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**Unformatted text preview: **M Qﬁfgmasgg Sggag éfgggg Iaa “5% {E} : mag gmégg 2g fgg‘ggggggg ggggg
{2} Egg: ggggg m gag; §gaagé W31 3532: agga gggggg g “8?; 3 a; ‘3 €><2gaga$g % ES {:23ng gaggss, mm f3; “2an 5%?“ ésaéW’? 4%; sigma; {Emmy 25k; {533% gag”? é f
$ {a ’1 , z»
kw}? wag/3?: $2.,» 5; 384 . CHAPTER 7 Transverse Shear A steel Wide-ﬂange beam has the dimensions shown in Fig. 7~11a. If it
is subjected to a shear of V = 80 kN, (a) plot the shear-stress distribution
acting over the beam’s cross-sectional area, and (b) determine the shear
force resisted by the web. (b) (a) Solution _ Part (a). The shear—stress distribution will be parabolic and varies in
the manner shown in Fig. 7411b. Due to symmetry, only the shear stresses
at points 8', B, and C have to be computed. To show how these values
are obtained, we must first determine the moment of inertia of the cross-
sectional area about the neutral axis. Working in meters, we have 0.02m ‘ 1 =_ [117: (0.015 m)(_0.200 my] _ k—e—oaoomlw—di
" T i 3’ 0.100m ‘ + 2[‘1—12-(0.300 m)(0.02 m)3 + (0.300 m)(0.02 m)(0.110 mm] = 1556(10’6) m4 For point B’, (3’ = 0.300 m, and A’ is the dark shaded area shown in
Fig. 7—11c. Thus, 0 ' QB: = 7’A’ = {0.110 m](0.300 m)(0.02 m) = 0.660(10‘3) m3 ' (0) Fig. 7—11
so that
my _ 80 kN(-O.660(10“3) m3) .. = 1.13 MPa
1th 155.6(10’6) m4(0.300 m) TB! 2 For point B, t3 = 0.015 m and Q B = Q35 Fig. 7~11c. Hence VQB 80 kN(0.660(1-0f3) m3) _
= —-——- = W = 22.6 MP
73 1:3 155.6(10’6) m4(0.015m) a SECTION 7.3 Shear Stresses in Beams . 385 (”Note from the discussion of “Limitations on the Use of the Shear
Formula” that the calculated value for both 73 and 1-3 will actually be very misleading. Why? ,. 002m
For point C Q; = 0.015 m and A’ is the dark shaded area shown 1111 1-‘-_——O.'300m—‘—‘”‘
, Fig. 7~11d. Considering this area to be composed of two rectangles we “L : have A’
QC = Ey’A’ = [0.110 m}(0.300 m)(0.02 m)
‘+ [0.05 m](0.015 m)(0.100 m)
= 0.735(10‘3) m3
Thus, (d)
" VQC _ so kN[0.735(10’3) m3] '
E: ” 155.6(10‘6) m4(0.o15 m) = 25.2 MP3 TC=7 = 'fyr’rPart (b). The shear force in the web will be determined by ﬁrst
3 “formulating the shear stress at the arbitrary location y within the web,
‘_ Pig. 7-11e. Using units of meters, we have
I = 155. 6(10'6) m4 w ‘
t = 0.015 m , ‘ ’ = (0.300 m)(0.02’m) + (0.015 m)(0.1 m - y)
Q = Ey’A’ = (0.11 In)(0.300 m)(0.02 m) + [y + am 111 —- y)](0.015 m)(0.1 m —- y) = (0.735 — 7.50 y2)(1073) m3 . so that I i ‘ (e)
" __ 2 ~3 3 ‘
7 = 1/2 = so kN(O.735 7.50 y )(10 )m mg. 7.11
1: ((155.6(10‘6) m4)(0.015 m) = (25.192 — 257.07 yz) Me This stress acts on the area strip dA = 0.015 dy shown at Fig. 7~11e,
and therefore the shear force resisted by the web is 0.1m
Vw = J “1.4 = I (25192 ~ 257.07 y2)(106)(0.015 m) dy
A", .O.1in Vw = 73.0 kN Ans. Note that by comparison, the web supports 91% of the total shear (80 kN),
whereas the ﬂanges support the remaining 9%. Try solving this problem by
ﬁnding the force in one of-th'e flanges (3. 496 kN) using the same method.
Then Vw = V - 2V,» = 80 kN — 2(3. 496 kN) = 73.0 kN. ...

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