Unformatted text preview: Problem 5.5“! A compound beam ABCD (see ﬁgure) is supported at
points A. B, and D and has :1 splice (represented by the pin connection)
at point C. The distance a = 6.0 ft and the beam is a W 16 X 57
wideﬂange shape with an allowable bending stress of 10,800 psi. Find the allowable uniform load 9'le that may be placed on top of
the beam. taking into account the weight of the beam itself. MW: 2 =61]th _. 20.,th I
I gm = 5a: qrallow = gm — (weight of beam) fim— din—of DATA: a = 6 ft = 72 in. alum = [0,800 psi
.114“! RE: 459:1 RD: 2% W 16 x 57 S = 92.2 in.3 34—1— Aunwnnua UNIFORM LOAD u 200,800 psi)(92.2 in?)
gm" 502 in.)2
= 9221bift quill"W = 9221b/ﬁ — 571blft = 8651blft +— Pin connection at point C. = 76.833 lbr‘in. Prnhlem 5.35 A steel beam of length L = 16 in. and crosssectional __ 3;: 2393??!9; dimensions b = 0.6 in. and h = 2 in. (see ﬁgure) supports a uniform load 1 i i E I of intensity q = 240 lbfin., which includes the weight of the beam. _
Calculate the shear stresses in the beam (at the cross section of maximum shear force} at points located 1M in.. “2 in.. 3M in.. and l in. I I I ' ' ' b = D 6 in from the top surface of the beam. From these calculations. plot a graph — ' ' ' showing the distribution of shear stresses from top to bottom of the beam. Snlulinn 5.05 Shear stresses in a simple beam Distance from the yl 1
top surface (in) (in) (psi)
0 1.00 0
0.25 0.75 l050
0.50 0.50 1800
0.75 0.25 2250
1.00 (NA) 0 2400
v 9:2
54:]. (5—39): 1' = —_y31)
GRAPH or SHEAR STRESS 'r
L 3
V=%=l9201b f=%=0.4in.‘ o
1050
UNTts: pounds and inches 1800
2250
192“ (2)2 2 NA 1' 4400 '
= _. : 4 _ . . .. p5:
T 2(o.43[ 4 n] (2 00m y?) 233:3
(r = psi; y. = in.) 1353? Shear SlfBSSES ill the WOIJS [If Beams Willi Flanges Problem 5.101 Il‘ll‘lluﬂll 5.195 A wide—ﬂange beam (see ﬁgure) having
the cross section described below is subjected to a shear force V. Using the
dimensions of the cross section, calculate the moment of inertia and then
determine the following quantities: (a) The maximum shear stress TM in the web.
(13) The minimum shear stress 7min in the web.
(c) The average shear stress 1' (obtained by dividing the shear 8"“ force by the area of the web) and the ratio Twirlm. (d) The shear force th carried in the web and the ratio th1 V. Note: Disregard the ﬁllets at the junctions of the web and ﬂanges and
detemﬁne all quantities. including the moment of inertia. by considering Probe. 5.101 through 5.105
the cross section to consist of three rectangles. Prohle 5.101 Dimensions of cross section: b = 6 in., r = 0.5 in.,
h =12in.. h1=10.5in., and V = 30 kt Solution 5.101 Wideﬂange beam (b) MINIMUM SHEAR STRESS IN THE wen (Eq. 548b) Vb
Tmin = — h: " Hz) = Si ‘—
b = 6.0 in. 8k< 1 p
I = 0.5 in. .
g, = 12 o in (c) AVERAGE SHEAR mess IN THE wen (Eq. 550)
. I. v
hl = 10.5 In. Tm, = __ = psi ‘—
V = 30 k zhl
7"“ = 1.0M «—
MOMENT OF INERTIA (Eq. 547) Tam I = Lam: _ bk? + m?) : 333 4 in 4 (d) Sam FORCE IN THE WEB (Eq. 549)
' ' If!
[2 vm = f (2%,. + rm...) = 28.25 R 4— (aJ MAXIMUM SHEAR STRESS IN THE wen (Eq. 548a) m
—— = 0.942 «— Tmln = _ + = psi 4— v BuilIUp Beams Prnhlem 5.111 A prefabricated wood Ibeam serving as a floor joist
has the cross section shown in the ﬁgure. The allowable load in shear
for the glued joints bﬂween the web and the ﬂanges is 65 lbﬁn. in the longitudinal direction. Determine the maximum allowable shear force Vm for the beam. Solution 5.111 Wood lbeam All dimensions in inches. Find Vm baSed upon shear in the glued joints.
Allowable load in shear for the glued joints is 65 lbfin.
'. flu“ = 65 Ibfm. f=ﬂ V’m =‘(ﬂ
1 Q
_bh"' (la—ma _ 1 3 1 ,
 12 I2 r~ 12 (5)(9.5) 12 (4.375K3)
= 170.57 in.4 Q = Qnmﬂ = Afdf= (5)(0.75)(4.375) = 16.406 in.3 Jun“! _ (651b/in.)(]70.57 in.‘)
' Q ' 16.4mm} Vm =6761b 4— ...
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This note was uploaded on 03/15/2012 for the course CIVL 000 taught by Professor Kk during the Spring '10 term at HKUST.
 Spring '10
 KK

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