearing stress in member AB at B, (c) the corresponding average
bearing stress in each of the support brackets at B.
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle
P
FAB
FAC
=
=
sin 20° sin110° sin 50°
P sin110°
FAB =
sin 20°
(9)sin110°
=
= 24.73 kN
sin 20°
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PROBLEM 1.24 (Continued)
(a)
Allowable pin diameter.
τ =
FAB
F
2 FAB
= πAB 2 =
where FAB = 24.73 × 103 N
2
2 AP
πd
24d
d2 =
2 FAB
πτ
=
(2)(24.73 × 103 )
= 131.18 × 10−6 m 2
6
π (120 × 10 )
d = 11.45 × 10−3 m
(b)
11.45 mm
Bearing stress in AB at A.
Ab = td = (0.016)(11.45 × 10−3 ) = 183.26 × 10−6 m 2
σb =
(c)
FAB
24.73 × 103
=
= 134.9 × 106
Ab
183.26 × 10−6
134.9 MPa
Bearing stress in support brackets at B.
A = td = (0.012)(11.45 × 10−3 ) = 137.4 × 10−6 m 2
σb =
1
2
FAB
A
=
(0.5)(24.73 × 103 )
= 90.0 × 106
137.4 × 10−6
90.0 MPa
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PROBLEM 1.25
Determine the largest load P which may be applied at A when θ = 60°,
knowing that the average shearing stress in the 10-mm-diameter pin at
B must not exceed 120 MPa and that the average bearing stress in
member AB and in the bracket at B must not exceed 90 MPa.
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle
P
FAB
FAC
=
=
sin 30° sin 120° sin 30°
P=
FAB sin 30°
= 0.57735 FAB
sin 120°
P=
FAC sin 30°
= FAC
sin 30°
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PROBLEM 1.25 (Continued)
If shearing stress in pin at B is critical,
A=
π
4
d2 =
π
4
(0.010) 2 = 78.54 × 10−6 m 2
FAB = 2 Aτ = (2)(78.54 × 10−6 )(120 × 106 ) = 18.850 × 103 N
If bearing stress in member AB at bracket at A is critical,
Ab = td = (0.016)(0.010) = 160 × 10−6 m 2
FAB = Abσ b = (160 × 10−6 )(90 × 106 ) = 14.40 × 103 N
If bearing stress in the bracket at B is critical,
Ab = 2td = (2)(0.012)(0.010) = 240 × 10 −6 m 2
FAB = Abσ b = (240 × 10−6 )(90 × 106 ) = 21.6 × 103 N
Allowable FAB is the smallest, i.e., 14.40 × 103N
Then from Statics
Pallow = (0.57735)(14.40 × 103 )
= 8.31 × 103 N
8.31 kN
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PROBLEM 1.26
Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support
the end of a horizontal beam. Knowing that the average normal stress in
the link is − 140 MPa, and that the average shearing stress in each of the
two pins is 80 MPa, determine (a) the diameter d of the pins, (b) the
average bearing stress in the link.
SOLUTION
Rod AB is in compression.
A = bt
where b = 50 mm and t = 6 mm
A = (0.050)(0.006) = 300 × 10−6 m 2
P = −σ A = −(−140 × 106 )(300 × 10−6 )
= 42 × 103 N
For the pin,
Ap =
π
Ap =
(a)
P
τ
=
P
Ap
42 × 103
= 525 × 10−6 m 2
80 × 106
Diameter d
d =
(b)
4
and τ =
d2
Bearing stress
σb =
4 Ap
π
=
(4)(525 × 10 −6 )
π
= 2.585 × 10−3 m
P
42 × 103
=
= 271 × 106 Pa
dt
(25.85 × 10−3 )(0.006)
d = 25.9 mm
σ b = 271 MPa
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PROBLEM 1.27
For the assembly and loading of Prob. 1.7, determine (a) the average
shearing stress in the pin at B, (b) the average bearing stress at B in
member BD, (c) the average bearing stress at B in member ABC, knowing
that this member has a 10 × 50-mm uniform rectangular cross section.
PROBLEM 1.7 Each of the four vertical links has an 8 × 36-mm uniform
rectangular cross section and each of the four pins has a 16-mm diameter.
Determine the maximum value of the average normal stress in the links
connecting (a) points B and D, (b) points C and E.
SOLUTION
Use bar ABC as a free body.
ΣM C = 0 : (0.040) FBD − (0.025 + 0.040)(20 × 103 ) = 0
FBD = 32.5 × 103 N
(a)
Shear pin at B
τ =
where
A=
τ =
(b)
Bearing: link BD
Bearing in ABC at B
π
4
d2 =
π
4
(0.016) 2 = 201.06 × 10−6 m 2
32.5 × 103
= 80.8 × 106
(2)(201.06 × 10−6 )
τ = 80.8 MPa
A = dt = (0.016)(0.008) = 128 × 10−6 m 2
σb =
(c)
FBD
for double shear,
2A
1
2
FBD
A
=
(0.5)(32.5 × 103 )
= 126.95 × 106
−6
128 × 10
σ b = 127.0 MPa
A = dt = (0.016)(0.010) = 160 × 10−6 m 2
σb =
FBD
32.5 × 103
=
= 203 × 106
−6
A
160 × 10
σ b = 203 MPa
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PROBLEM 1.28
The hydraulic cylinder CF, which partially controls the position of rod
DE, has been locked in the position shown. Member BD is 58 in. thick
and is connected to the vertical rod by a 83 -in.-diameter bolt.
Determine (a) the average shearing stress in the bolt, (b) the bearing
stress at C in member BD.
SOLUTION
Use member BCD as a free body, and note that AB is a two force member.
l AB =
82 + 1.82 = 8.2 in.
ΣM C = 0: (4 cos 20°)
8
1.8
FAB − (4sin 20°)
FAB
8.2
8.2
−(7 cos 20°)(400sin 75°) − (7sin 20°)(400 cos 75°) = 0
3.36678FAB − 2789.35 = 0
∴ FAB = 828.49 lb
1.8
FAB + Cx + 400cos 75° = 0
8.2
(1.8)(828.49)
− 400cos 75° = 78.34 lb
Cx =
8.2
ΣFx = 0: −
8
FAB + C y − 400sin 75° = 0
8.2
(8)(828.49)
+ 400sin 75° = 1194.65 lb
Cy =
8.2
ΣFy = 0: −
C =
Cx2 + C y2 = 1197.2 lb
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PROBLEM 1.28 (Continued)
(a)
(b)
Shearing stress in the bolt: P = 1197.2 lb
2
A=
π 2 π 3
d =
4
4 8
τ =
1197.2
P
=
= 10.84 × 103 psi =
A 0.11045
Bearing stress at C in member BCD: P = 1197.2 lb Ab = dt =
σb =
= 0.11045 in 2
3
8
10.84 ksi
5
= 0.234375 in 2
8
P
1197.2
=
= 5.11 × 103 psi = 5.11 ksi
0.234375
Ab
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PROBLEM 1.29
The 1.4-kip load P is supported by two wooden members of uniform cross section
that are joined by the simple glued scarf splice shown. Determine the normal and
shearing stresses in the glued splice.
SOLUTION
P = 1400 lb
θ = 90° − 60° = 30°
A0 = (5.0)(3.0) = 15 in 2
σ =
P cos 2 θ
(1400)(cos 30°)2
=
A0
15
σ = 70.0 psi
τ =
P sin 2θ
(1400)sin 60°
=
2 A0
(2)(15)
τ = 40.4 psi
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PROBLEM 1.30
Two wooden members of uniform cross section are joined by the simple scarf
splice shown. Knowing that the maximum allowable tensile stress in the glued
splice is 75 psi, determine (a) the largest load P that can be safely supported, (b)
the corresponding shearing stress in the splice.
SOLUTION
A0 = (5.0)(3.0) = 15 in 2
θ = 90° − 60° = 30°
σ =
(a)
(b)
P=
P cos 2 θ
A0
σ A0
(75)(15)
=
= 1500 lb
2
cos θ
cos 2 30°
τ =
P sin 2θ
(1500)sin 60°
=
2 A0
(2)(15)
P = 1.500 kips
τ = 43.3 psi
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PROBLEM 1.32
Two wooden members of uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that the
maximum allowable shearing stress in the glued splice is 620 kPa,
determine (a) the largest load P that can be safely applied, (b) the
corresponding tensile stress in the splice.
SOLUTION
θ = 90° − 45° = 45°
A0 = (150)(75) = 11.25 × 103 mm 2 = 11.25 × 10−3 m2
τ = 620 kPa = 620 × 103 Pa
P sin 2θ
τ =
2 A0
(a)
P=
2 A0τ
(2)(11.25 × 10−3 )(620 × 103 )
=
sin2θ
sin 90°
= 13.95 × 103 N
(b)
σ =
P = 13.95 kN
P cos 2 θ
(13.95 × 103 )(cos 45°)2
=
A0
11.25 × 10−3
= 620 × 103 Pa
σ = 620 kPa
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PROBLEM 1.33
A steel pipe of 12-in. outer diameter is fabricated from 14 -in.-thick plate by
welding along a helix that forms an angle of 25° with a plane perpendicular
to the axis of the pipe. Knowing that the maximum allowable normal and
shearing stresses in the directions respectively normal and tangential to the
weld are σ = 12 ksi and τ = 7.2 ksi, determine the magnitude P of the
largest axial force that can be applied to the pipe.
SOLUTION
1
d o = 6 in.
2
ri = ro − t = 6 − 0.25 = 5.75 in.
do = 12 in. ro =
A0 = π (ro2 − ri2 ) = π (62 − 5.752 ) = 9.228 in 2
θ = 25°
Based on σ = 12 ksi: σ =
P
cos 2 θ
A0
P=
Based on τ = 7.2 ksi: τ =
P=
A0σ
(9.228)(12 × 103 )
=
= 134.8 × 103 lb
cos 2 θ
cos 2 25°
P
sin 2θ
2 A0
2 A0τ
(2)(9.288)(7.2 × 103 )
=
= 174.5 × 103 lb
sin 2θ
sin 50°
The smaller calculated value of P is the allowable value.
P = 134.8 × 103 lb
P = 134.8 kips
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PROBLEM 1.34
A steel pipe of 12-in. outer diameter is fabricated from 14 -in.-thick plate by
welding along a helix that forms an angle of 25° with a plane perpendicular
to the axis of the pipe. Knowing that a 66 kip axial force P is applied to the
pipe, determine the normal and shearing stresses in directions respectively
normal and tangential to the weld.
SOLUTION
1
d o = 6 in.
2
ri = ro − t = 6 − 0.25 = 5.75 in.
do = 12 in. ro =
A0 = π (ro2 − ri2 ) = π (62 − 5.752 ) = 9.228 in 2
θ = 25°
Normal stress:
Shearing stress:
σ =
τ =
P cos 2 θ
(66 × 103 ) cos 2 25°
=
= 5875 psi
A0
9.228
σ = 5.87 ksi
P sin 2θ
(66 × 103 ) sin 50°
=
= 2739 psi
2 A0
(2)(9.228)
τ = 2.74 ksi
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PROBLEM 1.35
A 1060-kN load P is applied to the granite block shown. Determine the
resulting maximum value of (a) the normal stress, (b) the shearing stress.
Specify the orientation of the plane on which each of these maximum
values occurs.
SOLUTION
A0 = (140 mm)(140 mm) = 19.6 × 103 mm 2 = 19.6 × 10 −3 m 2
P = 1060 × 103 N
σ =
(a)
P
1060 × 103
cos 2 θ =
cos 2 θ = 54.082 × 106 cos 2 θ
−3
A0
19.6 × 10
Maximum tensile stress = 0 at θ = 90°.
Maximum compressive stress = 54.1 × 106 at θ = 0°.
(b)
|σ |max = 54.1 MPa
Maximum shearing stress:
τ max =
P
1060 × 103
=
= 27.0 × 106 Pa at θ = 45°.
2 A0
(2)(19.6 × 10−3 )
τ max = 27.0 MPa
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PROBLEM 1.36
A centric load P is applied to the granite block shown. Knowing that the
resulting maximum value of the shearing stress in the block is 18 MPa,
determine (a) the magnitude of P, (b) the orientation of the surface on
which the maximum shearing stress occurs, (c) the normal stress exerted on
that surface, (d) the maximum value of the normal stress in the block.
SOLUTION
A0 = (140 mm)(140 mm) = 19.6 × 103 mm 2 = 19.6 × 10−3 m 2
τ max = 18 MPa = 18 × 106 Pa
θ = 45° for plane of τ max
(a)
Magnitude of P. τ max =
|P|
so P = 2 A0 τ max
2 A0
P = (2)(19.6 × 10−3 )(18 × 106 ) = 705.6 × 103 N
sin 2θ is maximum when 2θ = 90°
(b)
Orientation.
(c)
Normal stress at θ = 45°.
σ =
(d)
P = 706 kN
θ = 45°
P cos 2 θ
(705.8 × 103 ) cos 2 45°
=
= 18.00 × 106 Pa
A0
19.6 × 10−3
Maximum normal stress: σ max =
σ max =
σ = 18.00 MPa
P
A0
705.8 × 103
= 36.0 × 106 Pa
19.6 × 10−3
σ max = 36.0 MPa (compression)
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PROBLEM 1.38
Link BC is 6 mm thick and is made of a steel with a 450-MPa ultimate
strength in tension. What should be its width w if the structure shown is
being designed to support a 20-kN load P with a factor of safety of 3?
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
ΣM A = 0:
(480) FBC − 600 P = 0
FBC =
600 P (600)(20 × 103 )
=
= 25 × 103 N
480
480
For a factor of safety F.S. = 3, the ultimate load of member BC is
FU = (F.S.)( FBC ) = (3)(25 × 103 ) = 75 × 103 N
But FU = σ U A
∴A =
For a rectangular section
FU
σU
=
75 × 103
= 166.67 × 10−6 m 2
450 × 106
A = wt or w =
A 166.67 × 10−6
=
t
0.006
w = 27.8 × 10−3 m or 27.8 mm
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PROBLEM 1.39
A 34 -in.-diameter rod made of the same material as rods AC and AD in
the truss shown was tested to failure and an ultimate load of 29 kips was
recorded. Using a factor of safety of 3.0, determine the required diameter
(a) of rod AC, (b) of rod AD.
SOLUTION
Forces in AC and AD.
Joint C:
ΣFy = 0:
Joint D:
Ultimate stress. From test on
ΣFy = 0:
3
4
σU =
-in. rod:
σ all =
Allowable stress:
(a)
Diameter of rod AC.
(b)
Diameter of rod AD.
σ all =
FAC
1 πd2
4
d2 =
d2 =
1
FAC − 10 kips = 0
5
FAC = 22.36 kips T
1
FAD − 10 kips = 0
17
FAD = 41.23 kips T
PU
29 kips
= 1 3 2 = 65.64 ksi
A
π (4)
4
σU
F .S .
4FAC
πσ all
4 FAD
πσ all
65.64 ksi
= 21.88 ksi
3.0
=
=
4(22.36)
= 1.301
π (21.88)
d = 1.141 in.
=
4(41.23)
= 2.399
π (21.88)
d = 1.549 in.
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