Upload
nisarg-shah
View
226
Download
1
Embed Size (px)
Citation preview
8/18/2019 Mechanics:-Sample Problems Chapter 8
http://slidepdf.com/reader/full/mechanics-sample-problems-chapter-8 1/5
Two forces PI and Pz, of magnitude PI::::: 15kN and
P2 ::::: 8 eN, are applied as shown to the end A of bar AB,
which is welded to a cylindrical member BD of radius
c
:::::
20 mm (Fig. 8.21). Knowing that the distance rom A to
the axis of memberBD is a
:::::
50 mm and assuming hat all
slressesemainbelow the propOrtional imit of the material,
determine a) the normal and shearingstresses t point K of
the transverse ection of memberBD located at a distance
b
:::::
60 mm from end B, (b) the principal axesand principal
stresses t K, (c) the maximumshearing tressat K.
rt'I jf()~ ~
F J'( J.- 6
([)
~t
fo P,
.
,s K N
15 t O. 0
.5
~ 'IS () (VrO
~ cljf"tct/ct')
Du~
PJ..
..
,-
".
.,.
-
,
G
0
(t ,,~
-"vt
v
,
...
.".
.,.
,.
0
12,
(-,<-It
ffy-IA
....
,.
-'"
I ft 1<N
".
:. 7 , () IV ""
-
I
- : -
- - - _: - - - - - - ::-: - - -- -- - -: --
-.';- :- -".', -::-::-.' /"':.'.-_:->:' . :', :.;:-;--:':-',"':\:';~~-:'..fU;;,:,:,;~;{~,:",/~;-':i:'~) 1'-;~:::.I'~;:;.. :~'.~~{ :.~'Jd<)"~}:::.'i;:;::;':~;j/i.~i1.i:_:~u.S'YH ~:~\~~.~)~"i"<i.'U~.)Y'i~ '~Ik»'~\t:~,~;?~~~~ :,t~_;:~;i~:~~(Nt{,
i
Fig. 8.21 P;2 '" H) kN
My
~T
a
0
F
-
M.
v
#./ftO.D6
1081) N'rn
(5
d.iff dlUt')
,r- KN
",gxO.O.5
Fig. 8.22
,...
...
y/
-
-
::< 7.50 N.II)
My
2.-
.,.
-
N~
,so
".
...
)l
v = IS kN
Fig. 8.23
N
. 75. 0
~k
)
(vA
t &'jY)( e.-
N ft- do~
J\Jr()
)
f.. IJ. t<
(' D J- p(JJA
8/18/2019 Mechanics:-Sample Problems Chapter 8
http://slidepdf.com/reader/full/mechanics-sample-problems-chapter-8 2/5
8/18/2019 Mechanics:-Sample Problems Chapter 8
http://slidepdf.com/reader/full/mechanics-sample-problems-chapter-8 3/5
~
Ne}-
~x ::
() Wdij
(J,;::. + 107
,-..
""
- l, .
y ~ -- r()rA((f(\
+ Q i J
.4 f'fPd..
f "6 heCJ..I'
-
52.5
~ '[ "'y
&p ~
~
f~ - fr:J
(6'7C-Q'~
o'(>z e s :.
2.-
75.1 MP",--
(2~. t 1'1PrA.
-2./.1i. MP'1
g
.,.
-
Of
0
-
..
2-
-<910u eA
r."'J=
-52.5 MPa
Fig. 8.24
I~ PIA
t07
MPa..
8p
==
22.2°
15kN
(
;11in= -21.4 MPa.
Fig. 8.26
)
= 75.1 MP~l
'T,n,,~
as=22.8°
8/18/2019 Mechanics:-Sample Problems Chapter 8
http://slidepdf.com/reader/full/mechanics-sample-problems-chapter-8 4/5
A
~
r
Ii
.Sin. ~
5001b
n. -.. .. .. ... ..
4.5 n.
4.5 in.
A
A == 2501b
B = 2.50b
= 62.51bill.
M f= 625
Ii
~/V:250lb
..-T =9001b.in.
~O.9-in. diameter
G
T:: 6290 psi
--LH
T = 6290 psi
,L
(0)
T =6290psi
K
= 62-90 si
H
T = 524 psi
"l
(b)
T=524psi
:,.
- (T = 8730 psi
.,.:::'0
K
(1';:0
(c)
u = 8730 psi
-u = 8730 psi
L
..,. =
6810 psi
.u =" 8730psI
, "" ~'"~ ""~;";, ',;~i;;:.;{;:.1~.i; ;f,<",'i:kv:, ,; J';':' ~>:,;i;\:i.:;:' ;,./i;'~~iD~iifi~t~;ii.fiii~Xl~At#wkMi.{~;~.&.~;~;.,{~;
SAMPLE PROBLEM 8.4
A horizontal 500-lb force acts at point D of crankshaft AB which is held in
static equilibrium by a twisting couple T and by reactions at A and B. Know-
ing that the bearings are self-aligning and exer~no couples on the shaft, de-
termine the normal and shearing stressesat points H, J, K, and L located at the
ends of the vertical and horizontal diameters of a transverse section located
2.5 in. to the left of bearing B.
SAMPLE PROBLEM 8.4
SOLUTION
Free Body. Entire Crankshaft. A =
B
= 250 Ib
+ ~2:M~= 0: -(500 Ib)(1.8 n.) + T = 0 T = 9001b
.
n.
Internal Forces n TransverseSection. We replace he reactionB and
the twisting couple T by an equivalent orce-couple ystemat the centerC of
the transverse ectioncontainjngH. J, K, and L.
%
v = B = 250 Ib T = 900 Ib . n.
My =
(250 b)(2.5 n.) = 6251b in.
The geometric properties of the O.9-in.-diameter section are
A = 7T(O.45n.)2
=
0.636 n2 1= 7T(O.45n.)4 = 32.2 X 10-3 n4
J = 7T(O.45n.)4= 64.4X 10-3 n4
Stresses Produced by Twisting Couple T. Using Eq. (3.8), we
determine he shearingstresses t points H, J, K, and L and show them in
Fig. (a).
Tc (900 b . n.)(0.45 n.)
=
6290psi
'T = J = 64.4 X 10 3 n4
Stresses Produced by Shearing Force V. The shearing force V pro-
duces no shearing stressesat points J and L. At points Hand K we first com-
pute Q for a semicircle about a vertical diameter and then determine the shear-
ing stress produced by the shear force V = 250 lb. These stressesare shown
in Fig. (b).
Q =
(
.1Tc2
)( 4C)= ~c3 = ~(O45 n)3 =60 7 X 10-3 in3
,2 31T 33' . .
VQ (250 Ib)(60.7 X 1O-3in3)
.
T
= - ::::: = 524pSi
It (32.2 X 10-3 in4)(O.9 n.)
,
StressesProduced by the Bending Couple M,.. Since the bending cou-
ple My acts in a horizontal plane, it produces no stressesat Hand K. Using Eq.
(4.15), we detennine the normal stressesat points J and L and show them in
Fig. (c).
IMylc (625lb . in.)(O.45 in.) .
(J'
= - = =
8730
PS
I 32.2 X 10-3 in4
Summary. We add the stressesshown and obtain the total normal and
shearingtressest pointsH. J, K, andL
8/18/2019 Mechanics:-Sample Problems Chapter 8
http://slidepdf.com/reader/full/mechanics-sample-problems-chapter-8 5/5
4O~mm
y
~~
MJ :; t).,'5 kN. III
G
x
T'"m
I ~;:~8°
O~
I~
O"'"iL1
SAMPLE PROBLEM 8.5.
Three orces are appliedas shownat pointsA, B, and Jj of a short steelpost.
Knowing that the horizontal crosssectionof the post s a 40 X 140-mm ec-
tangle, determine he principal stresses, rincipal planesand maximumshear-
ing stressat point H. .
SOLUTION
Internal Forces n Section EFG. We replace he three applied orces
by an equivalent orce-couple ystemat the centerC of the rectangular ection
EFG. We have. .
Vx
=
-30 kN P
=
50kN V~
=
-75 kN
Mx =
(50 kN)(0.130m) - (75 kN)(0.200m)
=
-8.5 kN
.m
My =
0
M~ =
(30kN)(0.1O0 ) = 3 kN . m
We note that there s no twisting couple about he y axis. The geometric
propertiesof the rectangular ectionare
A= (0.040m)(0.140m) = 5.6X 10-3 m2
Ix= i2(0.040m)(0.140m? = 9.15
X 10-6 m4
lz = i2(0.140m)(0.040m? = 0.747 X 10-6 m4
Normal Stress at H. We note that normal stresses J'yare produced by
the centric force P and by the bending couples Mx and M<;.We determine the
sign of each stress by carefully examining the sketch of the force-couple sys-
tem at C.
P IMtla IMx1b
(J' = +-+---
Y A I I
,t x
50 kN (3 kN
.
m)(0.020 m) (8.5 kN
.
m)(O.O25m)
= + -
5.6X 10-3 m2 0.747 X 1O-6m4 9.15 X 10-6 m4
Uy = 8.93MPa + 80.3MPa - 23.2MPa cry=
66.0 MPa. <1
Shearing Stress at H. Considering first the shearing force Vx, we note
that Q = 0 with respect to the z axis,. since H is on the edge of the cross sec-
tion. Thus V
x
produces no shearing stressat H. The shearing orce Vz does pro-
duce a shearing stress at H and we write
Q
=
A1YJ ~ [(0.040 m)(0.045m)](0.0475m)
=
85.5 X 10-6 m3
VzQ (75 kN)(85.5 X 10-6 m3) .
Tyz=
Ixl= (9.15X 1O-6m4)(0.040m) Tyz= 17.52MPa <J
Principal Stresses~ rincipal Planes, and Maximum Shearing Stress
at H. We draw Mohr's circle for the stresses t point H