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Unformatted text preview: Prob. 6.56. A steelreinforced timber beam has the cross
section shown in Fig. P6.5~6a (see Fig. P655a) and is used
as a simply supported beam to carry a uniformly distributed
105333; at intensity w = 2.0 kN/m, as shown in 16351563
6b. Determine the maximum normal stresses in the steel and the wood, (on), and (am)... under this loading condition.
The ratio of elastic moduli of the steel and wood is 5,! E.= 20. 352%»: 1'4: fie max/m um f/rerrff, W
f M“ / ” ” 
(ml, =0: a ,4} = (M 9,
(ml, ro:'—> a, = rm mm = m6 .154” my». )
— {5m} //.me) = F/fyum/ r/V/r/c'r.’ [7:51Af61? 5
M} Wand/Z 40W”): /o /, 26’ //0‘)mm" 1:362; = //Z)/4 )1 ‘ : /40m~ [6 M“ , 1?" /Z ’1; /z 1
= 5r. BOW/00mm" 5’51 I‘M]; )5 Eu [Biff/09)!»ij :50 zzﬂ/ﬂbn‘]
{axe/11%” (9, an»)
'A227K/0")m‘ = 6/03 M: Aggy/lo’a‘jm4 : Mp“ Prob. 6510. A simply supported beam spans a length of
L = 4 m and canies a unifome distributed load of magni tude w. The cross section of the beam consists of a timber 240 mm
beam (E, = 10 GPa) to which is attached a steel bottom plate (E, = 210 GPa). The z axis is the NA. (8) If the distrib uted load is w = 4 kN/m (including beam weight), ﬁnd the 6 mm
maximum stresses in the wood and in the steel, (am), and "F (mm), (b) Determine the radius of curvature, p, at the
midspan. (b) {a} Mix/mum r/n'rrtf, w[ filo/I érr'am / A ' ML: MAM, Wm ) 1 MW" 6’7: T— 9%
= gé/U‘M LOté’ﬁ'ﬂn a/A/ﬂ.’ [M m, +5734=o w/ZL 5/[ffw 'lzann){/40mm)(l¢0ktm) _
fil/(w'34i’mmil/40mml/émmn ‘0 E (u : 6, = Z¢6mm va‘ 53.656“. WC/ﬂi/a/ ﬂail/a] mire/fizz!
.57: 51L; fﬂfg 3 z, MM 3
[w = / 0/21“ fﬁu/O/u )= Wfﬂ40mm)/Z¢O “Mammal
= ZZ/.5))’//0“jm1i J
I = 635114047; "Wm/J6“) ,//¢ommJ/6mnl/5o,crmm]‘
3’ / 6 4
= 5467//0 )"M (canﬁ 0r) nen‘ p476} #45. 6,f—/0 max .57 = 53/1,, m1; ) = 51334. 337/0 ")M]
Mm E,“ a,  (gmmMa, 42:4,") MIN)” 5 Ugo/a = 5? ‘ 336,33I/0")M‘ ‘ 156/37“
WWE, (—4) —MM§§&)
(dz"(L E 0170f : 7—: 39’633 (10“ )M"
= 4/, 756/17/4
(0:14)” " (an): : 415M/jé
{K} ﬂit/Mr 0/ Ca/Va/ﬂ/C’J/ f.
ﬂan? ZZ’
’4 _ ﬂ; _ 5 _r J—
/? ' 5f " (mm )[if6,>’i//0"/M‘] aim/’0 ) ““
ﬂ —' 420. 42m lﬂ=420m I""Prob. 6522. The cross section of a steelreinforced con Y Y
crete beam of width b is shown in Fig. P6.522a. The allow
able compressive stress in the concrete is a", and the allow
able tensile stress in the steel is (7,. The diameter of each of
the three steel bars is d, and the ratio of the moduli of z
elasticity is EJE,. Assume that the area of the steel bars is
concentrated along a horizontal line at a distance of h from
the top of the beam, and assume that the concrete is only
effective in compression. That is, tension is carried solely by
the steel bars. Calculate the maximum allowable bending
moment for the beam. Let b = 12 in., h = 16 in., d = 0.875 area effective
in compression area effective
in tension in., (am), = 1.5 ksi, (am), = 18 ksi, and 5,15, = 10. 96522
'l' ri
'. Cross Gecizion (Xij Plane)
tjI Itaamxi $25“? F5 ' Pa = 0
M (lg/5 F6 = Fr.
x 2% her/ﬂow: M FJZ—é’) nighck o
h‘cc
for iinwltj elasiJc material behavior,
.5 Es h—CJ
0": TV" 1 E c 65: r @011th (511?!"
Fe = SiOZiciA" Z( L? (Mch “136:433': f 4
iz(E‘C magi: é'7Q( If )
E5 3 2 2h ﬂ; = 5
cf+ «EEHMC; KEY ‘f i o T
Cf + (sooneimcc 4%, lob inz' = o ‘ C; T
—E Earl—ca
(093M: ’TC E (69mm=—_f in L
m=gﬂ_ 13—53 1,2
Noam! Eci c:c i‘ loibmémgeisn. m) = l_<3_.Jag_'L (0’) ' ow k ‘ . .
Since iO’SZiiow = l.6l:i 2' l2 < l8,bo4, {he Sized ranrorccmenize Wili reach allowable stress limits helixbike, Connerelm will , So
MAJIOW. 3 g; (ll—Ca;
= (minnow i (h— $33119, )
= \8 ksi + (um. 5.5‘15lpinﬂ Mallow = 45‘! klp' In. Prob. 6.7 A rectangular box beam with height h = 12 in. and
width b = 8 in. has a constant wall thickness 1 = 0.50 in. It is made of structural steel with 0,, = 36 ksi and E = 30 X 10’
ksi. Calculate numerical values for the yield moment My, the
plastic moment MP, and the shape factor f: P6.73, P6.74, and P6.718 m5 lbzl:Xln2l:)5 ﬁll/1.02M); 7ln.(llin,73
I = '2 “ ' = I2 ‘ _2 =515,_5_8_Lni 2
(:11 _ abksus? .58'm‘n _ . .
MT: C ' 12in. ‘ﬂm Since. ‘l'Jae CroSS aeolian is GlOl/Llollj eﬂmrndmic, ‘Hne NA docsn‘l: move. 80
M ( «Wt/gig + (“Elk—2m
dt= do = 2A‘ ‘ ht/zl'z + (la29+.
_ 5in.§5§n222 + 5.15'1n.§1in.)(o.6fn )
' (Sirlle + Win(0.5M) =El . 510’8'.(l'.'+'.',
P: Cﬁ—éA—(ﬁic+ at) = kg m 2; ( m "m (4.o52in. “i (1.052an
= 2155 kjgln.
13: %FY ’45 kl m __ ’2'8 " 2255.5kiPin. M My = 2.250 kip' in. MP = 2750 kipﬂﬂ
1*? = Leis Prob. 6.79. The structural tee section shown in Fig. P6.79 has
the dimensions d = 5.05 in., bf = 8.02 in., I, = 0.350 in., g, =
0.620 in., and it is made of steel with 0'}! = 36 ksi and E = 30 X
103 ksi. Calculate numerical values for the yield moment My, the
plastic moment MP, and the shape factor f. P6.79 and P6.710 _ _2_7‘L_A_ (dtF/zlbtl‘f dim; a—H
7t= 2A —' 'Jﬁtfi' Wei—W F _ =47+4in14a 2 i 2) +z.zusin. .‘I'+2 in2 + o, m. . '51m'
= 4,153? in:
bg‘trs {aid3‘15 0144
I: I2 + b;tg(d%'7’j+ 12 +£«(cl—‘ql( 2 “7&2
_ 3.921n.(o.o2m.)3 + _ _
 :2 8.02 \n. (o. u21n)(4.‘—H in. — 7Q?“
0,35in.(‘l.’~5'm.)5 _ _ _
+ :2 + 0.35m.(4/~}3in)(2.215IVI.— n)‘ (31: ’ , ’ ‘l My = 7'» = 5b 31.124125“. )‘ = mama [gigin, In Jche ¥Knlj plasLic, %&,the Plasizic NA is locaieed b5
I A: = A1; = éfbftf + tw (0i 2E2 .02‘m.(o. bzin.) + 0,65 Mill/45 Inﬂ . 3.gg£35’mz
i v3 PM AC: 215$? 52lp45‘
z __r__ .
25L, d? dc: 2b,». = 2(%.02’m.) = 0.2Q555 m,
255 ., (MM —2a + t— +432 _
dt= ZAt i: At
=L24illm
A 5bk5’(5,2lpl453@2, _ ' ’
MP: gé’idu A; = I 2, (0.2035501, + I.2o4 m) = “25.61% klg'ln. Me lb5ﬁ‘ﬁ kigin. _
i9" MY = i‘ZFlep 'P'm. ‘ I.%le5% My = “6‘40 kipin. MP = Hobo klp'in.
1? = .3ola *Prob. 6.721. A beam with rectangular cross section of width
b and height h, as shown in Fig. P6.721a, is made of material
whose stressstrain curve (in tension or compression) is approx
imately the form shown in Fig. P6.721b. Let on = a). on =
$03,, an = 5,, and 5" = $6,. (a) Determine an expression for
the moment that causes ﬁrst yielding, M“. (b) Determine an ex
pression for the moment Mn that causes the outer ﬁbers to reach
the second yield stress, on = 03,. (c) Determine an expression
for the fully plastic moment, M p, the moment that causes all ﬁbers
(exfoept an inﬁnitesimally small core near the neutral axis) to
reach the second yield stress. (a) (b)
P6.721 a) — ' MY. is Jche moment Jgnod: Causes the outermost/ﬁber tojusl: tjleldﬁo
.I 3&(%l Grth Mm = C = _7772—— = ‘3 b) ~ ’
' Myg is the mom em: that wees the owLzrmosle fiber to read/1 6L sizress
o? 0’72, = CY. Note: strain distribution remélems linear, ‘lla
C
moment hm}, wees all ﬁbers to reach analyzes 0F 5Y1 = Gy.
ﬂ Wamﬂt Howrth
——
c) '
' P is the
GYzA
Mp= 2 (dude) = ...
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 Spring '06
 XI,YUNPING

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