Kansas University quarterly . -x2)k .....,(, I) <iM. PI. II _Pia,p (,,_,.!(_Ar,f^i = ° (I2) Simplifying (12 we have 30] ^\ -75|Xi +4o|-i-^- +^5\-Y-\ -MJ-i-j-+2=o (13) One root of (13), found by trial is —j-=.22-|-- This root makesthe second derivative of (13) negative and makes Mx a this value of -j- in (11) we have (Mx)max=.o86Pl (14) Differentiating (5) with respect to x to find for what position of theload the moment of the load is a maximum we have =0 (15) From (15) we find 2I 3 X x] -nti + Differentiating (5) with respect to z we havef^lMr I 5k dz 1 2l3^ ^ ^6) 148
Kansas University quarterly . -x2)k .....,(, I) <iM. PI. II _Pia,p (,,_,.!(_Ar,f^i = ° (I2) Simplifying (12 we have 30] ^\ -75|Xi +4o|-i-^- +^5\-Y-\ -MJ-i-j-+2=o (13) One root of (13), found by trial is —j-=.22-|-- This root makesthe second derivative of (13) negative and makes Mx a this value of -j- in (11) we have (Mx)max=.o86Pl (14) Differentiating (5) with respect to x to find for what position of theload the moment of the load is a maximum we have =0 (15) From (15) we find 2I 3 X x] -nti + Differentiating (5) with respect to z we havef^lMr I 5k dz 1 2l3^ ^ ^6) 148 KANSAS UNIVERSITY QUARTERLY. Solving this equation we find Placing the value of z in (16) eqnal to that in (17) and simplifyingwe have ^°]xj -7°i-Ti +^°n-f -^^\i-\ -441x1 +3 =-0 (t8) One root of (18) is --^=.^16. Substituting this value of— in (17) we have z=r. 724I Substituting these values of x and z in (5) we have (Mr)max=:—.044PI (19) Hence the maximum moment occurs under the load; is positive,and equals . FIG. II. Case II. The moving load uniformly distributed per horizontalfoot. Let the notation be as in case one as far as applicable with w equalto the intensity of loading and x the portion of the span covered withthe load, measured from A. Proceeding as in case one we find that wx2 \ - X Vn=—7— and V=wx a i 1- 2I ( 2I both of which are independent of the shape of the rib. The moment about any point of the rib to the right of the load is ( x I Mr=VjZ— - z r wx;—Hy (20) The moment about any point of the rib under the load is Mi=V J z— w-—- —Hy (21) murphy: maximum bending moments. 149 From the equation of the x displacement we have rMiydz+ I Mrydz=o (22)O * X Substituting for Mj and Mj. their values and simplifying we have——I z2ydz-|-Vj I zydz—wx I -. z— ydz—H I y^dz^o 2»/q ^ o •^x 2) •^o (23) Substituting for y its value in (23), integrating and simplifying wehave wx^P wx^l wx5 =t51^H13 (24) 24 24 60 From (24) we find H: —5
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Keywords: ., bookcentury1800, bookdecade1890, bookpublisherlawre, bookyear1892
