# 8 - Prob 6.5-6 A steel-reinforced timber beam has the cross...

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Unformatted text preview: Prob. 6.5-6. A steel-reinforced timber beam has the cross section shown in Fig. P6.5~6a (see Fig. P65-5a) 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.’ [7:51Af-61? 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/ﬂb-n‘] {axe/11%” (9, an») 'A227K/0")m‘ = 6/03 M: Aggy/lo’a‘jm4 : Mp“ Prob. 65-10. 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. 65-22. The cross section of a steel-reinforced con- Y Y crete beam of width b is shown in Fig. P6.5-22a. 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. 965-22 'l' ri '. Cross Gecizion (Xij Plane) tjI Itaamxi \$25“? F5 ' Pa = 0 M (lg/5 F6 = Fr. x 2% her/ﬂow: M- FJZ—é’) nigh-ck o h‘cc for iinwltj elasi-Jc 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 helix-bike, Conner-elm 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: P6.7-3, P6.7-4, and P6.7-18 m5 lb-zl:Xln-2l:)5 ﬁll/1.02M); 7-ln.(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 + (la-29+. _ 5in.§5§n222 + 5.15'1n.§1-in.)(o.6fn ) ' (Sirlle + Win-(0.5M) =El . 510’8'.(l'.-'+'.', P: Cﬁ—éA—(ﬁic+ at) -= kg m 2; ( m "m (4.o|52in. “i (1.052an = 2155 kjg-ln. 13: %FY ’45 kl m __ ’2'8 " 2255.5kiP-in. M My = 2.250 kip' in. MP = 2750 kipﬂﬂ- 1*? = Leis Prob. 6.7-9. The structural tee section shown in Fig. P6.7-9 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. P6.7-9 and P6.7-10 _ _2_7‘L_A_ (d-tF/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 {aid-3‘15 01-44 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 [gig-in, 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\$? 52lp|45‘ z __r__ . 25L, d? dc: 2b,». = 2(%.02’m.) = 0.2Q555 m, 255- ., (MM —2a + t— +432 _ dt=| ZAt i: At =L2|4illm A 5bk5’(5,[email protected], _ ' ’ MP: gé’idu A; = I 2, (0.2035501, + I.2|o4 m) = “25.61% klg'ln. Me lb5ﬁ‘ﬁ kigin. _ i9" MY = i‘ZFlep 'P'm. ‘ I.%le5% My = “6‘40 kip-in. MP = Hobo klp'in. 1? = |.3|ola *Prob. 6.7-21. A beam with rectangular cross section of width b and height h, as shown in Fig. P6.7-21a, is made of material whose stress-strain curve (in tension or compression) is approx- imately the form shown in Fig. P6.7-21b. 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) P6.7-21 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 ow-Lzrmosle 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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## This note was uploaded on 04/28/2008 for the course CVEN 3161 taught by Professor Xi,yunping during the Spring '06 term at Colorado.

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8 - Prob 6.5-6 A steel-reinforced timber beam has the cross...

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