Mechanics of engineeringComprising statics and dynamics of solids: and the mechanics of the materials of constructions, or strength and elasticity of beams, columns, arches, shafts, etc . that for a prismatic beam thevalue of ZQ is proportional to J, the total shear, and henceto the ordinate of the shear diagram for any particularcase of loading. The utility of such a diagram, as obtain- SHEAR IN FLEXURE. 289 edin Figs. 234-237 inclusive, is therefore evident, for bylocating the greatest shearing stress in the beam itenables us to provide proper relations between the load-ing and the form and
Mechanics of engineeringComprising statics and dynamics of solids: and the mechanics of the materials of constructions, or strength and elasticity of beams, columns, arches, shafts, etc . that for a prismatic beam thevalue of ZQ is proportional to J, the total shear, and henceto the ordinate of the shear diagram for any particularcase of loading. The utility of such a diagram, as obtain- SHEAR IN FLEXURE. 289 edin Figs. 234-237 inclusive, is therefore evident, for bylocating the greatest shearing stress in the beam itenables us to provide proper relations between the load-ing and the form and material of the beam to secure safetyagainst rupture by shearing. The table in § 210 gives safe values which the j*-3—maximum Z0 in any case should not exceed. It isonly in the case of beams with thin webs (see and 240) however, that Z0 is likely to need at-tention. For a Rectangle we have, Fig. 253, (see eq. 5, § Fig. 253. 254) b0=b, I=l/l2bh\ and fzdF=y2bh . j (Jo :.Zq=Xq=^ i , =|. (total shear)-4-(whole area) Hence the greatest intensity of shear in the cross-sectionis ~ as great per unit of area as if the total shear wereuniformly distributed over the Fig. 254. Fig. 235. e •-. ?fj -Hbt-\ Fig. 256. Fig. 257. 11 For a Solid Circular section Fig. 254 Ib0eJo nr2 4:r 4 J %7;r*.2r 2 3>r 3 tit2 [See § 26 Prob. 3]. For a Hollow Circular section (concentric circles) , we have similarly, 290 MECHANICS OF ENGINEERING. z- J Yr* 4r2 ~r22 4r2l ° ^(r/-r/)2(r-r2)L 2 3- 2 * 3ttJ _4 J(r^—ri) ~~3 7r(r14—rf)^—r2) Applying this formula to Example 2 § 252, we first have-as the max. shear Jm = y£P =1,735 lbs., this being the abut-ment reaction, and hence (putting tt = (22 -4- 7)) v ™o^ 4x7xl735[] OQ, „ A maX* = 3x22[256-i50]()-294 ** p6r B* ^ which cast iron is abundantly able to withstand in shear-ing. For a Hollow Rectangular Beam, symmetrical about itsneutral surface, Fig. 256 (box girder) 7== J#(Mi2-iW) =_3 _WjHV]_ The same equa
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