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154 - img‘e beam with inclined load ‘ Ft P angle...

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Unformatted text preview: img‘e beam with inclined load ‘ . Ft P . angle between the z axis and fire neutral 0, - Jziq lb/m‘t » - We batman the yam; and the \m g = s m p mm a I 9 - 4w 130° tan e = tancm+I-so") -tan_og v) finding moments L— My= ‘21—'- = Q(9fnd21.2 . 8 (sq..- 5-45): temps}; tons a. ’- tmo =1: Baeo lb—in. 1’ 13' M; -046 =4(cosd)!3 T T— ‘05)“) =% JThe neutral axis lies along the other diaflOna . GED (— HM -h_b5=l4aln.12= bins :25!» in Neutral'2 our-'5 nn (Eq.;_____ 6- 23) tanp: g tane 21;; tanok flab-92 tam at {0.5425 p: 303°4— Marimum, tensile stress (fl point A ._ {12(4) =77zpsi4— Q3391: beam with fueling lggd L , 9.5% P- 5' k d - 9.6.51" Nfde “Flange beam: w 10:30 Zy216. 7 m? 1;: 170 In.‘ (1 =10 41 in b - 5.910 in. Mamet: of inertia. I, a h 3 = 5,213 i Io’mm.‘ I1 I, =bh’ :1: oq4tlo’mm‘ "Ir ’ Nggtml 9155 an gauges: tanp: : tune =33 tanox mmm 1 1y My: 43%;); = 51,0 so lb in. =f§f tafld. =4 {”190 a ’2. $044 M2 = M. a Hit/ego lb- in p a 55 5' ‘__ - Neutm} axis n'n (Eq. 6-132 LUMML m‘nin} (figs-ts) “"6 = g tats-1: mud , s 049:; 5-”; ’31:“) ~ ”If- (M 79.34 = W», 0ng "1307722 57.73 s «9‘ mm“ 11,0014: m... 6m = HA MM 4—... 1. 6'4 - *6”. = 5,450 psi +— Pomt 5 ‘33 ~--- bl: = -7..fl0§ tn. rant/2 =—£23$Ia. dis-rfi-‘M-to "(4-— 15'4- ...
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