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Solutions5 - 4.3 Using an allowable stress of 155 We...

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Unformatted text preview: 4.3 Using an allowable stress of 155 We, determine the largest bending moment M that can be applied to the wide flange beam shown. Neglect the efi‘ect of fillets. MDWB-Id' 04‘ inter-ha. «Juau‘l‘ x-m‘s I, r {gamma}: + (magnum? = ZSflSTZ-xfo‘ mm" Ix = 7'; (3)040“ = 510m 910‘ m." I3 2 I, = zs'flsrzxro‘m’ I = I. + Iz+la -‘ xiv»: xlo‘ m? = gaflwxro“ M" wH-L c.- llama] = no mm = 0,1l0 w: wf'i'ln' (3d: 1'55“}0" Po. _ (54.944 no"\(|55'x 10‘) 0 no 2 80.1H03 N-m 0.5 in. 0.5 in. 0.5 in. “WWW?— A9,, ‘9 2.8m: 0.5615 5.375 v -: 34225 7 0.75%: The heIJ'l'N'o axis 19285 0.75 in. above lpo‘H'om. j“? '1'“ 2.0 "0.75- = [.25 in a jhf = ‘- 0.75- M. I. :- fi-Lfif-r-AA,‘ = {50.530.513-4- (1.253(05)‘ '-' 0.384351”? 1: = fibflhfiu Atop: -. fiz- (4.5)(0513 + (2.ul(o.s)‘ = 0.60%375 in" I = I.+ Ia .- 1.5”:3753...“ M 1614—le M = [$1 Tofu: QDMflr£55lon Mrw '4' 20."! Idle-m Baa-Hm: +ensrm M= LEW = 25,5 K94}. Problem 4.16 4.16 Knowing that for the extruded beam shown the allowable stress is 12 ksi in tension and 16 ksi in compression, determine the largest couple M that can be applied. Gomez He Shaffer as M414 Mm = 20.4 kip. in. «a Problem 4.18 X15? : 30-l7-5 :I?.S may; 9 C3. OILS-M I1 = T‘ihhn 9.01.2 = 7%(901(15)3+(.eoa)(3)2 = 25.25 #03 ma" I2 = 7172 bah}? A24: = 7‘5 (20305334 (300)00)‘ = 352619.605 m," I = I' + I; =- sr..31swo" my.“ = 91.3?5 x as)"‘1 m ‘* T' IG|= 11%] M = 5a ‘ _ 13F: +ensfou and: M .- L—M: 0522375340") 1' 113,3 NH”? 9 Bflfi'o" .: COHPmsl‘OH M = WEI) -_" 106;, Ndm .Dl choose Sawyer Val’ue The neaf‘my axis pics .I-lS‘mm above ‘Hze boflbm. .3 Yin? -'- 47.5 mm = -0.0I75'm M: [06.l N-M ‘ Problem 4.47 4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN - m. knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (5) the maximum stress in the concrete. E5 _ Zoo GPa. Y} = E. _ 2569a. 8.0 As 2 HL ital = (WEWRT = :.s_2c=z§><re:>3 m" mi5 = 12.154xto" mm‘ 5L_J:‘_;T 'Locwie 'Hne nw'irqf axn‘s ‘5 J50» 250 X-gf - (:2.:64x/03)(4-00—x3 = 0 “As—J 125x2+ 12.16'H10’x — 4.3657 “0‘ = o — ma; >403 + (IQ-144x16)" + (4)(:25)(4.36571g n96 SoiVinj ‘For x x -_— W (21025) x = $4.55“ M.) 'two—x = 2454.51... I = :55 250 X“ + (ILIC-hzoSquo—n" = 3 (25°)! L’sikssia + (:2. :64 x:03)(145.45 )1 = 1.04m we" mm” = Lollrotlxlod’ mi' 5 2 ;m l (0.) Shel]: = - 245.45 Mm = *0.2'~i$‘4$"m d 6: _ (8.0)(I7SXI0‘) —o.24545) = 330x10“ “Pa. = 330 MPcL 4 (_ 01‘04 x [0-3 (bi Cohcr-ez‘IeI “V: 13.9.55 MM = CLUE-455 he =__ (t-oitr75xm3)(o.rfiss) [.0404 >6 {0'3 = —2£,O x106 Pa, =- -26.0 “Pa, «a ...
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