components, namely, ¢Fx, ¢Fy, and ¢Fz, which are taken tangent,
tangent, and normal to the area, respectively. As ¢A approaches zero, so
do ¢F and its components; however, the quotient of the force and area
will, in general, approach a finite limit. This quotient is called stress, and
as noted, it describes the intensity of the internal force acting on a specific
plane (area) passing through a point.
Gerilme:
z
!Fz
z
!Fx
!F
tyz
z
!Fy
x
z
tyx
y
sy
sx
txy
!F
1.3
23
STRESS
!A
1.3
the force acting normal to ¢A is
).Stress.
Since ¢FzThe
is normal
to of
thethe
area
intensity
force acting normal to ¢A is
F1
F2
the normal stress, s (sigma). Since ¢Fz is normal
to theF1area
0
txz
¢Fz
¢A
x
(1–4)
¢F
(a)z
y
1
1
1.3
x
(1–4)
! sz = lim
¢A : 0 ¢A
¢A
Stress.
The
intensity
of
the
force
acting
normal
to
is
n ¢A as shown in Fig. 1–10a, it is
23
STRESS
(b)
SyTRESS
23
x
y
(c)
Fig. 1–10
s (sigma).
normal
stress,
Since ¢Fz is normal to the area
ifthe
it force
“pushes”
on ¢A
it ison
1.3
Scalled
TRESS
¢A as shown 23
mal
or stress
“pulls”
in Fig. 1–10a, it is
Normal
Gerilme:
o as tensile stress, whereas if it “pushes” on ¢A it is called
ve
stress.
¢A
is
¢F
¢A is calledz
rce acting tangentsto
(1–4)
1
z = lim
he area
¢A : 0 ¢A
shear stress
!
tress.
Thecomponents,
intensity
of force acting tangent to ¢A is called
stress, t (tau). Here we have shear stress components, z
Kayma
Gerilmesi:
¢A as shown in Fig. 1–10a, it is
mal
force
or stress
“pulls” on
¢F
x
o0 (1–4)
as
tensile
stress,
whereas
if
it “pushes” on ¢A it isscalled
z
¢A
¢Fx
ve stress.
tzx = lim(1–5)
¢A : 0 ¢A
¢Fy
(1–5)
Tzx
0a,
it is The intensity of force¢F
0 ¢A
Tzy
y
tress.
acting
tangentx to ¢A is called
y
lim
calledt (tau). Heretzy
stress,
we=have
¢A :shear
0 ¢A stress components,
!
Fig. 1–11
z specifies the orientation of the
eatthe
axes
along
which
each
shear
¢F
Genel
Gerilme
x Hali:
in this subscript
tzxnotation
= lim z specifies the orientation of the
called
¢A : 0 the
¢A axes along which each
Fig.
1–11, and x and y indicate
z shear
(1–5)
.
¢Fy
z
he body is further
by
sz
tzysectioned
= lim
t
¢A :
0 ¢A
–10b,
and
the
y–z
plane,
Fig.
1–10c,
t
zx
State of Stress. If sthe body is further sectionedzy by
z
lement
thatFig.
represents
tyz
allel to of
thematerial
x!z plane,
1–10b, and the y–z plane,tFig.
1–10c,
xz
(1–5)
hosen
point
in
the
body.
Thiselement
at
in
this
subscript
notation
zstate
specifies
the orientation
txy oftyxthe
en “cut out” a cubic volume
of material
that
represents
sx
sy
e components
each face
Fig.
1–11,acting
and xacting
and yTon
indicate
the axes
which
zxthe chosen
of
stress
around
pointalong
in the
body.each
Thisshear
state
y
Tzy
y
s then characterized
x by three components
xacting on each face
!
!
ment, Fig. 1–12.
1–11
Fig. 1–12 by
of
the per
force
area,Fig.
the
State
of unit
Stress.
If inthe
body
is further
sectioned
Gerilmenin
Birimi
(SI birim
sisteminde):
h
shear
he
magnitudes
both Fig.
normal
allel
to stress
the x!zof
plane,
1–10b,
and per
the y–z
plane,
Fig.in
1–10c,
Since
represents
a and
force
unit
area,
the
2 of material that represents
units
of
newtons
per
square
meter
n
“cut
out”
a
cubic
volume
element
1
Pa
=
1
N/m
z
nal Standard
or
SI
system,
the
magnitudes
of
both
normal
and
2
Pa
= 1specified
N>m
2 is
small,
andof
f stress
acting
around
the
chosen
point
in theper
body.
Thismeter
state
ss
are
inrather
the basic
units
newtons
square
3
1
kPa
=
1000
Pa
2 acting on each face
110
2,
as
kilosymbolized
by
k,
then
characterized
by
three
components
ned
by
s
z
This unit,
called a pascal 11 Pa = 1 N>m 2 is rather small, and
tzy
3
2,work
-ering
1109Fig.
symbolized
by such
G, are
ment,
1–12.
1–10c,
tzx
2
prefixes
asused
kilosymbolized
by k,
1
MPa
=
1000
kPa110
= 2,
1 N/mm
6
9
alues
of
stress.*
Likewise,
in
the
resents
2, symbolized by M, or giga2, symbolized by G, are used
110 tyz
txz
ngineers
usually
express
stress
in stress.*
is
state
Since
stress
represents
unitLikewise,
area, inin the
ent
larger,
more
realistic
values
the
ta
tyx of per
xy force
s(ksi),
sy
x
ounds
per
square
inch
where
ch
face
nal Standard
or SI of
system,
magnitudes
of both
normal
and
nd-Second
system
units,the
engineers
usually
express
stress
in
1
z
sz
Tzx
Tzy
x
y
z
Fig. 1–11
sz
Tzx
z
T
sz zy
t
t zy
x
y
zx
Fig. 1–11
sx
txz
txy
z
x
tyx
tyz
sy
y
sz
tzy
Fig.tzx1–12
sx
txz
txy
tyz
tyx
sy
y
x
Fig. 1–12
y
ss
specified
in
units ofper
newtons
er are
square
inch (psi)
orbasic
kilopounds
squareper
inchsquare
(ksi), meter
where
x the
his
unit, =called
d 1kip2
1000 alb.pascal 11 Pa = 1 N>m22 is rather small, and
Fig.as
1–12
ering
kilo- 11032, symbolized by k,
n thework prefixes such
6 2
-3
2, symbolized
M, or
110in92, symbolized by G, are used
mal
and
mm
, where 1 mm by
= 10
m. gigaHowever,
nominator
of
a
fraction
and
therefore
it isof stress.* Likewise, in the
ntstress
larger,
more realistic values
esmeter
is expressed in units of N>mm2, where 1 mm = 10-3 m. However, in
2
m
=
1
MPa.
d-Second
of units,
engineers usually
express
stress itinis
all,
and aresystem
m,
prefixes
not allowed
in the denominator
of a fraction
and therefore
rthe
square
(psi) 2or= kilopounds
equivalent
1 MN>m2 = 1 per
MPa.square inch (ksi), where
by
k, inch1 N>mm
d 1kip2
re
used = 1000 lb.
in the
!8
n of
m
mation
24
CHAPTER 1
STRESS
1.4 Average Normal Stress in an
1
Axially Loaded Bar
Eksenel Yüklü Bir Çubukta Ortalama Gerilme
In this section we will determine the average stress distribu
the cross-sectional area of an axially loaded bar such as the
Fig. 1–13a. This bar is prismatic since all cross sections
P
throughout its length. When the load P is applied to the b
1.4 Average Normal Stress in an centroid of its cross-sectional area, then the bar will defo
throughout the central region of its length, as shown
Axially Loaded Bar
provided the material of the bar is both homogeneous and
Region of
In this section we will determine the average uniform
stress distribution Homogeneous
acting on
material has the same physical and mechan
deformation
the cross-sectional area of an axially loaded bar
such as the one shown in
throughout
its volume, and isotropic material has these sa
of barsections are
Fig. 1–13a. This bar is prismatic since all cross
the same
all directions.
Many engineering materials may be app
throughout its length. When the load P is applied to the bar in
through
the
being
both homogeneous and isotropic as assumed he
centroid of its cross-sectional area, then the bar will deform
uniformly
throughout the central region of its length, as shown in example,
Fig. 1–13b,contains thousands of randomly oriented crystals
provided the material of the bar is both homogeneous and isotropic.
millimeter of its volume, and since most problems involvin
Homogeneous material has the same physical and mechanical properties
have
a physical size that is very much larger than a sing
throughout its volume, and isotropic material has these same
properties
above assumption
regarding its material composition is qu
in all directions. Many engineering materials may be approximated
as
P
P
!
being both homogeneous and isotropic as assumed here. Steel,
for
Note that anisotropic materials such as wood have differ
(a)thousands of randomly
(b)
example, contains
oriented
crystals in
cubic directions, and although this is the case, lik
ineach
different
Prizmatik
çubuk (tüm enkesitlerin
geometrisi
aynı)
millimeter of its volume, and since most problems involving this material
anisotropy is oriented along the bar’s axis, then the bar wi
have a physical
that merkezinden
is very much larger
than a single crystal, the
P yüküsize
ağırlık
tesir ediyor
above assumption regarding its material composition is quiteuniformly
realistic. when subjected to the axial load P.
Çubuğun
yapıldığı
malzeme
homojen
ve izotropik.
Note that
anisotropic
materials
such as wood
have different
properties
in different directions, and although this is the case, like Average
wood if the Normal Stress Distribution. If we p
Homojen malzeme: Malzemenin kapladığıalso
tüm
hacim aynı fiziksel ve mekanik özelliklere
anisotropy is oriented along the bar’s axis, then the bar will through
deform
the bar, and separate it into two parts, then equilib
uniformlysahip
when ise
subjected
to the
axial load P.
malzeme
homojendir.
the resultant normal force at the section to be P, Fig. 1–1
uniform
deformation
of the
it is necessary that th
Averageİzotropik
Normal malzeme:
Stress Distribution.
we pass
a tüm
section
Malzemenin Ifözellikleri
doğrultuda
aynı
ise, material,
yani kuvvetlerin
through the bar, and separate it into two parts, then equilibrium
requires
be subjected to a constant normal stress distribution, Fig. 1–
etkime
doğrultusuna
özellikler
göstermiyorsa
the resultant
normal
force at thegöre
section
to be P,farklılık
Fig. 1–13c.
Due to the malzeme izotropiktir.
z
P
uniform deformation of the material, it is necessary that the cross section
Ortalama normal gerilme dağılımı
be subjected to a constant normal stress distribution, Fig. 1–13d.
z
P
P
!F " s!A
!F " s!A
s
P
x
Internal force
s
P
y
Internal force
!A
!A
Cross-sectional
area
Cross-sectional
area
y
x
x
External force
y
1.4
AVERAGE NORMAL STRESS IN AN AXIALLY LOADE
External force
P
P
P
!
As a result,
each small area ¢A onPthe cross section is subjected to a
(c)
(d)
force ! ¢F = s ¢A, kuvveti
and the tesir
sum of
these
forces
acting overbirbirlerinden
the entire
(c) ve
Herbir küçük delta A
ediyor
bu kuvvetler
Fig.alanına
1–13
cross-sectional area must be equivalent to the internal resultant force P
Fig.
1–13
farklı. Ağırlık merkezine doğru
kuvvetlerin
büyük
beklenir,¢F
çünkü
eksenel
at the bu
section.
If we daha
let ¢A
and therefore
, then, dış
: dAolması
: dF
recognizing s is constant, we have
yük ağırlık merkezinden tesir ediyor.
+ c FRz = ©Fz;
L
dF =
s dA
LA
P = sA
!9
(d)
AVERAGE
NORMAL
TRESS
AN AXIALLY
AR25
As a 1.4
result,
each1.4
theLINOADED
cross
is Bsubjected
to 25
a
¢A
AVERAGE
Nsmall
ORMAL
Sarea
TRESS
IN AN on
ASXIALLY
Bsection
AR LOADED
force ¢F = s ¢A, and the sum of these forces acting over the entire
cross-sectional
mustis
sult,
each¢A
small
the cross
section
subjected
to a to the internal resultant force P
¢A on
all area
on area
the
cross
section
isarea
subjected
tobe
a equivalent
theof section.
If acting
we entire
letover
¢A
dA and therefore ¢F : dF1 , then,
, and
theatsum
these
the:entire
s ¢A
d= the
sum
of these
forces
actingforces
over
the
onalbearea
must berecognizing
tosthe
internal
resultant
force P
ust
equivalent
toequivalent
the internal
resultant
force
P have
is constant,
we
ction.
If we
let ! and
and! ¢F
therefore
, then, normal gerilme:
¢A :
dA ve
¢F : dF
let ¢A
therefore
then,
: dA
: dF, olursa,
ortalama
g swe
is constant,
we have
ant,
have
L
+ c FRz = ©Fz;
©Fz; dF =
dF =
s dA
s dA
LA
L
LA L
P = sA
P = sA
P
s =
A!
P
s =
A
dF =
1
s dA
LA
P = sA
(1–6)
P
P
s(1–6)
=
A
(1–6)
P
Here
P
P the cross-sectional
age
at
point
on
area
tressnormal
at anystress
point s
onany
cross-sectional
area
s ! — the cross-sectional area
s! —
=theaverage
A
A normal stress at any point on
nal
resultant
which acts
normal
force,normal
which force,
acts through
the through
centroid the
of centroid of
P P=isinternal
normal
force,
which acts through the centroid of
ross-sectional
area.
determined
using
the
method
of
al
area. P is determined
using
theresultant
method
of
s
s
cross-sectional area. P is determined using the method of
ons and the
ofthe
equilibrium
equations
ofequations
equilibrium
the equations of equilibrium
-sectional
area
of thesbar
where sand
is determined
ea of the bar
where
issections
determined
A passes
= cross-sectional
area
of of
thethebar
where s is determined
he
load
P
crossoadinternal
P passes
through
the through
centroidthe
of centroid
the
crosse
uniform
stress
distribution
will
produce
zero
moments
about
ress distribution will produce zero moments about
Since the internal load P passes through the centroid of the crossy axes
passing
this1–13d.
point,To
Fig.
1–13d.
ng
through
thisthrough
point, Fig.
show
this,Toweshow this, we
section
the
uniform
stress
distribution
e
moment
of
P
about
each
axis
to
be
equal
to the
moment of will produce zero moments about
f P about each axis to Pbe equal the moment
of
the
andnamely,
y axes passing through this
point, Fig. 1–13d. To show this, we
P
distribution
about
thexaxes,
about the axes,
namely,
each axis to be equal to the moment of
Tension the moment of P about
Compression
!require
the stress distributionFig.
about
the axes, namely,
1–15
y dF
=© Mxy; dF0 ==
ys
dA== s ysydA
dA= s y dA
L
LA LA
LA
LA
L
AA
In other words, the two normal stress components on the element must
Normal Kuvvet Diyagramı
inxmagnitude
but
opposite
direction.
ThisdA
is referred
as
0= =- s in
yx dF
=
ys
= s yto
dA
R2xdA
x; dA
0equal
==-- 1M
My; be
dF
== -=©-M
xs
dA
=© x dF
xs
s
x dA
L
A
L
A
L
A
LAL
LA L
LA
LA uniaxial stress.
A
A
Eğer
eksenel
yüklü
bir elemanda
farklı yükleme bölgeleri var ise ortalama gerilmelerin
The previous analysis applies to members subjected to either tension
maksimum
olduğu
bölgenin
tespiti
için normal
kuvvet değişimin gösteren Normal Kuvvet
ationssatisfied,
arecompression,
indeed
satisfied,
definition
or
as
shown
Fig.
As
a graphical
ndeed
since
by
the
0centroid,
=1–15.
-of the
1M
=since
© Mby
; in
x centroid,
dF
= - interpretation,
xs dA = - s thex dA
R2ydefinition
yof
x dA
= 0.
(See
Appendix
A.)
A
=E S0.Smagnitude
(See
Appendix
A.)
LA P is equivalent
A
LA s to the volume
of the
internal
resultant
force
sL
1
T0Rand
Diyagramının
çizilmesi
gerekebilir.
28
C
H
A
P
T
E
R
1
S
T
R
E
S
S
base2. !A
under the stress diagram; that is, P = s A 1volume = height *
!A
ium.
Itapparent
should These
be
apparent
that
only
a
normal
stress
exists
ould be
that
a
normal
stress
exists
Furthermore,
asonly
a
consequence
of
the
balance
of
moments,
this
resultant
equations are indeed satisfied, since by definition of the centroid,
mall
volume
of
material
located
each
element
of element
material
located
at each
on point on
passes
through
the
of
thisat=
volume.
y
dA
=centroid
0 and
x point
dA
0. (See Appendix A.)
s
1
1
EXAMPLE
1.6
section
of
an
axially
loaded
bar.
If
we
consider
vertical
an
axially
loaded
bar.
If
we
consider
vertical
!
1
Although
we have developed this analysis
for prismatic bars, this
m
of theFig.
element,
Fig. 1–14,
apply theofequation of
lement,
1–14, then
apply then
the equation
assumption
can be
relaxed
somewhat
to include
bars
that ise
have
amm
slight
Şekilde
verilen
çubuğun
genişliği
35 mm,
kalınlığı
10normal
dir.stress
Şekilde
verilen
yükler altında
librium,
Equilibrium.
It
should
be
apparent
that
only
a
exists
The
bar
in
Fig.
1–16a
has
a constant
width
of 35 mm and a thickness
taper. ForThe
example,
it Fig.
can be
shown,
using
the morewidth
exact analysis
of the
bar
in
1–16a
has
a
constant
of
35
mm
and
a
thickness
on
any oluşacak
small
volume
element
ofofmaterial
located
at maximum
each point
on normal stress in the bar
10
mm. Determine
the
average
ortalama
gerilmeyi
hesaplayınız.
theory
of çubukta
elasticity,
that for
a tapered
bar
of rectangular
cross
section,
for
s1¢A2
= 0maksimum
s¿stress
s1¢A2
- s¿1¢A2
=-mm.
0s¿1¢A2
s¿ normal
of
10
Determine
the
maximum
the
bar
whenaverage
it isbar.
subjected
to the
loadingin
shown.
the
cross
section
of
an
axially
loaded
If
we
consider
vertical
which the angle between two adjacent sides is 15°, the average normal
s
=
s¿
1–14
s =ass¿when
Fig. A
1–14
it is subjected
to
the
loading
shown.
equilibrium
element,
Fig. 1–14,
then
apply
the
of
B equation
C 4 kN
9 kN
D
stress,
calculated
by of
s =the
P>A,
is
only
2.2%
less
than
its
valueFig.
found
12 kN
22 kN
force
equilibrium,
from the theory of elasticity.
B
C 4 kN
9 kN
A
D
to
as a
12 kN
mm the
Maximum
Normals1¢A2
Stress.
In our analysis
= 0;
© FzAverage
- s¿1¢A2
= 0 35both
9 kN
22 kN
4 kN
internal force P and the cross-sectional area9A
along the (a)
kNwere constant
4
kN
s =normal
s¿ stress s = P>A is
35 mm
longitudinal
as a result the
! axis of the bar, and
12
(a) kN
also constant throughout the bar’s length. Occasionally,
however, the bar PAB ! 12 kN
Çözüm: to several external loads along its axis, or a change in its
may be subjected
9 kN
cross-sectional
area
mayyükler
occur.altında
As a result,
the midir?
normal stress within the
Çubuk
verilen
dengede
12
kN
PBC ! 30 kN
12different
kN
PAB
! 12and,
kN if the maximum
bar could be
from one section to the
next,
9 kN
Maksimum
hesabıthen
niçinitgereklidir?
average normal
stressortalama
is to begerilme
determined,
becomes important
9
kN
to find the location where the ratio P!A is a maximum. To do this it PisCD ! 22 kN
necessary to
the internal force P at various sections
along
the
12determine
kN
PBC !
30 kN
(b)
bar. Here it may be helpful to show this variation
9 kN by drawing an axial or
normal force diagram. Specifically, this diagram is a plot
of the normal
P (kN)
force P versus its position x along the bar’s
length.
As
a
sign
convention,
PCD ! 22 kN
22 kN
P will be positive if it causes tension in the member, and
30 negative if it
22
causes compression. Once the internal loading
the bar is
(b) throughout
s¿
Fig. 1–14
22 kN
!10
!A
(a)
12 kN
12 kN
12 kN
B
A
35 mm
C
9 kN
9 kN
PAB
! 12 kN
4 kN
4 kN
9 kN
(a)
35 mm
9 kN
12 kN
12 kN
P (kN)
22 kN
P (kN)
PAB ! 12 kN
9 kN
30
PAB ! 12 kN
22
9 kN
PCD ! 22 kN
12 22 kN
9 kN PBC ! 30 kN
(b)
9 kN
PBC ! 30 kN
12 kN
PCD ! 22 kN
22 kN
(b)
4 kN
(a)
PBC ! 30 kN
12 kN
12 kN
22 kN
D
x
(c)
9 kN
SOLUTION
22 kN
Internal
Loading.
By inspection,
the internal axial forces in regions
22 kN
(b) PCD ! 22 kN
!
30
AB, BC, and CD are all constant yet have different magnitudes. Using
22
(b) the method of sections, these loadings are determined in Fig. 1–16b;
12P (kN)
and the normal
force diagram which represents these results graphically
x
P (kN)
is shown in Fig. 1–16c. The largest loading is in region BC, where
(c)
30
PBC = 30 kN. Since the cross-sectional area of the bar is constant, the
SOLUTION 22
30
largest average normal stress also occurs within this region of the bar.
12
22
Internal Loading.
By
inspection, the internal axial forcesxin regions
!
12
Average
Normal
Stress. Applying Eq. 1–6, we have
(c)have different
AB, BC, and CD are all constant yet
magnitudes.
Using
PCD ! 22 kN
x
10 mm these loadings are determined in Fig. 1–16b;
the method of sections,
(c)
SOLUTION
3011032 N
PBC
and the normal force diagram which represents these results graphically
=
=
= 85.7 MPa
sinBCregions
Ans.
SOLUTION
Loading.
ByThe
inspection,
internal
forces
10.035 m210.010 m2
isInternal
shown in
Fig. 1–16c.
largest30the
loading
is inaxial
region
BC,
whereA
kN
BC, and
are
allcross-sectional
constant
yet have
magnitudes.
Using
Loading.
By inspection,
axial forces
PAB,
kN. CD
Since
the
areadifferent
ofthe
theinternal
bar
is constant,
the in regions
BC = 30 Internal
NOTE:
The
stress
distribution
the method
sections,
these
loadings
are
determined
in
Fig.
1–16b;
35of
mm
AB,
BC,
and CD
are
all
constant
yet
have
different
magnitudes.
Using acting on an arbitrary cross section of
largest
average
normal
stress
also
occurs
within
this
region
of
the
bar.
85.7 MPa
the
bar
within
region
BC is
shown in Fig. 1–16d. Graphically the volume
and the normal
force diagram
which
represents
these
results
graphically
the method
of (d)
sections,
these
loadings
are
determined
in Fig.
1–16b;
! the
Average
Stress.
Applying
1–6,represents
we
“block”)
represented
by this distribution of stress is equivalent to
is shownNormal
in
Fig.normal
1–16c. force
The
largest
loading
is(or
inhave
region
BC,
where
and
diagramEq.
which
these
results
graphically
AkN;
VERAGE
Nwhere
ORMAL
TRESSMPa2135
IN AN Amm2110
XIALLY LOADED
Fig.
1–16
the
30region
that
is,
30
kN =S185.7
mm2. BAR
PBC = 30is kN.
Since
area ofloading
theload
bar1.4
constant,
the
shown
inthe
Fig.cross-sectional
1–16c. The3 largest
isisofin
BC,
2
N
30110
P
largest average
normal
alsocross-sectional
occurs within this
region
of
the
bar.
BC stress the
PBC
the1.4
bar
is
constant,
the STRESS IN AN AXIALLY LOADED BAR
AVERAGE
NORMAL
=
= area
85.7of
MPa
sBC= =30 kN. Since
Ans.
A
10.035
m210.010
m2
largest
average
normal
stress
also
occurs
within
this
region
of
the
bar.
!
Average Normal
Stress. Applying Eq. 1–6, we have
EXAMPLE
1.7 Normal
Average
Stress.
Eq. 1–6,
we section
have of
NOTE: The
stress
distribution
acting Applying
on
an arbitrary
cross
3
EXAMPLE
1.7in
2 N Graphically the volume
30110
PBCis shown
the bar within
BC
Fig. 1–16d.
! sregion
=
Ans.
BC =
3 = 85.7 MPa
2 N is equivalent to
30110
A by this
10.035
m210.010
m2
PBCdistribution
“block”)
represented
of
stress
The(or
80-kg
lamp
is
supported
by
two
rods
AB
and
BC
asasshown
in
The 80-kg
lamp
supported
by two
rodsAB
BC
shown
in
=kN
=ağırlığındaki
=AB
85.7
MPa
sis,BC30
Ans. vasıtasıyla
Şekilde
verilen
80Aiskg
lamba
veand
BC
çubukları
asılı bir şekilde
the load of 30
kN; that
= 185.7
MPa2135
mm2110
mm2.
10.035
m210.010
m2
Fig.
1–17a.
If AB has
a diameter
of 10
mm
andsection
BC has
a diameter of
30 kN
Fig.
1–17a.
AB
has
a diameter
ofon10
and
BC
has
aofdiameter
of
NOTE: IfThe
stress
distribution
acting
anmm
arbitrary
cross
8region
mm, determine
theinaverage
normal
in
each
rod.
taşınmaktadır.
çubuğunun
çapı
10stress
ve
BC
çubuğunun çapı 8 mm ise herbir çubuktaki
thedetermine
bar within
BC
is AB
shown
Fig. 1–16d.
Graphically
the
volume
8 mm,
the
average
normal
stress
inmm
each
rod.
NOTE:
The
stress
distribution
acting
on an
arbitrary
cross section of
Pa
(or “block”)
represented
by
this
distribution
of
stress
is
equivalent
to volume
the
bar within
region BC
is shownhesaplayınız.
in Fig. 1–16d. Graphically the
ortalama
normal
gerilmeyi
the load (or
of 30
kN;
that
is,
30
kN
=
185.7
MPa2135
mm2110
mm2.
“block”) represented by this distribution of stress is equivalent to
y
the load of 30 kN; that is,A30 kN = 185.7 MPa2135 mm2110
mm2.
C
FBA
FBC
y
A
C
FBA
FBC
5
5
60!
60!
5
B
60!
4
B
5
3
4
60!
3
x
4
B
x
B
80(9.81) " 784.8 N
(a)
(b)
Fig. 1–17
!
SOLUTION
(a)
Internal Loading. We must first determine the axial force in each
rod. A free-body diagram of the lamp is shown in Fig. 1–17b. Applying
Fig. 1–17
the equations of force equilibrium,
+ ©F = 0;
:
x
80(9.81) " 784.8 N
(b)
FBC A 45 B - FBA cos 60° = 0
SOLUTION + c ©F = 0; F A 3 B + F sin 60° - 784.8 N = 0
y
BC 5
BA
Internal Loading. We must
determine
axial
FBC first
= 395.2
N,
= 632.4
N force in each
FBA the
rod. A free-body diagram of the lamp is shown in Fig. 1–17b. Applying
Byforce
Newton’s
third law of action, equal but opposite reaction, these
the equations of
equilibrium,
+ ©F = 0;
:
x
!11
forces subject the rods to tension throughout their length.
Average Normal
Stress. Applying Eq. 1–6,
4
FBC A 5 B - FBA cos 60° = 0
FBC
395.2 N
1
3
3
4
8.05 MPa
29
rod. A free-body diagram of the lamp is shown in Fig. 1–17b. Applying
the equations of force equilibrium,
SOLUTION
+ ©F = 0; Internal
We60°
must
FBC Loading.
:
= 0first determine the axial force in each
A 45 B - FBA cos
x
rod. A3 free-body diagram of the lamp is shown in Fig. 1–17b. Applying
FBCequations
+ c ©Fy = 0; the
60° equilibrium,
- 784.8 N = 0
A 5 B + FBAofsin
force
FBC = 395.2 N,
FBA = 632.4 N
+
FBC A 45 B - FBA cos 60° = 0
:
©Fx = 0;
4 AVERAGE NORMAL STRESS IN AN AXIALLY L3OADED BAR
29
By Newton’s third
law =
of 0;action,
FBCequal
+ c ©F
FBAopposite
sin 60° -reaction,
784.8 N these
= 0
A 5 B + but
y
forces subject the rods to tension throughout their length.
FBC = 395.2 N,
FBA = 632.4 N
! Stress. Applying
Average Normal
Eq. 1–6,
1
8.05 MPa
By Newton’s
law of
FBC third395.2
N action, equal but opposite reaction, these
sBC
=the rods to tension
= 7.86
MPa
and BC as shown
in =subject
forces
throughout
theirAns.
length.8.05 MPa
ABC
p10.004 m22
BC has a diameter
of
Average Normal Stress. Applying Eq. 1–6,
each rod.
FBA
632.4 N
=
MPa
sBA =
Ans.
FBC 2 = 8.05
395.2
N
ABA sp10.005
m2 =
= 7.86 MPa
Ans.
BC =
!
2
ABC
p10.004 m2
y
acting over a cross
C NOTE: The average normal stress distribution
FBA
632.4 N
FBA is shown in Fig.
section of rod AB
and
this
cross
BC
=
= at a point on
=
8.05
MPa
sBAF1–17c,
Ans.
ABAas shown
p10.005
m221–17d.
(d)
section, an element of material is stressed
in Fig.
32
8.05 MPa
8.05 MPa
632.4 N
(c)
C H A P T E R 51 3 S T R E S S
NOTE: The 4average normal stress distribution acting over a cross
B
60!
section of rod AB is shown in
x Fig. 1–17c, and at a point on this cross
section, an element of material is stressed as shown in Fig. 1–17d.
32
CHAPTER 1
632.4 N
(d)
(c)
1.5 Average Shear Stress
STRESS
Shear stress has been defined in Section 1.3 as the stress compo
!
acts in the plane of the sectioned area. To show how this s
1.5
Average Shear Stress
1
develop, consider the effect of applying a force F to the bar in F
A
Fig. 1–17
If stress
the
are considered
rigid,
and
F iscomponent
large enough,
1.5supports
Shear
Stress
Shear
has Average
been defined
in Section
1.3
as the
stress
that it w
FC
1
Ortalama
Kayma Gerilmesi
material
ofthe
thesectioned
bar to deform
failhow
along
planes
acts the
in the
plane of
area. Toand
show
thisthe
stress
can iden
B
Shear
stress
has
been
defined
in
Section
1.3
as
the
stress
component
that
F
develop,
consider
of applying
a force
to the
bar in Fig. 1–20a.
AB
and
CD.the
A effect
free-body
diagram
of Fthe
unsupported
center se
D
A
acts in the plane of the sectioned area. To show how this stress can
he axial force in each
C
If the
supports
are
considered
rigid,
and
F
is
large
enough,
it
will
cause
thedevelop,
bar, consider
Fig. 1–20b,
indicates
the
V = F>2
the effect
of applyingthat
a force
F toshear
the bar force
in Fig. 1–20a.
n Fig. 1–17b. Applying
A
the
material
of
the
bar
to
deform
and
fail
along
the
planes
identified
by
C
If the supports
aresection
considered
rigid,
and
F is
large enough,
it will cause
applied
at
each
to
hold
the
segment
in
equilibrium.
The
B
1.5 Average
Shear Stress
AB andthe
CD.
A free-body
diagram
of and
the fail
unsupported
center
segmentbyof
D
material
of the bar
to deform
along the planes
identified
stress distributed
overthe
each
sectioned
area
that
develops
t
(a) B
the shear
bar,
indicatesdiagram
that
shear
force Vcenter
= F>2
must
ABFig.
and 1–20b,
CD. A free-body
of the
unsupported
segment
ofbe
D
force
isstress
defined
by
Shear stress
has been defined in Section
1.3
as
the
component
applied
at
each
section
toindicates
holdthat
thethat
segment
in equilibrium.
Themust
average
the
bar,
Fig.
1–20b,
the shear
force V = F>2
be
e Shear
Stress
32
F
80(9.81)
C H A"P T784.8
E R 1N S T R E S S
(b)
N = 0
F
applied
at
eachthis
section
to
hold
the segment
in equilibrium.
Thethis
average
acts in the
show
how
stress
cansectioned
shearTostress
distributed
over
each
area
that develops
shear
(a) of the sectioned area.
! plane
shear
stress
distributed
over
each sectioned area that develops this shear
2.4
N
(a)
develop,
consider
the
effect
of
applying
a
force
F
to
the
bar
in
Fig.
1–20a.
force
is
defined
by
defined in Section 1.3 as the stress component that
is defined by
Fare considered rigid, and F isforce
If the supports
he sectioned
area. To
show Fhow this stress canlarge enough, it will cause
pposite
reaction, these
theapplying
materialaof
the bar
deform
along the planes identified by
effect
of
force
F totothe
bar inand
Fig.fail
1–20a.
eir length.
AB and CD. A free-body diagram of the unsupported center segment of
nsidered rigid, and F is large enough, it will cause
V
the bar, Fig. 1–20b, indicates
that the shear force V = F>2 must be
tavg =
MPa identified by
r to deform and fail along the8.05
planes
applied atAns.
each
section to hold the segment in equilibrium. The average
A
VV
V 8.05
7.86 diagram
MPa
MPa
V
ody
of
the
unsupported
center
segment
of
t
=
(1–7)
t
=
(1–7)
avg
avg
shear stress distributed
AA
V over each sectioned area that develops this shear
V
V
ndicates
that
(b)
! the shear
!V = F>2 must be
force
is defined
byV force
(b)
(b) in equilibrium. The average
n8.05
to MPa
hold the segment
Ans.
d over each sectionedFFarea thatF develops this shear
on acting over a cross
t a point on this cross
own in Fig. 1–17d.
tavg
Here
Here
Here
(d) t tavg
avg
tavg
Here
632.4
N
(c)V
tavg = average shear stress
(1–7)at the section, which is assumed to be the
=
average
shear
at
section,
whichtoisbe
assumed
to
tavg t
=avg
average
shear
stress
atstress
the section,
is assumed
the
A
same
at each
point
located
on the
thewhich
section
(c)
!
!
same
at
each
point
located
on
the
section
V = same
internalat
resultant
shear force
on theon
section
each point
located
the determined
section from
(c)
(c)
Fig. 1–20
the
equations
of
equilibrium
V
V
=
internal
resultant
shear
force
on
the
section
determined
from
V: Kesitte tesir eden kesme kuvveti
=
(1–7)V = internal resultant shear force on the section determine
Fig.
1–20
area at the
A =
the
equations
ofsection
equilibrium
A
Fig. 1–20
the equations of equilibrium
A: Kesit
alanı
A = area at the section
tavg =
area at
the
section
A
The=distribution
of average
shear stress acting over the sections is
Ortalama kayma gerilmesinin yönü kesitteki
kesme
kuvveti
ile aynıdır.
shown in Fig. 1–20c. Notice that tavg is in the same direction as V, since
The which
distribution
average
shear
stress
acting
sectionstois
tavg = average shear stress at the section,
is assumed
tocreate
be the
the
shear
stressofmust
associated
forces
all ofover
whichthe
contribute
in
Fig.
1–20c.
Notice
that
t
is
in
the
same
direction
as
since
the
internal
resultant
force
V
at
the
section.
same at each point located onshown
the section
The distribution of average
shear stress acting V,
over
the se
avg
!12 or directto
The
loading
case
discussed
here
is
an
example
of
simple
the
shear
stress
must
create
associated
forces
all
of
which
contribute
V = internal resultant shear force on shown
the section
determined
in Fig.
1–20c.from
Notice that tavg is in the same direction as
shear, resultant
since the shear
caused
the direct action of the applied load F.
forceisV
at thebysection.
the equations of equilibrium the internal
the
shear
stress
must
create
allconnections
of which cont
This
type
of
shear
often
occurs
in
various
types forces
ofofsimple
r stress at the section, which is assumed to beThe
theloading case discussed here isassociated
an example
simple
or direct
A = area at the section
t
1.5
1.5
z
t
tyz !z
z
!
Kayma Gerilmesi
t Dengesi
ytzy
Section planetzy
z
33
AVERAGE SHEAR STRESS
1.5
1.5
Section plane
z
AVERAGE SHEAR STRESS
AVERAGE SHEAR STRESS
33
1
t
Section plane
t
Section
tzy plane
Her ne kadar yukarıdaki
başlıktan
kayma gerilmelerine aitt bir denge
denklemi yazılacağı
t
t !z
!x
yz
t
tzy
1.5
t
1
33
AVERAGE SHEAR STRESS
anlaşılabilirse
de aslında yazılacak denge
için olacaktır.
!tdenklemi kayma kuvvetleri
!z
t
tyz
z
t¿yz
t¿zy
y
!y
(a)
!x
(b)t
Pure shear
tyz !z
Fig. 1–21 Fig.!1–21
x
y
(b) 1–21
(b)
! Fig.
t
1
t
Pure shear
t t
t
Pure shear
(a)
(a)
t¿yz
t
y
! Section plane
t
(a)
!y x
Fig.
1–21
x
t
t
t¿tzy
! xzy
zy
!y
!
y yz t¿zy
!y
!x
t¿ x
t
y
Pure
shear!
!x
tyz !z
t¿
t¿yz
tyz !z
t¿yz
Pure shear
(b)
t
(b)
Fig. 1–21
Shear Stress
Equilibrium.
Figure 1–21a shows a volume element
rium. Figure
1–21a
a located
volume
element
!y takenshows
of
material
at
a
point
on1–21a
the
surface
a sectioned
Pure
shear aof
Shear Stress Equilibrium. Figure
shows
volume
elementarea
t¿zy
t
Shear
Stress
Equilibrium.
Figureshows
1–21a shows
a volume
element
x
Stress
Equilibrium.
Figure
a volume
element
int
located
on
the
surface
ofthe
asurface
sectioned
area
!of
is
to
aonshear
stress
andarea
equilibrium
tzythe
material
taken
at1–21a
a point
located
on
surface
ofmoment
a sectioned
area
of material which
taken
at
asubjected
point
located
of
a. Force
sectioned
(a)
(b)
rial taken at a requires
point islocated
on to
thestress
surface
of aton
sectioned
area
the
shear
acting
this
face
of
the
element
to be
which
subjected
a
shear
stress
.
Force
and
moment
equilibrium
zy
which
is subjected
to a shear and
stress tmoment
equilibrium
hear
stress
Force
equilibrium
tshear
zy. Force and moment
zy.gerilmesi
s subjected
to
a
stress
.
Force
and
moment
equilibrium
t
!
kesit
düzleminde
tesir
eden
kayma
gerilmesidir.
Dengeleyici
zy
Fig.
1–21
requires
the
shear
stress
acting
on
this
face
of
the
element
to
be
accompanied
by
shear
stress
acting
on
three
other
faces.
To
show
this
we kayma
requires the shear stress acting on this face of the element to be
ss the
shear
stress
acting
on
this
face
of
the
element
to
be
acting
on
this
face
of
the
element
to
be
accompanied
byacting
shear
stress
acting
faces.
show this we
accompanied
byfirst
shearconsider
stress
on
three
otheron
faces.
To yother
show
this
weTo
will
force
equilibrium
inthree
the
direction.
Then
gerilmelerinin
şiddet
ve
yönleri
ise denge
denklemleri
ile
belirlenecektir.
aniedwill
byfirst
shear
stress
acting
on
three
Tothe
show
this
we Then
consider
force
equilibrium
inother
theFigure
y faces.
direction.
Then
will
first
consider
force
equilibrium
in
y
direction.
Shear
Stress
Equilibrium.
1–21a
shows
a
volume
element
ess acting on
three other
faces.
To show this we
force on Then
t consider force
equilibrium
y direction.
of material
taken atina the
point
located
the surface of a sectioned area
quilibrium
inis the
yforce
direction.
Then
which
subjected
to a shear force
stress
tzy. Force and moment equilibrium
rce
force
requires
thestress
sheararea
stress stress
actingarea
on this face of the element to be
stress area
accompanied by shear stress acting on three other faces. To show this we
stress
œ 1¢x
force¢y2
equilibrium
in¢y
the=y0-direction.
©Farea
= 0;tzy1¢x
ttzy
¢y2
tœ ¢x Then
¢y = 0
yconsider
©Fy = will
0; first
¢x
œ zy
zy
©Fy = 0;
tzy1¢x ¢y2 - tzyœ ¢x ¢y = 0
œ
tzy = œ tzy
œ tforce
zy = tzy
!
tzy1¢x ¢y2 - tzy ¢x ¢y = 0tzy = tzy
œ
œ
œ
stress
area
tforce
=manner,
tzy
In a similar In
manner,
equilibrium
inkuvvet
the
z direction
= tyz
. bulunur.
zy
a similar
force
equilibrium
inyields
the z!tyzdirection
yields tyz =
œ tyz .
zIn
ekseni
doğrultusundaki
dengesinden
a
similar
manner,
force
equilibrium
in
the
z
direction
yields
t
=
t
.
yz
yz
Finally, taking
moments
about
the x axis,
Finally,
taking
moments
aboutœthe
the
x axis,
Finally,
taking
moments
about
x axis,
œ
œ
©F
=
0;
t
1¢x
¢y2
t
¢x
¢y
=
0
x
eksenine
göre
moment
denkleminden:
zy
y
zy
ilar manner,
in=the
yields tyz = tyz .
¢x
¢y2 -force
tzyequilibrium
¢x ¢ymoment
0 z direction
œ
tzy =moment
tmoment
zy
taking moments about the x axis,
œ
force
arm
tzy = tzy
œ
force in the
arm
force
arm
In a similarmoment
manner,
force
equilibrium
z direction yields tyz = tyz
.
stress area
Finally, taking moments about
the
x
axis,
stress
area
stress area
arm
œ
©Mx = 0; force
-tzyz
1¢x
¢y2 ¢z + tyz1¢x
¢z2 ¢y = 0
equilibrium
in the
direction
moment yields tyz = tyz .
stress
©Mxxarea
0;
1¢x¢y2
¢y2¢z¢z+ +tyzt1¢x
¢z2
tzy =-t
== 0;
ttzyyzzy
1¢x
¢z2
¢y¢y
= 0= 0
yz1¢x
bout
the
x
axis,
force
arm
tzytzy= =tyztyz
so that
stress
0;
-! tzy1¢x ¢y2 ¢z
+ tarea
1¢x ¢z2 ¢y = 0
yz
œ
œ
so that
that tzy = tzy
= tyz = tyz
= t
tzy = tyz
moment
©Mx = 0;
-tzy1¢x ¢y2 ¢zœ œ+ tyz1¢x ¢z2
œ œ¢y = 0
tzy
==tzy
= =tequal
= t= t and
In other words, all four shear stresses
must
yztyz
yz
tzy
thave
tyz= =tmagnitude
zy
olarak:or away from
tzyeach
= tyzother at opposite edges of
be directed Sonuç
either toward
orcethe element,
arm
In
other
words,
four
shear
stresses
must
have
equal
magnitude
andand
In
other
all
shear
stresses
must
have
equal
magnitude
œ words,
œ
1–21b.
This
isall
referred
to
the
complementary
property
so that
tFig.
=
tzy
=either
tyz
=toward
tfour
tasaway
yz = or
zydirected
be
from
each
other
at
opposite
edges
of of
!
of shear, and
the either
conditions
shown
Fig.œ 1–21,
material
is opposite edges
beunder
directed
toward
or in
away
from the
each
other at
œ
sr words,
area
t
=
t
=
t
=
t
=
t
the
element,
Fig.
1–21b.
This
is
referred
to
as
the
complementary
property
zy
yz
zy
yz
subjected
tothe
pure
shear.
element,
Fig. 1–21b.
This equal
is referred
to as theand
complementary property
all four
shear
stresses
must have
magnitude
of shear,
and
under
the stresses
conditions
shown
in Fig.
1–21, the
material is
In
other
words,
all
four
shear
must
have
equal
magnitude
of shear,
andfrom
under
theother
conditions
shownedges
in Fig.
1–21,and
the
material is
ted either toward
or away
each
at opposite
of
subjected
to toward
pure shear.
be directed
either
or
away
from each other property
at opposite edges of
subjected
to
pure
shear.
ment,
Fig.
1–21b.
This
is
referred
to
as
the
complementary
¢y2
¢z +Fig.
tyz
1¢x
¢yto=as the
0
yr,1¢x
thethe
element,
1–21b.
This¢z2
isin
referred
and under
conditions
shown
Fig. 1–21,
thecomplementary
material is property
of shear, and under the conditions shown in Fig. 1–21, the material is
ed to pure
shear.
tzy
= tyz
0;
33
AVERAGE SHEAR STRESS
area
subjected to pure shear.
œ
œ
= tzy
= tyz = tyz
= t
hear stresses must have equal magnitude and
or away from each other at opposite edges of
!13
36
36
CHAPTER 1 STRESS
CHAPTER 1 STRESS
S T RCEHS SA P T E R 136 S T R E S S
1
CHAPTER 1
EXAMPLE 1.11
1
EXAMPLE 1.11
1 1.11
AMPLE
1
EXAMPLE 1.11
1 !
STRESS
If the wood joint in Fig. 1–23a has a width of 150 mm, determine the
If the wood
Fig. 1–23a along
has a width
of 150 mm,
determine
the
average
shearjoint
stressindeveloped
shear planes
a–a and
b–b. For
the
wood
joint
inaaverage
Fig.
1–23a
hasmm,
a width
of 150ofthe
mm,
determine
the a–a
If the wood jointIfin
Fig.
1–23a
haseach
width
of
150
determine
shear
stress
developed
along
shear
planes
and
b–b.
For
plane,
represent
the
state
stress
on
an
element
of
the
material.
If the wood
joint 150
in Fig.mm
1–23aolduğuna
has a width göre,
of 150 mm,
Aşağıda
verilenaverage
şekildeki
kalınlığı
a-a determine
ve b-b the
shear elemanların
stress
developed
along shear
planes
and
For
average shear stress
developed
along
shear
planes
and
b–b.
For
each
plane,
represent
thedeveloped
state
ofa–a
stress
onb–b.
an element
theand
material.
average
sheara–a
stress
along
shear
planesofa–a
b–b. For
each
plane,
represent
the
state
of
stress
on
an
element
of
the
material.
each
plane,
represent
the
state
of
stress
on
an
element
of
the
material.
düzlemlerinde meydana gelen ortalama
kaymarepresent
gerilmelerini
düzlemlerdeki
each plane,
the statehesaplayınız.
of stress on anBu
element
of the material.
gerilme durumlarını malzemeye ait bir elemandaki gerilme hali ile gösteriniz.F
a
1
SSTTRREESSSS
6 kN
a
a
a
6 kN
6 kN a
6 kN
6 kN
6 kN
b
b
b mm, determine the
11ig. 1–23a has a width of 150
a
a
a
6 kN
b
b
a
b
b
F b
6 kN
6 kN 6 kN
6 kN F
b
0.1 m 0.125 bm
developed along shear planes a–a and b–b. For
(b)
(b)
0.1 m of0.125
If the
the
wood joint
joint
in Fig.
Fig.
1–23a has
has aa width
width
150mmm,
mm, determine
determine the
the
wood
in
1–23a
of
150
the state of stress ! If
on
an
element
of
the
material.
0.1 m 0.125 m
0.1 m 0.125 m
(a)0.1 m 0.125 m
average shear
shear stress
stress developed
developed along
along shear
shear
planes a–a
a–a and
and b–b.
b–b. For
For
average
planes
(a)
each
plane,
represent
the
state
of
stress
on
an
element
of
the
material.
Fig.
1–23 çizelim:
each plane,
represent
the
state of stressçizerek
on an element
the material.
(a) diyagramını
Çözüm:
cisim
kuvvetofdağılımını
(a) Serbest
(a)
F
Fig. 1–23
Fig. 1–23
F
F
F
a
F
6 kN
F
6 kN
6 kN
6 kN
F
(b) F
(b)
(b)
Fig. 1–23
Fig. 1–23
SOLUTION
FF
SOLUTION
Internal Loadings. Referring to the free-body diagram of the
6 kN
SOLUTION
aa SOLUTION
aa
F
SOLUTION
Internal
Loadings.
Referring
to the free-body diagram of the
member,
Fig.
1–23b,
!
kN
66of
kN
Internal
Loadings.
Referring
to diagram
the free-body
of the
6
kN
Internal
Loadings.
Referring
to
the
free-body
thediagram
member,
Fig.
1–23b,
6
kN
6
kN
Internal
Loadings.
Referring
to the free-body diagram of the
6 kN
F
F
member,
Fig.
1–23b,
+
member,
Fig.
1–23b,
Yatay
kuvvet
dengesini
yazalım:
kN - F - F = 0
F = 3 kN
(b)
: ©F
x = 0; Fig.61–23b,
member,
bb
+ ©Fx = 0;
bb
6 kN - F - F = 0
F = 3 kN
:
+6 ©F
+ ©Fx = 0;
6= kN
- F - FF
3 kN cut across shear planes a–a
kN x-=F0;- F Now
0 consider
==
30kN6 kN -FofF=segments
(b)
:
:
+
the
equilibrium
(b)
©F
=
0;
F
=
0
F = 3 kN
:
!
x
Now
consider
equilibrium
of segments
and
b–b,
shown the
in Figs.
1–23c and
1–23d. cut across shear planes a–a
0.1 m
m 0.125
0.125 m
m
0.1
Now
consider
thesegments
of segments
cut
across
planes a–a
Now
consider the
equilibrium
of
cut
across
planes
a–ashear
and
b–b,
shown
inshear
Figs.
1–23c
and
1–23d.
a-a
kesitindeki
kuvvet
durumu
veequilibrium
gerilme
hali:
Now
consider
the
equilibrium
of
segments cut across shear planes a–a
Fig. 1–23
and
b–b,
shown
in :
Figs.
1–23c
and
1–23d.
+ ©F
and (a)
b–b, shown in
Figs.
1–23c
and
1–23d.
(a)
0; shown
Vain-Figs.
3 kN
= 0 and V1–23d.
and
b–b,
1–23c
x =
a = 3 kN
3 kN
+ ©F = 0;
Va = 3 kN
:
Va - 3 kN = 0
x
3
kN
Fig. 1–23
1–23
+V
+ ©F
+Va©F
Fig.
Va =
:
©F
0; = 0:
= 0 3 kN
3 kN
V
:
3 kN
=kN30;kN
0 = 0Vb = V3 kN
+ax3 ©F
x =
3 kN
x = 0;
a 3 kN
:
V-a V
-b 3=kN
x = 0;
a = 3 kN
3 kN
ta ! 200 kPa
Va
+ ©F
:
=
0;
3
kN
V
=
0
Vb = 3 kN
x
b
+
+ta ©F
! 200 kPa
Va3©F
:
=V
0;b = 0Average
3 kNV+b- =V
= 0Stress.
Vb = 3 kN
:
kNx 3b kN
x = 0;
Shear
: ©F
3 kN - Vb = 0
Vb = 3 kN
x = 0;
V! a
Vtaa ! 200 kPa
t
!
200
kPa
V
(c)
a
a
Referring to the free-body diagram of the
Average Shear Stress.
SOLUTION
3
SOLUTION
Average
Shear Stress.
b-b
kesitindeki
kuvvet
ve gerilme
(c)
Average
Shear
Stress.durumu
3110
2N
Vhali:
a Shear
Average
Stress.
(c)
1t
2
=
=
Ans.
Internal
Loadings.
Referring
to
the
free-body
diagram
of
the= 200 kPa
a
avg
3110
N the
Internal Loadings.
Referring to the 3free-body
diagram32of
(c)
AV
a a =10.1 m210.15 m2
3
=
200
1t
2
=
kPa
Ans.
3110
member,
Fig.
1–23b,
3
a avg2 N
V
3=
kN
a
1–23b,
a3
N - F - F = 0 member,
FFig.
kN 3110 2VN
3110
2Ans.
N
m2
200 m210.15
= = 200 kPa Aa V=a 10.1
kPa
Ans.
1ta2avg =
= 1ta2avg =
3
= 200 kPa
1ta2avg
=
= 3110 2 N
Ans.
Vbm2
Aa m210.1 m210.15
A3 kN 10.1 m210.15
10.1 m210.15
m2
3
= 160 kPa
1t0b2avg = VF=A
Ans.
+ ©Fx = 0;a
a3 kN
3 kN
6
kN
F
F
=
=
+
2
N
3110
:
©F across
= 0; shear
6 Vplanes
kN - F
3 kN m210.15 m2
ilibrium of segments
cut
a–a- F = 0
3 kN
:
10.125
3 =AbFb =
tb = 160xkPa
b 31103V
= 160 kPa
1t3110
Ans.
N =
2 bN
3
b2avg 2
s. 1–23c and 1–23d.
2N
3110Ans.
V
A
10.125
m2
! 1ttbb2=avg160=kPaVb = 1tb2Vavg
b
b
=
=
= 160 m210.15
kPa
Ans.
=
160
kPa
b
2avgofacross
=stress
=on elements
= 160
kPa a–a and b–b is Ans.
1t
Now
consider
the equilibrium
equilibrium
of segments
segments
cut
across
shear
planes a–a
a–a
The state
located on
sections
Ab of
10.125
m210.15
m2 shear
bcut
A(d)
m210.15
m2
Now
planes
Vb consider
b the10.125
tb =
Vb160 kPa
Ab
10.125
m210.15 m2
Vb
tb = shown
160 kPa in Figs. 1–23c
and
b–b,
and
1–23d.
The
state
of
stress
on
elements
located
on
sections
a–a and b–b is
shown
in
Figs.
1–23c
and
1–23d,
respectively.
a-a
ve
b-b
kesitindeki
kuvvetleri
yatay
kuvvet
dengesinden
hesaplayalım:
(d)
and
shown in Figs. 1–23c and 1–23d.
V
- 3 kN = 0
3 kN
a =b–b,
Theon
state
of stress
on elements
located
on
sections
a–a and b–b is
The state of stress
elements
located
on
sections
a–a
and
b–b
is
(d)
shown
in
Figs.
1–23c
and
1–23d,
respectively.
The state of stress on elements located on sections a–a and b–b is
(d)
shown
in Figs.
1–23c
and
1–23d,
respectively.
shown
in==Figs.
respectively.
++b ©F
V
:
0; 1–23c
Vaaand
kN ==
= 3 in
kNFigs.
N - V
=©F
3xxkN
V
shown
1–23c and 1–23d, respectively.
V
:
0;
V
-- 331–23d,
kN
00
bkN= 0
aa = 3 kN
33 kN
ss.
Va
a
6 kN
+ ©F = 0;
:
! : ©Fxx = 0;
+
kN -- V
Vbb == 00
33 kN
kN
Vbb == 33 kN
V
Average Shear
Shear
Stress.
Ortalama
gerilmeler:
Average
Stress.
= 200 kPa
Ans.
m210.15 m2
3
3
3110
2
N
Va
3110 2 N
= 200 kPa
1taa22avg
= Va ==
Ans.
3
avg
1t
=
Ans.
3110 2 N
A
10.1 m210.15 m2
m2 = 200 kPa
= 160 kPa Aaa 10.1 m210.15
Ans.
5 m210.15 m2
N
31103322 N
Vbb
3110
V
2
=
=
= 160 kPa
1t
Ans.
b
avg
2
=
=
1t
Ans.
n elements
located bonavgsections
and m210.15
b–b is m2
Abb a–a
10.125
m210.15
m2 = 160 kPa
Vb
A
10.125
V
b
!
nd 1–23d, respectively.
The state
state of
of stress
stress on
on elements
elements located
located on
on sections
sections a–a
a–a and
and b–b
b–b is
is
The
shown
in
Figs.
1–23c
and
1–23d,
respectively.
shown in Figs. 1–23c and 1–23d, respectively.
11032 N
!14
1.5
EXAMPLE 1.12
AVERAGE SHEAR STRESS
1.5
37
AVERAGE SHEAR STRESS
1
37
The inclined member in Fig. 1–24a is subjected to a compressive force
of 600 lb. Determine the average compressive stress along the smooth
1.5 AVERAGE SHEAR S
1.12
! EXAMPLE
1
areas of contact
defined by AB and BC, and the average shear stress
along the horizontal plane defined by DB.
1.5 AVERAGE SHEAR STRE
The inclined member in Fig. 1–24a is subjected to a compressive force
EXAMPLE
1.12
600 lb
of 600 lb. Determine the average
compressive stress along the smooth
5 4
ahşapshear
birleşim
areas of
contact
by
AB
BC, andverilen
the average
stress alt yüzeyinde zemin üzerine
1.5 AVERAGE
SHEAR
STRESSdefined
37 andŞekilde
member in Fig. 1–24a is subjected to a compressive force
3 The inclined
EXAMPLE
1.12
along the horizontal plane defined by DB.
of 600 lb.
Determine
the averageDiyagonal
compressive eleman
stress along
the smooth
1.5 AVERAGE SHEAR STRESS
37
ankastre
mesnetlidir.
üzerine
600 lb
1.5 AVERAGE SHEAR STRESS
37
600
lb
of contact
defined
ABisand
BC, and
average shear
stress
Theareas
inclined
member
in Fig. by
1–24a
subjected
tothe
a compressive
force
lb birleşim
5 4
1.5 AVERAGE SHEAR STRESS
37600
kuvvet
uygulanmaktadır.
ABDB.
vestress
BC along
çizgilerindeki
along
the
horizontal
defined
by
1.5
AVERAGE
SHEAR STRESS
of
lb.1Determine
theplane
average
compressive
the600
smooth
3
5
4
1.5
A
VERAGE
S
HEAR
S
TRESS
37
areas of contact defined by AB and BC, and the
average
shear
stress
600 lb
1 enkesit alanlarında meydana gelen
3
essive force
ortalama
gerilmeleri ve
along the1 horizontal
plane defined by DB. 5 4
A
the
EXAMPLE
1.12
3 lb
sivesmooth
force
1
600
600 lb
DB hattındaki yatay düzlemdeki
ortalama kayma
ssive force
shear
stress
1.12 C
1EXAMPLE
in.
he smooth
5 4
1
5 4
B
heforce
smooth
ve
Fig. 1–24a ishesaplayınız.
subjected to a compressive
force
3
hear
stress
3
D The inclined member ingerilmesini
hear
stress
smooth
The inclinedofmember
1–24a
subjected
to a compressive
force the smooth
2in
in.Fig.
600 lb. Determine
theisaverage
compressive
stress along
A
r stress
the average
compressive
stress
along
theaverage
smoothshear stress
Cby AB and
3 in.of contact
areas
defined
BC,
and
the
1.5ofin.600 lb. Determine
1 in.
areas of contact
defined
by AB
BC,
and
the
shear stress
B and
along
the horizontal
plane
defined
by average
DB.
(a)
Fig. 1–24
A
FAB
Çözüm:
along the horizontal plane
defined by DB.
D
600 lb
C
2 in.
SOLUTION
1 in.600 lb
5 4
600
lb
B
3 in.
in.
A
FBC
Internal Loadings. The 1.5
free-body
diagram
of the inclined
member
5 4
D 3
5 4
C
(a)
600
lb
Fig. 1–24
3
is shown in Fig. 1–24b. The compressive
forces
acting
on
the
areas
of
1
in.
(b)
2
in.
FAB
3 600 lb
B
5 4
contact are SOLUTION
3 in.
1.5
in.
360 lb
5600
lb
D
4
3
600 lb
+ ©F = 0;
in.
:
FAB Loadings.
- 600
lb A 3 B = 0
FAB =diagram
360 lb of the inclined
(a)2 member
Fig.
x
FBC1–24
Internal
5 4 3 5 The free-body
600 lbF 5 4
AB
3 in.
1.5 in.
acting on the areas of
5 4
(b)
3
SOLUTION
+ c ©Fy = 0; is shown
FBC in
- Fig.
6003 1–24b.
lb A 45 B =The
0 compressive
FBC = 480forces
lb
(a)
Fig. 1–24
3
contact are
V
A
F
FBC
Loadings.
The
Also, from the free-body diagram of theInternal
top segment
ABD of
thefree-body diagram of the inclined member360 lb AB
C= 360 lb
+ ©F = 0;
A lb A 3 B = 0
:
-SOLUTION
600
600areas
lb
AB 1
AB
in.
is shown
Fig.
1–24b.
The compressive forces(c)acting on the
of
bottom member,
Fig.x 1–24c, the Fshear
force
acting
theF
sectioned
5
(b
Cinon
B
1 in.
5 member
4
4 Loadings.
g. 1–24horizontal plane
F
contact
are
DB
is
Internal
The
free-body
diagram
of
the
inclined
BC
c
B
+ ©Fy = 0;
FBC - 600 lb A 5DB = 0
FBC = 480 lb
FAB
3
+ ©F
is shown
inx Fig.
acting
the
2The
in. compressive
(b)
+ ©F = 0; Also, from the free-body
:
= the
0;1–24b.
F
lb A 35 Bof=forces
0
FAB V=on
360
lbareas of
D
. 1–24 :
AB - 600
V
=
360
lb
diagram
of
top
segment
ABD
the
x
F
2
in.
g. 1–24
AB
contact
are
3 in.
(c)
1.5
in.
600 lb
FABbottom member,
Fig. compressive
1–24c,+
shear
force along
acting
the3lbsectioned
FBC average
ed member
c ©F
= 0;
F -on
600
= 480 lb
A=45 B 0= 0 F FBC
3the
in.
y stresses
Average !Stress.
The
the
+ ! ©F
–24
1.5 in.
(a)FAB BC
:
=
0;
600
lb
=
360
lb
A
B
Fig.
1–24
5 4
x
AB
F
5
horizontal
plane
DB
is
AB
areas
of
V
(b)
FAB of the
horizontal
and vertical
of the inclined
member
are free-body diagram
Also,
from the
of the top segment ABD
FBCplanes
(a)
dhemember
Fig.
3
4 1–24
FAB lb
FBCF SOLUTION
d member
(c)
c
+
+
©F
=
0;
F
600
lb
=
0
F
=
480
A
B
360
lb
DB
yüzeyine
etki
eden
yatay
kuvvet:
y
BC
BC
AB
bottom
member, Fig. 1–24c, the
shear force acting on the sectioned
e areas of
:s©Fx= =(b)0;
V =
Ans. 5
= 360 lb
= 360
240 lb
psi
SOLUTION
AB
he areas of
FBC
member
V
(b)
240
psi
F
Loadings.
Thethe
free-body
of the
inclined
member ABD of the
horizontal
plane
DBdiagram
is diagram
AInternal
11 in.211.5
in.2
Also,
from
free-body
of the
top segment
BC
AB
Average
Stress.
compressive
stresses
along
the
360 The
lbThe average
areas of
(c)
FBC
Internal
free-body
diagram
of
the
inclined
member
(b)Loadings.
is
shown
in
Fig.
1–24b.
The
compressive
forces
acting
on
the
areas
of
bottom
member,
Fig.
1–24c,
the
shear
force
acting
on
the
sectioned
(b)
360
lb
+
Fand
lb :of©F
BC vertical480
horizontal
planes
thex inclined
member
are
= 0;forces
V =
360 of
lb
is shown
in
Fig.
1–24b.
The
compressive
acting
on
the
areas
(b)
Ans.
s
=
=
=
160
psi
contactlbare Fhorizontal plane
BC
160 psi
V
360 lbDB is
360 lb
ABC 36012
in.211.5
AB in.2
BD of the
contact are
3 = The
Average
Stress.
average
compressive
stresses
along the360 lb
+
Ans.
s
=
=
240
psi
AB
:
©F
=
0;
F
600
lb
=
0
F
=
360
lb
+
A
B
(c)
x
AB
AB
240
psi
600
lb
5
:
©F
=
0;
V
=
360
lb
A
11
in.211.5
in.2
3
e sectioned
are shown
AB horizontal
+ ©F = 0;
:
FABin! -Fig.
6001–24d.
lbxA 5 B = 0and vertical
F
= planes
360 lb of the inclined member are
BD
of These
the stress! VVdistributions
x
4 AB
(d)
5=
4 0;
BD of the
The
average
shear
stress
acting
on
the
horizontal
plane
defined
by
(c) + c
©F
F
600
lb
=
0
= 480
A
B
F
360lb
lb
F
600
lb
480
lb
y
BC
Average
Stress.
The
average
stresses along the
ABFBC compressive
5
BC
sectioned
4
(c)0;
3 =600
V
Ans.
s
=
=
240 psi
lb
c
Ans.
s
=
=
160
psi
+
©F
=
F
600
lb
=
0
F
=
480
lb
A
B
AB
sectioned
DB is
of the
y
160 psi
5 4BC A
240 psi
horizontal
anddiagram
vertical
planes
of
member
are 360 lb V
Also, BC
from
the
the
topthe
segment
ABD
AAB
11inclined
in.211.5
in.2 of the
125 in.211.5
in.2BCof
BCfree-body
(c)
V
5
600
lb
4
360
lb
(c)
ctioned
3
Also, from
the
free-body
diagram
of
the
top
segment
ABD
of
the
600 lb
F
360
lb
member,
Fig.
the1–24d.
shearAB
force
acting 480
on the
tavg =bottom
Ans.
801–24c,
psiin Fig.
These stress
distributions
are =
shown
FBC
3
lb sectioned
(c)Ans.
sAB
=
= the sectioned
= 240 psi
5 in.211.5
4Fig. 1–24c,
along the
600 lb
13
in.2
bottom member,
the
shear
force
acting
on
5
4
240
psi
Ans.
s
=
=
=
160
psi
horizontal
planeacting
DB ison the horizontal
(d)
11 in.211.5 by
in.2
BC AABplane
The average
shear stress
ABC defined
12 in.211.5 in.2
5 4
3
horizontal
plane3DB
is
uniformly
distributed
over the sectioned area
in
along This
the stress is shown
80
psi
+
DB
is
FBClb are shown
480 lb
3
: ©Fx = 0; These stress distributions
V = 360
along Fig.
the1–24e.
in
Fig.
1–24d.
360
lb
Ans.
s
=
=
=
160
psi
+ ©F = 0;
(e)
Ans.
lb lb BC
V360
= average
360
x
(
A
12 in.211.5
in.2horizontal
240:
psi
ng the
The
stress
acting
onAns.
the
BC
=
=shear
80compressive
psi
Averagetavg
Stress.
The
average
stresses
along
theplane defined by
13
in.211.5
in.2
Ans.
DBstress
is planes
Stress.
average
compressive
stresses
along
the
These
distributions
are shown
in Fig.
1–24d.
horizontalThe
and
vertical
of the inclined
member
are
240Average
psi
Ans.
is shown
uniformly
distributed
over
the lb
sectioned
area
in
(d)
240This
psi stress and
Ans.
360
lbhorizontal
horizontal
vertical
planes
of
the
inclined
member
are
The
average
shear
stress
acting
on
the
plane defined80bypsi
F
360
160
psi
AB
Ans.
tavg =
Ans.
= 80 Ans.
psi
s
=
=
=
240
psi
(e)
F
240 psiFig. 1–24e.
AB
360
lb
DB
AB is
13 in.211.5 in.2
240 psi
Ans.
AAB
11 in.211.5
in.2
Ans.
sAB
=
= 240 psi
160
psi =
Ans.
360distributed
lb
240 psi
A
11
in.211.5
in.2
160
psi
This
stress
is
shown
uniformly
over
the
sectioned
area in
AB
(d)
tavg
Ans.
= 80 psi
FBC
480= lb
defined
Ans. by
13
in.211.5
in.2
Ans.
s
=
=
=
160
psi
Fig.
1–24e.
160 psi
(e
FBC BC
480 lb 12 in.211.5 in.2
160 psi
(d)
!
Ans.
s360
= This! =stress A
= 160 psidistributed
defined by
BC lb
isBC
shown
uniformly
over the sectioned area in 160 psi
(d)
80 p
A
12
in.211.5
in.2
defined by
BC
These stress distributions
are shown in Fig. 1–24d.
Ans.
Fig.
1–24e.gerilmesi:
(e)
(d) distributions
360 lb
ined by
DBThese
yüzeyindeki
ortalama
kayma
(d)
stress
are
shown
in Fig.
1–24d.
The360
average
shear
stress
acting
on the horizontal plane defined by
lb
(d)
Ans.
The average
shear
ned area
in
DB
islb stress acting on the horizontal plane defined by
80360
psi
Ans.
360 lb
DB
is
360 lb
(e)
Ans. in
ed area
360
lb
80 psi
t
Ans.
=
=
80
psi
avg lb
360
ed area in
80 psi
13 in.211.5
in.2
tavg =
Ans.
= 80 psi
(e)
13
in.211.5
in.2 distributed over the sectioned area in
area in
(e)
80 psi
This stress
!
! is shown uniformly
80 psi
(e) is
This stress
shown
uniformly distributed over the sectioned area in
80 psi
Fig.
1–24e.
(e)
Fig. 1–24e.
(e)
!15
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