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LAAJ inn! in LA L._U_J’ L, :I l ngj L _i__t i _1 Lnnnti L. _.. .] L _.U L L_4._n 7 5.117 through 5.120 (a) Using singularity ﬁmctions, write the equations deﬁning the
PROBLEM 5'“ shear and bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam. ‘25 kN/m SOLUTION 33 SJMML'NJ 1'?A '' 9., 4mm
toil—F: co P,+F?o  40(i.8)(2$)«‘¥o = o 0.6m ' 0.61:: j?“ _.. 12° .. G2.5 W W = mot0.6)”  2502."? =" "if
V 62.5  25<xo.c)‘ + “(hawwanes)“tact—2.9)“ kN ‘3
M = 62.5x  i2.5<xo.t)‘+ t2.s<n—2.'4>zI+o<2«My Ho <¥~2Q>l Ridn an
LOCdePoia‘i vixen V e 0 . Assume. 0.6 < x‘ 4 L9 :0 = ézts masQuad + o  4c:  o x‘= 1.5 M M = (enema—owes? +o  worm) o = 41:25 wm «
5.117 through 5.129 (a) Using singularity functions, write the equations deﬁning the PROBLEM 5113 shear and bending moment for the beam and loading shown. (1‘)) Determine the
maximum value of the bending moment in the beam. SOLUTION By Symme'i‘rj Ra ’ 9:
+7 2: ,0 RA+QL(L2)(L2S) 1.8 —(t2)C:.zr)= 0
RA ' Q; = 2.“! LIN L25 — L25<x~L2>° + 1.25 02.4? —5!,‘:’ I o
v=—t.25x +l.2$<x.L2)' —1.25<x_2.q) + 2.4 [.8 (X‘L3> w 4 M =‘O.£ZS x' + 0.625 (x—m)‘ « o.szs'(x2;q >‘l
+ 23 x  1.8 <x L8) kidm M...“ occurs a"! X= 18». M... ”(cashier+(oez§)(o.s)‘ + o +(2.4)(:.9)  o = 2.52 w.M 43's? am: ...
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