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Unformatted text preview: 4.43 and 4.44 Wooden beams and steel plates are securely bolted together to form
PROBLEM 4.44 the composite members shown. Using the data given below, determine the largest
permlsmble bendmg moment when the oomposnte beam 1s bent about a horizontal 5 X % in. 3X13. \ Wood Steel
’_ Modulus of elasticity: 2 X 106 psi 30 X 10‘ psi
Allowable stress: 2000 psi 22 ksi Use Wood as ‘He reference md‘ew‘aJ i SOLUTION
0 in. ____ n: L0 in wooJ Ljﬁii h " Es/Ew = 30/1 =15 in s+eeﬂ
For H: +V‘amS‘FOV’meci secHon
I' : %—b.k.3 + nlA. A'Z
iii(53%? +(’5X~’>’)(J=z3<5.25‘)z = 1054.4 2N = ibk1=JL9—(Gﬁ(lo)s= 500 in“ l WWI: n=r.0, yrs—in J 5:2000fsi
M = W = I.028 we" J19 3v. Stet)? n = IS; 3: 5.5 .'n ) 6‘ = 21 la: = 22v/o3 )osi
M=——————(‘(f;'§’:5)_€:i‘q) = 635w A»; Cheese ‘He slam/Her \Mjue M = 685*“): IHn = €85 kipin «I \ PROBLEM 4.47 4.47 and 4.48 A 6 X lOin. timber beam has been strengthened by bolting to it the
steel straps shown. The modulus of elasticity is 1.5 X 106 psi for the wood and 30
X 10‘ psi for the steel. Knowing that the beam is bent about a horizontal axis by a
couple of moment 200 kip'in., determine the maximum stress in (a) the wood, (b)
the steel. SOLUTION Use vocal as Hie V‘E‘reJ‘Ech manic/3:.)
Y): LC in M1009 h: Es/Ew = 30/15 = 20 RV 5‘53?) mes‘rofmea Seoﬁ 0‘4 The neu‘iva) axi's jigs 3.][4 in «hoof +b€ bo‘H’om.
1. : gfwf i mag}: ‘Ti9(e )(Jo)3+ (I.o)(4c»)(2.3izc)l =
11 '4 Trix353%: + nzAzo’a.‘ : 72:0.(5)(_;.)3+ (20)(2.5)<2_864)1 I= I.+ .121 = 12528 2.1+ (o3 wood: n = [.0 3/ = 10.5 3.114 = 7.336 .‘n : '1‘wa t _(I.05(200)S738‘) : __ ll7q “S’ I. (H S'l'ee) 1 n:
um
I l252.8 29 j = — 3.”“f in : — 20 20°) '3‘,” : 97L} ksf l2$28 \‘ PROBLEM 4.53 centers as shown. The modulus of elasticity is 20 GP; for concrete and 200 GPA for steel, determine the largest allowable positlve bending moment in a portion of slab l m wide.
16—mm diameter SOLUTION Considevéfec'f’IOn 180 MM male WW one sﬁJraA.
As : col = %(ls)" : 201.06 mm" HA5 r Z.OI06 No” my?
Locate. H12 nea‘f‘v‘ﬁi) ands
I30 x—z— — ('IOOx)(2.0I06xto3) = o Clo x1+ 2.0:04xlo3x ~~ 201.06 V103 =0
X _ g.o40(—:.>(lq3 +5]£2.0ng10332+(‘+)(7o)(20:.oe x10 ) (23070)
x = 37.377 Mm ) /00x = é2£03 mm Sch/{HS 4m x I = Jg,(l§3¢:~)><'s + (2.0106woﬁﬂoonx)z = 'MlsOXBZ 37.7)3 +(2.0106 “03)(92303 )2 = “.012 was mm” = 11.0/8x/o"m”
 nM . ‘ g;
lam—£11 .. M w Comorz’l’t : V) =I J y = 37. 3?? mm '3 0.0373‘77 m J 6":qVIO‘Pd. M = (6/ wo‘Xmoxs x10“)
(1.0)(033'2397) shag}: n : to J 3,: 62,603 m. = 0.0mm m) 62 Izox/O‘ Po. t 2.65:; x103 NM M _ (no 110°)(H.0l3xlo—e)
 (10)(O.062g03) ~ 2!120 1103 NM \CLooSe Ht: sound/er Vague M = 2.!!20 x103 NM We above is Hm MIOWaUa Posih've momen‘l’ For a. IZO 1mm wt‘Je .seo'l'u‘ou. For a. [m t 1000 m MAM) MUJHPL, L7 ’33)" = 5.554 M 2 (5.556)(2~11Q0x103) = 11".73><Io3 vi = H.73kHm «a ...
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 Fall '09
 MOORTHY

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