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kw}? wag/3?: $2.,» 5; 384 . CHAPTER 7 Transverse Shear A steel Wideﬂange beam has the dimensions shown in Fig. 7~11a. If it
is subjected to a shear of V = 80 kN, (a) plot the shearstress distribution
acting over the beam’s crosssectional 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(10f3) 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 13 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. 711e. 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 ofth'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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 Winter '06
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