Unformatted text preview: 71m 7—1
g r“: [:1 5.33 Determine (a) the distance a for which the maximum absolute value of the PROBLEM 5.33 bending moment in the beam is as small as possible, (b) the com: 11 . xim
normal mes due to bending. 8‘” dmg ”‘3 “m (
20 IcN 40 IN (Hint: Draw the bending—amen: diagram and then equate the absolute values of the largest positive and negative bending moments obtained) I SOLUTION g W360X64 Readi'ion 4+ 8 'QZMC : 0
20a. (2.4)(uo)+(1+.o)8 = o
8* 24  5' :1. Bending manner} 0:} 'D ‘
GM 3 93MB : O 3
53 ~ + 1.
L‘ 8 MD: La. 8 r 33.4  8a,  Beneiinﬂ momewT ad' C 2° M DZM¢=O '
D 20a+M=O ‘ j
[*0 v M==20a. { E30538 "" Me 1' Mg (
20 Q : 33.“ " 80.
Q. '' L37!” m 4
Mc 7 47120: mm MDa 21mm kNM ; For W 360: 6‘! “Meal size! sec‘l'i'on 5 = IoSox [03 m3
= loaowlo“ m‘ i . ‘ 3 .\
Normu? dress 6‘= lgl ‘4' W ‘' 26.61:!0‘91 ‘' 26.6 “Pa. 4 l ...
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