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Unformatted text preview: Emblem 5.44 A cantilever beam AB is loaded by a couple M0 at
its free end (see ﬁgure). The length of the beam is L = 1.5 m and the
longitudinal normal strain at the top surface is 0.001. The distance from the top surface of the beam to the neutral surface is 75 mm. Calculate the radius of curvature p, the curvature K, and the vertical
deﬂection 8 at the end of the beam. Snlutlnn 5.44 Cantilever beam tnhlem 5.51ll A railroad tie (or sleeper) is subjected to two rail
loads, each of magnitude P = 175 ltN, acting as shown in the ﬁgure.
The reaction q of the ballast is assumed to be uniformly distributed over the length of the tie. which has crosssectional dimensions b=300mmandh=250mm Calculate the maximum bending stress am in the tie due to the loads P, assuming the distance L = 1500 mm and the overhang
length a = 500 mm. l_.;4 M. ASSume that the deﬂection curve is nearly flat.
Then the distance BC is the same as the length L
of the beam. , _£_l.5n‘t=
.smB—p —?5m 0.02
8 = arcsin 0.02 = 0.02rad
5 = p(l  cos 9) = (75 m}(l — cos (0.02 rad»
=l5.0mm 4 L
NOTE: 5 = 100. which conﬁrms that the deflection curve is nearly ﬂat. 3%};1. #aa' Fly—1
1“: Sululinn 5.5.10 Railroad tie (or sleeper) DATA P=17SkN
L=1500mm
_ 2P bhl
q L+2a s
BENDINGMOMENT DIAGRAM
Ml
0 A
my
“2
z 2
P
Mean.“
2 L+2a
qL )1 PL
l1,1:— —+ _—_.
2 2(2 " 2
— P (5 «f—ﬂ
+21 2 2 b=300mm
a=500mm s = — = 3.125 x10" m3 h: 250mm Substitute numerical values: M1 = 17.500 N m
Mm M2 = —21,375 N  m
= 21.875Ntn MAmIM BENDING STRESS a,“ M 21,875N'm =—m‘““=—"—m~—“—=7.UW 4—
s 3.125x twin3 3 (Tension on top: compression on bottom] Problem 5.51'9 A beam ABC with an overhang from B to C supports a
uniform load of 160 lhfft throughout its length (see ﬁgure). The beam is
a channel section with dimensions as shown in the figure. The moment of
inertia about the z axis (the neuual axis) equals 5.14 111.4 Calculate the maximum tensile su'ess a", and maximum compressive
stress a} due to the uniform load. Solution 5.519 Beam with an overhang AT CROSS SECTION OF MAXIMUM PomNE M] BENDING MOMENT
M 13,500 lb: . 2.496 ' . _
0 _ a: = 1‘72 ___ W = 6560 p51
1._, ‘7’ 1a 5.14 m.
335 n M l3 5001b: . 0.674 ' .
“1 m= [‘C'=———( ’ 53.”. mar/70951
1: = 5.14 in.‘ = ' ""
Cl = 0‘6“ “1' 02 = 2'496 111' AT CROSS sacnon 0F MAXIMUM mam
RA. = 600 lb R8 = 1300 "3 SENDING MOMENT
M = 11251bﬁ=135001bin. . .
1 ' M 24, ll: . 0.674 . ,
M2 = 20001011 = 24,000 10111. 0', = 10' = = 3,150 1331
I, 5.14 tn.
M262 (24.000 lbin.)(2.496 in.) ,
0‘ 11 5.14 'm.‘ pm MAXIMUM 511153555
0" = 6,560 psi 0} = 11,650 psi 4— Pfﬂlllﬂl'l‘l 5.53 A cantilever beam of length L = 6 ft supports a uniform
load of intensity q = 200 lblft and a concean load P = 2500 lb
(see ﬁgure). Calculate the required Section modulus 5 if awn“, = 15,000 psi. Then
Select a suitable wideﬂange beam (W shape) from Table 51, Appendix
E. and recalculate 5' taking into account the weight of the beam. Select a
new beam size if necessary. Solution 5.63 Cantilever beam P=25001b q=2001blft L=6ft TRIALSECT‘ION W8><21
Cam 2 15'0"” Psi 5 = 13.2 in.3 qo = 21 1er
2
REQUIRED SECTION MODULUS Mo = ‘30 L = 373 "311 = 45351b1n_
(#3 2
M = + — = . ~
m“ PL 2 15‘000 lb h + 3'6“) !b R Mm = 223.200 + 4.536 = 227.7001bin. =18,600 lb—fl = 223,2001bin. R ’ d S _ Mm _ 227,700 “rm _ 15 2 I j s = Mm: = 223v2m1h'in=14 88 in 3 alum “um... 15.000 psi _ I m
Vﬂhw '5'090 PSi 15.2 1.1.3 < 13.2 in.3 Beam is satisfactory. Use WSXZI ‘— ...
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This note was uploaded on 03/15/2012 for the course CIVL 000 taught by Professor Kk during the Spring '10 term at HKUST.
 Spring '10
 KK

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