F1–19. If the54
eyebolt is made
a yield
F1–22. The pin is made of a material having a failure
C H Aof
P TaE material
R 1
S T having
RESS
required
shear stress of tfail = 100 MPa. Determine the minimum
1 stress of sY = 250 MPa, determine the minimum
P
diameter d of its shank. Apply a factor of safety F.S. = 1.5
required diameter of the pin to the nearest mm. Apply a
F1–19. If the eyebolt is made of a material havingfactor
a yield
is made
a material having a failure
against yielding.
of safetyF1–22.
of F.S. The
= 2.5pin
against
shearoffailure.
shear stress of tfail = 100 MPa. Determine the minimum
1 stress of sY = 250 MPa, determine the minimum required
F1–16
diameter d of its shank. Apply a factor of safety F.S. = 1.5
required diameter of the pin to the nearest mm. Apply a
against yielding.
factor of safety of F.S. = 2.5 against shear failure.
F1–17. ÖDEV
The strut NO
is glued
3. to the horizontal member at
surface AB. If the strut has a thickness of 25 mm and the glue
d
80 kN
Soruan average
1. Diyagonal
AB
yüzeyi boyunca, yatay duran elemana yapıştırılmıştır.
can withstand
shear stresseleman
of 600 kPa,
determine
30 kN
the maximum
force
P
that
can
be
applied
to
the
strut.
Diyagonal elemanın kalınlığı 25mm ise ve yapıştırıcının taşıyabileceği ortalama kayma
d
gerilmesi 600 kPa ise diyagonal
elemana uygulanabilecek en büyük P kuvvetini hesaplayınız.
P
n at
ear,
80 kN
kNSIMPLE CONNECTIONS
DESIGN30OF
1.7 DESIGN OF SIMPLE55
CONNECTIONS
F1–22
1.7
55
F1–19
PROBLEMS
50 mm
PROBLEMS
F1–23. If the bolt head and the supporting
1 bracket are F1–22
1
made of the same material having a failure shear stress of
t
,
determine
the
maximum
allowable
force
P
=
120
MPa
F1–19
fail
F1–23.
If the
the bolt
bolt
head
and
the supporting
bracket are
F1–20.
If the to
bara assembly
is made
Member B
is subjected
compressive
force of
of atomaterial
*1–76.having
The lapbelt
to The
be subjected
to aso
force
can*1–76.
be isapplied
to
that
it be
does
not pull
•1–73. Member B is 3subjected
of assembly
lapbelt
assembly
is to
subjected
to a force
60! a compressive force that
made
of
the
same
material
having
a
failure
shear
stress of
a yield
stress
of
ksi,
determine
the
minimum
s
=
50
3
A and B are
both made
of
wood
and
are
thick,
in.
of
800
N.
Determine
(a)
the
required
thickness
t
of
Y
through of
the800
plate.
a factor
safety
of F.S.thickness
= 2.5
8
N. Apply
Determine
(a)ofthe
required
t of
1800 lb. If A and B are both made of wood and are 8 in. thick,
t
,
determine
the
maximum
allowable
force P
=
120
MPa
required
dimensions
1>8
in.
Apply
a
h
1 nearest
ne54
to the nearest
h
of
in.
the
belt
if
the
allowable
tensile
stress
for
the
material
fail
C H4determine
A
P T the
E R 1smallest
S
T1Rand
E dimension
SA
Sh2 to the
B
shear
failure.
to the
nearest
dimension
h against
of
the belt
if the allowable tensile stress for the material
4 in. the smallest
F1–20.
If1.5
the
bar
assembly
isEach
made
of
a material
having
that
can be lap
applied
to the
bolt so that it does not pull
factorsoofthat
safety
F.S.
=
against
yielding.
bar
has
a
zontal segment
it
does
not
fail
in
shear.
The
is
(b)
the
required
length
(s
)
=
10
MPa,
d
t
allow
l
the horizontal
segment
that it does not
fail in shear.
The
is (st)allow = 10 MPa, (b) the required lap length dl
a yield
stress
of so
determine
thecanminimum
s
Y = 50 ksi, if
through the
plate.stress
Applyofa factor of safety of F.S. = 2.5
thickness
of 0.5
stress
forF1–22.
theaverage
is tallow
= 300
psi.
the=aglue
sustain ifanthe
allowable
shear
# segment
gshear
a yield
Thein.
pin
is made
a materialishaving
failure
forof
the
tallow
300
psi.in. Apply
glue
sustainhaving
an allowable
80
mm
required
dimensions
and h2 atoyield
the
nearest
1>8
a
h1segment
F1–19. If the eyebolt
is shear
made stress
of a F1–17
material
having
F1–22.
The
pin
is the
made
of shear
acan
material
a failure shear stress of
against
failure.
and
(c)
required
diameter
(t
)
=
0.75
MPa,
dr of
g
equired
shear stress of tfail = 100 MPa. Determine allow
the minimum
and (c)
theminimum
required diameter dr of
(tallow
)100
0.75.MPa,
g = MPa
factor
of
safety
F.S.
=
1.5
against
yielding.
Each
bar
has
a
stress of sY = 250
MPa,
determine
the
minimum
required
shear
stress
of
t
Determine
the
=
failshear stress for
Soru
2. Şekildeki
kayma
gerilmesi
# the
olan
themm.
pinApply
if the
thebirpinmalzemeden
is
S. = 1.5
required
diameter
of the pinbulon
to the göçme
nearest
a allowable
pin
if
the
allowable
shear
stressa forimal
the pin is
thickness
0.5 in.of safety F.S. = (t1.5 ) =required
diameter d of its shank.
Apply aoffactor
diameter of the pin to the nearest mm. Apply
30
MPa.
allow
p
factor
of
safety
of
F.S.
=
2.5
against
shear
failure.
edilmiştir.
Bulon
için
gerekli
en
küçük
çapı
en
yakın
mm
ölçüsünde
hesaplayınız.
Kayma
80
mm
(t
)
=
30
MPa.
mm
75
allow
p
F1–18.
Determine
the
maximum
average
shear
stress
against yielding.
factor of safety of F.S. = 2.5 against shear failure.
15 kip
h1
göçmesine
karşı F.S.=2
kullanınız.
developed
in the 30-mm-diameter
pin.emniyet
30katsayını
kip
B h2
800 lb
B 15 kip
30 kN
h2
C
13
12
30 kN
ress
5
d
C
h
A
F1–20
AB
15 kip
13
12
h
5
B
15 kip
30 kN
800 lb
h1 800 N
30 kip
80 kN
A
t
45 mm
45 mm
t
dl
A
75 mm
30 mm
800 N
40 mm80 kN
30 mm
dl
F1–20
P
F1–2340 mm
F1–21. Determine the maximum force P that can be
applied to the rod if it is made of material having a yield
dr
F1–22
dr
# Y 1–73
F1–22
F1–24. Six nails are used to 800
hold
at A against
stress Prob.
of s
the possibility
that failure
= 250 MPa. ConsiderProb.
1–73
N the hanger
P
F1–23
F1–21.
Determine
the
maximum
force
P
that
can
be
The lever isoccurs
attached
to
the
shaft
A
using
a
key
that
the
column.
Determine
the
minimum
required
diameter800
of N
in1–74.
the rodThe
andlever
at section
a–a. Apply
a shaft
factorAofusing
safety
is
attached
to
the
a
key
that
p
F1–19
Soru
3.
Şekildeki
çubuğa
uygulanabilecek
en
büyük
P
kuvvetini
hesaplayınız.
Çubuk
akma
applied
to
the
rod
if
it
is
made
of
material
having
a
yield
F1–23.
If
the
bolt
head
and
the
supporting
bracket
are
dth d and of
length
25
Ifyielding.
the
shaft
is fixed
and
each
to head
the nearest
if it is
made are
of material
in. Prob.
F.S. of
=has
2 against
Prob.
1–76
If thenail
bolt
and the1>16
supporting
bracket
amm.
width
d and
length
of 25
mm. If the
shaft
isF1–23.
fixed and
1–76
Six
nails
are used
to hold
the hanger at A against
stress
of
smaterial
MPa.
Consider
the
possibility
failuret = F1–24.
of
the
same
aolan
failuremalzemeden
shear
stress ofthathaving
Y = 250having
al force of 200made
N ais
applied
perpendicular
to
the
gerilmesi
#
yapılmıştır.
a-a
kesitinde
göçme
ihtimalini
.
Apply
a
factor
of
safety
of
16
ksi
fail
made
of
the
same
material
having
a
failure
shear
stress
ofF.S. = 2
vertical
force
ofrod
200and
N at
is section
applieda–a.
perpendicular
to the
the
column.
Determine
the
minimum
required diameter of
occurs
in
the
Apply
a
factor
of
safety
t
,
determine
the
maximum
allowable
force
P
=
120
MPa
40
kN
fail
determine the dimension
ddetermine
if thebulundurunuz.
allowable
shear Akmaya
against
failure.
tfailF.S.=2
,shear
determine
the maximum
allowable force P
=shear
120
MPa
gözönünde
karşı
emniyet
katsayısını
kullanınız.
handle,
the dimension
d if the allowable
having
each
nail totothethe
nearest
1>16 in. if it is made of material
of F.S.
= made
2 against
yielding.
thatbar
can
be
applied
to the
bolt
so that
it•1–77.
does not
pull
The
wood
specimen
is
subjected
pull
of
r the
key isIftallow
=stress
35assembly
MPa.
F1–20.
the
is
of
a
material
having
that can be applied
to the
so specimen
that it does
not pull to the pull of
•1–77.
Thebolt
wood
is subjected
fora the key isF1–18
tallow = 35 MPa.
inimum
having
tfail
. Apply a factor of safety of F.S. = 2
= 16 ksinormal
through
plate.
a factor
safety
of F.S.
=
2.5 testing machine.
10 kN
in athrough
tension
If
the
allowable
a yield
stress
of sYthe
ksi,Apply
determine
theofminimum
= 50
the
plate.
Apply
a
factor
of
safety
of
F.S.
2.5allowable normal
10
kN
in
a
tension
testing
machine.
If=the
300
lb/ft
d
Apply a
shear
failure. the
d nearest 1>8 in. Apply
against shear
failure.
stress
foragainst
the shear
woodfailure.
isstress
)allow
12 MPa
(stagainst
required dimensions
h2 to the
a
h1 and
for = the
woodand
is (st)allow = 12 MPa and the
P
ar has a a
allowable
shear stress is tallowable
determine
the = 1.2 MPa, determine the
MPa, stress
factor of safety F.S. = a1.5 againstayielding.
a
allow = 1.2shear
40amm Each bar has
is tallow
A
required
dimensions
b
and
t
so
that
the
specimen
reaches
300
lb/ft
thickness of 0.5 in.
required dimensions b and t so that the specimen reaches
50 mm80 mm
20 mm
80
mm
20 mm
these stresses simultaneously.
Thestresses
specimen
has a width The
of specimen has a width of
P
these
simultaneously.
500 mm a
40
mm
25
mm.
500 mm
B
25 mm.
mm
75mm
60
mm
200
N 50
200 N
ha1
30
kip
Prob. 1–74
Section a-a Prob. 1–74
# h2
120
60 mm
A mm
30 mm
The joint is fastened
together
using
bolts.together
1–75. The
joint
is two
fastened
using two bolts.
kip F1–21
B 15
C
ne the required diameter
of
the
bolts
if
the
failure
Soru
4.
Aşağıdaki
birleşim
iki
bulon
kullanılarak
Determine the required diameter
of
the
bolts
if the failure
Section
a-a
ess for the bolts is shear
t
Use
a
factor
of
=
350
MPa.
fail
40
mm
#
olduğuna
gerekli
bulon
çapını
stress for the
bolts is tfailgöre
a factor
of
= 350
MPa. Use
A
120 mm
15 kip
F1–20
r shear of F.S. = 2.5.
safety
for shear
of F.S.kullanınız.
= 2.5. F1–21
emniyet
katsayısı
30 mm
80 kN
75 mm
10 kN
B
10 kN
A
9 ft
30 mm
F1–24
gerçekleştirilmiştir. Akma göçme gerilmesi
t Kaymaya karşı F.S.=2.5
belirleyiniz.
A
t
40 mm
A
9 ft
80 kN
30 mm
P
F1–23
can be
b
P
F1–23 b
F1–21. Determine the maximum force P that can be
a yield
30
mm
applied to the rod if it is made of material having a yield
30 mm
F1–24. Six nails are used to hold the hanger at A against
t failure
F1–24. Six nails are used to hold the hanger at A against
stress of sY = 250 MPa. Consider the possibility that failure
the column. Determine the minimum required diameter of
of safety
the column. Determine the minimum required diameter of
occurs in the rod and at section a–a. Apply a factor of safety
each nail to the nearest 1>16 in. if it is made of material
each nail to the nearest 1>16 in. if it is made of material
of F.S. = 2 against yielding.
having tfail = 16 ksi. Apply a factor of safety of F.S. = 2
having tfail = 16 ksi. Apply a factor of safety of F.S. = 2
against shear
40 kN failure.
against shear failure.
kN
a
#
kN
30040
lb/ft
Prob. 1–75
300 lb/ft
40 mm
10 kN
10 kN
Prob. 1–77
P
Prob. 1–75
Prob. 1–77
50 mm
a
A
120 mm
B
B
A
60 mm
#22
Section a-a
F1–21
9 ft
F1–24
9 ft
F1–24
F1–24
Download

Ödev3