Unformatted text preview: 5.97 Determine the largest permissible Imifonnly distributed load w for the beam
PROBLEM 593 shown, knowing that the allowable normal stress is +i2 ksi in tension and  19,5 ksi
in compression. 5. 98 Solve Prob. 5.97, assuming that the cross section of the beam is reversed, with the
ﬂange of the beam resting on the supports at B and C. SOLUTION
Rem/haw; E+C~3éw=0 B=C=12w Shear? VA=O
V5: 0 —8w e 8w
V3=3w+l8w=loiu
‘19 r iOvu 20w * JOw
V3 —iow+l?w= 8w V°=~SWv8w= 0 Areas?‘ A +o B (HUN12w) wszw
'3 +0 5 (“Gt)(toiu') = 50w Bewoir'nj momen‘l’s: MA = 0 MB = O‘3Qw 232“!
Ma r3QW+SOw = iﬁw Cevi'i'mi‘A and manned a‘Pincr'Hm. _ Gas) A (if!) Ad‘tin‘) I Cw“ 0.7”"!
Q07‘1\ Bend in: momen+ (pin'+S Tension ad' and C ~(I2)(I.H3H33
Comp. d Band C (Hq;$a( 2.3mm
Tension a5“ E —(23(:2..3<l05)
Compression ad‘E (‘l?‘.5‘)(l.1l3'l3)  mm. mm. ._.
'%.6 ku'fin
'234387 k:r‘ in
27.4w tap3.. — :1 II 1 I] Alﬁoudh 1"“ W‘ 3‘ C '32“: =  17.2”» wa 0.531 "‘r ’i.
E '3 U "’ 27. qéq W LSSI} kip/fa Smﬂesi w . 0.531. kilo/in = us In}, In 4 ...
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