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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. MDWBId' 04‘ interha. «Juau‘l‘ xm‘s I, r {gamma}: + (magnum?
= ZSﬂSTZxfo‘ mm" Ix = 7'; (3)040“ = 510m 910‘ m." I3 2 I, = zs'ﬂsrzxro‘m’
I = I. + Iz+la ‘ xiv»: xlo‘ m? = gaﬂwxro“ M"
wHL 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 Nm 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. : ﬁLﬁfrAA,‘ = {50.530.5134 (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."! Idlem
BaaHm: +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? : 30l75 :I?.S may; 9 C3. OILSM 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 .IlS‘mm above ‘Hze boﬂbm. .3 Yin? ' 47.5 mm = 0.0I75'm M: [06.l NM ‘ 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 Xgf  (:2.:64x/03)(400—x3 = 0
“As—J 125x2+ 12.16'H10’x — 4.3657 “0‘ = o — ma; >403 + (IQ144x16)" + (4)(:25)(4.36571g n96
SoiVinj ‘For x x _— W (21025)
x = $4.55“ M.) 'two—x = 2454.51... I = :55 250 X“ + (ILIChzoSquo—n"
= 3 (25°)! L’sikssia + (:2. :64 x:03)(145.45 )1 = 1.04m we" mm” = Lollrotlxlod’ mi'
5 2 ;m
l
(0.) Shel]: =  245.45 Mm = *0.2'~i$‘4$"m
d
6: _ (8.0)(I7SXI0‘) —o.24545) = 330x10“ “Pa. = 330 MPcL 4 (_ 01‘04 x [03 (bi Cohcrez‘IeI “V: 13.9.55 MM = CLUE455 he =__ (toitr75xm3)(o.rﬁss)
[.0404 >6 {0'3 = —2£,O x106 Pa, = 26.0 “Pa, «a ...
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 Spring '07
 nesteranko

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